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.