NCERT Solutions · Class 11 Chemistry Hydrocarbons

Aromaticity is not "any ring with alternating double bonds" — it needs three structural conditions plus a head-count rule, and all four must hold together.A compound is aromatic only if it satisfies every one of the following.1. The system must be cyclic. The carbon (or carbon+heteroatom) skeleton must form a closed ring — an open chain, however conjugated, cannot be aromatic because there is no loop for the electron cloud to close around.2. The ring must be planar. Every atom of the ring has to lie in (or very close to) one plane. This is the step people skip: planarity is not a minor detail, it is what lets the next condition work at all. If even one ring atom is pushed out of plane (as happens in strained or bridged rings), the orbital overlap around the ring is broken and the system cannot be aromatic.3. Each ring atom must supply a p-orbital, all of them parallel, to build one continuous \(\displaystyle \pi\) system. Every atom in the ring must be \(\displaystyle sp^2\) (or \(\displaystyle sp\)) hybridized, leaving one unhybridized p-orbital perpendicular to the ring plane. Because the ring is planar (condition $\displaystyle 2$), all these p-orbitals are parallel to each other, so they overlap side-by-side all the way around the ring. This continuous overlap delocalizes the \(\displaystyle \pi\) electrons into two doughnut-shaped clouds, one above and one below the ring plane, rather than leaving them as isolated double bonds. This is what "complete conjugation around the ring" means — a lone \(\displaystyle sp^3\) carbon anywhere in the ring (no p-orbital to contribute) breaks the loop and kills aromaticity at that point.4. Hückel's rule: the ring must contain exactly \(\displaystyle (4n+2)\) delocalized \(\displaystyle \pi\) electrons. Here \(\displaystyle n\) is a whole number \(\displaystyle 0, 1, 2, 3, \ldots\), and "\(\displaystyle \pi\) electrons" means the electrons sitting in that continuous ring \(\displaystyle \pi\) cloud from condition $\displaystyle 3$ (each double bond contributes $\displaystyle 2$, and a lone pair in a ring p-orbital, such as on O, N or a carbanion carbon, also contributes $\displaystyle 2$).\[\text{number of ring } \pi \text{ electrons} = 4n + 2 , \qquad n = 0, 1, 2, 3, \ldots \]Working out the allowed counts: \(\displaystyle n=0 \Rightarrow 2\), \(\displaystyle n=1 \Rightarrow 6\), \(\displaystyle n=2 \Rightarrow 10\), \(\displaystyle n=3 \Rightarrow 14\), and so on. A ring electron count that is NOT one of these numbers (for example $\displaystyle 4$ or $\displaystyle 8$) fails Hückel's rule — such a system is called antiaromatic (e.g., cyclobutadiene, with $\displaystyle 4$ ring \(\displaystyle \pi\) electrons) or simply non-aromatic if it also fails a structural condition.The everyday example is benzene, \(\displaystyle \text{C}_6\text{H}_6\): it is a six-membered ring (condition $\displaystyle 1$), it is planar (condition $\displaystyle 2$), all six carbons are \(\displaystyle sp^2\) with a p-orbital each, giving one continuous delocalized \(\displaystyle \pi\) system (condition $\displaystyle 3$), and it holds $\displaystyle 6$ \(\displaystyle \pi\) electrons from its three formal double bonds, which is \(\displaystyle 4n+2\) with \(\displaystyle n=1\) (condition $\displaystyle 4$) — so benzene satisfies all four and is aromatic. The same four checks (not just the electron count on its own) are what you apply to any other candidate ring, whether it is all-carbon or contains a heteroatom such as furan or pyridine.Answer: A ring system is aromatic only if it meets all of the following: (i) it is cyclic, (ii) it is planar, (iii) each ring atom is \(\displaystyle sp^2\)/\(\displaystyle sp\) hybridized so that its p-orbital can overlap continuously with its neighbours all around the ring, giving one delocalized \(\displaystyle \pi\) electron cloud, and (iv) that delocalized cloud contains \(\displaystyle (4n+2)\) \(\displaystyle \pi\) electrons, where \(\displaystyle n = 0, 1, 2, 3, \ldots\) (Hückel's rule) — e.g. benzene has $\displaystyle 6$ \(\displaystyle \pi\) electrons, \(\displaystyle n=1\).

A ring is aromatic only when all three Hückel conditions hold together: it is planar, every ring atom contributes an unhybridised p‑orbital so the π electrons are delocalised all the way round in one closed loop, and that loop holds \(\displaystyle 4n+2\) π electrons for some whole number \(\displaystyle n = 0, 1, 2, \dots\) (so $\displaystyle 2$, $\displaystyle 6$, $\displaystyle 10$, $\displaystyle 14$ … electrons). Breaking any one of the three is enough to lose aromaticity — and each of these three rings breaks a different one.(i) The six‑membered ring with two ring double bonds and an exocyclic \(\displaystyle =\text{CH}_2\)This ring has a double bond across the top, a double bond across the bottom, and — at the ring carbon between them on the other side — a third double bond that runs out of the ring to a \(\displaystyle \text{CH}_2\) group (an exocyclic methylene; the compound is a cyclohexa‑$\displaystyle 1,3$‑diene carrying an exocyclic methylene, sometimes called $\displaystyle 5$‑methylenecyclohexa‑$\displaystyle 1,3$‑diene). Three C=C bonds are present in total, which makes it tempting to expect six delocalised π electrons exactly as in benzene.The step people get wrong here: counting total π bonds in the molecule is not the same as counting π electrons circulating in the ring. Trace the ring itself, atom by atom, ignoring the branch: the ring carbon that carries the exocyclic double bond is joined to both its ring neighbours by single bonds. Its one p‑orbital is already spent forming the π bond to the outside \(\displaystyle \text{CH}_2\), so it has nothing left to hand on to either ring neighbour. That breaks the chain of overlapping p‑orbitals exactly at that atom — you cannot trace one continuous loop of overlap all the way round the six ring atoms and back to the start. The four π electrons of the two ring double bonds stop and start on either side of that carbon instead of closing into a ring current, and the remaining two π electrons belong to the exocyclic bond, entirely outside the ring. Condition (ii), complete delocalisation in the ring, fails, so the system is not aromatic even though six π electrons exist somewhere in the molecule.(ii) The five‑membered ring with two double bonds and one \(\displaystyle \text{CH}_2\)This is an ordinary five‑membered ring with two ring double bonds and one \(\displaystyle \text{CH}_2\) group between them — cyclopenta‑$\displaystyle 1,3$‑diene. That \(\displaystyle \text{CH}_2\) carbon is \(\displaystyle sp^3\) hybridised: all four of its bonds (two C–C single bonds to its ring neighbours, two C–H bonds) come from \(\displaystyle sp^3\) orbitals, so unlike every other ring atom here, it has no unhybridised p‑orbital left over at all.The mistake to watch for: one \(\displaystyle sp^3\) carbon anywhere in a ring is enough to stop the whole ring from being aromatic, no matter how the rest is drawn. A closed conjugated loop needs a p‑orbital sitting on every ring atom, edge to edge, so the π cloud can spread continuously around the circle. Here one atom offers nothing, so the loop of p‑orbitals has a physical gap in it — delocalisation cannot get past that carbon in either direction. Only the four π electrons of the two double bonds exist at all here, and \(\displaystyle 4n+2 = 4\) has no whole‑number solution for \(\displaystyle n\), so this ring fails both condition (ii) and condition (iii). (It is worth noting the near‑miss: pull one \(\displaystyle \text{H}^-\) off that same \(\displaystyle \text{CH}_2\) and the carbon left behind becomes \(\displaystyle sp^2\), with the leftover lone pair sitting in a p‑orbital instead of an \(\displaystyle sp^3\) orbital. The resulting cyclopentadienyl anion then has five continuous p‑orbitals carrying six π electrons and is aromatic — the entire difference between the non‑aromatic molecule drawn here and that aromatic ion is that one p‑orbital.)(iii) The eight‑membered ring with four alternating double bondsThis is cycloocta‑$\displaystyle 1,3,5,7$‑tetraene (COT): every ring carbon here is \(\displaystyle sp^2\) and does carry a p‑orbital, so — unlike (i) and (ii) — the ring as drawn looks fully conjugated all the way round, with no break in the chain of p‑orbitals.Count the π electrons: four C=C bonds give \(\displaystyle 4 \times 2 = 8\) π electrons.Applying Hückel's rule means solving \(\displaystyle 4n+2 = 8\) for a whole number \(\displaystyle n\), not just checking that $\displaystyle 8$ is an even count. \(\displaystyle 4n+2=8 \Rightarrow n = 1.5\), which is not a whole number, so $\displaystyle 8$ is not a member of the aromatic series \(\displaystyle 2, 6, 10, 14, \dots\). It falls instead on the series \(\displaystyle 4n\) \(\displaystyle (4, 8, 12, \dots)\), the electron count associated with destabilised, antiaromatic rings. Condition (iii) fails, so the ring is not aromatic. (This is also why the real molecule does not stay flat: cyclooctatetraene puckers into a non‑planar "tub" shape precisely to avoid forcing eight π electrons into one flat, fully delocalised ring, which would additionally cost it condition (i), planarity.)**Answer: (i) is not aromatic because its third double bond is exocyclic — the p‑orbital of that ring carbon is used up bonding to the outside \(\displaystyle \text{CH}_2\), so the ring's own π system cannot close into one continuous delocalised loop (fails complete ring delocalisation). (ii) is not aromatic because the ring contains an \(\displaystyle sp^3\) \(\displaystyle \text{CH}_2\) carbon with no p‑orbital at all, physically breaking the loop, and it holds only $\displaystyle 4$ π electrons, not \(\displaystyle 4n+2\) (fails both delocalisation and Hückel's rule). (iii) is not aromatic because, although it is a fully conjugated ring on paper, it carries $\displaystyle 8$ π electrons and \(\displaystyle 4n+2=8\) has no whole‑number solution for \(\displaystyle n\) — $\displaystyle 8$ belongs to the destabilising \(\displaystyle 4n\) series, not the aromatic \(\displaystyle 4n+2\) series (fails Hückel's rule; the real molecule also puckers non‑planar to escape this).

Benzene itself has nothing to direct anything — every answer below is built by choosing the ORDER in which two electrophilic substitutions happen, because whichever group goes onto the ring first is the one that decides where the second group lands.Each step is an electrophilic aromatic substitution: a catalyst generates an electrophile (a positively-polarised species) which replaces one ring hydrogen. Getting the target's two groups into the right relative positions (para, meta, or built as a single carbonyl-bearing group) is a question of picking the right first substituent, not of forcing the second one to go somewhere it doesn't want to.(i) p-Nitrobromobenzene, \(\displaystyle \text{O}_2\text{N–C}_6\text{H}_4\text{–Br} \) (para isomer)Put the bromine on first — bromine is an ortho/para-director, so a ring that already carries Br sends the next electrophile mainly to the para position. Nitrating first would be the mistake here: \(\displaystyle -\text{NO}_2\) is a meta-director, so a later bromination would land meta, never para.Step $\displaystyle 1$ — bromination: \[\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} + \text{HBr} \] benzene + bromine → bromobenzene + hydrogen bromide. Checking atoms: left side has \(\displaystyle 6\text{C}, 6\text{H}, 2\text{Br}\); right side \(\displaystyle \text{C}_6\text{H}_5\text{Br}\) (6C, 5H, 1Br) plus \(\displaystyle \text{HBr}\) (1H, 1Br) also gives 6C, 6H, 2Br — balanced as written, $\displaystyle 1$:$\displaystyle 1$:$\displaystyle 1$:1. \(\displaystyle \text{FeBr}_3\) (or Fe, which converts to \(\displaystyle \text{FeBr}_3\) in the reaction mixture) is the Lewis-acid catalyst that polarises \(\displaystyle \text{Br}_2\) into an electrophilic bromine.Step $\displaystyle 2$ — nitration (the electrophile here is \(\displaystyle \text{NO}_2^+\), the nitronium ion, generated from conc. \(\displaystyle \text{HNO}_3\) by conc. \(\displaystyle \text{H}_2\text{SO}_4\)): \[\text{C}_6\text{H}_5\text{Br} + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{O}_2\text{N–C}_6\text{H}_4\text{–Br (para)} + \text{H}_2\text{O} \] Atom count: left \(\displaystyle \text{C}_6\text{H}_5\text{Br} + \text{HNO}_3 = \text{C}_6\text{H}_6\text{BrNO}_3\); right \(\displaystyle \text{C}_6\text{H}_4\text{BrNO}_2 + \text{H}_2\text{O} = \text{C}_6\text{H}_6\text{BrNO}_3\). Balanced.Bromine's lone pairs feed into the ring by resonance, stabilising attack at ortho and para — but bromine is also a bulky atom, so it sterically blocks the position right next to itself. The result is that para attack dominates over ortho, and fractional distillation (or crystallisation) separates the major p-nitrobromobenzene from the minor ortho by-product.(ii) m-NitrochlorobenzenePut the nitro group on first — \(\displaystyle -\text{NO}_2\) is a meta-director, so chlorinating a nitrated ring sends the chlorine to the meta position. Doing it the other way round is the trap: chlorine is an ortho/para-director, so chlorinating benzene first and nitrating second would give ortho/para-nitrochlorobenzene, never meta.Step $\displaystyle 1$ — nitration: \[\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O} \] benzene + nitric acid → nitrobenzene + water. Left: \(\displaystyle \text{C}_6\text{H}_7\text{NO}_3\); right: \(\displaystyle \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O} = \text{C}_6\text{H}_7\text{NO}_3\). Balanced.Step $\displaystyle 2$ — chlorination: \[\text{C}_6\text{H}_5\text{NO}_2 + \text{Cl}_2 \xrightarrow{\text{anhyd. FeCl}_3} m\text{-ClC}_6\text{H}_4\text{NO}_2 + \text{HCl} \] nitrobenzene + chlorine → m-nitrochlorobenzene + hydrogen chloride. Left: \(\displaystyle \text{C}_6\text{H}_5\text{Cl}_2\text{NO}_2\); right: \(\displaystyle \text{C}_6\text{H}_4\text{ClNO}_2 + \text{HCl} = \text{C}_6\text{H}_5\text{Cl}_2\text{NO}_2\). Balanced.The reason the second group is forced to meta: \(\displaystyle -\text{NO}_2\) withdraws electron density from the ring by resonance, and it withdraws most strongly from the ortho and para positions (its resonance structures put positive charge there), leaving the meta position comparatively electron-richer. The incoming \(\displaystyle \text{Cl}^+\) electrophile therefore attacks meta to the nitro group.(iii) p-NitrotolueneAlkylate the ring before nitrating it — nitrobenzene is deactivated so strongly that a Friedel–Crafts reaction (alkylation or acylation) simply will not run on it. That rules out making nitrobenzene first and trying to attach a methyl group afterwards; the methyl group has to go on first, onto plain benzene.Step $\displaystyle 1$ — Friedel–Crafts alkylation: \[\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl} \] benzene + chloromethane → toluene (methylbenzene) + hydrogen chloride. Left: \(\displaystyle \text{C}_7\text{H}_9\text{Cl}\); right: \(\displaystyle \text{C}_6\text{H}_5\text{CH}_3 (\text{C}_7\text{H}_8) + \text{HCl} = \text{C}_7\text{H}_9\text{Cl}\). Balanced. Anhydrous \(\displaystyle \text{AlCl}_3\) is the Lewis acid that pulls \(\displaystyle \text{Cl}^-\) off \(\displaystyle \text{CH}_3\text{Cl}\), leaving an electrophilic \(\displaystyle \text{CH}_3^+\)-type species that attacks the ring.Step $\displaystyle 2$ — nitration: \[\text{C}_6\text{H}_5\text{CH}_3 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{C}_6\text{H}_4\text{NO}_2 + \text{H}_2\text{O} \] toluene + nitric acid → nitrotoluene + water. Left: \(\displaystyle \text{C}_7\text{H}_9\text{NO}_3\); right: \(\displaystyle \text{C}_7\text{H}_7\text{NO}_2 + \text{H}_2\text{O} = \text{C}_7\text{H}_9\text{NO}_3\). Balanced.Methyl is an ortho/para-director (it feeds electron density into the ring by hyperconjugation), so this step gives a mixture of ortho- and para-nitrotoluene, not one pure product. The step people skip is realising the reaction does not stop at "toluene → p-nitrotoluene" — it genuinely gives both isomers, and p-nitrotoluene must be picked out afterwards by fractional distillation (the para isomer, boiling near $\displaystyle 238$ °C, is separated from the lower-boiling ortho isomer, boiling near $\displaystyle 222$ °C).(iv) Acetophenone, \(\displaystyle \text{C}_6\text{H}_5\text{COCH}_3 \)Acetophenone is a ring with a \(\displaystyle -\text{COCH}_3\) group directly attached — that whole group is installed in one step by Friedel–Crafts ACYLATION, so there is no ordering puzzle here at all.\[\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} \] benzene + acetyl chloride → acetophenone (methyl phenyl ketone) + hydrogen chloride. Left: \(\displaystyle \text{C}_6\text{H}_6 + \text{C}_2\text{H}_3\text{OCl} = \text{C}_8\text{H}_9\text{OCl}\); right: \(\displaystyle \text{C}_6\text{H}_5\text{COCH}_3 (\text{C}_8\text{H}_8\text{O}) + \text{HCl} = \text{C}_8\text{H}_9\text{OCl}\). Balanced.Anhydrous \(\displaystyle \text{AlCl}_3\) reacts with acetyl chloride to generate the electrophilic acylium ion, \(\displaystyle \text{CH}_3\text{CO}^+\), which attacks the ring. (Acetic anhydride, \(\displaystyle (\text{CH}_3\text{CO})_2\text{O}\), works the same way in place of acetyl chloride, giving acetophenone plus acetic acid instead of HCl.) The aside worth remembering here: because the \(\displaystyle -\text{COCH}_3\) group that gets attached is strongly electron-withdrawing, it deactivates the ring the moment it forms — so, unlike Friedel–Crafts alkylation, acylation stops cleanly at mono-substitution with no risk of a second acyl group going on.Answer: (i) brominate benzene (\(\displaystyle \text{Br}_2/\text{FeBr}_3\)) then nitrate (\(\displaystyle \text{HNO}_3/\text{H}_2\text{SO}_4\)) — bromine's ortho/para-directing, sterically hindered ortho, gives p-nitrobromobenzene as the major product; (ii) nitrate benzene (\(\displaystyle \text{HNO}_3/\text{H}_2\text{SO}_4\)) then chlorinate (\(\displaystyle \text{Cl}_2/\text{FeCl}_3\)) — the meta-directing nitro group sends chlorine to the meta position, giving m-nitrochlorobenzene; (iii) Friedel–Crafts alkylate benzene with \(\displaystyle \text{CH}_3\text{Cl}/\text{AlCl}_3\) to get toluene, then nitrate with \(\displaystyle \text{HNO}_3/\text{H}_2\text{SO}_4\) and separate the resulting ortho/para mixture by fractional distillation to isolate p-nitrotoluene; (iv) Friedel–Crafts acylate benzene directly with \(\displaystyle \text{CH}_3\text{COCl}/\text{AlCl}_3\) (or \(\displaystyle (\text{CH}_3\text{CO})_2\text{O}/\text{AlCl}_3\)) to get acetophenone in a single step.

A carbon's "degree" — whether it counts as $\displaystyle 1$°, $\displaystyle 2$°, $\displaystyle 3$° or $\displaystyle 4$° — is decided purely by how many OTHER CARBON atoms it is bonded to, never by its position in the chain and never by how many hydrogens it happens to carry.Every carbon atom makes exactly four bonds. So once the bonds to other carbons are counted, whatever is left over must be bonds to hydrogen:\[\text{H atoms on a carbon} = 4 - (\text{number of carbon atoms bonded to it}) \]That single rule is the whole method here — no reaction, just careful bond-counting, so it pays to draw out every carbon rather than only the ones written on the main line.The compound\[\text{H}_3\text{C-CH}_2\text{-C(CH}_3)_2\text{-CH}_2\text{-CH(CH}_3)_2 \]has $\displaystyle 9$ carbon atoms in all: the five sitting on the chain, plus two \(\displaystyle \text{CH}_3\) groups hanging off the middle carbon and two more \(\displaystyle \text{CH}_3\) groups hanging off the carbon at the right end. Label them so none get missed:\[\underset{C_1}{\text{CH}_3}-\underset{C_2}{\text{CH}_2}-\underset{C_3}{\text{C}}(\underset{C_{3a}}{\text{CH}_3})(\underset{C_{3b}}{\text{CH}_3})-\underset{C_4}{\text{CH}_2}-\underset{C_5}{\text{CH}}(\underset{C_{5a}}{\text{CH}_3})(\underset{C_{5b}}{\text{CH}_3}) \]A step people get wrong here: reading \(\displaystyle -\text{CH(CH}_3)_2\) as one carbon. It is three — the \(\displaystyle \text{CH}\) itself plus its two methyl branches — exactly the way \(\displaystyle \text{C(CH}_3)_2\) in the middle is three carbons, not one.Now go through every carbon and count only its carbon neighbours:\(\displaystyle C_1\) (the leftmost \(\displaystyle \text{CH}_3\)) is bonded only to \(\displaystyle C_2\) → $\displaystyle 1$ carbon neighbour → $\displaystyle 1$° (primary) → \(\displaystyle 4-1=3\) H atoms.\(\displaystyle C_2\) (the first \(\displaystyle \text{CH}_2\)) is bonded to \(\displaystyle C_1\) and \(\displaystyle C_3\) → $\displaystyle 2$ carbon neighbours → $\displaystyle 2$° (secondary) → \(\displaystyle 4-2=2\) H atoms.\(\displaystyle C_3\) (the branched carbon in the middle) is bonded to \(\displaystyle C_2\), \(\displaystyle C_4\), \(\displaystyle \mathrm{C_{3a}}\) and \(\displaystyle \mathrm{C_{3b}}\) → $\displaystyle 4$ carbon neighbours → $\displaystyle 4$° (quaternary) → \(\displaystyle 4-4=0\) H atoms. It has no hydrogen at all, so it sits outside the $\displaystyle 1$°/$\displaystyle 2$°/$\displaystyle 3$° count the question asks for, but it is worth flagging: a carbon bonded to four other carbons simply has none left for H.\(\displaystyle \mathrm{C_{3a}}\) and \(\displaystyle \mathrm{C_{3b}}\) (the two methyls sitting on \(\displaystyle C_3\)) are each bonded only to \(\displaystyle C_3\) → $\displaystyle 1$ carbon neighbour each → $\displaystyle 1$° (primary) → $\displaystyle 3$ H atoms each.\(\displaystyle C_4\) (the second \(\displaystyle \text{CH}_2\)) is bonded to \(\displaystyle C_3\) and \(\displaystyle C_5\) → $\displaystyle 2$ carbon neighbours → $\displaystyle 2$° (secondary) → $\displaystyle 2$ H atoms.\(\displaystyle C_5\) (the \(\displaystyle \text{CH}\) at the right end) is bonded to \(\displaystyle C_4\), \(\displaystyle \mathrm{C_{5a}}\) and \(\displaystyle \mathrm{C_{5b}}\) → $\displaystyle 3$ carbon neighbours → $\displaystyle 3$° (tertiary) → \(\displaystyle 4-3=1\) H atom.\(\displaystyle \mathrm{C_{5a}}\) and \(\displaystyle \mathrm{C_{5b}}\) (the two methyls sitting on \(\displaystyle C_5\)) are each bonded only to \(\displaystyle C_5\) → $\displaystyle 1$ carbon neighbour each → $\displaystyle 1$° (primary) → $\displaystyle 3$ H atoms each.Collecting these by type: As a check, add up every hydrogen counted this way: \(\displaystyle 5(3) + 2(2) + 1(1) + 1(0) = 15 + 4 + 1 + 0 = 20\). With $\displaystyle 9$ carbons and $\displaystyle 20$ hydrogens the formula is \(\displaystyle \text{C}_9\text{H}_{20}\), which matches \(\displaystyle 2n+2 = 2(9)+2 = 20\) for a saturated, open-chain (no ring) alkane — so the bond-counting above is self-consistent.**Answer: In H₃C–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂, the $\displaystyle 1$° (primary) carbons are the five methyl carbons — the leftmost CH₃ and the four CH₃ branches (two on the middle carbon, two on the right-end carbon) — each bearing $\displaystyle 3$ H atoms; the $\displaystyle 2$° (secondary) carbons are the two chain CH₂ groups, each bearing $\displaystyle 2$ H atoms; the $\displaystyle 3$° (tertiary) carbon is the branched CH at the right end, bearing $\displaystyle 1$ H atom (the middle carbon itself is $\displaystyle 4$°/quaternary, bearing $\displaystyle 0$ H atoms).

Branching lowers the boiling point of an alkane, because it shrinks the area of contact between neighbouring molecules and weakens the only force holding them together.Alkanes are non-polar, so the sole intermolecular attraction between one molecule and the next is the van der Waals force (London dispersion force) — a weak, momentary pull between temporarily-induced dipoles. This force does not act at a single point; it acts wherever the surfaces of two molecules lie close to each other, so its total strength for a given molecule scales with how much surface area it can lay alongside a neighbour.Take the three isomers of pentane, all with the same molecular formula \(\displaystyle \text{C}_5\text{H}_{12} \), so the comparison isolates shape from mass:1. n-Pentane — a straight, unbranched chain of five carbons. A long chain lies flat against a neighbouring chain over its whole length, so contact area is largest. Boiling point: \(\displaystyle 36\ ^\circ\text{C} \).2. Isopentane ($\displaystyle 2$-methylbutane) — the same five carbons, but with one methyl branch on the chain. The branch makes the molecule bulge outward, so it can no longer lie flush against a neighbour along its full length; the contact area is reduced. Boiling point: \(\displaystyle 28\ ^\circ\text{C} \).3. Neopentane ($\displaystyle 2,2$-dimethylpropane) — a central carbon carrying four methyl groups, the most compact, nearly spherical arrangement possible for \(\displaystyle \text{C}_5\text{H}_{12} \). A sphere touches a neighbouring sphere at only a small patch, so contact area is smallest of the three. Boiling point: \(\displaystyle 9.5\ ^\circ\text{C} \).So the order is\[\text{n-pentane (unbranched)} \;>\; \text{isopentane (one branch)} \;>\; \text{neopentane (most branched)} \]This is a surface-area effect, not a molecular-weight effect — that is the point people miss. All three isomers have exactly the same mass and the same number and kind of bonds; only the shape differs. More branching packs the same set of carbon atoms into a rounder, more compact shape. A rounder shape presents less surface to a neighbouring molecule, so the van der Waals forces between molecules are weaker, and weaker intermolecular forces need less thermal energy to overcome — hence a lower boiling point.The general rule that follows: among isomeric alkanes, boiling point decreases as branching increases, because branching moves the molecule from an elongated shape (large contact area, stronger van der Waals attraction, higher boiling point) toward a compact, spherical shape (small contact area, weaker van der Waals attraction, lower boiling point).**Answer: Branching lowers the boiling point. Branching makes an alkane molecule more compact and closer to spherical, which reduces the surface area of contact between neighbouring molecules; since the only attraction between alkane molecules is the (surface-area-dependent) van der Waals force, less contact area means weaker attraction and hence a lower boiling point. For the \(\displaystyle \text{C}_5\text{H}_{12} \) isomers: n-pentane (unbranched) \(\displaystyle 36\ ^\circ\text{C}\) > isopentane (one branch) \(\displaystyle 28\ ^\circ\text{C}\) > neopentane (most branched) \(\displaystyle 9.5\ ^\circ\text{C}\) — boiling point falls steadily as branching increases.

Without benzoyl peroxide, \(\displaystyle \text{H}^+ \) attacks the double bond first and goes where it makes the more stable carbocation (Markovnikov addition); with benzoyl peroxide, a bromine atom attacks first and goes where it makes the more stable radical (the peroxide effect) — and because the roles of H and Br as "first attacker" are swapped, Br ends up on the opposite carbon.Propene is \(\displaystyle \text{CH}_2=\text{CH}-\text{CH}_3 \). Call the terminal \(\displaystyle =\text{CH}_2\) carbon \(\displaystyle \mathrm{C_{1}}\) (it already carries $\displaystyle 2$ hydrogens) and the middle \(\displaystyle =\text{CH}-\) carbon \(\displaystyle \mathrm{C_{2}}\) (it carries $\displaystyle 1$ hydrogen and is attached to the methyl group, C3).Mechanism $\displaystyle 1$ — no peroxide: ionic addition, gives $\displaystyle 2$-bromopropaneThis is ordinary electrophilic addition. \(\displaystyle \text{HBr} \) is polarized \(\displaystyle \overset{\delta+}{\text{H}}-\overset{\delta-}{\text{Br}} \), so the proton is the species that attacks the \(\displaystyle \pi\) bond first.Step $\displaystyle 1$ — protonation. \(\displaystyle \text{H}^+ \) bonds to C1. Look at the two possible outcomes before deciding which one happens:\[\text{CH}_2=\text{CH}-\text{CH}_3 + \text{H}^+ \rightarrow \text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \quad \text{(H adds to C1: secondary cation on C2)} \]versus H adding to \(\displaystyle \mathrm{C_{2}}\), which would leave the positive charge on \(\displaystyle \mathrm{C_{1}}\): \(\displaystyle \overset{+}{\text{C}}\text{H}_2-\text{CH}_2-\text{CH}_3 \), a primary cation.The secondary cation \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \) is flanked by two alkyl groups, so it is stabilised by hyperconjugation and the +I effect from both sides; the primary cation has only one. The system takes the lower-energy path, so H bonds to \(\displaystyle \mathrm{C_{1}}\) and the positive charge sits on C2. This is exactly Markovnikov's rule: the carbon that already has more hydrogens picks up the extra H. (The rule is often mis-stated as "the halogen goes to the more-substituted carbon" — that is just the consequence, not the reason; the reason is always which carbocation is more stable.)Step $\displaystyle 2$ — nucleophilic capture. \(\displaystyle \text{Br}^- \) (the leftover part of \(\displaystyle \text{HBr} \)) attacks the electron-poor \(\displaystyle \mathrm{C_{2}}\):\[\text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH}_3 + \text{Br}^- \rightarrow \text{CH}_3-\text{CHBr}-\text{CH}_3 \]Overall: \[\text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3-\text{CHBr}-\text{CH}_3 \] Balance check: left side has \(\displaystyle \text{C}_3\text{H}_6 \) (propene) plus \(\displaystyle \text{HBr} \), i.e. $\displaystyle 3$ C, $\displaystyle 7$ H, $\displaystyle 1$ Br; the product \(\displaystyle \text{CH}_3\text{CHBrCH}_3 \) is also $\displaystyle 3$ C, $\displaystyle 7$ H, $\displaystyle 1$ Br — H and Br from \(\displaystyle \text{HBr} \) have simply been distributed across the two carbons that were double-bonded, so nothing needs a coefficient.Product: $\displaystyle 2$-bromopropane (isopropyl bromide), \(\displaystyle \text{CH}_3\text{CHBrCH}_3 \) — bromine on the middle carbon, C2.Mechanism $\displaystyle 2$ — with benzoyl peroxide: free-radical addition, gives $\displaystyle 1$-bromopropaneBenzoyl peroxide, \(\displaystyle \text{C}_6\text{H}_5\text{CO}-\text{O}-\text{O}-\text{CO}-\text{C}_6\text{H}_5 \), has a weak O–O bond. On warming it breaks homolytically (one electron to each oxygen, not both to one), which is what starts a radical chain instead of an ionic path.Initiation. \[\text{C}_6\text{H}_5\text{COO}-\text{OOCC}_6\text{H}_5 \rightarrow 2\,\text{C}_6\text{H}_5\text{COO}^{\bullet} \] (Balanced by inspection: one molecule of formula \(\displaystyle \text{C}_{14}\text{H}_{10}\text{O}_4 \) splits into two identical fragments of formula \(\displaystyle \text{C}_7\text{H}_5\text{O}_2 \), and \(\displaystyle 2 \times \text{C}_7\text{H}_5\text{O}_2 = \text{C}_{14}\text{H}_{10}\text{O}_4\) — atoms conserved, so no coefficients beyond the "$\displaystyle 2$" are needed.)The benzoyloxy radical then pulls a hydrogen atom off \(\displaystyle \text{HBr} \), because forming the strong O–H bond of benzoic acid releases more energy than the H–Br bond broken: \[\text{C}_6\text{H}_5\text{COO}^{\bullet} + \text{H}-\text{Br} \rightarrow \text{C}_6\text{H}_5\text{COOH} + \text{Br}^{\bullet} \] (Check: left side \(\displaystyle \text{C}_7\text{H}_5\text{O}_2 + \text{HBr} = \text{C}_7\text{H}_6\text{O}_2\text{Br}\); right side \(\displaystyle \text{C}_6\text{H}_5\text{COOH} (=\text{C}_7\text{H}_6\text{O}_2) + \text{Br}^{\bullet}\), same atom count on both sides.)This is the step that matters: it manufactures a bromine free radical, \(\displaystyle \text{Br}^{\bullet} \), which is now the species that attacks the alkene first — not \(\displaystyle \text{H}^+ \) as in Mechanism 1.Propagation step $\displaystyle 1$ — Br• adds to the double bond. Again compare the two possible points of attack:\[\text{Br}^{\bullet} + \text{CH}_2=\text{CH}-\text{CH}_3 \rightarrow \text{BrCH}_2-\overset{\bullet}{\text{C}}\text{H}-\text{CH}_3 \quad \text{(Br adds to C1: secondary radical on C2)} \]versus Br adding to \(\displaystyle \mathrm{C_{2}}\), leaving the odd electron on \(\displaystyle \mathrm{C_{1}}\): \(\displaystyle \overset{\bullet}{\text{C}}\text{H}_2-\text{CHBr}-\text{CH}_3 \), a primary radical.Free radicals are stabilised the same way carbocations are — by hyperconjugation from neighbouring C–H bonds — so the secondary radical \(\displaystyle \text{BrCH}_2-\overset{\bullet}{\text{C}}\text{H}-\text{CH}_3 \) is lower in energy, and that is the path taken. Notice: it is the same rule (attack the terminal, more-hydrogenated carbon to put the odd electron/charge on the more substituted carbon) — but this time it is Br, not H, doing the attacking, so Br itself ends up on C1.Propagation step $\displaystyle 2$ — chain-carrying step. The carbon radical abstracts H from a fresh \(\displaystyle \text{HBr} \) molecule, finishing the product and regenerating \(\displaystyle \text{Br}^{\bullet} \) so the chain continues: \[\text{BrCH}_2-\overset{\bullet}{\text{C}}\text{H}-\text{CH}_3 + \text{H}-\text{Br} \rightarrow \text{BrCH}_2-\text{CH}_2-\text{CH}_3 + \text{Br}^{\bullet} \]Adding the two propagation steps together, the \(\displaystyle \text{Br}^{\bullet} \) that appears on both sides cancels, giving the same overall atom balance as before: \[\text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \] ($\displaystyle 3$ C, $\displaystyle 7$ H, $\displaystyle 1$ Br on each side, exactly as in Mechanism $\displaystyle 1$ — only which carbon carries the Br has changed.)(Termination steps such as \(\displaystyle 2\,\text{Br}^{\bullet} \rightarrow \text{Br}_2 \) or two carbon radicals combining also occur, but only rarely, since radical concentration stays low; they simply stop individual chains rather than change the product formed.)Product: $\displaystyle 1$-bromopropane (n-propyl bromide), \(\displaystyle \text{BrCH}_2\text{CH}_2\text{CH}_3 \) — bromine on the end carbon, C1.A short aside on why this happens only with HBr and not with HCl or HI: both propagation steps above must be exothermic for the chain to sustain itself, and that balance of bond energies works out only for the H–Br/C–Br/C–H combination. This peroxide-induced reversal is called the peroxide effect or Kharasch effect, and because it puts Br on the carbon that already had more hydrogens (opposite of the ionic case), it is also called anti-Markovnikov addition.Answer: Without peroxide, \(\displaystyle \text{H}^+ \) attacks propene first and adds to the terminal carbon \(\displaystyle \mathrm{(C_{1})}\) because that generates the more stable secondary carbocation \(\displaystyle \text{CH}_3\overset{+}{\text{C}}\text{HCH}_3 \); \(\displaystyle \text{Br}^- \) then bonds to \(\displaystyle \mathrm{C_{2}}\), giving $\displaystyle 2$-bromopropane, \(\displaystyle \text{CH}_3\text{CHBrCH}_3 \) (Markovnikov addition, ionic mechanism). With benzoyl peroxide present, the peroxide generates \(\displaystyle \text{Br}^{\bullet} \) radicals from HBr; \(\displaystyle \text{Br}^{\bullet} \) itself now attacks first and adds to the terminal carbon \(\displaystyle \mathrm{(C_{1})}\) because that generates the more stable secondary radical \(\displaystyle \text{BrCH}_2\overset{\bullet}{\text{C}}\text{HCH}_3 \), which then abstracts H from another HBr molecule to give $\displaystyle 1$-bromopropane, \(\displaystyle \text{BrCH}_2\text{CH}_2\text{CH}_3 \), and regenerate \(\displaystyle \text{Br}^{\bullet} \) for the chain (anti-Markovnikov addition, free-radical/peroxide effect).

Ozonolysis snips every C=C bond and turns each carbon end into a C=O; the C–C single bonds around the ring stay untouched. Because benzene's ring (Kekulé's picture) has three double bonds alternating with three single bonds, cutting all three double bonds does not leave one broken chain — it fully disconnects the ring into three separate two-carbon fragments, each fragment being the piece that was joined by a surviving single bond.Setting up o-xylene's ring. Number the ring carbons $\displaystyle 1$–$\displaystyle 6$, with the two methyl groups on \(\displaystyle \mathrm{C_{1}}\) and \(\displaystyle \mathrm{C_{2}}\) (that is what "$\displaystyle 1,2$-dimethyl" means). Kekulé's structure needs alternating double and single bonds, and there are exactly two ways to place them on this ring: These aren't two different compounds — o-xylene is one molecule — they are the two equally-valid ways of drawing where the alternating double bonds fall.The balanced transformation, one double bond at a time. Each C=C bond takes up one \(\displaystyle O_3 \) to form an ozonide, which is then split reductively. Zinc dust in the work-up is essential here — an aside worth remembering: without Zn, the initial split releases hydrogen peroxide, which would oxidise the aldehyde products on to carboxylic acids; Zn intercepts that \(\displaystyle H_2O_2 \) (as \(\displaystyle ZnO \)) so the aldehydes/ketones survive intact. Bookkeeping this out, atom for atom, per double bond:\[\text{R}_2\text{C}=\text{CR}_2 + O_3 + Zn \;\rightarrow\; \text{R}_2\text{C}=O + O=\text{CR}_2 + ZnO \](check: left side has $\displaystyle 1$ C=C's worth of C,H, plus $\displaystyle 3$ O from ozone and $\displaystyle 1$ Zn; right side has the same C,H split between two carbonyls, $\displaystyle 2$ O in those carbonyls + $\displaystyle 1$ O in ZnO = $\displaystyle 3$ O, and $\displaystyle 1$ Zn — balanced.) The whole ring, having three double bonds, therefore needs three \(\displaystyle O_3 \) and three Zn:\[C_8H_{10}\;(\text{o-xylene}) + 3\,O_3 + 3\,Zn \;\rightarrow\; \text{three carbonyl fragments} + 3\,ZnO \]Following Structure I. Track each carbon: \[C_8H_{10} + 3\,O_3 + 3\,Zn \;\rightarrow\; 2\,CH_3COCHO + OHC\text{-}CHO + 3\,ZnO \] Atom check: C: \(\displaystyle 3+3+2=8\) ✓, H: \(\displaystyle 4+4+2=10\) ✓, O (organic + ZnO): \(\displaystyle 2+2+2+3=9\), matching \(\displaystyle 3\times O_3=9\) O atoms ✓.Following Structure II. Track each carbon: \[C_8H_{10} + 3\,O_3 + 3\,Zn \;\rightarrow\; CH_3COCOCH_3 + 2\,OHC\text{-}CHO + 3\,ZnO \] Atom check: C: \(\displaystyle 4+2+2=8\) ✓, H: \(\displaystyle 6+2+2=10\) ✓, O: same total of $\displaystyle 9$ across products ✓.Why this supports Kekulé's structure of benzene. If the ring genuinely had one fixed set of alternating double bonds — say, permanently Structure I — ozonolysis of o-xylene could only ever give methylglyoxal and glyoxal, never diacetyl. If it were permanently Structure II, only diacetyl and glyoxal would appear, never methylglyoxal. What is actually found in the flask is a mixture of all three carbonyl compounds — glyoxal, methylglyoxal, and diacetyl — together. A short aside on the easy mistake here: this is not evidence of two different xylenes reacting; it is one o-xylene giving both product sets, which is only possible if the ring's double bonds are not frozen at either single arrangement. That is exactly Kekulé's own proposal for benzene: the ring oscillates between (is a resonance hybrid of) the two structures, so at any instant some molecules behave as if Structure I and others as if Structure II, and ozonolysis catches both — producing methylglyoxal, glyoxal, and diacetyl side by side.Answer: Ozonolysis of $\displaystyle 1,2$-dimethylbenzene (o-xylene) gives a mixture of three dicarbonyl compounds — glyoxal (ethanedial, \(\displaystyle OHC\text{-}CHO\)), methylglyoxal ($\displaystyle 2$-oxopropanal, \(\displaystyle CH_3COCHO\)), and diacetyl (butane-$\displaystyle 2,3$-dione, \(\displaystyle CH_3COCOCH_3\)) — because the balanced ring-cleavage \(\displaystyle C_8H_{10}+3O_3+3Zn\rightarrow\) products \(\displaystyle +\,3ZnO\) can go two ways depending on which Kekulé structure (double bond between the methylated carbons, or single bond between them) is drawn: one path gives \(\displaystyle 2\,CH_3COCHO + OHC\text{-}CHO\), the other gives \(\displaystyle CH_3COCOCH_3 + 2\,OHC\text{-}CHO\). Getting all three products together shows benzene's ring is not locked into either single Kekulé arrangement but is a hybrid/oscillation of both, which is precisely what Kekulé's structure (with this oscillation) proposes.

Acidity here means how easily a C–H bond releases \(\displaystyle \text{H}^+ \), and that depends on the hybridisation of the carbon holding the hydrogen.Step $\displaystyle 1$: Identify the hybridisation of the carbon atom bearing the acidic H in each molecule. Step $\displaystyle 2$: Compare the % s-character of these hybrid orbitals.An sp orbital is built from one s and one p orbital, so it is \(\displaystyle \tfrac{1}{2} \) s-character, i.e. $\displaystyle 50$%. An sp\(\displaystyle ^2\) orbital is built from one s and two p orbitals shared over three hybrid orbitals, so it is \(\displaystyle \tfrac{1}{3} \) s-character, i.e. about $\displaystyle 33$%. An sp\(\displaystyle ^3\) orbital is built from one s and three p orbitals shared over four hybrid orbitals, so it is \(\displaystyle \tfrac{1}{4} \) s-character, i.e. $\displaystyle 25$%.\[\text{sp (50\% s)} \;>\; \text{sp}^2 \text{(33\% s)} \;>\; \text{sp}^3 \text{(25\% s)} \]This is the step people skip: s-character is not a side detail, it is the whole reason. An s orbital is spherical and sits close to the nucleus, while a p orbital is directional and extends further out. A hybrid orbital with more s-character is therefore held closer to, and more tightly by, the carbon nucleus. The electron pair of a C–H bond built from such an orbital is pulled in toward carbon, which makes carbon relatively more electronegative toward that hydrogen. A more electronegative carbon polarises the C–H bond more strongly, so it lets go of that hydrogen as \(\displaystyle \text{H}^+ \) more readily — the bond is more acidic.Step $\displaystyle 3$: Rank the acidity in the same order as the % s-character.Since ethyne's carbon is sp ($\displaystyle 50$% s), benzene's is sp\(\displaystyle ^2\) ($\displaystyle 33$% s), and n-hexane's is sp\(\displaystyle ^3\) ($\displaystyle 25$% s), the acidity of the C–H bond falls in exactly that order:\[\text{ethyne} \;>\; \text{benzene} \;>\; \text{n-hexane} \]This also matches what is observed experimentally: ethyne is acidic enough to be deprotonated by strong bases and even displaces hydrogen from sodium metal or from ammoniacal \(\displaystyle \text{AgNO}_3 \)/\(\displaystyle \text{Cu}_2\text{Cl}_2 \) solutions (forming a precipitate), benzene is only very weakly acidic, and the fully saturated n-hexane shows essentially no acidic character at all under ordinary conditions.Answer: Decreasing order of acidic behaviour — ethyne > benzene > n-hexane. This follows the decreasing % s-character of the hybrid orbital on the C–H carbon (sp, $\displaystyle 50$% > sp\(\displaystyle ^2\), $\displaystyle 33$% > sp\(\displaystyle ^3\), $\displaystyle 25$%): more s-character pulls the bonding electrons closer to carbon, polarising the C–H bond more and making it easier to lose \(\displaystyle \text{H}^+ \).

Benzene's ring carries a delocalized cloud of \(\displaystyle \pi\) electrons that is electron-rich on both faces — it pulls in electron-seeking electrophiles but pushes away electron-rich nucleophiles.Start from benzene's structure. Every ring carbon is \(\displaystyle sp^2\) hybridised: each forms two \(\displaystyle \sigma\) bonds to its neighbouring carbons and one \(\displaystyle \sigma\) bond to hydrogen, all in the plane of the ring. That leaves one unhybridised \(\displaystyle p\) orbital on every carbon, standing perpendicular to the ring. These six \(\displaystyle p\) orbitals overlap sideways all the way around the ring, and the six electrons in them are not tied to any one carbon — they spread out into two continuous, doughnut-shaped clouds, one above the plane of the ring and one below it. This is what "delocalized \(\displaystyle \pi\) system" means: the negative charge of these six electrons is smeared evenly over the whole ring instead of sitting in three separate double bonds.Why electrophiles attack easily. An electrophile, symbol \(\displaystyle E^+\), is any electron-deficient or positively charged species (for example \(\displaystyle Br^+\), \(\displaystyle NO_2^+\), or \(\displaystyle SO_3H^+\), generated in the usual substitution reactions of benzene). Because the faces of the benzene ring are unusually rich in electron density, an approaching \(\displaystyle E^+\) is electrostatically attracted to that cloud from above or below the ring — the same way a positive ion is drawn toward any region of high electron density. The electrophile bonds to one ring carbon, pulling a pair of \(\displaystyle \pi\) electrons out of the delocalized system and forming a positively charged intermediate called the arenium ion (or benzenonium ion), in which that one carbon has become \(\displaystyle sp^3\) while the other five carbons still share the remaining four \(\displaystyle \pi\) electrons:\[C_6H_6 \;+\; E^{+} \;\longrightarrow\; \left[C_6H_6E\right]^{+} \;\longrightarrow\; C_6H_5E \;+\; H^{+} \]Check this is balanced: on the left, \(\displaystyle C_6H_6\) ($\displaystyle 6$ carbons, $\displaystyle 6$ hydrogens) plus \(\displaystyle E^+\) gives a total charge of \(\displaystyle +1\); the arenium ion \(\displaystyle [C_6H_6E]^+\) has the same $\displaystyle 6$ carbons, $\displaystyle 6$ hydrogens, one \(\displaystyle E\), and charge \(\displaystyle +1\) — nothing is lost forming it. In the second step this ion loses one proton, \(\displaystyle H^+\), from the \(\displaystyle sp^3\) carbon: \(\displaystyle C_6H_6E \to C_6H_5E + H\), so $\displaystyle 6$ hydrogens on the left match $\displaystyle 5$ + $\displaystyle 1$ on the right, and charge \(\displaystyle +1\) on the left matches \(\displaystyle 0 + 1\) on the right. Losing that proton lets the carbon that held it go back to \(\displaystyle sp^2\) and rejoin the ring's \(\displaystyle \pi\) system, restoring full delocalization over all six carbons — the aromatic sextet is regenerated. This is the step that is easy to miss: the reaction is a substitution, not an addition, precisely because losing \(\displaystyle H^+\) afterwards is what gives back the extra stability (the resonance/delocalization energy) that addition would otherwise destroy. An addition reaction, by contrast, would leave two ring carbons permanently \(\displaystyle sp^3\) and break the continuous \(\displaystyle \pi\) cloud, which is energetically unfavourable — so benzene resists addition and instead substitutes.Why nucleophiles attack with difficulty. A nucleophile, symbol \(\displaystyle Nu^-\), is an electron-rich species (for example \(\displaystyle OH^-\), \(\displaystyle CN^-\), or \(\displaystyle NH_2^-\)) that normally attacks a carbon bearing a partial or full positive charge, such as the carbon attached to a halogen in an alkyl halide \(\displaystyle R\text{–}X\), where the more electronegative \(\displaystyle X\) pulls electron density away from carbon and leaves it electron-deficient. Benzene offers no such site: every ring carbon sits inside the same electron-rich delocalized \(\displaystyle \pi\) cloud, with no carbon left short of electron density for a nucleophile to be drawn toward. An incoming \(\displaystyle Nu^-\) is therefore repelled by the ring's own electron cloud rather than attracted to it — an electron-rich species approaching an already electron-rich surface is simply working against electrostatic repulsion, so there is no low-energy pathway for the nucleophile to bond to a ring carbon. (Nucleophilic substitution on an aromatic ring only becomes feasible when strong electron-withdrawing groups are attached to the ring first, which pull electron density away from a particular carbon and make it electron-deficient enough for a nucleophile to attack — but that is a special, activated case, not how unsubstituted benzene behaves.)Answer: Benzene's six ring carbons are all \(\displaystyle sp^2\), each leaving one \(\displaystyle p\) orbital that overlaps continuously around the ring; the six \(\displaystyle \pi\) electrons in these orbitals are delocalized into electron-rich clouds above and below the ring plane. This electron-rich \(\displaystyle \pi\) system attracts electron-deficient electrophiles (\(\displaystyle E^+\)), which add to a ring carbon to give an arenium ion intermediate that then loses \(\displaystyle H^+\) to restore the aromatic sextet — an easy, favourable substitution (\(\displaystyle C_6H_6 + E^+ \rightarrow [C_6H_6E]^+ \rightarrow C_6H_5E + H^+\)) that also preserves the ring's extra delocalization (resonance) stability, which addition would destroy. The same electron-rich cloud repels electron-rich nucleophiles (\(\displaystyle Nu^-\)), and since no ring carbon is left electron-deficient the way the carbon in \(\displaystyle R\text{–}X\) is, there is no site for a nucleophile to be drawn to — so nucleophilic substitution on plain benzene proceeds only with great difficulty.

Three carbons short of a ring, or already a chain of six — you either join small pieces into the ring or strip hydrogens off a chain that's already the right length.Benzene, \(\displaystyle \mathrm{C_6H_6}\), needs six carbons arranged in a ring with three double bonds (its degree of unsaturation is $\displaystyle 4$: one ring + three π bonds). The three starting materials give three different routes to that same ring.(i) Ethyne \(\displaystyle \rightarrow\) benzeneEthyne (acetylene), \(\displaystyle \mathrm{HC \equiv CH}\), has two carbons — you need three molecules of it to make the six-carbon ring. When ethyne vapour is passed through a red-hot iron tube at \(\displaystyle 873\ \mathrm{K}\), three molecules undergo cyclic polymerisation (they join head-to-tail and close into a ring):\[3\ \mathrm{HC \equiv CH} \xrightarrow[\text{red-hot iron tube}]{873\ \mathrm{K}} \mathrm{C_6H_6\ (benzene)} \]Check the balance: left side has \(\displaystyle 3 \times 2 = 6\) carbon atoms and \(\displaystyle 3 \times 2 = 6\) hydrogen atoms; benzene has $\displaystyle 6$ carbons and $\displaystyle 6$ hydrogens. Nothing is added or removed — the three triple bonds simply rearrange into the aromatic ring's alternating framework, so this equation is already balanced by matching atom counts on both sides.(ii) Ethene \(\displaystyle \rightarrow\) benzeneEthene, \(\displaystyle \mathrm{CH_2=CH_2}\), is fully saturated on each carbon except for its one double bond — it will not trimerise directly the way ethyne does. The route is to first turn ethene into ethyne, then trimerise the ethyne exactly as in part (i). That takes two steps.Step $\displaystyle 1$ — addition of bromine. Ethene adds bromine across its double bond to give $\displaystyle 1,2$-dibromoethane (ethylene dibromide):\[\mathrm{CH_2=CH_2} + \mathrm{Br_2} \longrightarrow \mathrm{BrCH_2-CH_2Br\ (1,2\text{-}dibromoethane)} \]Carbon: \(\displaystyle 2=2\); hydrogen: \(\displaystyle 4=4\); bromine: \(\displaystyle 2=2\) — balanced as written, since this is a simple $\displaystyle 1$:$\displaystyle 1$ addition.Step $\displaystyle 2$ — double dehydrohalogenation. Heating $\displaystyle 1,2$-dibromoethane with excess alcoholic potassium hydroxide eliminates two molecules of \(\displaystyle \mathrm{HBr}\) (one from each carbon) to give ethyne:\[\mathrm{BrCH_2-CH_2Br} + 2\,\mathrm{KOH(alc)} \longrightarrow \mathrm{HC \equiv CH\ (ethyne)} + 2\,\mathrm{KBr} + 2\,\mathrm{H_2O} \]Balance check: carbon \(\displaystyle 2=2\); hydrogen, left \(\displaystyle 4+2=6\), right \(\displaystyle 2+4=6\); bromine \(\displaystyle 2=2\); potassium \(\displaystyle 2=2\); oxygen \(\displaystyle 2=2\). This is where it is easy to under-count — using only one equivalent of \(\displaystyle \mathrm{KOH}\) would remove just one \(\displaystyle \mathrm{HBr}\) and stop at a mono-bromoalkene, not the alkyne, so the "$\displaystyle 2$" in front of \(\displaystyle \mathrm{KOH}\) is essential, not a rounding choice.Step $\displaystyle 3$ — trimerise the ethyne, exactly as in part (i):\[3\ \mathrm{HC \equiv CH} \xrightarrow[\text{red-hot iron tube}]{873\ \mathrm{K}} \mathrm{C_6H_6\ (benzene)} \]So the full path is: ethene \(\displaystyle \to\) $\displaystyle 1,2$-dibromoethane \(\displaystyle \to\) ethyne \(\displaystyle \to\) benzene.(iii) Hexane \(\displaystyle \rightarrow\) benzenen-Hexane, \(\displaystyle \mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3}\) (\(\displaystyle \mathrm{C_6H_{14}}\)), already has all six carbons in a chain — nothing needs joining. It only needs to close into a ring and lose hydrogen atoms until three double bonds appear. This is done by passing hexane vapour over a chromium oxide–alumina (or platinum–alumina) catalyst at about \(\displaystyle 773\ \mathrm{K}\) under high pressure — a process called aromatisation (also called reforming), which is a combined cyclisation and dehydrogenation:\[\mathrm{C_6H_{14}} \xrightarrow[\ 773\ \mathrm{K},\ \text{high pressure}\ ]{\mathrm{Cr_2O_3/Al_2O_3}} \mathrm{C_6H_6\ (benzene)} + 4\,\mathrm{H_2} \]Balance check: carbon \(\displaystyle 6=6\); hydrogen, left \(\displaystyle 14\), right \(\displaystyle 6 + 4\times2 = 6+8=14\). This is the step where students often guess the wrong number of \(\displaystyle \mathrm{H_2}\) molecules — the shortcut is to use the degree-of-unsaturation formula \(\displaystyle \mathrm{DoU} = \dfrac{2C+2-H}{2}\), where \(\displaystyle C\) and \(\displaystyle H\) are the numbers of carbon and hydrogen atoms in the molecule. For hexane, \(\displaystyle \mathrm{DoU} = \dfrac{2(6)+2-14}{2} = 0\) (a fully saturated open chain). For benzene, \(\displaystyle \mathrm{DoU} = \dfrac{2(6)+2-6}{2} = 4\) (one ring plus three double bonds). The difference, \(\displaystyle 4-0=4\), is exactly the number of \(\displaystyle \mathrm{H_2}\) molecules that must be removed, because losing one \(\displaystyle \mathrm{H_2}\) raises the degree of unsaturation by exactly 1.Answer: (i) \(\displaystyle 3\,\mathrm{HC{\equiv}CH} \xrightarrow[873\text{ K}]{\text{red-hot Fe tube}} \mathrm{C_6H_6}\) — cyclic trimerisation of ethyne. (ii) Ethene \(\displaystyle \xrightarrow{\mathrm{Br_2}}\) $\displaystyle 1,2$-dibromoethane \(\displaystyle \xrightarrow{2\,\mathrm{KOH(alc)},\,-2\mathrm{HBr}}\) ethyne, then \(\displaystyle 3\,\mathrm{HC{\equiv}CH} \to \mathrm{C_6H_6}\) as in (i). (iii) \(\displaystyle \mathrm{C_6H_{14}} \xrightarrow[773\text{ K, high pressure}]{\mathrm{Cr_2O_3/Al_2O_3}} \mathrm{C_6H_6} + 4\,\mathrm{H_2}\) — catalytic aromatisation (reforming) of n-hexane.

Hydrogenation just adds one \(\displaystyle H_2\) molecule across the \(\displaystyle C=C\) double bond — so to reverse it, look at every C–C single bond in the saturated skeleton and ask "could this have been the double bond that got saturated?"First write out $\displaystyle 2$-methylbutane and number its carbons so nothing gets lost:\[\underset{C_1}{CH_3}-\underset{C_2}{CH}(-\underset{C_b}{CH_3})-\underset{C_3}{CH_2}-\underset{C_4}{CH_3} \]It is a $\displaystyle 4$-carbon chain (butane) carrying one methyl branch on \(\displaystyle C_2\). There are exactly four C–C single bonds where a double bond could sit: \(\displaystyle C_1\!-\!C_2\), \(\displaystyle C_2\!-\!C_b\), \(\displaystyle C_2\!-\!C_3\), and \(\displaystyle C_3\!-\!C_4\). Each alkene has the molecular formula \(\displaystyle C_5H_{10}\) — one degree of unsaturation, one \(\displaystyle C=C\) — so each equation below balances simply because adding one \(\displaystyle H_2\) turns \(\displaystyle C_5H_{10}\) into \(\displaystyle C_5H_{12}\): carbon count stays at $\displaystyle 5$, and hydrogen goes \(\displaystyle 10+2=12\), matching $\displaystyle 2$-methylbutane exactly.Double bond at \(\displaystyle C_1\!-\!C_2\) (or, equivalently, at \(\displaystyle C_2\!-\!C_b\)). \(\displaystyle C_1\) and \(\displaystyle C_b\) are both plain methyl groups hanging off \(\displaystyle C_2\), so moving the double bond from one to the other gives the same molecule — this is the step people miss, and it's why the count of distinct alkenes is $\displaystyle 3$, not 4. Removing one H from \(\displaystyle C_2\) and one H from \(\displaystyle C_1\) (or from \(\displaystyle C_b\)) gives:\[CH_2=C(CH_3)-CH_2-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH(CH_3)-CH_2-CH_3 \]This alkene is $\displaystyle 2$-methylbut-$\displaystyle 1$-ene.Double bond at \(\displaystyle C_2\!-\!C_3\). Remove one H from \(\displaystyle C_2\) and one H from \(\displaystyle C_3\):\[CH_3-CH=C(CH_3)-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH(CH_3)-CH_2-CH_3 \]Both carbons of this double bond carry a \(\displaystyle CH_3\) on the same side (\(\displaystyle C_2\) has \(\displaystyle C_1\)'s methyl and \(\displaystyle C_b\)'s methyl, which are identical), so there is no cis/trans pair to worry about — only one compound exists here. This is $\displaystyle 2$-methylbut-$\displaystyle 2$-ene.Double bond at \(\displaystyle C_3\!-\!C_4\). Remove one H from \(\displaystyle C_3\) and one H from \(\displaystyle C_4\):\[CH_2=CH-CH(CH_3)-CH_3 \;+\; H_2 \xrightarrow{\text{Ni}} CH_3-CH_2-CH(CH_3)-CH_3 \](written from the other end, this is the same target molecule, \(\displaystyle CH_3-CH(CH_3)-CH_2-CH_3\)). Since \(\displaystyle C_4\) becomes a terminal \(\displaystyle =CH_2\) with two identical H's, again no cis/trans isomer arises. This is $\displaystyle 3$-methylbut-$\displaystyle 1$-ene.Checking that nothing was missed: those four bonds are the only C–C single bonds in the skeleton, and the first two collapse into one product by symmetry, so three is the complete list.Answer: There are exactly three such alkenes — $\displaystyle 2$-methylbut-$\displaystyle 1$-ene, \(\displaystyle CH_2=C(CH_3)CH_2CH_3\); $\displaystyle 2$-methylbut-$\displaystyle 2$-ene, \(\displaystyle CH_3CH=C(CH_3)CH_3\); and $\displaystyle 3$-methylbut-$\displaystyle 1$-ene, \(\displaystyle CH_2=CHCH(CH_3)CH_3\) — each of which adds one molecule of \(\displaystyle H_2\) (Ni/Pt/Pd catalyst) to give $\displaystyle 2$-methylbutane, \(\displaystyle CH_3CH(CH_3)CH_2CH_3\).

Reactivity toward an electrophile \(\displaystyle E^+\) tracks how much electron density is sitting on the ring — a substituent that pushes electron density in speeds up attack, a substituent that pulls electron density out slows it down.Electrophilic aromatic substitution starts with \(\displaystyle E^+\) attacking the \(\displaystyle \pi\)-cloud of the ring. Whatever raises the electron density of that \(\displaystyle \pi\)-cloud above benzene's own level makes the ring a better nucleophile toward \(\displaystyle E^+\) (faster reaction); whatever lowers it makes the ring a poorer nucleophile (slower reaction). So to rank a set of substituted benzenes, look at every substituent already on the ring and ask whether it is a net electron-donor or a net electron-withdrawer, and how many of each are present.(a) Chlorobenzene, \(\displaystyle p\)-nitrochlorobenzene, $\displaystyle 2,4$-dinitrochlorobenzeneThese three are: \(\displaystyle \text{C}_6\text{H}_5\text{Cl}\) (one Cl on the ring), \(\displaystyle \text{C}_6\text{H}_4(\text{Cl})(\text{NO}_2)\) (one Cl and one \(\displaystyle \text{NO}_2\), para to each other), and \(\displaystyle \text{C}_6\text{H}_3(\text{Cl})(\text{NO}_2)_2\) (one Cl and two \(\displaystyle \text{NO}_2\) groups, at the $\displaystyle 2$- and $\displaystyle 4$-positions).Chlorine carries a lone pair that it can donate into the ring by resonance (this is exactly why Cl is an ortho/para director — that donation puts extra electron density, and so extra stability for the intermediate cation, at the ortho and para carbons). At the same time, chlorine is more electronegative than carbon and pulls electron density out of the ring through the sigma framework (its inductive, \(\displaystyle -I\), effect). For chlorobenzene the inductive pull slightly outweighs the resonance donation, so the ring ends up a little less reactive than plain benzene — but chlorine is still by far the mildest deactivator of the three substituents in this question.A short aside, because this is the step most people get backwards: "ortho/para-directing" and "activating" are not the same claim. Chlorine directs to ortho/para (resonance argument, about where the intermediate is most stable) while still net deactivating the ring relative to benzene (inductive argument, about how fast overall). Both are true of the same atom at once.\(\displaystyle \text{NO}_2\), by contrast, has no lone pair to donate. Its nitrogen is doubly bonded to one oxygen and carries a formal positive charge balanced by the other, so it withdraws ring electron density both by induction (\(\displaystyle -I\), through the sigma bonds) and by resonance (\(\displaystyle -M\): the ring's own \(\displaystyle \pi\) electrons can delocalize onto the electronegative oxygens, most strongly draining density from the ortho and para carbons — which is why \(\displaystyle \text{NO}_2\) is a meta director and a strong deactivator, not a mild one like Cl).So each successive compound adds a strongly-deactivating \(\displaystyle \text{NO}_2\) group on top of the same mildly-deactivating Cl: Each added \(\displaystyle \text{NO}_2\) drains more electron density from the ring, so each added \(\displaystyle \text{NO}_2\) makes the ring a slower target for \(\displaystyle E^+\):\[\text{Chlorobenzene} \;>\; p\text{-Nitrochlorobenzene} \;>\; 2,4\text{-Dinitrochlorobenzene} \](b) Toluene, \(\displaystyle p\)-nitrotoluene, \(\displaystyle p\)-dinitrobenzeneThese are: \(\displaystyle \text{C}_6\text{H}_5\text{CH}_3\) (one \(\displaystyle \text{CH}_3\) on the ring), \(\displaystyle \text{C}_6\text{H}_4(\text{CH}_3)(\text{NO}_2)\) (one \(\displaystyle \text{CH}_3\) and one \(\displaystyle \text{NO}_2\), para to each other, written in the question as \(\displaystyle p\text{-H}_3\text{C–C}_6\text{H}_4\text{–NO}_2\)), and \(\displaystyle \text{C}_6\text{H}_4(\text{NO}_2)_2\) (two \(\displaystyle \text{NO}_2\) groups, para to each other, written as \(\displaystyle p\text{-O}_2\text{N–C}_6\text{H}_4\text{–NO}_2\)).The methyl group has no lone pair either, but its \(\displaystyle \text{C–H}\) sigma electrons can overlap with the ring's \(\displaystyle \pi\) system (hyperconjugation), and this pushes electron density into the ring rather than pulling it out. So \(\displaystyle \text{CH}_3\) is a genuine electron donor — it makes toluene's ring more electron-rich than benzene's, hence more reactive toward \(\displaystyle E^+\), and it directs ortho/para for the same reason (extra density, and extra cation stability, at those positions).Now compare the three rings directly: \[\text{Toluene} \;>\; p\text{-Nitrotoluene} \;>\; p\text{-Dinitrobenzene} \]The two lists run on the same logic: count the electron-donors, count the electron-withdrawers (and how strong each one is), and the ring with the most net electron density reacts fastest with \(\displaystyle E^+\).Answer: (a) Chlorobenzene > \(\displaystyle p\)-Nitrochlorobenzene > $\displaystyle 2,4$-Dinitrochlorobenzene; (b) Toluene > \(\displaystyle p\)-Nitrotoluene > \(\displaystyle p\)-Dinitrobenzene — in both series, reactivity falls as strongly electron-withdrawing \(\displaystyle \text{NO}_2\) groups are added, since each one drains electron density from the ring and makes it a slower target for the electrophile.

Nitration is an electrophilic attack on the ring, so whichever ring has the most electron density reacts fastest.How the electrophile is made and what it doesNitration replaces a ring hydrogen with a nitro group, \(\displaystyle -NO_2\). The actual attacking species is not nitric acid itself but the nitronium ion, \(\displaystyle NO_2^+\), generated when concentrated nitric acid is protonated by concentrated sulfuric acid:\[HNO_3 + 2\,H_2SO_4 \rightarrow NO_2^{+} + H_3O^{+} + 2\,HSO_4^{-} \]Check the balance atom by atom: left side has \(\displaystyle H = 1 + 2(2) = 5\), \(\displaystyle N = 1\), \(\displaystyle O = 3 + 2(4) = 11\), \(\displaystyle S = 2\); right side has \(\displaystyle H = 0 + 3 + 2(1) = 5\), \(\displaystyle N = 1\), \(\displaystyle O = 2 + 1 + 2(4) = 11\), \(\displaystyle S = 2\). Every atom matches, and the charges balance too: the left side is neutral, and the right side is \(\displaystyle (+1) + (+1) + 2(-1) = 0\).Once \(\displaystyle NO_2^{+}\) is formed, it attacks a ring carbon. This is the slow, rate-determining step — the ring's electrons attack the electrophile and form a positively charged intermediate (the arenium ion, or sigma complex), which then loses \(\displaystyle \mathrm{H^{+}}\) to restore the aromatic ring and give the nitro compound:\[C_6H_6 + NO_2^{+} \rightarrow C_6H_5NO_2 + H^{+} \]This is already balanced: \(\displaystyle C = 6\) and \(\displaystyle H = 6\) on both sides once you count the \(\displaystyle H^+\) released, \(\displaystyle N = 1\), \(\displaystyle O = 2\), and the charge goes from \(\displaystyle +1\) on the left to \(\displaystyle +1\) on the right (the proton). The product, \(\displaystyle C_6H_5NO_2\), is nitrobenzene.Because the slow step is the electrophile attacking the ring, anything that makes the ring more electron-rich lowers the energy of that step and speeds up nitration; anything that makes the ring electron-poor does the opposite. This is the whole question.Toluene — the methyl group pushes electrons inToluene is \(\displaystyle C_6H_5CH_3\), benzene with one \(\displaystyle H\) replaced by a methyl group, \(\displaystyle -CH_3\). The \(\displaystyle C-H\) bonds of that methyl group can align with the ring's \(\displaystyle \pi\) system and donate a little electron density into it — a no-bond delocalization called hyperconjugation — with a small additional push from the inductive (\(\displaystyle +I\)) effect of the alkyl group. This raises the electron density of the ring, especially at the carbons ortho and para to the methyl group, so the arenium-ion intermediate formed when \(\displaystyle NO_2^+\) attacks there is extra stabilized. That is why toluene nitrates markedly faster than benzene itself.m-Dinitrobenzene — two nitro groups pull electrons outm-Dinitrobenzene, \(\displaystyle C_6H_4(NO_2)_2\), already carries two \(\displaystyle -NO_2\) groups. The nitro group is strongly electron-withdrawing in two ways at once: inductively (\(\displaystyle -I\)), because the electronegative \(\displaystyle N\) and \(\displaystyle O\) atoms pull electron density through the sigma bonds, and by resonance (\(\displaystyle -M\)), because the nitrogen can draw a ring electron pair into the \(\displaystyle N=O\) system, leaving the ring carbons ortho and para to it short of electrons. With two such groups already deactivating the ring, its electron density is far below that of plain benzene, so the arenium-ion intermediate for a third nitration is much less stable and forms much more slowly. (This is the step people get backwards: it is tempting to think a ring that already has nitro groups on it should accept another one just as easily — it does not, because a nitro group is a deactivator, not an activator, for further substitution.)Putting the three in orderRanking by how easily each accepts a further nitro group, from fastest to slowest:\[\text{toluene} > \text{benzene} > \text{m-dinitrobenzene} \]Toluene's electron-donating methyl group speeds up attack by \(\displaystyle NO_2^+\); benzene is the unactivated, undeactivated reference; and m-dinitrobenzene's two electron-withdrawing nitro groups strongly resist further attack by the same electrophile.Answer: Toluene undergoes nitration most easily, because its \(\displaystyle -CH_3\) group donates electron density into the ring (by hyperconjugation and the \(\displaystyle +I\) effect), stabilizing the arenium-ion intermediate formed by attack of \(\displaystyle NO_2^{+}\); benzene is intermediate; m-dinitrobenzene is hardest to nitrate because its two \(\displaystyle -NO_2\) groups withdraw electron density from the ring by both the \(\displaystyle -I\) and \(\displaystyle -M\) effects, so the order of ease of nitration is toluene > benzene > m-dinitrobenzene.

Ethylation of benzene is Friedel–Crafts alkylation, and the catalyst's real job is generating a carbocation — any strong enough Lewis acid can do that, not just \(\displaystyle \mathrm{AlCl_3} \).In this reaction, benzene, \(\displaystyle \mathrm{C_6H_6} \) (six \(\displaystyle \mathrm{CH}\) units joined in the aromatic ring), is treated with ethyl chloride, \(\displaystyle \mathrm{C_2H_5Cl} \) (chloroethane, the alkylating agent supplying the ethyl group), in the presence of a catalyst. Anhydrous \(\displaystyle \mathrm{AlCl_3} \) is the catalyst usually named in the textbook, but its role is purely that of a Lewis acid — a species with an empty orbital that can accept the lone pair on the chlorine of \(\displaystyle \mathrm{C_2H_5Cl} \) and pull the \(\displaystyle \mathrm{Cl} \) away as a chloride complex. Once that happens, the ethyl group is left behind as an ethyl carbocation, \(\displaystyle \mathrm{CH_3CH_2^+} \), which is the actual electrophile that attacks the ring.Because the job is just "accept a chloride ion strongly enough to ionise the C–Cl bond," several other metal halides do it equally well. The one most commonly used as a substitute, and the one NCERT accepts as the answer, is anhydrous ferric chloride, \(\displaystyle \mathrm{FeCl_3} \) (iron(III) chloride — the iron centre has the same kind of electron-deficient, empty-orbital character as aluminium in \(\displaystyle \mathrm{AlCl_3} \)). Anhydrous \(\displaystyle \mathrm{BF_3} \) (boron trifluoride), anhydrous \(\displaystyle \mathrm{SnCl_4} \) (tin(IV) chloride), and anhydrous \(\displaystyle \mathrm{ZnCl_2} \) (zinc chloride) work the same way and are also acceptable answers.The word "anhydrous" is doing real work here, not decoration. If the catalyst is hydrated, water molecules occupy the empty orbital on the metal first, satisfying its Lewis acidity before it ever meets the alkyl halide — the catalyst is effectively dead. This is a step people skip: naming the right metal halide is not enough; it must be moisture-free to ionise the \(\displaystyle \mathrm{C-Cl} \) bond at all.With \(\displaystyle \mathrm{FeCl_3} \) as the catalyst, the overall ethylation is\[\mathrm{C_6H_6 + C_2H_5Cl \xrightarrow{\text{anhydrous } FeCl_3} C_6H_5CH_2CH_3 + HCl} \]Balancing check: count atoms on each side — left side has \(\displaystyle \mathrm{C}: 6+2=8 \), \(\displaystyle \mathrm{H}: 6+5=11 \), \(\displaystyle \mathrm{Cl}: 1 \); right side, ethylbenzene \(\displaystyle \mathrm{C_6H_5CH_2CH_3} \) contributes \(\displaystyle \mathrm{C_8H_{10}} \) and the by-product \(\displaystyle \mathrm{HCl} \) contributes one more \(\displaystyle \mathrm{H} \) and one \(\displaystyle \mathrm{Cl} \), giving \(\displaystyle \mathrm{C}: 8 \), \(\displaystyle \mathrm{H}: 10+1=11 \), \(\displaystyle \mathrm{Cl}: 1 \). Both sides match with all coefficients equal to $\displaystyle 1$, so the equation is already balanced as written.Mechanistically: \(\displaystyle \mathrm{FeCl_3} \) coordinates to the chlorine of \(\displaystyle \mathrm{C_2H_5Cl} \), generating the ethyl carbocation \(\displaystyle \mathrm{CH_3CH_2^+} \) and the complex ion \(\displaystyle \mathrm{FeCl_4^-} \). The carbocation is attacked by the \(\displaystyle \pi \) electrons of benzene to give an arenium (sigma-complex) intermediate, which then loses a proton to reform the aromatic ring — giving ethylbenzene, \(\displaystyle \mathrm{C_6H_5CH_2CH_3} \) (benzene with one hydrogen replaced by an ethyl group). The lost proton combines with \(\displaystyle \mathrm{FeCl_4^-} \) to release \(\displaystyle \mathrm{HCl} \) and regenerate \(\displaystyle \mathrm{FeCl_3} \), so the catalyst is not consumed overall.**Answer: Anhydrous ferric chloride, \(\displaystyle \mathrm{FeCl_3} \), can be used in place of anhydrous \(\displaystyle \mathrm{AlCl_3} \) as the Lewis acid catalyst for ethylation of benzene (anhydrous \(\displaystyle \mathrm{BF_3} \), \(\displaystyle \mathrm{SnCl_4} \), or \(\displaystyle \mathrm{ZnCl_2} \) work the same way); it ionises \(\displaystyle \mathrm{C_2H_5Cl} \) into the electrophilic ethyl carbocation that attacks benzene, giving \(\displaystyle \mathrm{C_6H_6 + C_2H_5Cl \xrightarrow{FeCl_3} C_6H_5CH_2CH_3 + HCl} \).

The Wurtz reaction always joins two identical alkyl groups, so it can only build alkanes with an even number of carbons cleanly — an odd-carbon alkane forces you to mix two different halides, and that mixture reacts three ways at once.The Wurtz reaction is: \[2\,R\text{–}X + 2\,Na \xrightarrow{\text{dry ether}} R\text{–}R + 2\,NaX \] Here \(\displaystyle R\) is an alkyl group, \(\displaystyle X\) is a halogen (usually Br or Cl), and the two \(\displaystyle R\text{–}X\) molecules that come together must be identical — sodium removes the halogen from each and the two carbon fragments join to give a single symmetrical alkane \(\displaystyle R\text{–}R\), which always has an even number of carbon atoms (two identical halves always add to an even total).The step people get wrong: to make an odd-carbon alkane you cannot use one halide alone — you must react two different alkyl halides together, say bromomethane, \(\displaystyle CH_3Br\), and bromoethane, \(\displaystyle C_2H_5Br\), aiming for propane, \(\displaystyle C_3H_8\). But sodium does not know to pick one \(\displaystyle CH_3Br\) and one \(\displaystyle C_2H_5Br\) every time. All three possible pairings happen at once, statistically:Self-coupling of bromomethane gives ethane: \[2\,CH_3Br + 2\,Na \longrightarrow CH_3\text{–}CH_3 + 2\,NaBr \] (balance check: left side has \(\displaystyle 2\,C,\ 6\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_2H_6\) plus \(\displaystyle 2\,NaBr\))Self-coupling of bromoethane gives n-butane: \[2\,C_2H_5Br + 2\,Na \longrightarrow C_2H_5\text{–}C_2H_5 + 2\,NaBr \] (balance check: left side has \(\displaystyle 4\,C,\ 10\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_4H_{10}\) plus \(\displaystyle 2\,NaBr\))Cross-coupling, the one you actually wanted, gives propane: \[CH_3Br + C_2H_5Br + 2\,Na \longrightarrow CH_3\text{–}C_2H_5 + 2\,NaBr \] (balance check: left side has \(\displaystyle 3\,C,\ 8\,H,\ 2\,Br,\ 2\,Na\); right side has the same — \(\displaystyle C_3H_8\) plus \(\displaystyle 2\,NaBr\))So the flask ends up with a mixture of three alkane gases — ethane, propane, and n-butane — along with sodium bromide. Since these three hydrocarbons are all low-boiling gases with boiling points close together, separating pure propane from the ethane and butane formed alongside it is impractical, and the yield of the alkane you actually wanted is low. That is why the Wurtz reaction is a good route to even-carbon alkanes (where only one alkyl halide, reacting with itself, is needed) but a poor one for odd-carbon alkanes (where a second, unwanted, symmetrical coupling is unavoidable alongside the cross product).Answer: The Wurtz reaction couples two alkyl halide molecules into one symmetrical alkane \(\displaystyle R\text{–}R\), so a single halide always gives an even-carbon product. To reach an odd-carbon alkane like propane, two different halides — \(\displaystyle CH_3Br\) and \(\displaystyle C_2H_5Br\) — must be used together, but sodium then couples them in all three possible ways at once: \(\displaystyle 2CH_3Br+2Na\rightarrow C_2H_6+2NaBr\) (ethane), \(\displaystyle 2C_2H_5Br+2Na\rightarrow C_4H_{10}+2NaBr\) (n-butane), and the wanted \(\displaystyle CH_3Br+C_2H_5Br+2Na\rightarrow C_3H_8+2NaBr\) (propane). The three gaseous alkanes formed are hard to separate from each other, so the method gives a poor, impure yield of the odd-carbon alkane — which is why Wurtz synthesis is not preferred for odd-carbon-number alkanes.