NCERT Solutions · Class 11 Chemistry Organic Chemistry – Some Basic Principles and Techniques

A resonance structure never moves atoms — only electrons. A curved arrow always starts at a pair of electrons already sitting somewhere (a lone pair, or a π bond) and its head shows the new home those two electrons take up: a new bond, or a new lone pair. Every resonance structure of the same species must have the same atoms in the same positions, the same total number of electrons, and the same overall charge — only the way the π electrons and lone pairs are distributed changes.For a ring carbon bearing a substituent, call that carbon \(\displaystyle \mathrm{C_{1}}\), its two neighbours (ortho) \(\displaystyle \mathrm{C_{2}}\) and \(\displaystyle \mathrm{C_{6}}\), the next pair (meta) \(\displaystyle \mathrm{C_{3}}\) and \(\displaystyle \mathrm{C_{5}}\), and the carbon directly opposite (para) C4. This labelling is reused in parts (a), (b), (d) and (e) below.(a) Phenol, \(\displaystyle \mathrm{C_6H_5OH} \) — a benzene ring carrying one \(\displaystyle -\mathrm{OH} \) group on C1. Oxygen carries two lone pairs even though only one participates in each individual resonance structure.Structure I (starting point): the ring drawn with one Kekulé arrangement of alternating double bonds, \(\displaystyle \mathrm{C1{=}C2} \), \(\displaystyle \mathrm{C3{=}C4} \), \(\displaystyle \mathrm{C5{=}C6} \), and \(\displaystyle \mathrm{O} \) neutral with two lone pairs, single-bonded to C1.Arrow $\displaystyle 1$: one lone pair on O moves to form a new π bond between O and \(\displaystyle \mathrm{C_{1}}\) (arrow tail on the O lone pair, head landing between O and C1). To keep \(\displaystyle \mathrm{C_{1}}\)'s bonding count correct, the \(\displaystyle \mathrm{C1{=}C2} \) π electrons must simultaneously shift off \(\displaystyle \mathrm{C_{1}}\) and onto \(\displaystyle \mathrm{C_{2}}\) alone (second arrow, tail on the \(\displaystyle \mathrm{C1{=}C2} \) bond, head pointing to C2). Result — Structure II: \(\displaystyle \mathrm{O^+{=}C1} \), a negative charge sitting on \(\displaystyle \mathrm{C_{2}}\) (ortho), and the rest of the ring rearranged to \(\displaystyle \mathrm{C3{=}C4} \), \(\displaystyle \mathrm{C5{=}C6} \).Arrow set $\displaystyle 2$: starting again from Structure I but pushing the same O lone pair the other way — through the \(\displaystyle \mathrm{C1{=}C6} \) bond instead — gives Structure III: \(\displaystyle \mathrm{O^+{=}C1} \) with the negative charge on \(\displaystyle \mathrm{C_{6}}\) (the other ortho carbon).Arrow set $\displaystyle 3$: the negative charge doesn't stop at the ortho carbon — from Structure II, the lone pair now sitting on \(\displaystyle \mathrm{C_{2}}\) can push further around the ring: tail on that \(\displaystyle \mathrm{C_{2}}\) lone pair, head forming a new \(\displaystyle \mathrm{C2{=}C3} \) π bond, while the existing \(\displaystyle \mathrm{C3{=}C4} \) π electrons shift off \(\displaystyle \mathrm{C_{3}}\) onto C4. This gives Structure IV: \(\displaystyle \mathrm{O^+{=}C1} \) and the negative charge now on \(\displaystyle \mathrm{C_{4}}\) (para).So phenol has four resonance contributors: the neutral one, and three charge-separated ones with \(\displaystyle \mathrm{O^+} \) and a negative charge distributed over the two ortho carbons and the one para carbon — never on the meta carbons, because no arrow-pushing path reaches them. This is exactly why phenol undergoes electrophilic substitution preferentially at the ortho and para positions: those carbons already carry extra electron density in the real (hybrid) structure.Common trap: it's tempting to also push the ring's own double bonds toward the OH group so oxygen becomes negative — but oxygen already has a complete octet and two lone pairs in Structure I; it cannot accept more electron density, only donate. The arrow must originate on oxygen's lone pair, never end there.(b) Nitrobenzene, \(\displaystyle \mathrm{C_6H_5NO_2} \) — the nitro group \(\displaystyle -\mathrm{NO_2} \) on C1. First, the nitro group has its own internal resonance, independent of the ring: nitrogen is bonded to the ring carbon, double-bonded to one oxygen, and single-bonded to the other. Counting bonds and lone pairs, that gives N four bonds and no lone pair (formal charge \(\displaystyle \mathrm{N^+} \)), the doubly-bonded O neutral, and the singly-bonded O three lone pairs (formal charge \(\displaystyle \mathrm{O^-} \)). Swapping which oxygen carries the double bond gives an equivalent second structure — same \(\displaystyle \mathrm{N^+} \), same one \(\displaystyle \mathrm{O^-} \), just the other oxygen.Now bring the ring in. Arrow set (extending conjugation into the ring): tail on the ring's \(\displaystyle \mathrm{C1{=}C2} \) π bond, head forming a new π bond between \(\displaystyle \mathrm{C_{1}}\) and N; simultaneously, tail on the existing \(\displaystyle \mathrm{N{=}O} \) π bond, head moving those electrons fully onto that oxygen (giving it a third lone pair, so it too becomes \(\displaystyle \mathrm{O^-} \)). Electron and charge count still balances: the ring carbon that lost its share of the \(\displaystyle \mathrm{C1{=}C2} \) pair — \(\displaystyle \mathrm{C_{2}}\) — is left short an electron pair and becomes \(\displaystyle \mathrm{C2^+} \) (ortho), while N stays formally + (it still has four bonds: the new \(\displaystyle \mathrm{C1{=}N} \) double bond plus two single bonds to two now-negative oxygens). Net charge on this whole picture: \(\displaystyle +1 \,(\mathrm{C2}) +1\,(\mathrm{N}) -1\,(\mathrm{O}) -1\,(\mathrm{O}) = 0 \), matching the neutral starting molecule.Running the same arrow through the other Kekulé double bond, \(\displaystyle \mathrm{C1{=}C6} \), instead, puts the positive charge on \(\displaystyle \mathrm{C_{6}}\) (the other ortho carbon). And pushing on from the C2-cation structure — the \(\displaystyle \mathrm{C3{=}C4} \) π electrons shifting to fill in at \(\displaystyle \mathrm{C_{2}}\) — moves the positive charge to \(\displaystyle \mathrm{C_{4}}\) (para).So nitrobenzene's ring carries positive character at both ortho carbons and the para carbon in three resonance structures, while the nitro oxygens absorb the corresponding negative charge. No resonance path ever puts positive charge on a meta carbon — which is precisely why \(\displaystyle -\mathrm{NO_2} \) is a deactivating, meta-directing group: an incoming electrophile avoids attacking where the ring is already electron-poor.(c) But-$\displaystyle 2$-enal (crotonaldehyde), \(\displaystyle \mathrm{CH_3CH{=}CHCHO} \). Number the chain from the aldehyde carbon: \(\displaystyle \mathrm{C_{1}}\) is the \(\displaystyle -\mathrm{CHO} \) carbon, \(\displaystyle \mathrm{C_{2}}\) and \(\displaystyle \mathrm{C_{3}}\) carry the double bond, \(\displaystyle \mathrm{C_{4}}\) is the methyl carbon. This is a conjugated (alpha,beta-unsaturated) system, \(\displaystyle \mathrm{C3{=}C2{-}C1{=}O} \), because the C2–C3 double bond sits right next to the \(\displaystyle \mathrm{C_{1}}\)=O double bond with no intervening single-bonded sp3 carbon breaking the overlap.Structure I: \(\displaystyle \mathrm{CH_3{-}CH{=}CH{-}CH{=}O} \), carbonyl oxygen neutral with two lone pairs.Arrow: tail on the \(\displaystyle \mathrm{C1{=}O} \) π bond, head moving those electrons onto O (giving O a third lone pair, so \(\displaystyle \mathrm{O^-} \)); simultaneously, tail on the \(\displaystyle \mathrm{C3{=}C2} \) π bond, head forming a new π bond between \(\displaystyle \mathrm{C_{2}}\) and C1. Structure II: \(\displaystyle \mathrm{CH_3{-}\overset{+}{C}H{-}CH{=}CH{-}O^-} \) — the positive charge lands on \(\displaystyle \mathrm{C_{3}}\) (the carbon that was part of the alkene, now two bonds away from oxygen through the newly-conjugated chain), and the double bond has shifted to sit between \(\displaystyle \mathrm{C_{2}}\) and C1.The step people get wrong here: it looks like the positive charge should sit on \(\displaystyle \mathrm{C_{2}}\) (the carbon nearer the original double bond's other end), but tracing the arrow correctly, \(\displaystyle \mathrm{C_{2}}\) is the atom that gains the new π bond to \(\displaystyle \mathrm{C_{1}}\), so \(\displaystyle \mathrm{C_{2}}\) stays fully bonded — it is \(\displaystyle \mathrm{C_{3}}\), the far end of the original double bond, that is left electron-short. This is exactly why but-$\displaystyle 2$-enal (like all \(\displaystyle \alpha,\beta \)-unsaturated carbonyls) is electrophilic at that far ("beta") carbon and undergoes $\displaystyle 1,4$- (conjugate) addition there, not just at the carbonyl carbon.(d) Benzaldehyde, \(\displaystyle \mathrm{C_6H_5CHO} \) — the \(\displaystyle -\mathrm{CHO} \) group's carbon (call it \(\displaystyle \mathrm{C_a} \)) attached to ring carbon C1. This is the same pattern as nitrobenzene, because \(\displaystyle -\mathrm{CHO} \) is likewise an electron-withdrawing, conjugating group.Structure I: ring Kekulé form, \(\displaystyle \mathrm{C_a{=}O} \) neutral (two lone pairs on O), \(\displaystyle \mathrm{C_a{-}C1} \) single bond.Arrow: tail on \(\displaystyle \mathrm{C1{=}C2} \), head forming a new \(\displaystyle \mathrm{C1{=}C_a} \) π bond; tail on \(\displaystyle \mathrm{C_a{=}O} \), head moving those electrons fully onto O. Structure II: \(\displaystyle \mathrm{C2^+} \) (ortho), new double bond \(\displaystyle \mathrm{C1{=}C_a} \), and \(\displaystyle \mathrm{C_a{-}O^-} \) now a single bond to a negatively charged oxygen.Repeating through \(\displaystyle \mathrm{C1{=}C6} \) gives Structure III with \(\displaystyle \mathrm{C6^+} \) (the other ortho carbon); pushing on from Structure II through \(\displaystyle \mathrm{C3{=}C4} \) gives Structure IV with \(\displaystyle \mathrm{C4^+} \) (para). As with nitrobenzene, positive character never reaches a meta carbon — consistent with \(\displaystyle -\mathrm{CHO} \) also being a meta-directing, deactivating group.(e) The benzyl cation, \(\displaystyle \mathrm{C_6H_5CH_2^+} \) — an exocyclic \(\displaystyle \mathrm{CH_2} \) carbon carrying the positive charge (an empty p orbital), attached to ring carbon C1.Structure I: ring Kekulé form, \(\displaystyle \mathrm{C1{-}CH_2^+} \) single bond, the cationic carbon has no π bond of its own — its p orbital is empty, not filled with a lone pair, which is exactly what makes it able to accept electron density rather than donate it (the reverse role compared with the O lone pair in phenol).Arrow: tail on the \(\displaystyle \mathrm{C1{=}C2} \) π bond, head forming a new π bond between \(\displaystyle \mathrm{C_{1}}\) and the exocyclic carbon (filling that carbon's empty p orbital). Structure II: the exocyclic carbon becomes neutral, \(\displaystyle \mathrm{C1{=}CH_2} \), while \(\displaystyle \mathrm{C_{2}}\) — having given up its share of the original π pair — becomes the new cation, \(\displaystyle \mathrm{C2^+} \) (ortho).Repeating through \(\displaystyle \mathrm{C1{=}C6} \) gives Structure III, cation at \(\displaystyle \mathrm{C_{6}}\) (other ortho); pushing on from Structure II through \(\displaystyle \mathrm{C3{=}C4} \) gives Structure IV, cation at \(\displaystyle \mathrm{C_{4}}\) (para). Four resonance structures in total, with the positive charge shared between the exocyclic carbon and the two ortho and one para ring carbons — this delocalization into the ring is the reason benzylic cations are markedly more stable than an ordinary alkyl cation of the same size, in which the charge would have nowhere else to go.(f) But-$\displaystyle 2$-enyl (crotyl) cation, \(\displaystyle \mathrm{CH_3CH{=}CHCH_2^+} \). Number the chain: \(\displaystyle \mathrm{C_{1}}\) is the methyl carbon, \(\displaystyle \mathrm{C_{2}}\)=\(\displaystyle \mathrm{C_{3}}\) is the double bond, \(\displaystyle \mathrm{C_{4}}\) carries the positive charge (empty p orbital).Structure I: \(\displaystyle \mathrm{CH_3{-}CH{=}CH{-}CH_2^+} \).Arrow: tail on the \(\displaystyle \mathrm{C2{=}C3} \) π bond, head forming a new π bond between \(\displaystyle \mathrm{C_{3}}\) and \(\displaystyle \mathrm{C_{4}}\) (filling \(\displaystyle \mathrm{C_{4}}\)'s empty orbital). Structure II: \(\displaystyle \mathrm{CH_3{-}\overset{+}{C}H{-}CH{=}CH_2} \) — the double bond has shifted down to \(\displaystyle \mathrm{C_{3}}\)=\(\displaystyle \mathrm{C_{4}}\), and the positive charge has moved up to C2.This is the standard allylic-cation resonance: the charge and the double bond simply trade places, one carbon over, because there is no ring to spread the charge further — only the two termini of this three-carbon allylic system (C2 and C4) ever carry the positive charge; \(\displaystyle \mathrm{C_{3}}\), the middle carbon, never does, since it is always part of the moving double bond.Answer: (a) Phenol has $\displaystyle 4$ resonance structures: the neutral form, and three with \(\displaystyle \mathrm{O^+} \) and a negative charge on, respectively, one ortho carbon, the other ortho carbon, and the para carbon (never meta). (b) Nitrobenzene: the nitro group's own $\displaystyle 2$ internal resonance forms (\(\displaystyle \mathrm{N^+} \), one \(\displaystyle \mathrm{O^-} \), swapping which oxygen), plus $\displaystyle 3$ ring-conjugated forms with a positive charge on one ortho carbon, the other ortho carbon, or the para carbon, and both oxygens negative — explaining its meta-directing behaviour. (c) But-$\displaystyle 2$-enal: $\displaystyle 2$ structures, \(\displaystyle \mathrm{CH_3CH{=}CHCH{=}O} \leftrightarrow \mathrm{CH_3\overset{+}{C}HCH{=}CHO^-} \), showing the beta carbon is the electrophilic site. (d) Benzaldehyde: $\displaystyle 4$ structures, exactly analogous to nitrobenzene — ring positive charge at each ortho carbon and the para carbon, oxygen negative, once the arrow pushes ring π electrons into the carbonyl. (e) The benzyl cation \(\displaystyle \mathrm{C_6H_5CH_2^+} \): $\displaystyle 4$ structures, the positive charge delocalized onto the exocyclic carbon and the two ortho and one para ring carbons, which is why it is unusually stable. (f) The but-$\displaystyle 2$-enyl (crotyl) cation: $\displaystyle 2$ structures, \(\displaystyle \mathrm{CH_3CH{=}CHCH_2^+} \leftrightarrow \mathrm{CH_3\overset{+}{C}HCH{=}CH_2} \), the simplest allylic resonance, charge alternating between the two end carbons of the three-carbon allyl system.

Electrophiles are electron-seeking species that accept an electron pair; nucleophiles are electron-rich species that donate one — the two are simply Lewis acids and Lewis bases wearing organic-chemistry names.The Greek roots say it plainly: electro-phile = "electron-loving," nucleo-phile = "nucleus-loving." An electrophile is short of electron density somewhere and reaches out to grab a pair of electrons to complete a bond. A nucleophile has a surplus — a lone pair, or a \(\displaystyle \pi\)-electron cloud — and offers that pair to an electron-poor centre (the "nucleus," i.e., a positively polarised atom) to form a new bond.Electrophiles \(\displaystyle (E^+)\)An electrophile is any species that can accept an electron pair — equivalently, a Lewis acid. Electrophiles fall into two kinds, and the mistake students make is thinking only cations count:1. Positively charged species. These are missing electrons outright. 2. Neutral molecules that are still electron-deficient. A species does not need a positive charge to be an electrophile — it only needs an atom with an incomplete octet or one made electron-poor by an attached electronegative atom. \[\overset{\delta+}{C} = \overset{\delta-}{O} \] leaving the carbon partially positive \(\displaystyle (\delta+)\) even though the molecule carries no net charge. This carbon is the electrophilic site nucleophiles attack in addition reactions.Nucleophiles \(\displaystyle (Nu^-)\)A nucleophile is any species that can donate an electron pair — a Lewis base. It, too, comes in two kinds:1. Negatively charged species (anions). 2. Neutral molecules with a lone pair or a \(\displaystyle \pi\)-bond to give away. A short aside on where this is easy to get backward: charge alone does not decide the label. A neutral molecule (\(\displaystyle BF_3\), a carbonyl compound) can be the electrophile, and a neutral molecule (\(\displaystyle NH_3, H_2O\)) can be the nucleophile — what matters is whether the species is short of electrons or has them to spare at the reacting atom, not whether it happens to carry a formal charge.Putting the two together in a real reactionIn the hydrolysis of bromoethane, the nucleophile hydroxide ion attacks the electrophilic carbon (made \(\displaystyle \delta+\) because bromine, more electronegative, pulls electron density away from it), displacing bromide: \[CH_3CH_2Br + OH^- \; \rightarrow \; CH_3CH_2OH + Br^- \] Checking the balance: both sides carry $\displaystyle 2$ carbons, $\displaystyle 6$ hydrogens, $\displaystyle 1$ oxygen, $\displaystyle 1$ bromine, and a net charge of \(\displaystyle -1\) — so the equation is balanced as written, with no coefficients needed since one molecule of each species reacts with one of the other. The product, ethanol, forms because the nucleophile's electron pair builds the new C–O bond while the C–Br bond breaks, sending the pair of electrons off with the departing bromide ion.Answer: An electrophile is an electron-deficient species (a Lewis acid) that accepts an electron pair to form a bond — examples include the cations \(\displaystyle CH_3{}^+ \), \(\displaystyle NO_2{}^+ \), \(\displaystyle H^+ \), and the neutral electron-deficient molecules \(\displaystyle BF_3 \), \(\displaystyle AlCl_3 \), and the carbonyl carbon of \(\displaystyle CH_3CHO \). A nucleophile is an electron-rich species (a Lewis base) that donates an electron pair to form a bond — examples include the anions \(\displaystyle OH^- \), \(\displaystyle CN^- \), \(\displaystyle Cl^- \), \(\displaystyle CH_3O^- \), and the neutral electron-pair or \(\displaystyle \pi\)-electron donors \(\displaystyle NH_3 \), \(\displaystyle H_2O \), \(\displaystyle ROH \), and alkenes like \(\displaystyle CH_2{=}CH_2 \).

A nucleophile has an electron pair it wants to give away; an electrophile is short of electrons and wants to take one. "Nucleophile" literally means nucleus-loving — it seeks out a positively-polarised (electron-poor) carbon and donates a lone pair to it, forming a new bond. "Electrophile" means electron-loving — it is itself electron-deficient (a positive charge, or an atom with only six electrons around it) and accepts the electron pair a nucleophile offers. Once you spot which species is short of electrons and which one has extra to spare, the label follows.(a) \(\displaystyle \text{CH}_3\text{COOH} + \text{HO}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \)Count atoms and charge on each side before doing anything else: left side has \(\displaystyle \text{C}_2\text{H}_5\text{O}_3\) with a net charge of \(\displaystyle -1\) (four H and two O from \(\displaystyle \text{CH}_3\text{COOH}\), plus one H and one O from \(\displaystyle \text{HO}^-\)); right side has the same \(\displaystyle \text{C}_2\text{H}_5\text{O}_3\) with charge \(\displaystyle -1\) (three H, two O in \(\displaystyle \text{CH}_3\text{COO}^-\), plus two H, one O in \(\displaystyle \text{H}_2\text{O}\)). Atoms and charge already match, so the equation as printed is balanced.This is a simple proton transfer. The O–H hydrogen of the carboxylic acid is the electron-poor site here (the carboxyl oxygen pulls electron density away from that H, leaving it with a partial positive charge). \(\displaystyle \text{HO}^-\) brings a lone pair on its own oxygen and grabs that proton, forming \(\displaystyle \text{H}_2\text{O}\) and leaving the acetate ion \(\displaystyle \text{CH}_3\text{COO}^-\) behind. Because \(\displaystyle \text{HO}^-\) is the one donating its electron pair, it is the nucleophile.(b) \(\displaystyle \text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow (\text{CH}_3)_2\text{C}(\text{CN})(\text{OH}) \)Written exactly this way the equation will not balance: the left side carries a charge of \(\displaystyle -1\) (from \(\displaystyle \text{CN}^-\)) while the right side, the neutral cyanohydrin, carries charge \(\displaystyle 0\). One proton-transfer step from the solvent has been folded silently into that single arrow — this is the step students usually skate past. Writing it out in full:Step $\displaystyle 1$, the actual nucleophilic addition — cyanide's carbon end attacks the carbonyl carbon of acetone, and the electrons of the C=O double bond swing onto oxygen: \[\text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow (\text{CH}_3)_2\text{C}(\text{CN})\text{O}^- \]Step $\displaystyle 2$, the alkoxide oxygen (still carrying the extra electron pair and the negative charge) pulls a proton off a water molecule, regenerating hydroxide: \[(\text{CH}_3)_2\text{C}(\text{CN})\text{O}^- + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_2\text{C}(\text{CN})(\text{OH}) + \text{OH}^- \]Adding the two steps (the alkoxide intermediate cancels) gives the true balanced overall change: \[\text{CH}_3\text{COCH}_3 + \text{CN}^- + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_2\text{C}(\text{CN})(\text{OH}) + \text{OH}^- \] Check: C, $\displaystyle 4$ = $\displaystyle 4$; H, $\displaystyle 8$ = $\displaystyle 8$; N, $\displaystyle 1$ = $\displaystyle 1$; O, $\displaystyle 2$ = $\displaystyle 2$; charge, \(\displaystyle -1 = -1\). Balanced.\(\displaystyle \text{CN}^-\) is the species with the spare lone pair, and it is the one that attacks the electron-poor carbonyl carbon to form the new C–C bond, so \(\displaystyle \text{CN}^-\) is the nucleophile. (The product \(\displaystyle (\text{CH}_3)_2\text{C}(\text{CN})(\text{OH})\) is $\displaystyle 2$-hydroxy-$\displaystyle 2$-methylpropanenitrile, the cyanohydrin of acetone.)(c) \(\displaystyle \text{C}_6\text{H}_6 + \text{CH}_3\text{CO}^+ \rightarrow \text{C}_6\text{H}_5\text{COCH}_3 \)Again the printed equation drops a product to keep it short: the left side carries charge \(\displaystyle +1\), the right side (a neutral ketone) carries charge \(\displaystyle 0\). Electrophilic aromatic substitution always ends with the ring losing a proton to get its aromatic sextet back, so the balanced equation is \[\text{C}_6\text{H}_6 + \text{CH}_3\text{CO}^+ \rightarrow \text{C}_6\text{H}_5\text{COCH}_3 + \text{H}^+ \] Check: C, \(\displaystyle 6+2=8\) on the left, \(\displaystyle 8\) on the right; H, \(\displaystyle 6+3=9\) on the left, \(\displaystyle 8+1=9\) on the right; O, \(\displaystyle 1=1\); charge, \(\displaystyle +1=+1\). Balanced.\(\displaystyle \text{CH}_3\text{CO}^+\) is the acetyl (acylium) cation: its carbonyl carbon carries a positive charge and only six electrons around it, so it is electron-deficient. It is attracted to the electron-rich \(\displaystyle \pi\) cloud of benzene, which supplies the electron pair for the new C–C bond; the ring then ejects \(\displaystyle \text{H}^+\) to restore aromaticity. Because \(\displaystyle \text{CH}_3\text{CO}^+\) is the electron acceptor, it is the electrophile. (The product \(\displaystyle \text{C}_6\text{H}_5\text{COCH}_3\) is methyl phenyl ketone, commonly called acetophenone.)Answer: (a) \(\displaystyle \text{HO}^-\) is a nucleophile; (b) \(\displaystyle \text{CN}^-\) is a nucleophile; (c) \(\displaystyle \text{CH}_3\text{CO}^+\) (the acetyl/acylium cation) is an electrophile.

Classifying a reaction means asking what actually changed between reactant and product: was an atom/group swapped for another (substitution), did a multiple bond disappear (addition), did a new multiple bond appear (elimination), or did the carbon skeleton itself reorganise (rearrangement)? Check each equation atom-by-atom before deciding, because the answer is not always what the first glance suggests.(a) \[\mathrm{CH_3CH_2Br + HS^{-} \longrightarrow CH_3CH_2SH + Br^{-}} \]Atom count, to confirm the equation balances: left side has \(\displaystyle 2\) carbons, \(\displaystyle 6\) hydrogens (\(\displaystyle 5\) from the ethyl group, \(\displaystyle 1\) from \(\displaystyle \mathrm{HS^-}\)), \(\displaystyle 1\) sulfur, \(\displaystyle 1\) bromine, net charge \(\displaystyle -1\). Right side: \(\displaystyle \mathrm{CH_3CH_2SH}\) is \(\displaystyle 2\) carbons, \(\displaystyle 6\) hydrogens, \(\displaystyle 1\) sulfur, and \(\displaystyle \mathrm{Br^-}\) supplies the bromine and the \(\displaystyle -1\) charge. Both sides match.Bromoethane's \(\displaystyle \mathrm{C-Br}\) bond is polar — bromine pulls electron density toward itself, leaving that carbon electron-poor. The hydrosulfide ion \(\displaystyle \mathrm{HS^-}\) is a good nucleophile: it uses its lone pair to attack that carbon, and \(\displaystyle \mathrm{Br^-}\) leaves as the stable, weakly-basic leaving group. One group, \(\displaystyle -\mathrm{Br}\), is simply replaced by another, \(\displaystyle -\mathrm{SH}\), at the same carbon; nothing elsewhere in the molecule changes bond order.This is a substitution reaction (nucleophilic substitution): bromoethane is converted to ethanethiol.(b) \[\mathrm{(CH_3)_2C=CH_2 + HCl \longrightarrow (CH_3)_2CClCH_3} \]Atom count: left side is \(\displaystyle \mathrm{C_4H_8}\) ($\displaystyle 2$-methylpropene) plus \(\displaystyle \mathrm{HCl}\), total \(\displaystyle \mathrm{C_4H_9Cl}\). The product, written out, is the same \(\displaystyle \mathrm{C_4H_9Cl}\) — it is the same molecule as \(\displaystyle \mathrm{(CH_3)_3CCl}\), just with one methyl group written separately. Balanced.The step people get wrong here is assuming \(\displaystyle \mathrm{H}\) and \(\displaystyle \mathrm{Cl}\) can add to either carbon at random — they can't. The double bond's electrons attack \(\displaystyle \mathrm{H^+}\) from \(\displaystyle \mathrm{HCl}\), and the proton lands on the carbon that already carries more hydrogens (the \(\displaystyle \mathrm{=CH_2}\) end), because that leaves the positive charge on the other carbon — the one bearing two methyl groups — where three alkyl groups can stabilise it: \(\displaystyle \mathrm{(CH_3)_2C^+CH_3}\), a tertiary carbocation. Chloride then bonds to that carbon. The \(\displaystyle \mathrm{C=C}\) double bond is used up and two new single bonds (\(\displaystyle \mathrm{C-H}\), \(\displaystyle \mathrm{C-Cl}\)) form in its place, so the degree of unsaturation drops from one to zero.This is an addition reaction (electrophilic addition, following Markovnikov's rule): $\displaystyle 2$-methylpropene gives $\displaystyle 2$-chloro-$\displaystyle 2$-methylpropane.(c) \[\mathrm{CH_3CH_2Br + OH^{-} \longrightarrow CH_2=CH_2 + H_2O + Br^{-}} \]Atom count: left side, \(\displaystyle \mathrm{C_2H_5Br}\) plus \(\displaystyle \mathrm{OH^-}\), gives \(\displaystyle \mathrm{C_2H_6BrO}\), charge \(\displaystyle -1\). Right side, ethene (\(\displaystyle \mathrm{C_2H_4}\)) plus \(\displaystyle \mathrm{H_2O}\) plus \(\displaystyle \mathrm{Br^-}\), also gives \(\displaystyle \mathrm{C_2H_6BrO}\), charge \(\displaystyle -1\). Balanced.Here \(\displaystyle \mathrm{OH^-}\) does not attack the carbon holding bromine at all — it pulls a hydrogen off the neighbouring carbon instead. As that \(\displaystyle \mathrm{C-H}\) bond breaks, its electron pair swings in to form a new \(\displaystyle \mathrm{C=C}\) \(\displaystyle \pi\) bond, and in the same step the \(\displaystyle \mathrm{C-Br}\) bond breaks and \(\displaystyle \mathrm{Br^-}\) departs. Two atoms leave from two adjacent carbons, and a double bond appears where there was none: the degree of unsaturation rises from zero to one.This is an elimination reaction (dehydrohalogenation): bromoethane loses \(\displaystyle \mathrm{HBr}\) to give ethene.(d) \[\mathrm{(CH_3)_3C{-}CH_2OH + HBr \longrightarrow (CH_3)_2CBrCH_2CH_3 + H_2O} \]This is the one where it's tempting to assume the product is "just" the bromide swapped in for the hydroxyl, and that assumption is wrong. Left side: \(\displaystyle \mathrm{(CH_3)_3CCH_2OH}\) (neopentyl alcohol) is \(\displaystyle \mathrm{C_5H_{12}O}\), plus \(\displaystyle \mathrm{HBr}\), total \(\displaystyle \mathrm{C_5H_{13}BrO}\). Right side: the central carbon of \(\displaystyle \mathrm{(CH_3)_2CBrCH_2CH_3}\) carries two methyls (\(\displaystyle 2\mathrm{C}\), \(\displaystyle 6\mathrm{H}\)) and an ethyl group \(\displaystyle \mathrm{-CH_2CH_3}\) (\(\displaystyle 2\mathrm{C}\), \(\displaystyle 5\mathrm{H}\)), plus itself (\(\displaystyle 1\mathrm{C}\)) and one bromine — \(\displaystyle \mathrm{C_5H_{11}Br}\) — plus \(\displaystyle \mathrm{H_2O}\), giving \(\displaystyle \mathrm{C_5H_{13}BrO}\). The two sides match atom-for-atom, so the equation is balanced — but look at the skeleton: the reactant's longest carbon chain off the quaternary carbon is a single \(\displaystyle \mathrm{CH_2OH}\); the product has a four-carbon (butane) chain with a methyl branch. Carbons have moved. That rules out plain substitution.Following each bond shows why: 1. \(\displaystyle \mathrm{HBr}\) protonates the \(\displaystyle -\mathrm{OH}\) oxygen: \(\displaystyle \mathrm{(CH_3)_3C{-}CH_2OH + HBr \rightarrow (CH_3)_3C{-}CH_2{-}OH_2^{+} + Br^{-}} \). 2. Water leaves, generating a carbocation on the \(\displaystyle \mathrm{CH_2}\) carbon: \(\displaystyle \mathrm{(CH_3)_3C{-}CH_2^{+}} \). That carbon is bonded to only one other carbon (the bulky \(\displaystyle \mathrm{C(CH_3)_3}\) group) — a primary carbocation, the least stable kind, sitting right next to a carbon with three methyl groups to spare. 3. One of those methyl groups migrates, bringing its bonding electron pair with it (a $\displaystyle 1,2$-methyl shift), and bonds to the electron-poor \(\displaystyle \mathrm{CH_2}\) carbon. The positive charge moves to the carbon the methyl left behind, which is now bonded to two remaining methyls and to a \(\displaystyle \mathrm{-CH_2CH_3}\) (ethyl) group — three carbon substituents, a tertiary carbocation: \(\displaystyle \mathrm{(CH_3)_2C^{+}{-}CH_2CH_3} \). 4. The bromide ion released in step $\displaystyle 1$ bonds to this tertiary carbocation, giving \(\displaystyle \mathrm{(CH_3)_2CBrCH_2CH_3} \), $\displaystyle 2$-bromo-$\displaystyle 2$-methylbutane.Why bother with the shift: a primary carbocation is high-energy because only one adjacent carbon can push electron density toward the empty orbital; sliding a methyl group over turns it into a tertiary carbocation, which three adjacent carbons stabilise. The molecule rearranges its own skeleton to reach that lower-energy intermediate before bromide ever gets a chance to attack the original carbon.This is a rearrangement reaction — a substitution of \(\displaystyle -\mathrm{OH}\) by \(\displaystyle -\mathrm{Br}\) that is accompanied by a change in the carbon skeleton, so it is classified separately from a plain substitution: neopentyl alcohol gives $\displaystyle 2$-bromo-$\displaystyle 2$-methylbutane, not $\displaystyle 1$-bromo-$\displaystyle 2,2$-dimethylpropane.Answer: (a) substitution reaction (bromoethane → ethanethiol via attack of \(\displaystyle \mathrm{HS^-}\)); (b) addition reaction (Markovnikov addition of \(\displaystyle \mathrm{HCl}\) to $\displaystyle 2$-methylpropene, giving $\displaystyle 2$-chloro-$\displaystyle 2$-methylpropane through a tertiary-carbocation intermediate); (c) elimination reaction (dehydrohalogenation of bromoethane by \(\displaystyle \mathrm{OH^-}\) to give ethene); (d) rearrangement reaction (the primary carbocation formed from neopentyl alcohol undergoes a $\displaystyle 1,2$-methyl shift to a tertiary carbocation before bromide attacks, giving $\displaystyle 2$-bromo-$\displaystyle 2$-methylbutane).

Two structures are isomers only if you would have to break a bond and remake it somewhere else to turn one into the other. If every atom stays exactly where it is and only electrons (and a formal charge) shift, the two structures are resonance contributors of one species, not two different compounds. That single test settles all three pairs.(a) Read both skeletal drawings carbon by carbon — they turn out to be a ketone drawn with the \(\displaystyle \mathrm{C{=}O} \) group in two different positions.Left structure: \(\displaystyle \mathrm{CH_3{-}CH_2{-}CO{-}CH_2{-}CH_3} \) — the carbonyl carbon is C-$\displaystyle 3$ of a five-carbon chain, with an ethyl group, \(\displaystyle \mathrm{-CH_2CH_3} \), on each side. This is pentan-$\displaystyle 3$-one (diethyl ketone).Right structure: \(\displaystyle \mathrm{CH_3{-}CH_2{-}CH_2{-}CO{-}CH_3} \) — the carbonyl carbon is C-$\displaystyle 2$, flanked by a propyl group, \(\displaystyle \mathrm{-CH_2CH_2CH_3} \), on one side and a methyl group on the other. This is pentan-$\displaystyle 2$-one (methyl propyl ketone).Check the molecular formula on each side, the way you'd check that an equation balances: Left: $\displaystyle 5$ carbons, H count \(\displaystyle 3+2+0+2+3=10\) → \(\displaystyle \mathrm{C_5H_{10}O} \). Right: $\displaystyle 5$ carbons, H count \(\displaystyle 3+0+2+2+3=10\) → \(\displaystyle \mathrm{C_5H_{10}O} \).Same molecular formula, same functional group (a non-terminal \(\displaystyle \mathrm{C{=}O} \)), but the carbonyl carbon sits at a different point along the chain, so the two carbon skeletons are genuinely different — a bond has moved from C-$\displaystyle 2$/C-$\displaystyle 4$ to C-3. That is structural (constitutional) isomerism — here, more precisely, position isomerism (also called metamerism, since it is the same functional group with a different split of alkyl groups on either side of it). It is not geometrical isomerism (there is no restricted-rotation double bond deciding the shape) and not resonance (the two skeletons are not even the same skeleton).(b) Both drawings sit around the same rigid \(\displaystyle \mathrm{C{=}C} \) double bond, each carbon carrying one H and one D (deuterium, the hydrogen isotope of mass number $\displaystyle 2$), so both have the formula \(\displaystyle \mathrm{C_2H_2D_2} \) and identical connectivity.This is the step that trips people up: a \(\displaystyle \mathrm{C{=}C} \) double bond is one \(\displaystyle \sigma \) bond plus one \(\displaystyle \pi \) bond, and rotating the two carbons relative to each other means breaking that \(\displaystyle \pi \) bond — it simply does not happen at ordinary temperatures. So which side of the double bond a group sits on is a real, fixed fact about the molecule, not just an artist's choice in how it was drawn.Left drawing: D is on the upper-left carbon and D is on the lower-right carbon — the two D atoms are diagonally placed, i.e. on opposite sides of the double bond. That is the trans arrangement. Right drawing: D is on the upper-left carbon and D is on the upper-right carbon — both D atoms are on the same side. That is the cis arrangement.Same atoms, same bonds, but a different arrangement in space around a double bond that cannot rotate freely — that is exactly geometrical (cis–trans) isomerism: trans-$\displaystyle 1,2$-dideuterioethene (left) and cis-$\displaystyle 1,2$-dideuterioethene (right).(c) Write out exactly what the central carbon is bonded to in each drawing.Left: the carbon is bonded to one H, one plain \(\displaystyle \mathrm{-OH} \), and one oxygen joined by a double bond that itself carries an H and a \(\displaystyle + \) charge, i.e. \(\displaystyle \mathrm{=O^+H} \). Right: the carbon itself carries the \(\displaystyle + \) charge (\(\displaystyle \mathrm{C^+} \)) and is singly bonded to one H and two plain \(\displaystyle \mathrm{-OH} \) groups.Formal charge has to be checked exactly the way atoms are checked in an equation — count bonding and non-bonding electrons on each atom using formal charge \(\displaystyle = (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons}) \):Left, the doubly-bonded O: $\displaystyle 1$ double bond to C ($\displaystyle 4$ bonding \(\displaystyle e^-\)) + $\displaystyle 1$ single bond to H ($\displaystyle 2$ bonding \(\displaystyle e^-\)) + $\displaystyle 1$ lone pair ($\displaystyle 2$ \(\displaystyle e^-\)) → \(\displaystyle 6 - 2 - \tfrac{6}{2} = +1 \). Matches the drawn \(\displaystyle ^+\mathrm{OH} \). Left, the central C: $\displaystyle 4$ bonds, $\displaystyle 0$ lone pairs → \(\displaystyle 4 - 0 - \tfrac{8}{2} = 0 \). Right, the central C: $\displaystyle 3$ single bonds, $\displaystyle 0$ lone pairs (empty orbital, a carbocation) → \(\displaystyle 4 - 0 - \tfrac{6}{2} = +1 \). Matches the drawn \(\displaystyle \mathrm{C^+} \). Right, each O: $\displaystyle 2$ single bonds + $\displaystyle 2$ lone pairs → \(\displaystyle 6 - 4 - \tfrac{4}{2} = 0 \) each.Both drawings come out to a net charge of \(\displaystyle \mathrm{ +1 \) on the whole species, and in both drawings the skeleton is identical — one carbon carrying one H and two oxygens, each oxygen carrying one H. Nothing has moved to a different atom; only a lone pair became a \(\displaystyle \pi \) bond (or vice versa) and the \(\displaystyle + \) charge relocated from O to C. That is precisely the definition of resonance contributors (canonical structures) of a single species — here, the two canonical forms of the protonated methanoic (formic) acid cation, \(\displaystyle [\mathrm{HC(OH)_2}]^+ ^{-}}\) and not two different compounds at all.Answer: (a) structural (position) isomers — pentan-$\displaystyle 3$-one and pentan-$\displaystyle 2$-one, both \(\displaystyle \mathrm{C_5H_{10}O} \), with the carbonyl at a different chain position; (b) geometrical (cis–trans) isomers — trans-$\displaystyle 1,2$-dideuterioethene and cis-$\displaystyle 1,2$-dideuterioethene; (c) resonance contributors, not isomers — the two canonical structures of the protonated methanoic acid cation \(\displaystyle [\mathrm{HC(OH)_2}]^+ \), differing only in whether the \(\displaystyle \pi \) bond and the \(\displaystyle + \) charge sit on oxygen or on carbon.

A curved arrow always tracks electrons, not atoms — and the TYPE of arrowhead tells you whether the bond splits evenly (homolysis) or unevenly (heterolysis).Two conventions, and mixing them up is the single most common mistake on this question: Now the four cleavages.(a) Dimethyl peroxide splitting at its weak O–O bond — homolysis, giving two free radicals.\[\text{CH}_3\text{–O–O–CH}_3 \;\xrightarrow{\text{homolysis}}\; \text{CH}_3\text{O}^{\bullet} \;+\; {}^{\bullet}\text{OCH}_3 \]The O–O bond is the weakest bond in the molecule. Two fish-hook arrows start from the middle of that O–O bond — one curls up onto the left oxygen, the other curls up onto the right oxygen — so each oxygen walks away with exactly one of the two bonding electrons. That single unpaired electron is exactly what the dot over each O in the product represents. Because the split is perfectly symmetric (both fragments are identical methoxy radicals, \(\displaystyle \text{CH}_3\text{O}^{\bullet} \)), this has to be homolysis: nothing here explains why one oxygen would grab both electrons over the other.(b) Propanone losing an alpha-hydrogen to hydroxide — heterolysis, giving a carbanion.\[\text{CH}_3\text{COCH}_3 \;+\; {}^{-}\text{OH} \;\xrightarrow{\text{heterolysis}}\; {}^{-}\text{CH}_2\text{COCH}_3 \;+\; \text{H}_2\text{O} \](Check the atoms before trusting the equation: left side is \(\displaystyle \text{C}_3\text{H}_7\text{O}_2^{-} \) — propanone \(\displaystyle \text{C}_3\text{H}_6\text{O} \) plus \(\displaystyle \text{OH}^{-} \); right side is the carbanion \(\displaystyle \text{C}_3\text{H}_5\text{O}^{-} \) plus \(\displaystyle \text{H}_2\text{O} \), which is also \(\displaystyle \text{C}_3\text{H}_7\text{O}_2^{-} \). It balances.)Here two full arrows move together, not one: ($\displaystyle 1$) a lone pair on the negatively charged oxygen of \(\displaystyle ^{-}\text{OH} \) swings over and lands on the hydrogen of one of propanone's methyl groups (the one alpha to the carbonyl), making the new O–H bond of water; ($\displaystyle 2$) simultaneously, the C–H bond that hydrogen leaves behind does NOT split evenly — both of its electrons stay on the carbon, because the hydrogen is departing as a bare proton, \(\displaystyle \text{H}^{+} \), not as a hydrogen radical. This is the step people mistake for two separate reactions (an acid-base step, then a separate bond-breaking step) — it's one continuous piece of electron-pushing, which is exactly why it's heterolysis and not homolysis. The carbon that keeps both electrons is now a carbanion, \(\displaystyle {}^{-}\text{CH}_2\text{COCH}_3 \) — it is stabilised by the neighbouring carbonyl group (its more familiar name is the enolate ion), which is why hydroxide can pull this proton off at all.(c) tert-Butyl bromide losing bromide — heterolysis, giving a carbocation.\[(\text{CH}_3)_3\text{C–Br} \;\xrightarrow{\text{heterolysis}}\; (\text{CH}_3)_3\text{C}^{+} \;+\; \text{Br}^{-} \]One full curved arrow, tail on the C–Br bond, head landing on the bromine atom: both bonding electrons travel to Br, never the other way. The arrow always points toward the more electronegative atom — drawing it from Br toward carbon is the reverse mistake, and it would put the negative charge on the wrong atom. Bromine can hold the extra electron pair comfortably as a stable, isolated bromide ion, \(\displaystyle \text{Br}^{-} \); carbon, left with only six electrons around it, becomes the carbocation \(\displaystyle (\text{CH}_3)_3\text{C}^{+} \) — the tert-butyl cation, a tertiary carbocation (three alkyl groups feeding electron density into the empty orbital by the inductive effect, which is why this bond breaks this way so readily compared to a primary halide).(d) Benzene attacking an electrophile — heterolysis of a π bond, giving a (resonance-stabilised) carbocation.\[\text{C}_6\text{H}_6 \;+\; \text{E}^{+} \;\xrightarrow{\text{heterolysis}}\; [\,\text{C}_6\text{H}_6\text{E}\,]^{+} \]This one isn't a sigma bond splitting into two separate pieces — it's one of benzene's \(\displaystyle \pi \) bonds breaking so its electron pair can form a brand-new \(\displaystyle \sigma \) bond to the electrophile. The arrow is still the full, double-barbed kind: tail on one ring C=C double bond, head pointing straight at \(\displaystyle \text{E}^{+} \). That pair leaves the ring entirely to bond with E, which is why it's classed with heterolysis rather than homolysis (no single electron is left orphaned on each atom — the whole pair goes to make the new bond). The ring carbon that used to share that pair is now short of it, so the positive charge that was on E is now spread over the ring instead: the product is the arenium ion (also called the cyclohexadienyl cation, or sigma complex) — a carbocation, just one whose positive charge is delocalised over three ring carbons instead of sitting on one atom, which is exactly why benzene can tolerate reacting with an electrophile at all.Answer: (a) Homolysis — the O–O bond of dimethyl peroxide splits symmetrically (two fish-hook arrows) into two methoxy free radicals, \(\displaystyle \text{CH}_3\text{O}^{\bullet} \). (b) Heterolysis — an alpha C–H bond of propanone breaks unevenly as hydroxide removes the proton, giving the carbanion \(\displaystyle {}^{-}\text{CH}_2\text{COCH}_3 \) (plus \(\displaystyle \text{H}_2\text{O} \)). (c) Heterolysis — the C–Br bond of tert-butyl bromide breaks with both electrons going to Br, giving the tertiary carbocation \(\displaystyle (\text{CH}_3)_3\text{C}^{+} \) (plus \(\displaystyle \text{Br}^{-} \)). (d) Heterolysis of a π bond — benzene's π electrons attack \(\displaystyle \text{E}^{+} \), giving the resonance-stabilised carbocation known as the arenium ion.

The Inductive effect is a permanent pull of \(\displaystyle \sigma\)-bond electrons toward a more electronegative atom, transmitted along the carbon chain; the Electromeric effect is a temporary, complete shift of a \(\displaystyle \pi\)-electron pair that appears only when an attacking reagent approaches a multiple bond.Inductive effect (I effect). When an atom or group more electronegative than carbon — a halogen, for instance — is attached to a chain, it draws the shared pair of the C–X \(\displaystyle \sigma\) bond slightly toward itself. That small polarisation is relayed to the next \(\displaystyle \sigma\) bond, and the next, growing weaker with each bond it crosses. Two things mark it out: it needs no external reagent (it exists as long as the molecule does — hence "permanent"), and it always travels through \(\displaystyle \sigma\) bonds only. Electromeric effect (E effect). This only exists in a molecule that already has a multiple bond, and only at the instant a reagent attacks it. The whole \(\displaystyle \pi\)-electron pair swings over to one of the two atoms of the multiple bond, and the moment the reagent is taken away, the electrons swing back — nothing about it is permanent. (This is the effect you would invoke for something like HBr adding across \(\displaystyle \text{CH}_2=\text{CH}_2\), not for comparing the resting acid strength of two molecules sitting in a bottle.)Both acidity orders in this question are decided before any reagent turns up — they are comparisons of how readily each acid ionises on its own. So the effect at work in both (a) and (b) is the Inductive effect, not the Electromeric effect.Why the Inductive effect controls acid strength. A carboxylic acid ionises as \[\text{RCOOH} \rightleftharpoons \text{RCOO}^- + \text{H}^+ \] How readily this goes forward depends on how comfortably the negative charge sits on the carboxylate ion \(\displaystyle \text{RCOO}^-\): the more that charge is spread out (stabilised), the more willingly the acid gives up \(\displaystyle \text{H}^+\), so the stronger the acid. A \(\displaystyle -\text{I}\) group next to \(\displaystyle -\text{COO}^-\) pulls some of that negative charge away and disperses it, stabilising the anion and strengthening the acid. A \(\displaystyle +\text{I}\) group does the opposite: it pushes more electron density onto an already-negative oxygen, piling up charge in one place, destabilising the anion, and weakening the acid.(a) \(\displaystyle \text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH}\). Chlorine is far more electronegative than carbon, so each C–Cl bond is a site of \(\displaystyle -\text{I}\) pull, relayed through the \(\displaystyle \sigma\) framework onto the \(\displaystyle -\text{COO}^-\) group. More electron-withdrawing chlorines directly on the carbon next to \(\displaystyle -\text{COOH}\) means more anion stabilisation, and that is the entire order: it tracks the number of \(\displaystyle -\text{I}\) chlorine atoms, nothing else.A trap worth naming: it is tempting to think piling atoms onto the molecule near the acidic \(\displaystyle \text{H}\) should make that \(\displaystyle \text{H}\) "harder to remove." The opposite is true here — what matters is not the acid molecule's own stability, but the stability of the ion left behind after \(\displaystyle \text{H}^+\) leaves.(b) \(\displaystyle \text{CH}_3\text{CH}_2\text{COOH} > (\text{CH}_3)_2\text{CHCOOH} > (\text{CH}_3)_3\text{C}\cdot\text{COOH}\). Here the substituents on the \(\displaystyle \alpha\)-carbon are methyl groups, and alkyl groups are electron-releasing: they show a \(\displaystyle +\text{I}\) effect, feeding electron density toward the chain instead of pulling it away. Notice the order runs the opposite way from (a) for exactly the same reason stated in reverse: in (a), more electron-withdrawing groups stabilise the anion and raise acid strength; in (b), more electron-donating groups destabilise the anion and lower acid strength. Both are the Inductive effect — one is \(\displaystyle -\text{I}\) acting to strengthen, the other is \(\displaystyle +\text{I}\) acting to weaken.Answer: Both orders are explained by the Inductive effect, not the Electromeric effect (which needs a multiple bond and an attacking reagent, and is only temporary). In (a), the \(\displaystyle -\text{I}\) effect of chlorine withdraws electron density and stabilises the carboxylate anion, so acid strength rises with the number of chlorines: \(\displaystyle \text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH}\). In (b), the \(\displaystyle +\text{I}\) effect of methyl groups pushes electron density onto the anion and destabilises it, so acid strength falls as the number of methyl groups on the \(\displaystyle \alpha\)-carbon rises: \(\displaystyle \text{CH}_3\text{CH}_2\text{COOH} > (\text{CH}_3)_2\text{CHCOOH} > (\text{CH}_3)_3\text{CCOOH}\).

Purification is not one technique — the right method depends on what's mixed with what: a solid with solid impurities, a liquid with a non‑volatile solid, or several similar‑looking liquids/solutes together. Each of the three techniques below exploits a different physical difference between the wanted substance and everything mixed in with it.(a) Crystallisation — separates a solid from soluble/insoluble impurities using the fact that solubility changes with temperature. The impure solid is dissolved in the minimum volume of a solvent in which it is readily soluble hot but only sparingly soluble cold. The hot solution is filtered (this removes insoluble impurities) and the filtrate is then cooled slowly. As the solution cools, it becomes supersaturated with respect to the desired compound, which is present in much larger amount than the soluble impurities, so it reaches its solubility limit first and crystallises out in a regular lattice that only accommodates molecules of the same substance — the soluble impurities stay behind, dissolved in the small amount of mother liquor, and are removed when the crystals are filtered off and washed.A step people skip: filtering the hot solution before cooling. Skip that step and any insoluble dust or grit gets trapped inside the crystals as they grow, instead of staying in the liquid where filtration removes it.Example: an impure sample of copper sulphate, \(\displaystyle \text{CuSO}_4\cdot 5\text{H}_2\text{O} \), is dissolved in the least amount of hot water, the hot solution is filtered to remove insoluble dirt, and the filtrate is left to cool undisturbed. Well-formed blue crystals of pure copper sulphate separate out, leaving the soluble coloured impurities in the mother liquor.(b) Distillation — separates a volatile substance from a non-volatile one, or two miscible liquids whose boiling points differ enough, by evaporating and then condensing. The liquid is heated in a distillation flask; the substance with the lower boiling point vaporises first, the vapour travels through a condenser where it is cooled back to a liquid, and this distillate is collected separately from what is left behind in the flask. Simple distillation is used when the liquid is stable at its boiling point and the impurity is either non-volatile (a dissolved solid) or a second liquid whose boiling point is at least about \(\displaystyle 25\ \text{K} \) higher; the vapour that comes off first is then already the almost-pure lower-boiling component. When two liquids have boiling points closer together than this, plain distillation cannot separate them cleanly because each fraction of vapour still contains a substantial amount of the higher-boiling liquid — that case needs fractional distillation instead, where a fractionating column full of surface area lets the vapour condense and re-evaporate many times on its way up, enriching it in the more volatile liquid at each stage.The step that trips people up: distillation is not the same as simply boiling something away — the point is that the vapour is caught and condensed as a separate, purified liquid, not just driven off into the air.Example: an aqueous solution of common salt, \(\displaystyle \text{NaCl} \), is distilled to recover pure water — water (boiling point \(\displaystyle 100\,^{\circ}\text{C} \)) evaporates and is condensed and collected, while the non-volatile salt stays behind as a solid residue in the flask.(c) Chromatography — separates the components of a mixture by how strongly each one clings to a stationary phase versus how readily it moves with a mobile phase. A stationary phase (a solid adsorbent packed in a column, or the solvent held in the pores of paper) is set up, and a mobile phase (a liquid or gas) is allowed to flow over or through it, carrying the dissolved mixture along. Different components of the mixture are adsorbed onto the stationary phase, or dissolve into it, to different extents — a component held more strongly by the stationary phase travels more slowly, and one that prefers the mobile phase travels faster — so as the mobile phase keeps moving, the components separate into distinct, spatially separated zones or bands that can then be collected or identified individually. This mode, where separation is governed by differing strength of adsorption on the solid stationary phase, is called adsorption chromatography.Where the mix-up happens: the mobile phase does not react with or dissolve the stationary phase — it only carries the sample components over it, and separation happens purely because those components differ in how much they are held back by the stationary phase.Example: column chromatography is used to separate the pigments of a plant leaf extract — the extract is loaded onto a column packed with alumina (the stationary phase) and washed through with a suitable solvent (the mobile phase); the different pigments, such as the carotenes, xanthophylls, and chlorophylls, are held with different strengths by the alumina and so travel down the column at different rates, emerging separately and letting each pigment be collected on its own.Answer: (a) Crystallisation purifies a solid using the difference in how solubility changes with temperature — dissolve hot, filter hot, cool slowly so the pure compound crystallises out while soluble impurities stay in the mother liquor (e.g., purifying impure copper sulphate, \(\displaystyle \text{CuSO}_4\cdot 5\text{H}_2\text{O} \)). (b) Distillation purifies a volatile liquid using the difference in boiling points — the liquid is vaporised and the vapour condensed and collected separately, leaving non-volatile or higher-boiling impurities behind (e.g., recovering pure water from a salt solution). (c) Chromatography separates components of a mixture using the difference in how strongly each is adsorbed by a stationary phase relative to a moving mobile phase, so they travel at different rates and separate into distinct zones (e.g., separating leaf pigments by column chromatography on alumina).

The technique is fractional crystallization: it separates two solutes dissolved in the same solvent by using the fact that a less soluble solute reaches saturation — and begins crystallizing out — before a more soluble one does, as the solution is cooled or concentrated. Aside: cooling must be slow and the solvent must not be driven off too far — cooling too fast or over-concentrating forces both compounds out of solution together, and the separation is lost. Answer: Fractional crystallization — dissolve the mixture in solvent \(\displaystyle S\), concentrate the solution and cool it slowly so the less soluble compound crystallizes out first and can be filtered off, leaving the more soluble compound in the mother liquor; repeating dissolution, cooling and filtration on each fraction purifies both compounds further.

The three methods separate liquids by exploiting different tricks to make a compound boil at a lower, safer temperature — plain distillation, reduced pressure, or a partner liquid that carries it over.1. Simple distillationThis is heating a liquid mixture until the more volatile component turns to vapour, then cooling and condensing that vapour back to liquid in a separate receiver.It works only when: Example: separating water \(\displaystyle \text{(b.p. } 373\ \text{K)}\) from a dissolved salt, or separating two miscible liquids whose boiling points differ by more than about \(\displaystyle 25\ \text{K}\), such as acetone \(\displaystyle \text{(b.p. } 329\ \text{K)}\) and water.2. Distillation under reduced pressureA liquid boils when its vapour pressure becomes equal to the external (atmospheric) pressure. Lowering the external pressure with a vacuum pump means the liquid's vapour pressure needs to climb to only that lower value — so it reaches that value, and starts boiling, at a lower temperature.The idea people miss: you are not changing the liquid, you are lowering the bar it has to clear. Distillation under reduced pressure is used for liquids that: By pumping the system down to a low pressure, the liquid distils at a temperature well below its normal boiling point, so it comes over intact instead of breaking down. This method is used, for example, to purify glycerol, which decomposes near its normal boiling point of \(\displaystyle 563\ \text{K}\) but can be distilled safely at about \(\displaystyle 453\ \text{K}\) under reduced pressure.3. Steam distillationThis is used for a liquid that: Steam is passed into the mixture (or the mixture is heated with water). Because the liquid and water do not mix, each contributes its own full vapour pressure independently, and the total vapour pressure above the mixture is simply the sum of the two: \[p_{\text{total}} = p_{\text{liquid}}^{\circ} + p_{\text{water}}^{\circ} \] where \(\displaystyle p_{\text{liquid}}^{\circ}\) and \(\displaystyle p_{\text{water}}^{\circ}\) are the vapour pressures of the pure liquid and of pure water at that temperature.This is the step that trips people up: because the two pressures simply add, the mixture reaches the boiling condition — total vapour pressure equal to atmospheric pressure — at a temperature lower than the boiling point of EITHER pure component on its own, including a liquid whose own boiling point is above $\displaystyle 373$ K. So the liquid distils over, carried along with steam, at a temperature below $\displaystyle 373$ K, without ever being heated to its own (possibly much higher, decomposition-prone) boiling point. The vapour of the two components condenses together and is then separated using a separating funnel, since the liquid and water do not mix.This is why steam distillation is the standard way to isolate essential oils (e.g. from flowers) and other high-boiling, water-insoluble, heat-sensitive organic liquids from plant material.Summary of the difference Answer: Simple distillation heats a thermally stable liquid to its own boiling point at normal pressure to separate it from less volatile material. Distillation under reduced pressure lowers the external pressure with a vacuum pump so a high-boiling, heat-sensitive liquid boils well below its normal boiling point (e.g. glycerol, boiling near $\displaystyle 453$ K under reduced pressure instead of $\displaystyle 563$ K at $\displaystyle 1$ atm), avoiding decomposition. Steam distillation passes steam into a water-immiscible, steam-volatile liquid; since the vapour pressures of the two immiscible liquids add \(\displaystyle (p_{\text{total}} = p_{\text{liquid}}^{\circ} + p_{\text{water}}^{\circ})\), the mixture boils below $\displaystyle 373$ K and below the liquid's own boiling point, letting it distil over intact, after which the water and liquid layers are separated with a separating funnel.

Dumas and Kjeldahl's methods both find the percentage of nitrogen in an organic compound, but they measure it in completely different ways — one measures a gas by volume, the other measures an acid by titration — and that difference decides which compounds each method can even be used on.(i) Dumas method — nitrogen is turned into \(\displaystyle N_2\) gas and its volume is measured.The organic compound is mixed with dry copper(II) oxide, \(\displaystyle CuO\), and heated strongly in a stream of carbon dioxide, \(\displaystyle CO_2\). The \(\displaystyle CuO\) oxidises the carbon and hydrogen of the compound to \(\displaystyle CO_2\) and water, while the nitrogen of the compound is set free mostly as nitrogen gas, \(\displaystyle N_2\), along with small amounts of nitrogen oxides:\[\text{Organic compound} + CuO \xrightarrow{\ \Delta\ } CO_2 + H_2O + N_2 \ (\text{+ some } N_xO_y) \]Any oxides of nitrogen formed are not left as they are — the gas stream is passed over hot copper gauze, which reduces them back to \(\displaystyle N_2\), so that all the nitrogen ends up as the free element:\[2NO + 2Cu \xrightarrow{\ \Delta\ } N_2 + 2CuO \]Check the balance: $\displaystyle 2$ N atoms on each side (in \(\displaystyle 2NO\) and in \(\displaystyle N_2\)), $\displaystyle 2$ O atoms on each side (in \(\displaystyle 2NO\) and in \(\displaystyle 2CuO\)), and $\displaystyle 2$ Cu atoms on each side — the copper is the reducing agent and is itself oxidised.The mixed gas stream (\(\displaystyle CO_2 + N_2\)) is then passed into a graduated tube standing over concentrated \(\displaystyle KOH\) solution. \(\displaystyle KOH\) absorbs \(\displaystyle CO_2\) completely (\(\displaystyle CO_2 + 2KOH \rightarrow K_2CO_3 + H_2O\)) but does not touch \(\displaystyle N_2\), so the gas that collects and can be read off the graduations is pure \(\displaystyle N_2\), measured at known temperature and pressure and then converted to STP.This is a measurement of moles of gas, not of mass — so the working goes through the gas volume, not through weighing anything. From the volume \(\displaystyle V\) (in mL, at STP) of \(\displaystyle N_2\) collected:\[\text{moles of } N_2 = \frac{V}{22400} \]using \(\displaystyle 22400\ \text{mL}\) as the volume occupied by $\displaystyle 1$ mole of any gas at STP. The mass of nitrogen this represents uses the molar mass of \(\displaystyle N_2\), which is \(\displaystyle 28\ \text{g mol}^{-1}\) (two nitrogen atoms, each of atomic mass $\displaystyle 14$):\[\text{mass of } N = \frac{V}{22400}\times 28 \]Dividing by the mass \(\displaystyle m\) (in g) of the organic compound taken and multiplying by $\displaystyle 100$ gives the percentage:\[\%\,N = \frac{28 \times V \times 100}{22400 \times m} \](ii) Kjeldahl's method — nitrogen is turned into ammonium sulfate, then \(\displaystyle NH_3\), and the \(\displaystyle NH_3\) is found by titration.The organic compound is heated with concentrated sulfuric acid, \(\displaystyle H_2SO_4\) (a small amount of \(\displaystyle K_2SO_4\) is added to raise the boiling point, and \(\displaystyle CuSO_4\) or selenium acts as a catalyst). This digestion converts the nitrogen present into ammonium sulfate, while the carbon and hydrogen are oxidised away as \(\displaystyle CO_2\) and water:\[\text{Organic compound } (N) + H_2SO_4 \xrightarrow{\ \Delta,\ \text{catalyst}\ } (NH_4)_2SO_4 + CO_2 + H_2O \]The digested mixture is then made strongly alkaline with excess sodium hydroxide, which displaces ammonia gas from the ammonium sulfate:\[(NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3\uparrow + 2H_2O \]Balance check: $\displaystyle 2$ N atoms, $\displaystyle 8$ H atoms ($\displaystyle 4$ in each \(\displaystyle NH_4^+\)) matched by $\displaystyle 2$ in \(\displaystyle NaOH\) and $\displaystyle 2$ more released as \(\displaystyle 2NH_3\) plus the water, $\displaystyle 1$ S and the O's all accounted for in the two sulfate salts — the equation is balanced.This liberated ammonia is distilled over into a known volume of a standard (accurately known strength) acid taken in excess, where it is absorbed and neutralised. The acid left over after neutralising the ammonia is then found by back-titrating it against standard \(\displaystyle NaOH\); the volume of acid actually consumed by the ammonia is obtained by subtracting the leftover from the total taken. This is the step people get wrong — you are not titrating the ammonia directly, you are finding how much of a known excess of acid it used up.From the volume \(\displaystyle V\) (mL) of acid of normality \(\displaystyle N\) that reacted with the ammonia: $\displaystyle 1000$ mL of a $\displaystyle 1$ N acid is chemically equivalent to $\displaystyle 14$ g of nitrogen (one gram-equivalent of \(\displaystyle N\), since the acid neutralises the ammonia mole-for-mole of nitrogen). So the mass of nitrogen present is:\[\text{mass of } N = \frac{V \times N \times 14}{1000} \]and the percentage of nitrogen is:\[\%\,N = \frac{\text{mass of } N}{m}\times 100 = \frac{V \times N \times 14 \times 100}{1000 \times m} = \frac{1.4\times N \times V}{m} \]where \(\displaystyle m\) is the mass (g) of the organic compound taken. The constant is $\displaystyle 1.4$, not $\displaystyle 14$ — the extra factor of $\displaystyle 10$ comes purely from converting the $\displaystyle 1000$ mL basis into a percentage (÷$\displaystyle 1000$ × $\displaystyle 100$); it is not a separate chemical fact to memorise.The difference that actually matters — which compounds each method can be used on.Dumas' method converts all the nitrogen in the compound to \(\displaystyle N_2\) gas by oxidation–reduction, so it works for every nitrogen-containing compound, including nitro compounds, azo compounds, and compounds with nitrogen in a ring (such as pyridine).Kjeldahl's method depends on the nitrogen being convertible to ammonium sulfate by hot concentrated \(\displaystyle H_2SO_4\). Nitrogen that is directly bonded to another nitrogen or to oxygen (as in nitro, azo, or diazo groups) or is locked in an aromatic ring (as in pyridine) does not get reduced to ammonium sulfate under these conditions, so Kjeldahl's method fails for these compounds and gives a low result, while Dumas' method still succeeds.Answer: Dumas' method oxidises the compound with \(\displaystyle CuO\) in a \(\displaystyle CO_2\) atmosphere, reduces any nitrogen oxides back to \(\displaystyle N_2\) over hot copper, and measures the volume \(\displaystyle V\) of \(\displaystyle N_2\) collected over \(\displaystyle KOH\): \(\displaystyle \%N = \dfrac{28\times V\times100}{22400\times m}\). Kjeldahl's method digests the compound with conc. \(\displaystyle H_2SO_4\) to \(\displaystyle (NH_4)_2SO_4\), liberates \(\displaystyle NH_3\) with excess \(\displaystyle NaOH\), absorbs it in a known excess of standard acid, and finds the amount consumed by back-titration: \(\displaystyle \%N = \dfrac{1.4\times N\times V}{m}\). Dumas measures nitrogen as a gas and works for every nitrogen compound (nitro, azo, pyridine-type included); Kjeldahl measures nitrogen by acid–base titration and fails for compounds whose nitrogen is not converted to ammonium sulfate, i.e. nitro, azo, diazo compounds and ring nitrogen such as in pyridine.

Halogens, sulphur and phosphorus are all estimated the same way: convert the element into one particular insoluble salt, weigh that salt, and scale back to find what fraction of the original compound was that element. This is the Carius method (for halogens and sulphur) and a closely related oxidation-and-precipitation method for phosphorus. In every case you need two masses — the mass of organic compound you started with, and the mass of the final weighable salt — because the percentage formula is built entirely from that ratio and the known molar mass of the salt.1. Estimation of halogens (Carius method)A known mass of the organic compound is heated with fuming nitric acid, \(\displaystyle \text{HNO}_3 \), in the presence of silver nitrate, \(\displaystyle \text{AgNO}_3 \), inside a sealed hard-glass tube (the Carius tube). The carbon and hydrogen in the compound are oxidised away as carbon dioxide and water, while the halogen present is converted straight into the insoluble silver halide:\[\text{Organic compound} + \text{HNO}_3 \xrightarrow{\text{AgNO}_3,\ \Delta} \text{CO}_2 + \text{H}_2\text{O} + \text{AgX}\ (\text{precipitate}) \]Here \(\displaystyle X \) stands for the halogen (Cl, Br or I) present in the compound. This equation is not meant to be balanced atom-for-atom the way a normal reaction is, because "organic compound" is not one fixed formula — it is a bookkeeping statement of what happens to each type of atom: carbon and hydrogen end up oxidised, and the halogen ends up locked into \(\displaystyle \text{AgX} \), a solid that will not redissolve in the acidic solution.The silver halide precipitate is filtered, washed, dried and weighed. Let \(\displaystyle m \) be the mass of organic compound taken and \(\displaystyle m_1 \) be the mass of \(\displaystyle \text{AgX} \) obtained. Since every mole of \(\displaystyle \text{AgX} \) contains exactly one mole of the halogen, the mass of halogen inside \(\displaystyle m_1 \) grams of \(\displaystyle \text{AgX} \) is found from the fraction of \(\displaystyle \text{AgX} \)'s molar mass that belongs to \(\displaystyle X \):\[\%\ \text{of halogen} = \frac{\text{atomic mass of } X}{\text{molar mass of AgX}} \times \frac{m_1}{m} \times 100 \]The step people get wrong here is treating \(\displaystyle m_1 \), the mass of AgX, as if it were the mass of halogen itself. It is not — \(\displaystyle \text{AgX} \) also contains the mass of silver, so you must always multiply by the atomic-mass fraction \(\displaystyle \dfrac{\text{atomic mass of }X}{\text{molar mass of AgX}} \) before you can call the number "mass of halogen."Using atomic masses \(\displaystyle \text{Ag} = 108,\ \text{Cl} = 35.5,\ \text{Br} = 80,\ \text{I} = 127 \):\[\%\ \text{Cl} = \frac{35.5}{143.5}\times\frac{m_1}{m}\times100,\quad \%\ \text{Br} = \frac{80}{188}\times\frac{m_1}{m}\times100,\quad \%\ \text{I} = \frac{127}{235}\times\frac{m_1}{m}\times100 \](the denominators $\displaystyle 143.5$, $\displaystyle 188$ and $\displaystyle 235$ are the molar masses of \(\displaystyle \text{AgCl} \), \(\displaystyle \text{AgBr} \) and \(\displaystyle \text{AgI} \) respectively).2. Estimation of sulphur (Carius method)The same Carius-tube procedure is used: a known mass of the compound is heated with fuming nitric acid, but this time the sulphur present is oxidised all the way to sulphuric acid:\[\text{S (in compound)} + \text{HNO}_3 \xrightarrow{\Delta} \text{H}_2\text{SO}_4 \]The resulting sulphuric acid is then precipitated as barium sulphate by adding excess barium chloride solution:\[\text{H}_2\text{SO}_4 + \text{BaCl}_2 \longrightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \]Checking this balances: $\displaystyle 1$ Ba, $\displaystyle 1$ S and $\displaystyle 4$ O appear on each side, and the $\displaystyle 2$ chlorines on the left reappear as $\displaystyle 2$ HCl on the right — so the coefficient $\displaystyle 2$ in front of HCl is what makes it balance, not 1.The white \(\displaystyle \text{BaSO}_4 \) precipitate is filtered, washed, dried and weighed. With \(\displaystyle m \) = mass of compound and \(\displaystyle m_1 \) = mass of \(\displaystyle \text{BaSO}_4 \) formed, and using \(\displaystyle \text{Ba}=137,\ \text{S}=32,\ \text{O}=16 \) so that the molar mass of \(\displaystyle \text{BaSO}_4 = 137+32+64 = 233 \):\[\%\ \text{of sulphur} = \frac{32}{233}\times\frac{m_1}{m}\times100 \]The same aside applies again: \(\displaystyle m_1 \) is the mass of the whole \(\displaystyle \text{BaSO}_4 \) salt, not the mass of sulphur — the factor \(\displaystyle \dfrac{32}{233} \) is what strips out the barium and oxygen to leave just the sulphur's share.3. Estimation of phosphorusA known mass of the organic compound is heated with fuming nitric acid, which oxidises the phosphorus present to phosphoric acid:\[\text{P (in compound)} + \text{HNO}_3 \xrightarrow{\Delta} \text{H}_3\text{PO}_4 \]This phosphoric acid is precipitated as magnesium ammonium phosphate by adding magnesia mixture (a solution of \(\displaystyle \text{MgCl}_2 \) and \(\displaystyle \text{NH}_4\text{OH} \)):\[\text{MgCl}_2 + \text{NH}_3 + \text{H}_3\text{PO}_4 \longrightarrow \text{MgNH}_4\text{PO}_4\downarrow + 2\text{HCl} \](Balance check: $\displaystyle 1$ Mg and $\displaystyle 1$ P on each side; hydrogens total \(\displaystyle 3+3=6 \) on the left and \(\displaystyle 4+2=6 \) on the right; the $\displaystyle 2$ chlorines on the left match the $\displaystyle 2$ in \(\displaystyle 2\text{HCl} \).)The \(\displaystyle \text{MgNH}_4\text{PO}_4 \) precipitate, once filtered and washed, is ignited (strongly heated), which converts it to magnesium pyrophosphate, \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) — this is the compound actually weighed, because \(\displaystyle \text{MgNH}_4\text{PO}_4 \) itself is not stable enough on heating in air to give a reliable mass:\[2\text{MgNH}_4\text{PO}_4 \xrightarrow{\Delta} \text{Mg}_2\text{P}_2\text{O}_7 + 2\text{NH}_3\uparrow + \text{H}_2\text{O} \](Balance check: Mg \(\displaystyle 2=2\); N \(\displaystyle 2=2\); P \(\displaystyle 2=2\); H on the left \(\displaystyle 2\times4=8\), on the right \(\displaystyle 2\times3 + 2 = 8\); O on the left \(\displaystyle 2\times4=8\), on the right \(\displaystyle 7+1=8\).)With \(\displaystyle m \) = mass of compound and \(\displaystyle m_1 \) = mass of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) finally weighed, and using \(\displaystyle \text{Mg}=24,\ \text{P}=31,\ \text{O}=16 \), so molar mass of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 = 2(24)+2(31)+7(16) = 48+62+112 = 222 \), and noting that each mole of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) carries two phosphorus atoms:\[\%\ \text{of phosphorus} = \frac{2\times31}{222}\times\frac{m_1}{m}\times100 = \frac{62}{222}\times\frac{m_1}{m}\times100 \]Here the mistake to avoid is using only one atomic mass of phosphorus ($\displaystyle 31$) in the numerator instead of \(\displaystyle 2\times31=62\) — the weighed salt \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) has two phosphorus atoms per formula unit, so leaving out that factor of $\displaystyle 2$ would understate the phosphorus content by half.Answer: In all three estimations, a known mass \(\displaystyle m \) of the organic compound is oxidised (heated with fuming \(\displaystyle \text{HNO}_3 \), for halogens in the presence of \(\displaystyle \text{AgNO}_3 \)) so that the element of interest is converted into one specific insoluble, weighable salt of mass \(\displaystyle m_1 \): the halogen as \(\displaystyle \text{AgX} \), sulphur as \(\displaystyle \text{BaSO}_4 \) (via \(\displaystyle \text{H}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \)), and phosphorus as \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) (via \(\displaystyle \text{H}_3\text{PO}_4 \rightarrow \text{MgNH}_4\text{PO}_4\downarrow \xrightarrow{\Delta} \text{Mg}_2\text{P}_2\text{O}_7 \)). The percentage of each element is then \(\displaystyle \%\ \text{element} = \dfrac{\text{mass of element in one mole of the salt}}{\text{molar mass of the salt}} \times \dfrac{m_1}{m} \times 100 \), giving \(\displaystyle \%\text{X} = \dfrac{\text{at. mass of X}}{\text{molar mass of AgX}}\times\dfrac{m_1}{m}\times100 \), \(\displaystyle \%\text{S} = \dfrac{32}{233}\times\dfrac{m_1}{m}\times100 \), and \(\displaystyle \%\text{P} = \dfrac{62}{222}\times\dfrac{m_1}{m}\times100 \).

Paper chromatography separates a mixture because its components split differently between two phases that do not mix — a film of water held inside the paper (stationary phase) and a solvent that flows through the paper (mobile phase).Chromatography as a family of techniques works because different components of a mixture get distributed unequally between a stationary phase and a mobile phase: a component that clings more to the stationary phase lags behind, and a component that dissolves more readily in the mobile phase races ahead. Paper chromatography is the specific technique where this distribution happens by partition, not adsorption — the mixture keeps re-dissolving into and out of two liquid phases rather than sticking to a solid surface.The two phases in paper chromatography How the separation is set upA small spot of the sample solution is applied near one end of a paper strip, on a line called the origin, and allowed to dry. The paper is then suspended vertically inside a closed chamber with the lower edge dipping into the solvent, keeping the origin spot above the solvent surface so the spot itself never touches the liquid directly. As the solvent rises up the dry paper by capillary action, it reaches the spot and carries the dissolved components upward along with it. This whole process of letting the solvent travel up the paper is called developing the chromatogram.Why the components separateAs the mobile phase moves through the region of the spot, each component of the mixture continuously partitions itself between the stationary water film and the moving solvent — dissolving into the mobile phase, being carried a little further, then re-partitioning back toward the stationary phase, over and over as the solvent front advances. A component that is more soluble in the mobile solvent (relative to how strongly it is held by the stationary water) spends more of its time in the moving phase and travels further up the paper in a given time; a component that partitions more strongly into the stationary water film is held back and travels only a short distance. Because different substances have different relative solubilities in the two phases, they end up travelling different distances in the same time — this is what separates one spot on the origin line into several distinct spots at different heights.After the solvent front has moved a convenient distance, the paper is removed and dried, and the spots are located (directly if coloured, or by spraying with a developing reagent, or under UV light if fluorescent). Each spot's position is described using the retardation factor, \(\displaystyle R_f \):\[R_f = \dfrac{\text{distance moved by the substance from the origin}}{\text{distance moved by the solvent front from the origin}} \]Here "distance moved by the substance" is measured to the centre of its spot, and "distance moved by the solvent front" is measured to the farthest point the mobile phase reached — both measured from the same origin line, along the same straight path up the paper. Because \(\displaystyle R_f \) is a ratio of two distances measured under the same run, it is a fixed, characteristic property of a given substance for a given solvent system, temperature, and type of paper — it does not depend on how far that particular run happened to develop. This is the property that lets an unknown spot be identified: its \(\displaystyle R_f \) is compared against the \(\displaystyle R_f \) of a known reference substance run under the same conditions.A step people get wrong: paper chromatography is a partition technique, so the separating agent is the invisible water trapped in the cellulose, not the visible paper fibres — describing the paper itself as the "stationary phase" misses the actual principle being tested.Answer: Paper chromatography works on the principle of partition (differential distribution) of the components of a mixture between two phases — a stationary phase of water held in the pores of cellulose paper, and a mobile phase of an organic solvent that flows past it by capillary action. A component that is more soluble in the mobile phase moves further; one that is held more strongly by the stationary water film moves less. This unequal movement separates the original mixed spot into distinct spots, each located by its retardation factor \(\displaystyle R_f = \dfrac{\text{distance moved by substance}}{\text{distance moved by solvent}} \), which is characteristic of that substance under fixed conditions and is used to identify it.

Silver nitrate does not only test for halogens — sulphur and nitrogen in the same fusion mixture also give precipitates with it, and nitric acid's job is to destroy those before the test even starts.When an organic compound is fused with sodium metal to make the sodium extract (Lassaigne's test), every heteroatom in the molecule gets converted to a sodium salt in the same solution: All of these can sit together in the one extract. If silver nitrate, \(\displaystyle \text{AgNO}_3\), were added straight to this mixture, it would not react with the halide ion alone — sulphide and cyanide ions are also precipitated by \(\displaystyle \text{Ag}^+\):\[\text{Na}_2\text{S} + 2\text{AgNO}_3 \rightarrow \text{Ag}_2\text{S}\!\downarrow + 2\text{NaNO}_3 \] (silver sulphide, a black precipitate)\[\text{NaCN} + \text{AgNO}_3 \rightarrow \text{AgCN}\!\downarrow + \text{NaNO}_3 \] (silver cyanide, a white precipitate, and it can also form soluble complexes with excess CN⁻)Both equations are already balanced: count sodium, silver, nitrate, and the S or CN group on each side and they match one-for-one. Notice the black \(\displaystyle \text{Ag}_2\text{S}\) and the white \(\displaystyle \text{AgCN}\) would appear right alongside the halide precipitate — you would have no way to tell, by colour, whether the precipitate you see is telling you about a halogen or about leftover sulphur/nitrogen from the fusion.This is the step people skip: they treat "add dilute HNO₃" as a generic acidifying habit, when its real, specific job here is to remove sulphide and cyanide before silver nitrate ever touches the solution.Boiling the sodium extract with dilute nitric acid decomposes both interfering ions and drives the products off as gases, so they leave the test tube instead of sitting in solution:\[\text{Na}_2\text{S} + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{S}\!\uparrow \] (hydrogen sulphide gas escapes)\[\text{NaCN} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{HCN}\!\uparrow \] (hydrogen cyanide gas escapes)Check the balance on each: sodium, nitrogen, oxygen and hydrogen atoms match on both sides in the first equation ($\displaystyle 2$ Na, $\displaystyle 2$ N, $\displaystyle 6$ O, $\displaystyle 2$ H, $\displaystyle 1$ S throughout), and in the second ($\displaystyle 1$ Na, $\displaystyle 2$ N, $\displaystyle 3$ O, $\displaystyle 1$ H, $\displaystyle 1$ C throughout) — one \(\displaystyle \text{HNO}_3\) is exactly enough to protonate one \(\displaystyle \text{CN}^-\) and free one \(\displaystyle \text{NO}_3^-\). Any carbonate picked up from atmospheric \(\displaystyle \text{CO}_2\) during fusion is destroyed the same way, \[\text{Na}_2\text{CO}_3 + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{O} + \text{CO}_2\!\uparrow, \] so that carbonate cannot form a silver precipitate either.The halide ion itself is untouched by this treatment — \(\displaystyle \text{NaX}\) and \(\displaystyle \text{HNO}_3\) do not react to remove \(\displaystyle X^-\), because the hydrohalic acids stay fully ionised in water rather than escaping as gas under gentle boiling. So after boiling, the solution still carries all its original halide ion, but no more sulphide, cyanide, or carbonate. Only now is silver nitrate added:\[\text{NaX} + \text{AgNO}_3 \rightarrow \text{AgX}\!\downarrow + \text{NaNO}_3 \]giving a precipitate that is unambiguously the silver halide — white and curdy for \(\displaystyle \text{AgCl}\) (chloride), pale yellow for \(\displaystyle \text{AgBr}\) (bromide), or yellow for \(\displaystyle \text{AgI}\) (iodide).A second reason sits underneath the same acidification, and it is worth naming separately: silver halides are insoluble even in dilute nitric acid, while silver carbonate, silver phosphate and similar silver salts of other anions are not. So working in acidic medium is a built-in filter — even if some other anion had survived, any precipitate it formed would simply redissolve in the acid, while a true halide precipitate would not. That is why the medium for this test is kept acidic with \(\displaystyle \text{HNO}_3\) throughout, not just at the start.**Answer: Nitric acid is added first because the sodium extract can also contain sodium sulphide and sodium cyanide (from sulphur or nitrogen in the compound), and these give their own precipitates with silver nitrate (black \(\displaystyle \text{Ag}_2\text{S}\), white \(\displaystyle \text{AgCN}\)) that would be mistaken for, or mask, the halide test. Boiling with dilute \(\displaystyle \text{HNO}_3\) decomposes \(\displaystyle \text{Na}_2\text{S}\) and \(\displaystyle \text{NaCN}\) into gaseous \(\displaystyle \text{H}_2\text{S}\) and \(\displaystyle \text{HCN}\) (and any carbonate into \(\displaystyle \text{CO}_2\)), which escape from solution, while the halide ion is unaffected. Adding \(\displaystyle \text{AgNO}_3\) after this leaves the silver halide, \(\displaystyle \text{AgX}\), as the only precipitate formed — and because \(\displaystyle \text{AgX}\) (unlike silver salts of carbonate or phosphate) does not dissolve back into dilute nitric acid, the acidic medium also guards against any other silver precipitate being wrongly read as a positive halogen test.

Fusion converts covalently bound N, S and halogens into free ions, because only ions give the simple precipitation and colour tests of qualitative analysis.In an organic compound, nitrogen, sulphur and the halogens are not present as separate \(\displaystyle \text{N}^{3-} \), \(\displaystyle \text{S}^{2-} \) or \(\displaystyle \text{X}^- \) ions — they are covalently bonded to carbon (as in \(\displaystyle \text{C-N} \), \(\displaystyle \text{C-S} \), \(\displaystyle \text{C-X} \) bonds). A covalent bond does not ionise in solution, so none of the standard inorganic spot tests can see these elements directly: This is the step students get wrong: testing the original organic compound straight away with \(\displaystyle \text{AgNO}_3 \) or lead acetate and concluding "no halogen/sulphur present" when in fact the element is present but covalently bound, so it simply cannot react in that form.Why sodium specifically. Metallic sodium is an extremely reactive reducing agent. When the organic compound is fused (strongly heated) with sodium metal, the sodium is reactive enough to break the covalent \(\displaystyle \text{C-N} \), \(\displaystyle \text{C-S} \) and \(\displaystyle \text{C-X} \) bonds and combine directly with these atoms, converting them into simple, water-soluble, ionic sodium salts. This molten mixture, plunged into distilled water, gives the "sodium fusion extract" (Lassaigne's extract), which now contains free ions that behave exactly like an ordinary inorganic salt solution.The fusion reactions (elements combine one-to-one, so each is already balanced by atom count on both sides):For nitrogen, sodium combines with the carbon and nitrogen of the compound to give sodium cyanide: \[\text{Na} + \text{C} + \text{N} \longrightarrow \text{NaCN} \]For sulphur, sodium combines directly with sulphur to give sodium sulphide: \[2\text{Na} + \text{S} \longrightarrow \text{Na}_2\text{S} \]For a halogen (\(\displaystyle \text{X} = \text{Cl}, \text{Br}, \text{I} \)), sodium combines directly with the halogen to give the sodium halide: \[\text{Na} + \text{X} \longrightarrow \text{NaX} \]If both nitrogen and sulphur are present in the same compound, sodium preferentially forms sodium thiocyanate instead of separate cyanide and sulphide: \[\text{Na} + \text{C} + \text{N} + \text{S} \longrightarrow \text{NaSCN} \]This is another common pitfall: when a compound has both N and S, the ordinary Prussian-blue test for nitrogen can fail (because the \(\displaystyle \text{CN}^- \) has been tied up as \(\displaystyle \text{SCN}^- \) instead), which is why a separate confirmatory test is needed for thiocyanate in such cases.Once the fusion extract is boiled and filtered, it contains free \(\displaystyle \text{Na}^+ \), \(\displaystyle \text{CN}^- \), \(\displaystyle \text{S}^{2-} \), \(\displaystyle \text{X}^- \) or \(\displaystyle \text{SCN}^- \) ions in aqueous solution — ordinary inorganic ions that now respond normally to \(\displaystyle \text{AgNO}_3 \) (halides), sodium nitroprusside or lead acetate (sulphide), and the iron(II)/iron(III) Prussian-blue test (cyanide).Answer: Fusion with metallic sodium is done because N, S and halogens exist in an organic compound as covalent bonds, not as free ions, so they cannot give the usual precipitation/colour tests of qualitative inorganic analysis. Reactive sodium metal, on strong heating, breaks these covalent bonds and combines with the atoms to form water-soluble, ionic sodium salts — \(\displaystyle \text{NaCN} \) (\(\displaystyle \text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN} \)) for nitrogen, \(\displaystyle \text{Na}_2\text{S} \) (\(\displaystyle 2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S} \)) for sulphur, \(\displaystyle \text{NaX} \) (\(\displaystyle \text{Na} + \text{X} \rightarrow \text{NaX} \)) for a halogen, or \(\displaystyle \text{NaSCN} \) when both N and S are present. These ionic salts, dissolved in the sodium fusion extract, now give the standard inorganic tests directly.

Sublimation separates a solid that vaporises directly (without melting) from one that does not — and camphor is exactly that kind of solid.To pick a separation technique, ask what physical property differs between the two components. Here that property is volatility of the solid itself: Aside — the mistake to avoid: it is tempting to reach for filtration or a solvent-based separation (like solvent extraction) because the two solids "look" like they should be separated by dissolving one away. But nothing in the question dissolves either component in a liquid; the property being exploited is volatility on heating, which is what makes sublimation the right technique — not solubility.Procedure. Heat the mixture gently in a china dish covered with an inverted funnel (its stem plugged with cotton wool). Camphor sublimes, rises as vapour, and re-solidifies as pure crystals on the cooler inner wall of the funnel, where it can be scraped off. Calcium sulphate, being non-volatile, is left behind as the residue in the dish.Answer: Sublimation. Camphor sublimes on gentle heating (solid → vapour → solid, collected on a cold surface) while the non-volatile calcium sulphate is left behind as residue, so the two are separated.

A mixture boils when the sum of the vapour pressures of its components equals the atmospheric pressure — not when either liquid alone reaches its own boiling point.In steam distillation you have two liquids that do not mix into a true solution and do not react with each other: water, and the organic liquid (which must be steam-volatile and virtually insoluble in water). Because the two are immiscible, each one is free to build up its own vapour pressure above the mixture exactly as if the other were not there. Water contributes its vapour pressure \(\displaystyle p_{\text{water}} \) and the organic liquid contributes its vapour pressure \(\displaystyle p_{\text{organic}} \), and — this is the step that is easy to get wrong — these are not weighted by mole fraction the way they would be in an ideal solution (Raoult's law). Since the two liquids are immiscible, each behaves as a pure substance occupying its own vapour space, so the total vapour pressure above the mixture is simply the sum of the two separate vapour pressures:\[p_{\text{total}} = p_{\text{water}}^{\circ} + p_{\text{organic}}^{\circ} \]Here \(\displaystyle p_{\text{water}}^{\circ} \) is the vapour pressure that pure water would exert at that temperature, and \(\displaystyle p_{\text{organic}}^{\circ} \) is the vapour pressure that the pure organic liquid would exert at that same temperature.A liquid (or a liquid mixture) boils at the temperature where its total vapour pressure becomes equal to the atmospheric pressure:\[p_{\text{total}} = p_{\text{atm}} \]Now compare this with what each liquid would need on its own. Pure water boils when \(\displaystyle p_{\text{water}}^{\circ} \) alone reaches \(\displaystyle p_{\text{atm}} \) (at \(\displaystyle 373\ \text{K} \), i.e. \(\displaystyle 100\,^{\circ}\text{C} \), at $\displaystyle 1$ atm). Pure organic liquid boils when \(\displaystyle p_{\text{organic}}^{\circ} \) alone reaches \(\displaystyle p_{\text{atm}} \), which for most steam-volatile organic liquids happens at a temperature well above \(\displaystyle 100\,^{\circ}\text{C} \).But in the mixture, water is already contributing part of the total pressure. So the organic liquid does not have to supply the full \(\displaystyle p_{\text{atm}} \) by itself — it only has to supply the leftover amount:\[p_{\text{organic}}^{\circ} = p_{\text{atm}} - p_{\text{water}}^{\circ} \]Because \(\displaystyle p_{\text{organic}}^{\circ} \) needed to reach boiling is now smaller than \(\displaystyle p_{\text{atm}} \), it is reached at a lower temperature than the organic liquid's own normal boiling point. In fact, since \(\displaystyle p_{\text{water}}^{\circ} \) is also less than \(\displaystyle p_{\text{atm}} \) below \(\displaystyle 100\,^{\circ}\text{C} \), the two partial contributions add up to \(\displaystyle p_{\text{atm}} \) at some temperature that is below \(\displaystyle 100\,^{\circ}\text{C} \) as well — so the mixture as a whole boils below the boiling point of either pure component. This is exactly why steam distillation is used: it lets a high-boiling organic liquid (or one that would decompose if heated all the way to its own boiling point) be carried over and vaporised at a much lower, safer temperature, together with steam.Answer: In steam distillation the organic liquid and water are immiscible, so each exerts its own vapour pressure independently and the mixture's total vapour pressure is the sum \(\displaystyle p_{\text{total}} = p_{\text{water}}^{\circ} + p_{\text{organic}}^{\circ} \) rather than a mole-fraction-weighted average. The mixture boils as soon as this sum equals the atmospheric pressure, which happens well before either pure liquid's own vapour pressure alone could reach atmospheric pressure. Hence the organic liquid vaporises (distils over with the steam) at a temperature below its normal boiling point.

A white AgCl precipitate needs free Cl⁻ ions available to combine with Ag⁺ — and CCl₄ has none to give up, before or after heating.The silver nitrate test detects halide ions already present in solution: \[\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\downarrow \] Here \(\displaystyle \text{Ag}^+ \) is the silver ion (supplied by silver nitrate, \(\displaystyle \text{AgNO}_3 \)) and \(\displaystyle \text{Cl}^- \) is a chloride ion free in solution. The equation is already balanced — one silver ion combines with one chloride ion to give one formula unit of silver chloride, \(\displaystyle \text{AgCl} \), a white insoluble solid, and the charges balance too: \(\displaystyle +1\) and \(\displaystyle -1\) cancel to a neutral precipitate.This test only works when chlorine is present as a free ion. That happens for an ionic chloride such as \(\displaystyle \text{NaCl} \), which dissociates completely in water, or for a compound that hydrolyses in water to release \(\displaystyle \text{HCl} \), which then ionises: \[\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq) \]In tetrachloromethane, \(\displaystyle \text{CCl}_4 \) (carbon tetrachloride), neither route is open. All four chlorine atoms are joined to the central carbon by ordinary covalent bonds — one C–Cl bond, repeated four times — so there is no ionic \(\displaystyle \text{Cl}^- \) sitting free in the liquid to react with \(\displaystyle \text{Ag}^+ \) directly.The remaining question is whether heating with water could still break those C–Cl bonds and generate \(\displaystyle \text{HCl} \) on the spot. Compare \(\displaystyle \text{CCl}_4 \) with its close relative \(\displaystyle \text{SiCl}_4 \) (silicon tetrachloride), which hydrolyses readily in moist air: \[\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl} \] Checking the balance: Si, $\displaystyle 1$ on each side; Cl, $\displaystyle 4$ on each side; H, \(\displaystyle 4\times2=8\) on the left matches \(\displaystyle 4\) (in \(\displaystyle \text{Si(OH)}_4 \)) \(\displaystyle +\ 4\) (in \(\displaystyle 4\,\text{HCl}\)) \(\displaystyle =8\) on the right; O, $\displaystyle 4$ on the left matches $\displaystyle 4$ (in \(\displaystyle \text{Si(OH)}_4 \)) on the right. In this reaction, water's oxygen lone pair attacks the silicon atom directly, because silicon (period $\displaystyle 3$) has empty valence \(\displaystyle 3d\) orbitals available to briefly hold five groups at once, in a five-coordinate transition state, before a chloride ion leaves and \(\displaystyle \text{HCl} \) is expelled.The trap is assuming carbon can do the same thing just because it carries the same four chlorine atoms as silicon — it cannot. Carbon is a period-$\displaystyle 2$ element: its valence shell is only \(\displaystyle 2s\) and \(\displaystyle 2p\), with no \(\displaystyle 3d\), or any other empty, orbital available. There is no orbital on the carbon atom for water's lone pair to attack in forming that five-coordinate intermediate, so the C–Cl bond in \(\displaystyle \text{CCl}_4 \) has no pathway to break by hydrolysis, no matter how strongly the mixture is heated. (The four bulky Cl atoms crowding the small carbon add steric protection on top of this, but the decisive reason is the missing d-orbital.)Since \(\displaystyle \text{CCl}_4 \) neither ionises nor hydrolyses, heating it with aqueous \(\displaystyle \text{AgNO}_3 \) never produces a free \(\displaystyle \text{Cl}^- \) ion for \(\displaystyle \text{Ag}^+ \) to capture, so the characteristic white precipitate of \(\displaystyle \text{AgCl} \) never forms.Answer: No. \(\displaystyle \text{CCl}_4 \) does not give a white precipitate of \(\displaystyle \text{AgCl} \) with silver nitrate, even on heating, because its C–Cl bonds are covalent, not ionic, and — unlike \(\displaystyle \text{SiCl}_4 \) — carbon has no vacant d-orbitals for water to attack, so \(\displaystyle \text{CCl}_4 \) cannot be hydrolysed and no free \(\displaystyle \text{Cl}^- \) ion is ever released to react with \(\displaystyle \text{Ag}^+ \).

Potassium hydroxide is chosen because it reacts with carbon dioxide completely and irreversibly — the "how much extra it weighs" trick for measuring carbon only works if every last molecule of \(\displaystyle \text{CO}_2 \) actually gets caught.The setup. In estimating carbon (Liebig's method), the organic compound is burnt in a stream of oxygen. Every carbon atom in the compound ends up as carbon dioxide gas, \(\displaystyle \text{CO}_2 \), and every hydrogen atom ends up as water vapour, \(\displaystyle \text{H}_2\text{O} \). These two gases are passed, one after the other, through separate weighed absorption tubes: water vapour is caught first (usually in anhydrous \(\displaystyle \text{CaCl}_2 \)), and the remaining gas stream — now carrying only \(\displaystyle \text{CO}_2 \) — is passed into a tube containing potassium hydroxide solution.The whole method depends on one idea: mass gained by the absorbing tube = mass of the gas it absorbed. So whatever absorbs \(\displaystyle \text{CO}_2 \) must take up all of it, not just some, or the measured mass undercounts the carbon in the sample.Why KOH specifically. Carbon dioxide is an acidic oxide — dissolved in water it behaves as a weak acid. Potassium hydroxide is a strong base. An acid meeting a strong base does not partially react and stop; it reacts until one of them is used up. Since KOH is kept present in excess in the tube, every molecule of \(\displaystyle \text{CO}_2 \) that passes through finds a hydroxide ion waiting for it, and the reaction runs to completion:\[2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]Here \(\displaystyle \text{KOH} \) is potassium hydroxide, \(\displaystyle \text{CO}_2 \) is carbon dioxide gas, \(\displaystyle \text{K}_2\text{CO}_3 \) is potassium carbonate (the salt formed), and \(\displaystyle \text{H}_2\text{O} \) is water.Balancing it: start by counting potassium. The product \(\displaystyle \text{K}_2\text{CO}_3 \) carries $\displaystyle 2$ potassium atoms, so the left side needs $\displaystyle 2$ KOH to match. Carbon: one \(\displaystyle \text{CO}_2 \) gives the one carbon in \(\displaystyle \text{K}_2\text{CO}_3 \) — already even. Hydrogen: $\displaystyle 2$ KOH supplies $\displaystyle 2$ hydrogen atoms, and the right side's \(\displaystyle \text{H}_2\text{O} \) also has $\displaystyle 2$ — matched. Oxygen as a check: left side has \(\displaystyle 2(1) \) from KOH plus \(\displaystyle 2 \) from \(\displaystyle \text{CO}_2 \), giving $\displaystyle 4$; right side has \(\displaystyle 3 \) from \(\displaystyle \text{K}_2\text{CO}_3 \) plus \(\displaystyle 1 \) from \(\displaystyle \text{H}_2\text{O} \), also 4. Every atom balances, so the equation is correct as written.Why this matters for the measurement, not just the chemistry. If a weaker or more volatile absorbent were used instead, some \(\displaystyle \text{CO}_2 \) could pass through unabsorbed, or the absorption could be an equilibrium that doesn't go to completion — either way the tube would gain less mass than the true amount of \(\displaystyle \text{CO}_2 \) evolved, and the calculated percentage of carbon would come out too low. Because the acid–base reaction between \(\displaystyle \text{CO}_2 \) and KOH goes essentially $\displaystyle 100$% to \(\displaystyle \text{K}_2\text{CO}_3 \), the mass gained by the KOH tube is trusted as exactly the mass of \(\displaystyle \text{CO}_2 \) produced, which is then used to back-calculate the mass, and hence percentage, of carbon in the original organic compound.Answer: A solution of KOH is used because it is a strong base that reacts completely and quantitatively with the acidic gas \(\displaystyle \text{CO}_2 \), via \(\displaystyle 2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \), so the increase in mass of the KOH tube gives exactly the mass of carbon dioxide evolved, which is needed to calculate the percentage of carbon in the compound.

Sodium fusion extract (Lassaigne's extract) turns any sulphur in the compound into sodium sulphide, and the test for that sulphide only works if the precipitate you get back is unambiguously lead sulphide — sulphuric acid ruins that by making its own precipitate.Where the sulphide comes from. When the organic compound is fused with sodium metal, any sulphur present is converted to sodium sulphide, which dissolves in water along with sodium cyanide and sodium halide (if present) to give the sodium fusion extract:\[\text{Na} + \text{C} + \text{N} + \text{S} \;\xrightarrow{\text{fusion}}\; \text{NaCN}, \; \text{Na}_2\text{S}, \; \text{NaX} \]The test itself. To confirm sulphur, the extract is acidified and treated with lead acetate solution, \(\displaystyle (\text{CH}_3\text{COO})_2\text{Pb} \) — "lead acetate" is lead(II) acetate, \(\displaystyle \text{Pb}^{2+} \) paired with two acetate ions. If sulphide is present, it forms a black precipitate of lead sulphide, \(\displaystyle \text{PbS} \):\[\text{Na}_2\text{S} + (\text{CH}_3\text{COO})_2\text{Pb} \;\longrightarrow\; \text{PbS}\downarrow \;(\text{black}) + 2\,\text{CH}_3\text{COONa} \]Check the balance: $\displaystyle 2$ Na on the left (in \(\displaystyle \text{Na}_2\text{S} \)) match $\displaystyle 2$ Na on the right (in \(\displaystyle 2\,\text{CH}_3\text{COONa} \)); $\displaystyle 1$ Pb matches $\displaystyle 1$ Pb; $\displaystyle 1$ S matches the $\displaystyle 1$ S locked in PbS; the two acetate groups that started on lead end up on the two sodium atoms. Nothing needs any other coefficient.Why acetic acid, and not sulphuric acid, is used to acidify. Acetic acid, \(\displaystyle \text{CH}_3\text{COOH} \), is a weak acid that does not react with lead acetate at all — it simply supplies the acidic medium needed for the test without introducing any new precipitating anion. Sulphuric acid, \(\displaystyle \text{H}_2\text{SO}_4 \), does the opposite: it reacts with the lead acetate itself, because sulphate ions form an insoluble salt with \(\displaystyle \text{Pb}^{2+} \):\[\text{H}_2\text{SO}_4 + (\text{CH}_3\text{COO})_2\text{Pb} \;\longrightarrow\; \text{PbSO}_4\downarrow \;(\text{white, insoluble}) + 2\,\text{CH}_3\text{COOH} \]Balance check: $\displaystyle 2$ H on the left end up as the $\displaystyle 2$ H's carried by the two \(\displaystyle \text{CH}_3\text{COOH} \) molecules on the right; the one sulphate group goes onto lead; $\displaystyle 1$ Pb matches $\displaystyle 1$ Pb. This is balanced with no extra coefficients needed elsewhere.The aside people miss: this is not just "sulphuric acid also happens to react" — that reaction produces a precipitate (\(\displaystyle \text{PbSO}_4\), white) every single time you add sulphuric acid to lead acetate, whether or not the compound being tested contains sulphur. That precipitate would appear over the black \(\displaystyle \text{PbS} \) (if any sulphide were present) and mask it, or appear by itself and be mistaken for a positive result even when there is no sulphur in the sample at all — a false positive. Acetic acid gives no such precipitate with lead acetate, so any precipitate that does form when the extract is added must be \(\displaystyle \text{PbS} \), and only \(\displaystyle \text{PbS} \), making the test reliable.Answer: Acetic acid is used because it does not react with lead acetate and so cannot form any precipitate of its own; sulphuric acid, in contrast, reacts with lead acetate to give a white precipitate of lead sulphate \(\displaystyle (\text{PbSO}_4) \), which would mask the black lead sulphide \(\displaystyle (\text{PbS}) \) precipitate that is meant to indicate sulphur, or falsely mimic a positive test even in its absence — so sulphuric acid would make the test unreliable.

Every carbon atom in the sample turns into one molecule of \(\displaystyle CO_2 \), and every two hydrogen atoms turn into one molecule of \(\displaystyle H_2O \) — combustion analysis is just carrying the percentage composition through mole ratios into product masses.Step $\displaystyle 1$: Work out what fraction of the sample is carbon and what fraction is hydrogen.The compound is $\displaystyle 69$% carbon and $\displaystyle 4.8$% hydrogen by mass, so the rest is oxygen: \[\%O = 100 - 69 - 4.8 = 26.2\% \] Oxygen doesn't matter for this question — combustion adds its own \(\displaystyle O_2 \), so only the carbon and hydrogen masses carry through to the products.Step $\displaystyle 2$: Find the actual mass of C and H burned, in the $\displaystyle 0.20$ g sample.A common slip here is using $\displaystyle 69$ g and $\displaystyle 4.8$ g (the amounts in $\displaystyle 100$ g of compound) instead of scaling down to the $\displaystyle 0.20$ g actually burned. The percentages apply to this sample, so: \[m(C) = 69\% \times 0.20\ g = 0.69 \times 0.20 = 0.138\ g \] \[m(H) = 4.8\% \times 0.20\ g = 0.048 \times 0.20 = 0.0096\ g \]Step $\displaystyle 3$: Convert each mass to moles of atoms, using molar mass = mass ÷ (mass per mole).Molar mass of C is \(\displaystyle 12\ g/mol \) and of H is \(\displaystyle 1\ g/mol \): \[n(C) = \frac{0.138\ g}{12\ g/mol} = 0.0115\ mol \] \[n(H\ atoms) = \frac{0.0096\ g}{1\ g/mol} = 0.0096\ mol \]Step $\displaystyle 4$: Turn moles of atoms into moles of product, using the balanced combustion equations.Carbon burns one atom to one molecule of carbon dioxide — already balanced as written, one C on each side: \[C + O_2 \rightarrow CO_2 \] So \(\displaystyle 1 \) mol C gives \(\displaystyle 1 \) mol \(\displaystyle CO_2 \): \[n(CO_2) = n(C) = 0.0115\ mol \]Hydrogen burns two atoms to one molecule of water. Writing it as a balanced equation makes the $\displaystyle 2$:$\displaystyle 1$ ratio explicit — $\displaystyle 4$ H atoms and one \(\displaystyle O_2 \) ($\displaystyle 2$ O atoms) on the left match $\displaystyle 2$ molecules of \(\displaystyle H_2O \) ($\displaystyle 4$ H, $\displaystyle 2$ O) on the right: \[4H + O_2 \rightarrow 2H_2O \] This is the step people get wrong: it is tempting to treat H and \(\displaystyle H_2O \) as $\displaystyle 1$:$\displaystyle 1$, the way C and \(\displaystyle CO_2 \) are. But every water molecule needs two hydrogen atoms, so: \[n(H_2O) = \frac{n(H\ atoms)}{2} = \frac{0.0096}{2} = 0.0048\ mol \]Step $\displaystyle 5$: Convert moles of product back to mass, using mass = moles × molar mass.Molar mass of \(\displaystyle CO_2 = 12 + 2(16) = 44\ g/mol \). Molar mass of \(\displaystyle H_2O = 2(1) + 16 = 18\ g/mol \). \[m(CO_2) = 0.0115\ mol \times 44\ g/mol = 0.506\ g \] \[m(H_2O) = 0.0048\ mol \times 18\ g/mol = 0.0864\ g \]Answer: Combustion of the $\displaystyle 0.20$ g sample produces $\displaystyle 0.506$ g of carbon dioxide and $\displaystyle 0.0864$ g of water.

In Kjeldahl's method, the nitrogen escapes as ammonia and gets trapped by a KNOWN excess of acid — what you actually measure is not the ammonia itself, but how much of that acid the ammonia used up.The ammonia liberated from the compound is passed into a measured volume of sulphuric acid — more acid than is needed, so all the ammonia is absorbed: \[2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \] (ammonia + sulphuric acid → ammonium sulphate). This is already balanced: sulphuric acid has two acidic hydrogens, so it takes two molecules of ammonia to use them both up.Whatever acid is left over (the "residual acid") is then found by titrating it against a standard sodium hydroxide solution: \[H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \] (sulphuric acid + sodium hydroxide → sodium sulphate + water). Again balanced two-for-two, for the same reason.The step everyone trips on here: sulphuric acid is diprotic, so its molarity is NOT its normality. You cannot just subtract "volume of acid minus volume of base" when the acid is diprotic and the base is monoprotic — you have to convert both to milliequivalents (meq) first.Total acid taken, before any of it touches the ammonia: \[\text{mmol } H_2SO_4 = M \times V = 0.5\ \text{mol L}^{-1} \times 50\ \text{mL} = 25\ \text{mmol} \] Each \(\displaystyle H_2SO_4 \) molecule supplies $\displaystyle 2$ acidic \(\displaystyle H^+ \), so its normality is \(\displaystyle 2 \times 0.5 = 1.0\ \text{N} \), and \[\text{total meq } H_2SO_4 = 1.0\ \text{N} \times 50\ \text{mL} = 50\ \text{meq} \]Acid left over after the ammonia had its share, found from the NaOH titration: \[\text{mmol NaOH used} = 0.5\ \text{mol L}^{-1} \times 60\ \text{mL} = 30\ \text{mmol} \] NaOH is monoprotic, so its normality equals its molarity, and \(\displaystyle 30\ \text{mmol NaOH} = 30\ \text{meq NaOH} \). At the point of neutralisation, meq of acid reacting = meq of base added, so the residual (unreacted) acid was also \[30\ \text{meq } H_2SO_4 \]Acid actually consumed by the ammonia is what's left after removing the residual part from the total: \[50\ \text{meq (total)} - 30\ \text{meq (residual)} = 20\ \text{meq } H_2SO_4 \text{ reacted with } NH_3 \]One equivalent of acid always neutralises exactly one equivalent of base — that equivalence is what turns "acid consumed" into "ammonia present."\(\displaystyle NH_3 \) is a monoacidic base (it accepts a single \(\displaystyle H^+ \) to become \(\displaystyle NH_4^+ \)), so its milliequivalents equal its millimoles: \[n(NH_3) = 20\ \text{meq} = 20\ \text{mmol} \] Every \(\displaystyle NH_3 \) molecule carries exactly one nitrogen atom, so the moles of nitrogen equal the moles of ammonia: \[n(N) = 0.020\ \text{mol}, \qquad \text{mass of } N = 0.020\ \text{mol} \times 14\ \text{g mol}^{-1} = 0.28\ \text{g} \]Percentage composition is always (mass of the element ÷ mass of the WHOLE ORIGINAL SAMPLE) × $\displaystyle 100$ — here the $\displaystyle 0.50$ g of organic compound you started with, not the mass of acid or ammonia.Using the formula \(\displaystyle \%N = \dfrac{\text{mass of N}}{\text{mass of compound}} \times 100 \), where "mass of compound" is the weighed sample \(\displaystyle W \): \[\%N = \frac{0.28\ \text{g}}{0.50\ \text{g}} \times 100 = 56\% \]As a check, the same result follows from the compact Kjeldahl relation \(\displaystyle \%N = \dfrac{1.4 \times \text{meq}(NH_3)}{W(\text{g})} \), where the constant \(\displaystyle 1.4 = \dfrac{14}{10} \) folds together the atomic mass of nitrogen ($\displaystyle 14$) and the mmol-to-percentage conversion: \[\%N = \frac{1.4 \times 20}{0.50} = 56\% \]Both routes agree, confirming the residual-acid bookkeeping was done correctly.Answer: The compound contains $\displaystyle 56$% nitrogen by mass.

Carius chlorine estimation converts every atom of chlorine in the sample into silver chloride, so you scale by the fraction of AgCl's mass that is chlorine, then compare that chlorine mass to the mass of compound you started with.Step $\displaystyle 1$ — Find what fraction of AgCl's mass is chlorine.Silver chloride is \(\displaystyle \text{AgCl} \), one silver atom bonded to one chlorine atom. Its molar mass is\[M(\text{AgCl}) = 108 \ (\text{Ag}) + 35.5 \ (\text{Cl}) = 143.5\ \text{g mol}^{-1} \]Since every mole of AgCl carries exactly one mole of Cl, the mass fraction of chlorine locked inside AgCl is\[\frac{\text{mass of Cl}}{\text{mass of AgCl}} = \frac{35.5}{143.5} \]This is the step people skip past: you are not weighing chlorine directly — you are weighing AgCl and then asking "how much of that weight is the chlorine part?"Step $\displaystyle 2$ — Convert the AgCl mass obtained into a mass of chlorine.The Carius estimation gave $\displaystyle 0.5740$ g of AgCl from the sample. Using the fraction from Step $\displaystyle 1$,\[\text{mass of Cl} = 0.5740\ \text{g} \times \frac{35.5}{143.5} \]Work the arithmetic in one line:\[0.5740 \times 35.5 = 20.377 \] \[\text{mass of Cl} = \frac{20.377}{143.5} = 0.1420\ \text{g} \]Step $\displaystyle 3$ — Turn that into a percentage of the original compound.Percentage composition always means: (mass of the element) divided by (mass of the whole sample you weighed out), not by the mass of AgCl and not by any intermediate quantity. Here the sample taken was $\displaystyle 0.3780$ g of the organic chloro compound.\[\% \text{Cl} = \frac{\text{mass of Cl}}{\text{mass of compound}} \times 100 = \frac{0.1420}{0.3780} \times 100 \]\[\% \text{Cl} = \frac{14.20}{3.780} = 37.566\ldots \approx 37.57\% \]So, packing both steps into a single working formula (the one worth remembering for every Carius chlorine problem):\[\% \text{Cl} = \frac{35.5}{143.5} \times \frac{\text{mass of AgCl}}{\text{mass of compound}} \times 100 \]Answer: The compound contains $\displaystyle 37.57$% chlorine by mass.

In the Carius method, all the sulphur in the compound ends up locked inside a fixed amount of \(\displaystyle \text{BaSO}_4\), so you find the mass of sulphur by scaling down the mass of \(\displaystyle \text{BaSO}_4\) by the fraction of its molar mass that is sulphur.What actually happens in the tube. The organic sulphur compound is heated with fuming nitric acid in a sealed Carius tube. The nitric acid oxidises every sulphur atom in the compound to sulphate ion, \(\displaystyle \text{SO}_4^{2-}\). This sulphate is then precipitated completely as barium sulphate (a white solid) by adding excess barium chloride solution to the solution:\[\text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \]Check the balance: $\displaystyle 1$ barium atom and $\displaystyle 2$ chlorine atoms on each side, $\displaystyle 2$ hydrogen atoms on each side, and $\displaystyle 1$ sulphur with $\displaystyle 4$ oxygens (as the sulphate group) carried straight through onto \(\displaystyle \text{BaSO}_4\) — nothing needs a coefficient beyond the $\displaystyle 2$ already in front of HCl. Every sulphur atom that started in the organic compound is now sitting inside one formula unit of \(\displaystyle \text{BaSO}_4\), so weighing the barium sulphate precipitate tells you exactly how much sulphur was present.Setting up the mass relationship. One mole of \(\displaystyle \text{BaSO}_4\) (molar mass \(\displaystyle 137 + 32 + 4(16) = 233\ \text{g mol}^{-1}\), using \(\displaystyle \text{Ba} = 137\), \(\displaystyle \text{S} = 32\), \(\displaystyle \text{O} = 16\)) contains exactly one mole of S (atomic mass \(\displaystyle 32\ \text{g mol}^{-1}\)). So the mass of sulphur is:\[\text{mass of S} = \dfrac{32}{233} \times \text{mass of } \text{BaSO}_4 \]This is the step people skip past — the mass of the precipitate you weigh is not the mass of sulphur; you must scale it down by the ratio (atomic mass of S) / (molar mass of \(\displaystyle \text{BaSO}_4\)), because most of that precipitate's mass is barium and oxygen, not sulphur.Percentage of sulphur in the original compound is then this mass of S expressed as a percentage of the mass of compound taken:\[\%\ \text{S} = \dfrac{32}{233} \times \dfrac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100 \]Here "mass of compound" is the $\displaystyle 0.468$ g of organic substance actually weighed out for the Carius tube — not any other mass in the experiment, since percentage composition is always mass of the element per unit mass of the original sample.Substituting the numbers. Mass of \(\displaystyle \text{BaSO}_4\) formed = $\displaystyle 0.668$ g, mass of compound taken = $\displaystyle 0.468$ g:\[\%\ \text{S} = \dfrac{32 \times 0.668 \times 100}{233 \times 0.468} \]Working the numerator and denominator separately:\[32 \times 0.668 \times 100 = 2137.6 \] \[233 \times 0.468 = 109.044 \]So:\[\%\ \text{S} = \dfrac{2137.6}{109.044} = 19.60 \]Answer: The organic compound contains $\displaystyle 19.60$% sulphur by mass.

The hybridisation of a carbon atom is decided by how many σ (sigma) bonds surround it — a double bond counts as one σ + one π, a triple bond as one σ + two π — not by which functional group it happens to sit next to.Number the six carbons of the chain so each one can be pointed to individually:\[\underset{C_1}{CH_2} = \underset{C_2}{CH} - \underset{C_3}{CH_2} - \underset{C_4}{CH_2} - \underset{C_5}{C} \equiv \underset{C_6}{CH} \]Now go carbon by carbon and count only the σ bonds around each one (that count is what fixes the hybridisation): A step people get wrong here: it is tempting to think \(\displaystyle C_3\) and \(\displaystyle C_4\) must be "half-hybridised" because they sit between a double bond and a triple bond. They do not — neither carbon is itself part of a multiple bond, so each one is counted purely by its own four single-bond neighbours and comes out as an ordinary sp³ –CH2– carbon, exactly as in an alkane.The question asks about the \(\displaystyle C_2-C_3\) bond. This bond is a plain single bond that joins the last carbon of the double bond, \(\displaystyle C_2\) (sp²), to the first carbon of the –CH2–CH2– bridge, \(\displaystyle C_3\) (sp³). So it is formed by the end-on overlap of an sp² orbital on \(\displaystyle C_2\) with an sp³ orbital on \(\displaystyle C_3\) — an sp²–sp³ σ bond.Answer: (c) sp²–sp³

Prussian blue is not the sodium salt that forms first — it appears only once that sodium salt meets iron in the +$\displaystyle 3$ state, giving the compound \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\). Track the iron and the cyanide through each step of the test and the answer falls out of the bookkeeping.Step $\displaystyle 1$ — sodium fusion makes cyanide, not any iron compound yet. When the nitrogen-containing organic compound is fused with sodium metal, its carbon and nitrogen atoms combine with sodium to give sodium cyanide in the fusion extract: \[\text{Na} + \text{C} + \text{N} \longrightarrow \text{NaCN} \] There is no iron anywhere in this step — the extract is just a solution containing \(\displaystyle \text{Na}^+\) and \(\displaystyle \text{CN}^-\) (plus \(\displaystyle \text{NaOH}\), \(\displaystyle \text{Na}_2\text{S}\), etc., from other elements). Iron is added only in the next step, so option (a), which already has iron in it, cannot be what forms here.Step $\displaystyle 2$ — boiling with FeSO₄ builds the ferrocyanide ion. The alkaline extract is boiled with freshly prepared iron(II) sulphate, \(\displaystyle \text{FeSO}_4\) (iron here is \(\displaystyle \text{Fe}^{2+}\), the ferrous state). Excess \(\displaystyle \text{CN}^-\) wraps around each \(\displaystyle \text{Fe}^{2+}\) to give the complex hexacyanidoferrate(II) ion, isolated as sodium ferrocyanide: \[\text{FeSO}_4 + 6\,\text{NaCN} \longrightarrow \text{Na}_4[\text{Fe(CN)}_6] + \text{Na}_2\text{SO}_4 \] Check the balance: iron, \(\displaystyle 1=1\); sodium, \(\displaystyle 6\) on the left equals \(\displaystyle 4+2\) on the right; \(\displaystyle \text{CN}\), \(\displaystyle 6=6\); sulphate, \(\displaystyle 1=1\). In \(\displaystyle \text{Na}_4[\text{Fe(CN)}_6]\) the iron is still \(\displaystyle \text{Fe}^{2+}\) — this is option (a). It is pale/colourless in solution, not blue, because the blue colour needs a second, higher oxidation state of iron to react with it. This is the step people stop at and wrongly pick (a): forming \(\displaystyle \text{Na}_4[\text{Fe(CN)}_6]\) is necessary for the test, but it is an intermediate, not the coloured product asked for.Step $\displaystyle 3$ — some iron is already Fe³⁺. \(\displaystyle \text{FeSO}_4\) solution is never perfectly pure \(\displaystyle \text{Fe}^{2+}\); air oxidises part of it to iron(III) sulphate on standing and boiling, so the mixture always carries some \(\displaystyle \text{Fe}^{3+}\) alongside the \(\displaystyle \text{Fe}^{2+}\) that became ferrocyanide.Step $\displaystyle 4$ — acidifying brings Fe³⁺ and ferrocyanide together, and that reaction is Prussian blue. On cooling and acidifying with dilute sulphuric acid, any iron hydroxide precipitate dissolves back into solution as \(\displaystyle \text{Fe}^{3+}\), and these ferric ions combine with the sodium ferrocyanide made in Step $\displaystyle 2$ to throw down an intensely coloured, insoluble precipitate: \[3\,\text{Na}_4[\text{Fe(CN)}_6] + 2\,\text{Fe}_2(\text{SO}_4)_3 \longrightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 6\,\text{Na}_2\text{SO}_4 \] Balance check: sodium, \(\displaystyle 12=12\); iron, \(\displaystyle 3\) (inside the three ferrocyanide units) \(\displaystyle +\,4\) (from the ferric sulphate) \(\displaystyle =7\) on the left, and on the right \(\displaystyle 4\) (outside) \(\displaystyle +\,3\) (inside) \(\displaystyle =7\); \(\displaystyle \text{CN}\), \(\displaystyle 18=18\); sulphate, \(\displaystyle 6=6\).The product, \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\), in words is ferric ferrocyanide (iron(III) hexacyanidoferrate(II)) — this is exactly Prussian blue. Its charges make sense with real oxidation states: the four outer iron atoms are \(\displaystyle \text{Fe}^{3+}\) (\(\displaystyle 4\times(+3)=+12\)), and each \(\displaystyle [\text{Fe(CN)}_6]\) unit inside carries the inner iron as \(\displaystyle \text{Fe}^{2+}\) with a \(\displaystyle 4-\) charge (three units give \(\displaystyle 3\times(-4)=-12\)), so the compound is neutral — a genuine, stable substance.Aside — why the other options fail on this same charge check. Option (c), \(\displaystyle \text{Fe}_2[\text{Fe(CN)}_6]\), would need the outer iron to also be \(\displaystyle \text{Fe}^{2+}\) (ferrous), giving \(\displaystyle 2\times(+2)=+4\) against the ferrocyanide's \(\displaystyle -4\) — this is a real compound (ferrous ferrocyanide, pale/white, not blue), but it needs Fe²⁺ everywhere, and this test always oxidises some iron to Fe³⁺, so it isn't what forms in the Prussian-blue step. Option (d), \(\displaystyle \text{Fe}_3[\text{Fe(CN)}_6]_4\), does not even balance charge for any simple oxidation state: \(\displaystyle 3\times(+3)=+9\) against \(\displaystyle 4\times(-4)=-16\) — the numbers don't match, so this is not a real product of the reaction at all.Answer: (b) \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\) — ferric ferrocyanide, the Prussian blue compound.

Chromatography is the technique that works when every other separation method runs out of things it can tell apart.Look at what each option actually depends on, and where that dependence breaks down.The step people undervalue here: chromatography does not need the components to differ in a single fixed property like boiling point or solubility-at-one-temperature. Its separating power comes from a continuous, repeated partitioning between two phases (adsorption/desorption or dissolution/re-dissolution happening over and over as the mixture moves), so even compounds that are extremely close in behaviour — geometrical isomers, closely related compounds in a natural extract, trace impurities — get pulled apart given enough length of stationary phase (a longer column, a bigger sheet) to repeat the partitioning enough times.This is also why it is the "best and latest" of the four: Answer: (d) Chromatography.

A nucleophile is an electron-rich species that attacks an electron-poor (electrophilic) carbon — so you decide "nucleophilic vs electrophilic" by asking which atom is short of electrons, not which atom is attacking.First, check that the equation you're given is even balanced, since a wrong count of atoms is where a lot of mistakes start silently:\[\text{CH}_3\text{CH}_2\text{I} + \text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KI} \]Here \(\displaystyle \text{CH}_3\text{CH}_2\text{I}\) is ethyl iodide (iodoethane), \(\displaystyle \text{KOH}\) is potassium hydroxide, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\) is ethanol, and \(\displaystyle \text{KI}\) is potassium iodide. Counting atoms on each side: left side has $\displaystyle 2$ C, $\displaystyle 6$ H ($\displaystyle 5$ on the ethyl group + $\displaystyle 1$ on OH), $\displaystyle 1$ I, $\displaystyle 1$ K, $\displaystyle 1$ O; right side has $\displaystyle 2$ C, $\displaystyle 6$ H ($\displaystyle 5$ on the ethyl group + $\displaystyle 1$ on OH), $\displaystyle 1$ O, $\displaystyle 1$ K, $\displaystyle 1$ I. Every atom matches with a coefficient of $\displaystyle 1$ in front of each species, so the equation is already balanced as written — no multiplying needed.Now look at what is actually happening at the carbon that carries the iodine, since that carbon is where the reaction takes place.Step $\displaystyle 1$ — find the electrophilic centre. Iodine is more electronegative than carbon, so it pulls the bonding electron pair of the \(\displaystyle \text{C–I}\) bond toward itself. That leaves the carbon attached to I carrying a partial positive charge (\(\displaystyle \delta^{+}\)), while iodine carries a partial negative charge (\(\displaystyle \delta^{-}\)). A carbon that is short of electron density like this is an electrophilic carbon — it is the site that something electron-rich will be drawn to.Step $\displaystyle 2$ — identify the attacking species. The reagent supplying the reactive species here is aqueous \(\displaystyle \text{KOH}\), which is fully ionic in water and furnishes hydroxide ions, \(\displaystyle \text{OH}^{-}\). The oxygen of \(\displaystyle \text{OH}^{-}\) carries lone pairs of electrons and a full negative charge, so it is strongly electron-rich. A species that seeks out an electron-poor centre and donates its own electron pair to form a new bond is, by definition, a nucleophile.Step $\displaystyle 3$ — the mechanism in words. The oxygen of \(\displaystyle \text{OH}^{-}\) uses one of its lone pairs to attack the \(\displaystyle \delta^{+}\) carbon of ethyl iodide from the side opposite the C–I bond. As the new \(\displaystyle \text{C–O}\) bond forms, the \(\displaystyle \text{C–I}\) bond breaks heterolytically, with both electrons of that bond leaving on iodine. This kicks iodine out as iodide ion, \(\displaystyle \text{I}^{-}\), which then pairs with \(\displaystyle \text{K}^{+}\) to give the by-product potassium iodide (KI). The carbon ends up bonded to \(\displaystyle \text{OH}\) instead of \(\displaystyle \text{I}\), giving the product ethanol, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\).The step people mix up: it is tempting to call this "electrophilic" because you might think of the halogen-bearing carbon as the reactive part being replaced. But the classification of a substitution reaction is named after the incoming attacking species, not the carbon skeleton. Here the species doing the attacking — hydroxide — is electron-rich (a nucleophile), and it is substituting for a leaving group (iodide) at an electron-poor carbon. That combination is a nucleophilic substitution. An electrophilic substitution would instead need an electron-poor (electrophile) species attacking an electron-rich centre, such as the benzene ring in nitration — the opposite situation from what happens here.So option (a), electrophilic substitution, does not describe this reaction; option (b), nucleophilic substitution, does — \(\displaystyle \text{OH}^{-}\) is the nucleophile that displaces \(\displaystyle \text{I}^{-}\) from the electrophilic carbon of ethyl iodide to give ethanol and potassium iodide.**Answer: (b) nucleophilic substitution — hydroxide ion (\(\displaystyle \text{OH}^{-}\)), an electron-rich nucleophile, attacks the electron-poor carbon of \(\displaystyle \text{CH}_3\text{CH}_2\text{I}\) and displaces iodide ion, converting ethyl iodide into ethanol (\(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\)) with KI as the by-product; the balanced equation is \(\displaystyle \text{CH}_3\text{CH}_2\text{I} + \text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KI}\).