NCERT Solutions · Class 12 Chemistry Alcohols, Phenols and Ethers

Each part asks for a specific named reagent — the trick is matching the reagent's strength (and dryness) to exactly how far the oxidation, substitution, or reduction is meant to go.(i) Oxidation of a primary alcohol to a carboxylic acidReagent: acidified potassium permanganate, \(\displaystyle \text{KMnO}_4/\text{H}^+ \) (warm) — acidified potassium dichromate, \(\displaystyle \text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4 \), also works.A primary alcohol \(\displaystyle \text{R–CH}_2\text{OH} \) is oxidised first to the aldehyde \(\displaystyle \text{R–CHO} \). In the aqueous, strongly oxidising medium the aldehyde is not isolated — it hydrates to the gem-diol \(\displaystyle \text{R–CH(OH)}_2 \), which the excess oxidant dehydrogenates straight on to the carboxylic acid \(\displaystyle \text{R–COOH} \). Because water and excess oxidant are both present, the reaction cannot be stopped at the aldehyde stage.(ii) Oxidation of a primary alcohol to an aldehyde (stopping there)Reagent: pyridinium chlorochromate, PCC, \(\displaystyle \text{C}_5\text{H}_5\text{NH}^+\,\text{CrO}_3\text{Cl}^- \), in anhydrous dichloromethane.PCC is a mild chromium(VI) oxidant used in a dry, non-aqueous solvent. With no water in the medium, the aldehyde \(\displaystyle \mathrm{ \text{R–CHO} }\) formed cannot hydrate to \(\displaystyle \text{R–CH(OH)}_2\) and it is the hydrate, not the aldehyde itself, that gets oxidised further — so the reaction halts cleanly at the aldehyde.(iii) Bromination of phenol to $\displaystyle 2,4,6$-tribromophenol Reagent: bromine water, \(\displaystyle \text{Br}_2(aq) \), at room temperature — no catalyst.The –OH group of phenol donates its lone pair into the ring by resonance, making the ring strongly electron-rich and directing incoming electrophiles to the ortho and para positions. This activation is powerful enough that even dilute aqueous bromine substitutes at all three open ortho/para positions (C-$\displaystyle 2$, C-$\displaystyle 4$, C-$\displaystyle 6$) without needing a Lewis-acid catalyst such as \(\displaystyle \text{FeBr}_3 \) — unlike benzene, which needs one for bromination to occur at all.(iv) Benzyl alcohol to benzoic acid Reagent: alkaline \(\displaystyle \text{KMnO}_4 \), heated, then acidified with dilute HCl (hot acidified \(\displaystyle \text{KMnO}_4 \) directly also works).Benzyl alcohol, \(\displaystyle \text{C}_6\text{H}_5\text{–CH}_2\text{OH} \), has a benzylic \(\displaystyle \text{–CH}_2\text{OH} \) group. Permanganate oxidises this side-chain carbon all the way through the aldehyde to the carboxylic acid; the aromatic ring itself is not attacked. Acidification of the manganate salt after the oxidation liberates the free acid, \(\displaystyle \text{C}_6\text{H}_5\text{–COOH} \) (benzoic acid).(v) Dehydration of propan-$\displaystyle 2$-ol to propeneReagent: concentrated phosphoric acid, $\displaystyle 85$% \(\displaystyle \text{H}_3\text{PO}_4 \), heated to about $\displaystyle 440$ K (concentrated \(\displaystyle \text{H}_2\text{SO}_4 \) with heat is an alternative).The acid protonates the –OH oxygen of \(\displaystyle \text{CH}_3\text{–CH(OH)–CH}_3 \), converting a poor leaving group into a good one, water. Loss of water gives the secondary carbocation \(\displaystyle \text{CH}_3\text{–}\overset{+}{\text{C}}\text{H–CH}_3 \); loss of a proton from an adjacent \(\displaystyle \text{CH}_3 \) group (E1 elimination) then gives propene, \(\displaystyle \text{CH}_3\text{–CH=CH}_2 \).(vi) Butan-$\displaystyle 2$-one to butan-$\displaystyle 2$-ol This step is a reduction of a ketone to a secondary alcohol, not an oxidation.Reagent: sodium borohydride, \(\displaystyle \text{NaBH}_4 \), in methanol (lithium aluminium hydride, \(\displaystyle \text{LiAlH}_4 \), followed by aqueous work-up, or catalytic hydrogenation with \(\displaystyle \text{H}_2/\text{Ni} \), all work equally well).A hydride ion, \(\displaystyle \text{H}^- \), is delivered from the boron to the electrophilic carbonyl carbon of butan-$\displaystyle 2$-one, \(\displaystyle \text{CH}_3\text{–CO–CH}_2\text{–CH}_3 \). This gives the alkoxide \(\displaystyle \text{CH}_3\text{–CH(O}^-\text{)–CH}_2\text{–CH}_3 \), which is protonated on aqueous work-up to butan-$\displaystyle 2$-ol, \(\displaystyle \text{CH}_3\text{–CH(OH)–CH}_2\text{–CH}_3 \).Answer: (i) acidified \(\displaystyle \text{KMnO}_4 \) (or \(\displaystyle \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+ \)), warm; (ii) PCC in \(\displaystyle \text{CH}_2\text{Cl}_2 \); (iii) bromine water, \(\displaystyle \text{Br}_2(aq) \); (iv) alkaline \(\displaystyle \text{KMnO}_4 \) then acidification; (v) concentrated \(\displaystyle \text{H}_3\text{PO}_4 \) (or conc. \(\displaystyle \text{H}_2\text{SO}_4 \)), heat; (vi) \(\displaystyle \text{NaBH}_4 \) (or \(\displaystyle \text{LiAlH}_4 \), or \(\displaystyle \text{H}_2/\text{Ni} \)).

Ethanol can hydrogen-bond to itself; methoxymethane cannot -- that missing O–H is the whole reason for the gap in boiling point.Ethanol is \(\displaystyle \text{CH}_3-\text{CH}_2-\text{OH} \) and methoxymethane (dimethyl ether) is \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3 \). Both have the molecular formula \(\displaystyle \text{C}_2\text{H}_6\text{O} \) and the same molar mass, $\displaystyle 46$ g mol⁻¹ ($\displaystyle 2$×$\displaystyle 12$ + $\displaystyle 6$×$\displaystyle 1$ + $\displaystyle 16$ = $\displaystyle 46$). Being constitutional isomers with the same number of electrons and the same molar mass, the dispersion (van der Waals) forces between their molecules are of comparable strength in both liquids. So dispersion forces cannot be what separates their boiling points -- the difference must come from somewhere else, and it does: from what the oxygen atom is attached to.In ethanol the oxygen carries one hydrogen directly on it: \(\displaystyle -\text{O}-\text{H} \). Oxygen is far more electronegative than hydrogen, so this O–H bond is strongly polarised, leaving the H atom distinctly electron-poor (\(\displaystyle \text{O}^{\delta-}-\text{H}^{\delta+} \)). That electron-poor hydrogen, sitting right on an oxygen, is exactly what is needed to be drawn toward a lone pair on the oxygen of a neighbouring ethanol molecule. This is intermolecular hydrogen bonding: \[\text{CH}_3\text{CH}_2-\text{O}-\text{H}\ \cdots\ \text{O}(\text{H})-\text{CH}_2\text{CH}_3 \] Liquid ethanol is therefore not a set of independent molecules; each molecule is linked to several neighbours through O–H\(\displaystyle \cdots\)O bridges (each roughly $\displaystyle 15$-$\displaystyle 25$ kJ mol⁻¹), building a hydrogen-bonded network throughout the liquid.In methoxymethane the oxygen is bonded to two carbons, \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3 \), and to no hydrogen at all. Every hydrogen in this molecule is a C–H hydrogen, and a C–H bond is only weakly polarised -- far too weakly for that hydrogen to act as a hydrogen-bond donor. The ether oxygen still has lone pairs and could accept a hydrogen bond from an O–H, N–H, or F–H elsewhere, but no such donor exists on another methoxymethane molecule. So methoxymethane molecules cannot hydrogen-bond to each other; they are held together only by the same weak dipole-dipole and dispersion forces present in ethanol, with nothing extra on top.Boiling requires supplying enough energy to pull the molecules of the liquid completely apart into the gas phase. Ethanol's molecules carry the extra burden of the O–H\(\displaystyle \cdots\)O hydrogen-bond network, so more energy (a higher temperature) must be supplied before they can escape into the vapour; methoxymethane's molecules, lacking that network, separate at a much lower temperature. This is exactly what is observed: ethanol boils at $\displaystyle 351.5$ K ($\displaystyle 78.5$ °C) while methoxymethane boils at $\displaystyle 248$ K (-$\displaystyle 24.9$ °C) -- a gap of about $\displaystyle 103$ degrees between two molecules of identical molecular formula and identical molar mass.**Answer: Ethanol, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH} \), has an O–H bond, so its molecules associate through intermolecular hydrogen bonding (O–H\(\displaystyle \cdots\)O); methoxymethane, \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3 \), has no hydrogen bonded to oxygen and so cannot hydrogen-bond to itself. Breaking these extra hydrogen bonds in ethanol takes more energy, giving ethanol the much higher boiling point ($\displaystyle 351$ K vs $\displaystyle 248$ K) even though both have the same molecular formula \(\displaystyle \text{C}_2\text{H}_6\text{O} \) and the same molar mass, $\displaystyle 46$ g mol⁻¹.

The idea: an ether \(\displaystyle \mathrm{R\!-\!O\!-\!R'}\) is named as an alkoxyalkane — the smaller group together with the oxygen becomes an alkoxy substituent (\(\displaystyle \mathrm{CH_3O-}\) = methoxy, \(\displaystyle \mathrm{C_2H_5O-}\) = ethoxy), and the larger group supplies the parent chain or ring.Three rules do all the work here: (i) \(\displaystyle \mathrm{CH_3CH_2\!-\!O\!-\!CH_2CH(CH_3)CH_3}\)Compare the two sides of the oxygen. On the left is \(\displaystyle \mathrm{CH_3CH_2-}\), two carbons. On the right is \(\displaystyle \mathrm{-CH_2CH(CH_3)CH_3}\), four carbons in all. The bigger side is the parent, so the ethyl side becomes ethoxy.Now name the parent side on its own. Its longest chain is \(\displaystyle \mathrm{CH_2\!-\!CH\!-\!CH_3}\), three carbons, so the parent is propane, with a \(\displaystyle \mathrm{CH_3}\) branch on the middle carbon.Number from the end that carries the oxygen, because that gives the lower locant set:\[\overset{1}{\mathrm{C}}\mathrm{H_2}(\mathrm{OC_2H_5})-\overset{2}{\mathrm{C}}\mathrm{H}(\mathrm{CH_3})-\overset{3}{\mathrm{C}}\mathrm{H_3}\]Locants are ethoxy at $\displaystyle 1$ and methyl at $\displaystyle 2$, i.e. the set \(\displaystyle \{1,2\}\); numbering the other way would give \(\displaystyle \{2,3\}\), which is higher. Alphabetical order puts ethoxy before methyl.Name: $\displaystyle 1$-ethoxy-$\displaystyle 2$-methylpropane.(ii) \(\displaystyle \mathrm{CH_3\!-\!O\!-\!CH_2CH_2Cl}\)The methyl side is smaller, so it becomes methoxy. The other side is \(\displaystyle \mathrm{-CH_2CH_2Cl}\), two carbons, so the parent is ethane, carrying \(\displaystyle \mathrm{Cl}\) on one carbon and \(\displaystyle \mathrm{OCH_3}\) on the other.Both substituents sit on a two-carbon chain, so either direction gives the locant set \(\displaystyle \{1,2\}\) — the tie is not broken by locants. The rule for a tie is that the substituent cited first alphabetically gets the lower number. Chloro comes before methoxy, so \(\displaystyle \mathrm{Cl}\) is on C-$\displaystyle 1$:\[\overset{1}{\mathrm{C}}\mathrm{H_2Cl}-\overset{2}{\mathrm{C}}\mathrm{H_2}(\mathrm{OCH_3})\]Name: $\displaystyle 1$-chloro-$\displaystyle 2$-methoxyethane.(iii) \(\displaystyle \mathrm{O_2N\!-\!C_6H_4\!-\!OCH_3}\) (para)Here the parent is the benzene ring, because a ring outranks the small methyl chain. The ring carries two groups: \(\displaystyle \mathrm{-OCH_3}\), named methoxy, and \(\displaystyle \mathrm{-NO_2}\), named nitro. The label p (para) means the two groups are on opposite carbons of the ring, a $\displaystyle 1,4$-relationship.Give C-$\displaystyle 1$ to one substituted carbon and count round to the other; para puts the second at C-$\displaystyle 4$, so the locant set is \(\displaystyle \{1,4\}\) whichever group you start from. Alphabetical order (methoxy before nitro) decides that methoxy takes the $\displaystyle 1$:Name: $\displaystyle 1$-methoxy-$\displaystyle 4$-nitrobenzene, also written $\displaystyle 4$-nitroanisole (anisole being the retained name for methoxybenzene).(iv) \(\displaystyle \mathrm{CH_3CH_2CH_2\!-\!O\!-\!CH_3}\)The methyl is the smaller group, so it becomes methoxy; the three-carbon chain is the parent, propane. Number from the end nearer the oxygen so the substituent gets locant $\displaystyle 1$:\[\overset{1}{\mathrm{C}}\mathrm{H_2}(\mathrm{OCH_3})-\overset{2}{\mathrm{C}}\mathrm{H_2}-\overset{3}{\mathrm{C}}\mathrm{H_3}\]Name: $\displaystyle 1$-methoxypropane.(v) Cyclohexane ring with two \(\displaystyle \mathrm{CH_3}\) on one carbon and \(\displaystyle \mathrm{OC_2H_5}\) on the carbon oppositeThe ring is the parent: cyclohexane. Its substituents are two methyl groups on the same carbon (gem-dimethyl) and one ethoxy group, \(\displaystyle \mathrm{-OC_2H_5}\), on the carbon directly across the ring — three bonds away by either path, so a $\displaystyle 1,4$-relationship.Now pick the numbering. There are only two sensible starting points: Compare term by term: the first terms tie at $\displaystyle 1$, and at the second term \(\displaystyle 1 < 4\). So \(\displaystyle \{1,1,4\}\) is lower and the gem-dimethyl carbon is C-$\displaystyle 1$, putting ethoxy at C-4. Note that the lowest-locant rule settles this before alphabetical order is consulted; alphabetical order only decides the order of citation in the written name, where ethoxy precedes methyl.Name: $\displaystyle 4$-ethoxy-$\displaystyle 1,1$-dimethylcyclohexane.(vi) Benzene ring bearing a single \(\displaystyle \mathrm{-OC_2H_5}\)The ring is the parent, benzene, and the one group attached is \(\displaystyle \mathrm{-OC_2H_5}\) = ethoxy. A monosubstituted ring needs no locant, because every ring carbon is equivalent until a second substituent appears — writing "$\displaystyle 1$-ethoxybenzene" adds nothing.Name: ethoxybenzene (its common name is phenetole).Answer: (i) $\displaystyle 1$-ethoxy-$\displaystyle 2$-methylpropane; (ii) $\displaystyle 1$-chloro-$\displaystyle 2$-methoxyethane; (iii) $\displaystyle 1$-methoxy-$\displaystyle 4$-nitrobenzene ($\displaystyle 4$-nitroanisole); (iv) $\displaystyle 1$-methoxypropane; (v) $\displaystyle 4$-ethoxy-$\displaystyle 1,1$-dimethylcyclohexane; (vi) ethoxybenzene.

Williamson's synthesis builds an ether by one \(\displaystyle \mathrm{SN_{2}}\) step: an alkoxide's oxygen attacks a primary (or methyl) carbon and pushes out a halide, so the bulkier of the two alkyl groups must always come in as the alkoxide, never as the halide.The general reaction is \[\text{R-O}^-\text{Na}^+ + \text{R}'\text{-X} \rightarrow \text{R-O-R}' + \text{NaX} \] where \(\displaystyle \text{R-O}^-\text{Na}^+ \) is a sodium alkoxide (or phenoxide) — it supplies the oxygen and one of the two groups on it — and \(\displaystyle \text{R}'\text{-X} \) (X = Cl, Br, I) is an alkyl halide that supplies the other group. The lone pair on the alkoxide oxygen attacks the carbon bearing X from the side opposite X, so that carbon must be unhindered — primary or methyl. If instead a secondary or tertiary halide is used, the alkoxide (which is also a strong base) pulls off a β-hydrogen and gives an alkene by E2 instead of the ether. That single fact decides which reagent must be the alkoxide and which must be the halide in every part below.(i) $\displaystyle 1$-Propoxypropane, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3 \), has an n-propyl group on each side of the oxygen, so only one pairing of reagents is possible: sodium propoxide (the alkoxide) with $\displaystyle 1$-bromopropane (a primary halide).Reagents: sodium propoxide, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^+ \), and $\displaystyle 1$-bromopropane, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \).\[\text{CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3 + \text{NaBr} \]The propoxide oxygen attacks the primary carbon carrying Br, Br⁻ leaves, and the two propyl chains are joined through oxygen to give $\displaystyle 1$-propoxypropane.(ii) Ethoxybenzene, \(\displaystyle \text{C}_6\text{H}_5-\text{O}-\text{CH}_2\text{CH}_3 \) (common name phenetole), joins a phenyl ring and an ethyl group through oxygen. The ring can never be delivered as the halide half of the reaction: in a haloarene the C–X bond is shortened and strengthened because a lone pair on the halogen conjugates into the ring, and the ring's own electron density blocks approach from behind that carbon, so backside \(\displaystyle \mathrm{(SN_{2})}\) attack at an aryl carbon does not happen. The ring must therefore be delivered as the nucleophile — sodium phenoxide — and the ethyl group as the halide.Reagents: sodium phenoxide, \(\displaystyle \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ \), and bromoethane, \(\displaystyle \text{CH}_3\text{CH}_2\text{Br} \).\[\text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{C}_6\text{H}_5-\text{O}-\text{CH}_2\text{CH}_3 + \text{NaBr} \]The phenoxide oxygen's lone pair attacks the primary carbon of ethyl bromide, Br⁻ leaves, and ethoxybenzene is formed.(iii) $\displaystyle 2$-Methoxy-$\displaystyle 2$-methylpropane, \(\displaystyle (\text{CH}_3)_3\text{C}-\text{O}-\text{CH}_3 \) (tert-butyl methyl ether), joins a tertiary tert-butyl group and a methyl group through oxygen. Here the tertiary group must be delivered as the alkoxide, not the halide: if the reagents were reversed — methoxide ion attacking tert-butyl halide — the small, strongly basic methoxide would instead remove a β-hydrogen from the tertiary halide, and the product would be almost entirely $\displaystyle 2$-methylpropene, \(\displaystyle (\text{CH}_3)_2\text{C}=\text{CH}_2 \), by E2, not the ether.Reagents: sodium tert-butoxide, \(\displaystyle (\text{CH}_3)_3\text{C}-\text{O}^-\text{Na}^+ \), and iodomethane, \(\displaystyle \text{CH}_3\text{I} \).\[(\text{CH}_3)_3\text{C}-\text{O}^-\text{Na}^+ + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_3\text{C}-\text{O}-\text{CH}_3 + \text{NaI} \]The methyl carbon of \(\displaystyle \text{CH}_3\text{I} \) has no β-hydrogen at all, so there is no elimination pathway open to it; the tert-butoxide oxygen attacks it cleanly from the back side, I⁻ leaves, and $\displaystyle 2$-methoxy-$\displaystyle 2$-methylpropane is formed.(iv) $\displaystyle 1$-Methoxyethane, \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_2\text{CH}_3 \) (ethyl methyl ether), joins a methyl group and an ethyl group, both of which are unhindered, so either could in principle serve as the halide without triggering elimination. The conventional choice is to alkylate methoxide with the primary ethyl halide.Reagents: sodium methoxide, \(\displaystyle \text{CH}_3\text{O}^-\text{Na}^+ \), and bromoethane, \(\displaystyle \text{CH}_3\text{CH}_2\text{Br} \).\[\text{CH}_3\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3-\text{O}-\text{CH}_2\text{CH}_3 + \text{NaBr} \]The methoxide oxygen attacks the primary carbon of ethyl bromide, Br⁻ leaves, and $\displaystyle 1$-methoxyethane is formed.Answer: (i) sodium propoxide \(\displaystyle (\text{CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^+) \) + $\displaystyle 1$-bromopropane \(\displaystyle (\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}) \rightarrow \) $\displaystyle 1$-propoxypropane \(\displaystyle (\text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3) \) + NaBr; (ii) sodium phenoxide \(\displaystyle (\text{C}_6\text{H}_5\text{O}^-\text{Na}^+) \) + bromoethane \(\displaystyle (\text{CH}_3\text{CH}_2\text{Br}) \rightarrow \) ethoxybenzene \(\displaystyle (\text{C}_6\text{H}_5\text{OCH}_2\text{CH}_3) \) + NaBr; (iii) sodium tert-butoxide \(\displaystyle ((\text{CH}_3)_3\text{C-O}^-\text{Na}^+) \) + iodomethane \(\displaystyle (\text{CH}_3\text{I}) \rightarrow \) $\displaystyle 2$-methoxy-$\displaystyle 2$-methylpropane \(\displaystyle ((\text{CH}_3)_3\text{C-O-CH}_3) \) + NaI; (iv) sodium methoxide \(\displaystyle (\text{CH}_3\text{O}^-\text{Na}^+) \) + bromoethane \(\displaystyle (\text{CH}_3\text{CH}_2\text{Br}) \rightarrow \) $\displaystyle 1$-methoxyethane \(\displaystyle (\text{CH}_3\text{OCH}_2\text{CH}_3) \) + NaBr — in every case the alkoxide's oxygen displaces the halide from R'-X in one \(\displaystyle \mathrm{SN_{2}}\) step, and whichever group is bulkier is always introduced as the alkoxide, never as the halide.

Williamson synthesis only works when the alkoxide ion can reach the halide carbon from the back — anything that blocks that backside attack (a crowded tertiary carbon, or a flat sp² carbon) breaks the method.The reaction is an \(\displaystyle S_N2 \) displacement: an alkoxide ion, \(\displaystyle R\text{-}O^- \), attacks the carbon bearing the leaving halogen in \(\displaystyle R'\text{-}X \), pushing out the halide ion as the new C–O bond forms.\[R\text{-}O^-\,Na^+ \;+\; R'\text{-}X \;\longrightarrow\; R\text{-}O\text{-}R' \;+\; NaX \]Here \(\displaystyle R\text{-}O^- \) is the alkoxide (or aryloxide) nucleophile, \(\displaystyle R'\text{-}X \) is the alkyl halide supplying the second carbon chain, and \(\displaystyle X \) is the leaving halogen (Cl, Br, or I). This only succeeds when \(\displaystyle R' \) is a carbon that is open to backside attack — a primary (and usually secondary) sp³ carbon. Two situations break that requirement.Limitation $\displaystyle 1$ — a tertiary (or heavily branched) alkyl halide gives elimination, not the ether.An alkoxide ion is not just a nucleophile, it is also a strong, bulky base. At a tertiary carbon, three alkyl groups crowd the back side, so the alkoxide cannot get behind the carbon to do \(\displaystyle S_N2 \). Instead it does the only thing left available to it: it pulls off a hydrogen from a neighbouring (β) carbon while the C–X bond breaks at the same time — an \(\displaystyle E2 \) elimination — and a \(\displaystyle C=C \) double bond is formed instead of a \(\displaystyle C\text{-}O \) bond.Example: sodium ethoxide, \(\displaystyle CH_3CH_2O^-\,Na^+ \), with tert-butyl bromide , \(\displaystyle (CH_3)_3C\text{-}Br \):\[CH_3CH_2O^-\,Na^+ \;+\; (CH_3)_3C\text{-}Br \;\longrightarrow\; (CH_3)_2C=CH_2 \;+\; CH_3CH_2OH \;+\; NaBr \]The product is $\displaystyle 2$-methylprop-$\displaystyle 1$-ene (isobutylene), \(\displaystyle (CH_3)_2C=CH_2 \), not the expected ether ethyl tert-butyl ether, \(\displaystyle CH_3CH_2\text{-}O\text{-}C(CH_3)_3 \). So Williamson synthesis fails whenever the halide component is tertiary (or otherwise very bulky): the base character of the alkoxide wins out over its nucleophile character, and elimination becomes the major or even exclusive pathway.Limitation $\displaystyle 2$ — vinylic and aryl (haloarene) halides do not react at all.In a vinyl halide (\(\displaystyle X \) on an sp² carbon of a \(\displaystyle C=C \)) or in a haloarene such as chlorobenzene , the halogen's lone pair is delocalized into the adjacent \(\displaystyle \pi \) system by resonance. That gives the \(\displaystyle C\text{-}X \) bond partial double-bond character — it becomes shorter and stronger than an ordinary single \(\displaystyle C\text{-}X \) bond in an alkyl halide, and the carbon carrying the halogen is sp² and locked in the plane of the ring, so there is no accessible back side for a nucleophile to attack. As a result these halides essentially do not undergo \(\displaystyle S_N2 \) (or \(\displaystyle S_N1 \)) with an alkoxide.So you cannot make anisole (methyl phenyl ether) this way:\[CH_3O^-\,Na^+ \;+\; C_6H_5\text{-}Cl \;\longrightarrow\; \text{no reaction under normal conditions} \]The workaround is to swap which fragment supplies the aromatic ring: use sodium phenoxide (the aryloxide) together with a reactive primary alkyl halide, so that the nucleophile attacks a normal sp³ carbon instead:\[C_6H_5\text{-}O^-\,Na^+ \;+\; CH_3\text{-}I \;\longrightarrow\; C_6H_5\text{-}O\text{-}CH_3 \;(\text{anisole}) \;+\; NaI \]Here the oxygen of the phenoxide ion is the nucleophile and methyl iodide, \(\displaystyle CH_3I \), is a primary halide fully open to backside attack, so ordinary \(\displaystyle S_N2 \) proceeds smoothly. The limitation is specifically that the aryl or vinyl group can never be the part carrying the leaving halogen — it can only ever be supplied through the alkoxide (as an aryloxide) half of the reaction.**Answer: Williamson synthesis is limited by the requirement of a clean \(\displaystyle S_N2 \) attack on the halide carbon. ($\displaystyle 1$) When the alkyl halide is tertiary (e.g. \(\displaystyle (CH_3)_3C\text{-}Br \)), the bulky alkoxide cannot attack from behind, so it instead abstracts a β-hydrogen and drives \(\displaystyle E2 \) elimination, giving an alkene (e.g. \(\displaystyle (CH_3)_2C=CH_2 \)) as the major product instead of the ether. ($\displaystyle 2$) Vinylic and aryl halides (e.g. chlorobenzene) cannot be used at all, because their \(\displaystyle C\text{-}X \) bond has partial double-bond character from resonance and sits on an sp² carbon shielded from backside attack — no substitution occurs. To make an aryl alkyl ether such as anisole, the aryl group must be introduced as the aryloxide (sodium phenoxide) reacting with a reactive alkyl halide such as \(\displaystyle CH_3I \), never as the halide component itself.

A symmetrical ether is built by making one molecule of the alcohol attack a second, protonated molecule of the very same alcohol -- this is the acid-catalysed intermolecular dehydration route, and it works only because both halves of the ether come from the same alcohol, propan-$\displaystyle 1$-ol.Overall reaction: two molecules of propan-$\displaystyle 1$-ol are heated together with excess concentrated sulphuric acid at $\displaystyle 413$ K:\[2\,\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow[413\,\text{K}]{\text{conc. }\text{H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3 \;+\; \text{H}_2\text{O} \]Here \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) is propan-$\displaystyle 1$-ol (n-propyl alcohol), and the product \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3\) is $\displaystyle 1$-propoxypropane, also known as di-n-propyl ether -- a straight-chain propyl group on each side of the ether oxygen.Because the carbon that loses its leaving group here is primary, it cannot support a carbocation, so this dehydration cannot go by the E1/\(\displaystyle \mathrm{SN_{1}}\) route used for tertiary alcohols. Instead it goes by a bimolecular, SN2-type displacement, in three steps.Step $\displaystyle 1$ -- Protonation of the leaving group. One molecule of propan-$\displaystyle 1$-ol accepts a proton from \(\displaystyle \text{H}_2\text{SO}_4\): the lone pair on the alcohol oxygen attacks the acidic proton of the acid. This converts the hydroxyl into a protonated alcohol (an oxonium ion), \[\text{CH}_3\text{CH}_2\text{CH}_2-\overset{+}{\text{O}}\text{H}_2 \] This step is the one people skip, and it is the reason the reaction needs an acid catalyst at all: \(\displaystyle -\text{OH}\) by itself is a poor leaving group, while the neutral water it becomes after protonation is a good one.Step $\displaystyle 2$ -- Nucleophilic attack: the bond-breaking, bond-making step. A second, still-neutral molecule of propan-$\displaystyle 1$-ol now supplies the nucleophile. The lone pair on its oxygen attacks the electrophilic carbon of the protonated molecule from the side directly opposite the \(\displaystyle -\overset{+}{\text{O}}\text{H}_2\) group -- backside attack, exactly as in any \(\displaystyle \mathrm{SN_{2}}\) substitution. As the new C-O bond forms, the C-\(\displaystyle \overset{+}{\text{O}}\text{H}_2\) bond breaks and a neutral water molecule leaves. The immediate product is a protonated ether: \[\text{CH}_3\text{CH}_2\text{CH}_2-\overset{+}{\text{O}}\text{H}-\text{CH}_2\text{CH}_2\text{CH}_3 \] in which the ether oxygen still carries the extra proton and the positive charge.Step $\displaystyle 3$ -- Deprotonation. A base present in the mixture -- another molecule of propan-$\displaystyle 1$-ol, or the \(\displaystyle \text{HSO}_4^{-}\) ion -- removes this extra proton from the oxygen. This regenerates the acid catalyst and releases the neutral ether: \[\text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3 \] Since \(\displaystyle \text{H}_2\text{SO}_4\) is returned at the end rather than consumed, it is a true catalyst here, not a stoichiometric reagent.A short aside on the conditions, because this is where the method is easy to get backwards: the very same protonated alcohol, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\overset{+}{\text{O}}\text{H}_2\), can instead lose a proton from the adjacent carbon to eliminate water and give propene. Using a lower temperature ($\displaystyle 413$ K, not the ~$\displaystyle 443$ K used for elimination) and an excess of alcohol favours substitution over elimination, because the second alcohol molecule is then present in high enough concentration to intercept the protonated alcohol as a nucleophile before it has time to eliminate. This is also why the method only gives a good yield of a symmetrical ether -- both halves must come from the same alcohol pool.Answer: Propan-$\displaystyle 1$-ol is converted to $\displaystyle 1$-propoxypropane (di-n-propyl ether) by heating excess propan-$\displaystyle 1$-ol with concentrated \(\displaystyle \text{H}_2\text{SO}_4\) at $\displaystyle 413$ K (intermolecular acid-catalysed dehydration). Mechanism: ($\displaystyle 1$) \(\displaystyle \text{H}_2\text{SO}_4\) protonates one \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) to give \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\overset{+}{\text{O}}\text{H}_2\); ($\displaystyle 2$) a second, neutral \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) molecule attacks this carbon by backside \(\displaystyle \mathrm{SN_{2}}\) displacement of water, giving the protonated ether \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\overset{+}{\text{O}}\text{H}-\text{CH}_2\text{CH}_2\text{CH}_3\); ($\displaystyle 3$) loss of a proton to \(\displaystyle \text{HSO}_4^{-}\)/another alcohol molecule gives the neutral product, $\displaystyle 1$-propoxypropane, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3\).

Elimination wins once the carbon under attack is secondary or tertiary — the intermediate carbocation is too stable to wait around for a second alcohol molecule to bump into a crowded carbon, so it just throws off a proton and becomes an alkene instead.Ethers are made from alcohols by acid-catalysed dehydration ("Williamson's continuous etherification") only when the mechanism can go through a clean back-side attack — an \(\displaystyle S_N2 \) step. Look at how it actually has to happen for a primary alcohol, say ethanol , and then see why a secondary or tertiary alcohol cannot follow the same path.Step $\displaystyle 1$ — protonation. The alcohol oxygen's lone pair picks up a proton from the concentrated \(\displaystyle H_2SO_4 \):CH3-CH2-OH + \(\displaystyle \mathrm{H^{+}}\) → CH3-CH2-OH2+ (protonated ethanol)This is the point of the whole trick: \(\displaystyle -OH \) is a poor leaving group, but \(\displaystyle -OH_2^+ \) leaves as neutral water, a good leaving group.Step $\displaystyle 2$ — nucleophilic attack (the step that decides everything). A second, unprotonated ethanol molecule uses the lone pair on its own oxygen to attack the electrophilic carbon of the protonated ethanol from the side opposite the leaving water molecule — a back-side, \(\displaystyle S_N2 \)-type attack:CH3-CH2-OH2+ + CH3-CH2-OH → CH3-CH2-O(+H)-CH2-CH3 + \(\displaystyle \mathrm{H_{2}O}\)Step $\displaystyle 3$ — deprotonation. A base (hydrogensulphate ion or another alcohol molecule) removes the extra proton, giving the neutral ether and regenerating the acid catalyst:CH3-CH2-O(+H)-CH2-CH3 → CH3-CH2-O-CH2-CH3 (diethyl ether) + \(\displaystyle \mathrm{H^{+}}\)Overall, at about \(\displaystyle 413\ \text{K} \) with excess ethanol: \(\displaystyle 2\,CH_3CH_2OH \xrightarrow{\text{conc. } H_2SO_4,\ 413\text{K}} CH_3CH_2-O-CH_2CH_3 + H_2O \).Now put a secondary alcohol (propan-$\displaystyle 2$-ol , CH3-CH(OH)-CH3) or a tertiary one (tert-butyl alcohol , (CH3)3C-OH) through the same three steps and Step $\displaystyle 2$ breaks down for two compounding reasons.Reason $\displaystyle 1$ — steric hindrance to the \(\displaystyle S_N2 \) attack. In Step $\displaystyle 2$ the incoming alcohol oxygen has to approach the carbon that is losing water, directly from the back. In a secondary alcohol that carbon already carries two alkyl groups; in a tertiary alcohol it carries three. There simply is not enough room behind that carbon for a second, equally bulky alcohol molecule to get close enough to bond — the back-side approach is blocked by the alkyl groups themselves.Reason $\displaystyle 2$ — the carbocation is too stable to wait. Once water leaves a protonated secondary or tertiary alcohol, it does not need a nucleophile to arrive in a concerted step at all — it can simply ionise on its own, because secondary and (especially) tertiary carbocations are stabilised by hyperconjugation and the +I effect of the extra alkyl groups. A carbocation that stable has a much faster, much easier fate available than waiting to be attacked by a second alcohol molecule: a base (often another alcohol molecule, acting this time as a base rather than a nucleophile) simply pulls off a proton from a carbon next door (a β-hydrogen). That converts the carbocation directly into an alkene — an E1 elimination — and this is both kinetically faster and thermodynamically preferred (the alkene product is the stable, conjugated π-system Zaitsev product) compared to the slow, sterically blocked substitution that would be needed to make an ether.So instead of an ether, acid treatment of propan-$\displaystyle 2$-ol gives propene, and acid treatment of tert-butyl alcohol gives $\displaystyle 2$-methylprop-$\displaystyle 1$-ene (isobutylene) — in each case elimination, not ether formation, dominates:CH3-CH(OH)-CH3 --(H+, heat)--> CH3-CH=\(\displaystyle \mathrm{CH_{2}}\) (propene) + \(\displaystyle \mathrm{H_{2}O}\)(CH3)3C-OH --(H+, heat)--> \(\displaystyle \mathrm{CH_{2}}\)=\(\displaystyle \mathrm{C(CH_{3})_{2}}\) ($\displaystyle 2$-methylprop-$\displaystyle 1$-ene) + \(\displaystyle \mathrm{H_{2}O}\)This is exactly the opposite outcome from what the reaction is meant to achieve, which is why acid dehydration is used to make ethers only from primary alcohols (where there is no branching to block the back-side attack and no unusually stable carbocation to tempt elimination), and is described as unsuitable for secondary or tertiary alcohols.Answer: Acid dehydration of secondary or tertiary alcohols is not suitable for making ethers because the mechanism needs a back-side (\(\displaystyle S_N2\)) attack by a second alcohol molecule on the carbon bearing the protonated \(\displaystyle -OH_2^+\) group, and in secondary/tertiary alcohols that carbon is too crowded (steric hindrance) for this to happen; instead, the more stable secondary/tertiary carbocation formed after water leaves loses a β-hydrogen (E1 elimination) much faster than it can be attacked by a nucleophile, so the major product is an alkene (e.g., propan-$\displaystyle 2$-ol gives propene, and tert-butyl alcohol gives $\displaystyle 2$-methylprop-$\displaystyle 1$-ene) rather than the desired ether.

Ether cleavage by a hydrogen halide always breaks a carbon–oxygen bond, and it breaks the ONE bond that gives either the least hindered carbon for a backside attack or the most stable carbocation — never the other one, and never a bond to an aromatic ring carbon. The three ethers below look similar, but each one hands the iodide ion a different-shaped carbon to attack, so each gives a different pair of products.The mechanism has the same first step in all three cases. Hydrogen iodide, \(\displaystyle \text{HI}\), is a strong acid, so the first thing that happens is protonation: a lone pair on the ether oxygen attacks the acidic hydrogen of \(\displaystyle \text{H–I}\), the \(\displaystyle \text{H–I}\) bond breaks heterolytically, and iodide ion, \(\displaystyle \text{I}^-\), is released while the ether oxygen becomes a positively charged oxonium centre (an "-O(H)+-" bridging the two carbon groups). What happens next — which of the two C–O bonds around that oxonium breaks, and how — is what differs between the three ethers.(i) $\displaystyle 1$-propoxypropane, \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3\) (di-n-propyl ether)Here both carbons attached to oxygen are primary \(\displaystyle \text{CH}_2\) carbons of n-propyl groups — small, unhindered, and identical to each other. With nothing to favour one carbocation over another, the reaction goes by the direct route: iodide ion attacks one of these primary carbons from the side opposite the oxygen (an \(\displaystyle S_N2\) attack), so the new C–I bond forms exactly as the C–O bond on that carbon breaks. The other half of the molecule keeps the oxygen and the proton picked up in the first step, so it leaves as a neutral alcohol rather than as a cation.\[\text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3 + \text{HI} \;\rightarrow\; \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{I} \]The products are propan-$\displaystyle 1$-ol (\(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) and $\displaystyle 1$-iodopropane, also called n-propyl iodide (\(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{I}\)). One aside worth flagging: with excess hot \(\displaystyle \text{HI}\), the propan-$\displaystyle 1$-ol formed does not just sit there — it is itself converted to a second molecule of $\displaystyle 1$-iodopropane (\(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{I} + \text{H}_2\text{O}\)), but the equation asked for is the single-cleavage step above.(ii) methoxybenzene, \(\displaystyle \text{C}_6\text{H}_5\text{-O-CH}_3\) (anisole)This ether is not symmetrical at all: one side of the oxygen is a methyl carbon (sp3, primary), and the other side is a ring carbon of benzene (sp2, locked flat by the aromatic system). The aryl–oxygen bond can never be the one that breaks. It cannot break by \(\displaystyle S_N2\), because backside attack needs the nucleophile to approach from behind the leaving group, and there is no "behind" on a flat sp2 ring carbon that is also bonded to two other ring carbons. It cannot break by \(\displaystyle S_N1\) either, because the resulting aryl cation would have its empty orbital lying in the plane of the ring, unable to share the ring's π-electron cloud — so it gets none of the resonance stabilisation that makes carbocations elsewhere in this problem feasible. So the only carbon available to iodide is the methyl group: iodide attacks the methyl carbon in a straightforward \(\displaystyle S_N2\) step, breaking the O–CH3 bond. The oxygen keeps its bond to the ring and to the proton from step one, so it is left as phenol.\[\text{C}_6\text{H}_5\text{-O-CH}_3 + \text{HI} \;\rightarrow\; \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I} \]The products are phenol (\(\displaystyle \text{C}_6\text{H}_5\text{OH}\)) and methyl iodide/iodomethane (\(\displaystyle \text{CH}_3\text{I}\)) — never iodobenzene, because the ring–oxygen bond is the one bond that is off-limits.(iii) benzyl ethyl ether, \(\displaystyle \text{C}_6\text{H}_5\text{CH}_2\text{-O-CH}_2\text{CH}_3\)Both carbons attached to oxygen here are, strictly by substitution pattern, primary \(\displaystyle \text{CH}_2\) carbons — one belongs to the benzyl group, the other to the ethyl group. But they are not equally happy to lose the oxygen. When the benzylic C–O bond ionises, the developing positive charge on that carbon is not stuck on one atom: it delocalises into the attached benzene ring, spreading onto the ortho and para ring carbons by resonance. The ethyl carbon has no such ring to share charge with. So although the geometry would allow \(\displaystyle S_N2\) at either carbon, the pathway through the resonance-stabilised benzylic cation is favoured, and cleavage happens at the benzylic C–O bond: the ethyl side keeps the oxygen and proton and leaves as ethanol, while iodide then bonds to the benzylic carbon.\[\text{C}_6\text{H}_5\text{CH}_2\text{-O-CH}_2\text{CH}_3 + \text{HI} \;\rightarrow\; \text{C}_6\text{H}_5\text{CH}_2\text{I} + \text{CH}_3\text{CH}_2\text{OH} \]The products are benzyl iodide (\(\displaystyle \text{C}_6\text{H}_5\text{CH}_2\text{I}\)) and ethanol (\(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\)).Answer: (i) \(\displaystyle \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{I}\) (propan-$\displaystyle 1$-ol + $\displaystyle 1$-iodopropane, by \(\displaystyle S_N2\) at either equivalent primary carbon); (ii) \(\displaystyle \text{C}_6\text{H}_5\text{-O-CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I}\) (phenol + methyl iodide, since the aryl–O bond cannot break); (iii) \(\displaystyle \text{C}_6\text{H}_5\text{CH}_2\text{-O-CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{I} + \text{CH}_3\text{CH}_2\text{OH}\) (benzyl iodide + ethanol, cleavage at the resonance-stabilised benzylic carbon).

The alkoxy oxygen has a lone pair that can conjugate into the ring — this is the same resonance donation you see in phenol, and it does two things at once: it floods the whole ring with extra electron density (so the ring reacts faster with an electrophile than benzene does), and it deposits that extra density specifically at the ortho and para carbons (so an incoming electrophile is funnelled to exactly those two positions).Setting up the molecule. Take anisole (methoxybenzene), \(\displaystyle \text{C}_6\text{H}_5-\text{O}-\text{CH}_3 \), as the representative aryl alkyl ether. Call the ring carbon carrying the \(\displaystyle -\text{OCH}_3 \) group \(\displaystyle \mathrm{C_{1}}\); going round the ring, \(\displaystyle \mathrm{C_{2}}\) and \(\displaystyle \mathrm{C_{6}}\) are ortho, \(\displaystyle \mathrm{C_{3}}\) and \(\displaystyle \mathrm{C_{5}}\) are meta, and \(\displaystyle \mathrm{C_{4}}\) is para to C1. The oxygen in \(\displaystyle -\text{OCH}_3 \) carries two lone pairs, and one of these lone pairs is not confined to oxygen — it can delocalize into the adjacent \(\displaystyle \pi \) system of the ring.(i) Why the ring is activated.Write the resonance (canonical) structures of anisole:Structure I — the ordinary Kekulé structure, no charge separation: \(\displaystyle \text{CH}_3-\overset{\displaystyle ..}{\text{O}}-\text{C}_6\text{H}_5 \), oxygen holding both lone pairs, ring bonds alternating single/double in the usual way.Structure II — one lone pair on oxygen swings in to form a \(\displaystyle \pi \) bond between O and C1. Oxygen, now making one \(\displaystyle \sigma \) and one \(\displaystyle \pi \) bond to carbon plus one \(\displaystyle \sigma \) bond to \(\displaystyle \text{CH}_3 \) while keeping only one lone pair, carries a formal positive charge: \(\displaystyle \text{CH}_3-\overset{+}{\text{O}}=\text{C}_1 \). Pushing this new double bond forces the ring double bonds to shift one step around the ring, and the electron pair displaced this way comes to rest on \(\displaystyle \mathrm{C_{2}}\) (ortho), which now carries a formal negative charge.Structure III — the same electron shift carried the other way around the ring instead, so the negative charge lands on \(\displaystyle \mathrm{C_{6}}\) (the other ortho carbon).Structure IV — shifting the double bonds two steps around the ring puts the extra electron pair on \(\displaystyle \mathrm{C_{4}}\) (para).No such canonical structure can be drawn with the negative charge on \(\displaystyle \mathrm{C_{3}}\) or \(\displaystyle \mathrm{C_{5}}\): alternating double bonds starting from \(\displaystyle \mathrm{C_{1}}\) can only ever terminate on a carbon that is an even number of bonds away going around the ring, and those positions are exactly ortho (C2, C6) and para \(\displaystyle \mathrm{(C_{4})}\), never meta.The true structure of anisole is a hybrid of I–IV. Structures II–IV show that real electron density is pushed off oxygen and onto the ring — the ring, taken as a whole, is more electron-rich than benzene's ring. Electrophilic aromatic substitution is rate-limited by how easily the ring can act as a nucleophile toward the electrophile \(\displaystyle E^{+} \) to form the arenium-ion (Wheland) intermediate; a ring with extra electron density does this more readily, i.e., with a lower activation energy, than plain benzene. This resonance (mesomeric, \(\displaystyle +M \)) donation from oxygen outweighs the inductive electron-withdrawal that oxygen's high electronegativity would otherwise cause (\(\displaystyle -I \)), so the net, observed effect on the rate is activation — anisole undergoes nitration, sulphonation, halogenation and Friedel–Crafts reactions faster than benzene itself.(ii) Why substitution goes ortho/para, not meta.Now let \(\displaystyle E^{+} \) actually attack the ring. Attack at any ring carbon gives a resonance-stabilized carbocation (the arenium ion), and the question is which arenium ion is lower in energy, since the more stable intermediate is reached through the more stable (lower-energy) transition state, so that pathway dominates.If \(\displaystyle E^{+} \) attacks \(\displaystyle \mathrm{C_{2}}\) (ortho) or \(\displaystyle \mathrm{C_{4}}\) (para), one of the resonance structures of the resulting arenium ion places the positive charge directly on \(\displaystyle \mathrm{C_{1}}\) — the very carbon that carries the \(\displaystyle -\text{OCH}_3 \) group. At that point oxygen's lone pair can donate into the empty p-orbital on \(\displaystyle \mathrm{C_{1}}\), exactly as in structures II–IV above, giving an additional resonance contributor of the form \(\displaystyle \text{CH}_3-\overset{+}{\text{O}}=\text{C}_1 \) in which oxygen (not carbon) bears the formal positive charge and every atom still has a full octet. This is an unusually stable resonance form — the positive charge sits on an electronegative atom that is nonetheless left with a complete octet — and it gives the ortho/para arenium ion a fourth, extra-stable resonance structure that a plain arenium ion (from benzene) does not have.If \(\displaystyle E^{+} \) attacks \(\displaystyle \mathrm{C_{3}}\) or \(\displaystyle \mathrm{C_{5}}\) (meta) instead, the three resonance structures of that arenium ion place the positive charge only on \(\displaystyle \mathrm{C_{2}}\), \(\displaystyle \mathrm{C_{4}}\), and \(\displaystyle \mathrm{C_{6}}\) — never on C1. Because the positive charge can never reach the oxygen-bearing carbon, oxygen's lone pair cannot be donated in to stabilize it, so the meta arenium ion has no extra resonance form and is distinctly less stable than the ortho/para one.Since the ortho- and para-attack intermediates are stabilized by this extra resonance contributor and the meta one is not, attack at ortho and para proceeds through a lower-energy pathway and is strongly preferred. This is why the alkoxy group of an aryl alkyl ether is both ring-activating and ortho/para-directing.Answer: In an aryl alkyl ether such as anisole, the oxygen's lone pair conjugates with the ring (resonance structures placing negative charge only at the ortho and para carbons, never meta), which (i) raises the ring's overall electron density and so lowers the activation energy for electrophilic attack relative to benzene — activation — and (ii) lets the arenium ion formed by ortho or para attack place the positive charge on oxygen (as \(\displaystyle \text{CH}_3-\overset{+}{\text{O}}=\text{C} \), a full-octet, extra-stable resonance form) while the meta arenium ion cannot access this stabilization — hence ortho, para-direction.

Protonation turns a hard-to-leave alkoxide into an easy-to-leave neutral alcohol — only then can iodide ion attack the carbon.Methoxymethane (dimethyl ether) is \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3\): an oxygen atom carrying two lone pairs, bonded to two methyl carbons. Ethers are not attacked directly by a nucleophile, because the leaving group would have to be an alkoxide ion \(\displaystyle \text{CH}_3\text{O}^-\), which is a strong base and a poor leaving group. Hydrogen iodide gets around this in two steps.Step $\displaystyle 1$ — Protonation of the ether oxygen. HI is a strong acid. One of the two lone pairs on the ether oxygen of \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3\) attacks the electrophilic hydrogen of \(\displaystyle \text{H}-\text{I}\). A new O–H bond forms, and the H–I bond breaks heterolytically, with both electrons of that bond leaving on iodine. This produces the protonated ether (a dimethyloxonium ion), \[\text{CH}_3-\overset{+}{\text{O}}(\text{H})-\text{CH}_3\] and releases an iodide ion, \(\displaystyle \text{I}^-\), into solution. This step is fast — it is simply an acid–base (proton-transfer) equilibrium, not bond-breaking at carbon.Step $\displaystyle 2$ — Nucleophilic attack at carbon (\(\displaystyle S_N2\)). Protonation is the key trick: the leaving group attached to each methyl carbon is no longer the strongly basic \(\displaystyle \text{CH}_3\text{O}^-\) but the neutral, weakly basic molecule methanol, \(\displaystyle \text{CH}_3\text{OH}\), which is a far better leaving group. Because each carbon on the oxonium ion is a methyl carbon — primary, with no steric bulk and no possibility of forming a stabilized carbocation — the reaction goes by a concerted back-side attack, not by a carbocation pathway.The iodide ion generated in Step $\displaystyle 1$, \(\displaystyle \text{I}^-\), is a strong, highly polarizable nucleophile. It attacks one of the two methyl carbons from the side directly opposite the oxygen (backside attack). As the new C–I bond forms, the C–O bond to that same carbon breaks at the same instant, with the bonding pair of electrons going to oxygen. This is a single concerted step (an \(\displaystyle S_N2\) displacement) through a trigonal-bipyramidal transition state in which iodine and oxygen are both partially bonded to the central carbon on opposite sides: \[\text{I}^- + \text{CH}_3-\overset{+}{\text{O}}(\text{H})-\text{CH}_3 \longrightarrow \text{CH}_3-\text{I} + \text{CH}_3-\text{OH}\]The carbon under attack is a methyl carbon (three identical hydrogens on it), so the inversion of configuration that always accompanies \(\displaystyle S_N2\) attack is not observable here — there is no stereocentre to invert.Products. One methyl group ends up as iodomethane (methyl iodide), \(\displaystyle \text{CH}_3\text{I}\); the other, still carrying the oxygen and its hydrogen, is released as methanol, \(\displaystyle \text{CH}_3\text{OH}\).If HI is in excess, the methanol formed above is itself an alcohol, and it is attacked by a second equivalent of HI through the identical two-step sequence — protonation of the –OH oxygen to give \(\displaystyle \text{CH}_3-\overset{+}{\text{O}}\text{H}_2\), followed by back-side \(\displaystyle S_N2\) attack of a second \(\displaystyle \text{I}^-\) on that carbon, expelling water and forming a second molecule of methyl iodide: \[\text{CH}_3\text{OH} + \text{HI} \longrightarrow \text{CH}_3\text{I} + \text{H}_2\text{O}\] so that with excess HI the overall transformation is \[\text{CH}_3-\text{O}-\text{CH}_3 + 2\,\text{HI} \longrightarrow 2\,\text{CH}_3\text{I} + \text{H}_2\text{O}\]Answer: HI first protonates the ether oxygen of \(\displaystyle \text{CH}_3-\text{O}-\text{CH}_3\) to give the oxonium ion \(\displaystyle \text{CH}_3-\overset{+}{\text{O}}(\text{H})-\text{CH}_3\) (releasing \(\displaystyle \text{I}^-\)); iodide ion then attacks one methyl carbon from the back side in a concerted \(\displaystyle S_N2\) step, breaking the C–O bond as the C–I bond forms. This gives methanol \(\displaystyle (\text{CH}_3\text{OH})\) and iodomethane \(\displaystyle (\text{CH}_3\text{I})\) as the products; with excess HI the methanol is further converted to a second molecule of \(\displaystyle \text{CH}_3\text{I}\) plus water.

The −OCH₃ group feeds its oxygen lone pair into the ring by resonance, so anisole is a strongly activated, ortho/para-directing arene — every electrophile below lands mainly at the position para to −OCH₃, with a smaller amount at ortho.Anisole is \(\displaystyle \mathrm{C_6H_5-OCH_3} \) (methoxybenzene). One lone pair on the ether oxygen overlaps with the ring π-system, pushing electron density onto the carbons ortho and para to itself (draw the resonance structures mentally: negative charge appears only at ortho and para carbons, never at meta). That is why every electrophilic aromatic substitution on anisole gives an ortho/para mixture, almost always with the para isomer as the major product because it is free of steric clash with the bulky −OCH�$\displaystyle 3$ group.(i) Friedel–Crafts alkylation of anisoleAnhydrous \(\displaystyle \mathrm{AlCl_3} \) pulls \(\displaystyle \mathrm{Cl^-} \) off methyl chloride to generate the electrophile, the methyl carbocation \(\displaystyle \mathrm{CH_3^+} \) (as the \(\displaystyle \mathrm{AlCl_4^-} \) ion pair). This attacks the ring at the position ortho or para to −OCH₃, giving an arenium (sigma-complex) intermediate that then loses \(\displaystyle \mathrm{H^+} \) to restore aromaticity:\[\mathrm{C_6H_5OCH_3 + CH_3Cl \; \xrightarrow{\text{anhyd. } AlCl_3} \; o\text{-}CH_3C_6H_4OCH_3 \;+\; p\text{-}CH_3C_6H_4OCH_3 \;+\; HCl} \]Products: $\displaystyle 1$-methoxy-$\displaystyle 2$-methylbenzene (o-methylanisole, minor) and $\displaystyle 1$-methoxy-$\displaystyle 4$-methylbenzene (p-methylanisole, major).(ii) Nitration of anisoleConcentrated \(\displaystyle \mathrm{H_2SO_4} \) protonates \(\displaystyle \mathrm{HNO_3} \) and expels water to generate the nitronium ion, \(\displaystyle \mathrm{NO_2^+} \), the attacking electrophile. It substitutes at ortho and para to −OCH₃:\[\mathrm{C_6H_5OCH_3 + HNO_3 \; \xrightarrow{\text{conc. } H_2SO_4} \; o\text{-}O_2NC_6H_4OCH_3 \;+\; p\text{-}O_2NC_6H_4OCH_3 \;+\; H_2O} \]Products: $\displaystyle 1$-methoxy-$\displaystyle 2$-nitrobenzene (o-nitroanisole, minor) and $\displaystyle 1$-methoxy-$\displaystyle 4$-nitrobenzene (p-nitroanisole, major).(iii) Bromination of anisole in ethanoic acid mediumBecause the ring is so strongly activated by −OCH₃, no Lewis-acid catalyst (no \(\displaystyle \mathrm{FeBr_3} \)) is needed at all — the electron-rich ring itself polarises the \(\displaystyle \mathrm{Br-Br} \) bond enough for one bromine to act as the electrophile \(\displaystyle \mathrm{Br^+} \), released as \(\displaystyle \mathrm{HBr} \) once the sigma-complex loses a proton. Carrying the reaction out in glacial ethanoic acid \(\displaystyle \mathrm{(CH_3COOH)} \) as solvent (rather than in water or CS₂) is exactly what pushes the reaction cleanly to the para product:\[\mathrm{C_6H_5OCH_3 + Br_2 \; \xrightarrow{\text{CH}_3\text{COOH}} \; p\text{-}BrC_6H_4OCH_3 \;(\text{major}) \;+\; o\text{-}BrC_6H_4OCH_3\;(\text{minor}) \;+\; HBr} \]Product named: $\displaystyle 1$-bromo-$\displaystyle 4$-methoxybenzene (p-bromoanisole) as the major product, with a little $\displaystyle 1$-bromo-$\displaystyle 2$-methoxybenzene (o-bromoanisole).(iv) Friedel–Crafts acetylation of anisoleAnhydrous \(\displaystyle \mathrm{AlCl_3} \) activates acetic anhydride (or acetyl chloride) by pulling off the leaving group, generating the acylium ion \(\displaystyle \mathrm{CH_3CO^+} \), a bulky electrophile. Because of its size, ortho attack is heavily disfavoured by the steric clash with the adjacent −OCH₃ group, so the reaction is essentially selective for the para position:\[\mathrm{C_6H_5OCH_3 + (CH_3CO)_2O \; \xrightarrow{\text{anhyd. } AlCl_3} \; p\text{-}CH_3COC_6H_4OCH_3 \;(\text{major product}) \;+\; CH_3COOH} \]Product named: $\displaystyle 1$-($\displaystyle 4$-methoxyphenyl)ethan-$\displaystyle 1$-one, i.e., $\displaystyle 4$-methoxyacetophenone (p-methoxyacetophenone), as the overwhelmingly major product.Answer: (i) \(\displaystyle \mathrm{CH_3Cl/anhyd.\,AlCl_3} \) gives o- and p-methylanisole (para major). (ii) \(\displaystyle \mathrm{HNO_3/conc.\,H_2SO_4} \) gives o- and p-nitroanisole (para major). (iii) \(\displaystyle \mathrm{Br_2} \) in ethanoic acid (no catalyst needed) gives p-bromoanisole as the major product with some o-bromoanisole. (iv) \(\displaystyle \mathrm{(CH_3CO)_2O/anhyd.\,AlCl_3} \) gives p-methoxyacetophenone as the essentially exclusive product. In every case the −OCH₃ group's resonance electron donation makes it a strong ortho/para director, and steric bulk of the incoming electrophile/group pushes the balance further toward the para isomer.

Each target alcohol has its \(\displaystyle -OH\) sitting on a carbon that, in the starting alkene, was one end of a C=C double bond — find that carbon, mentally strip off the \(\displaystyle -OH\) and a hydrogen from the next-door carbon, and the double bond falls back into place. Then the question is only which carbon protonates first, and Markovnikov's rule (protonation goes wherever it makes the more stable carbocation) tells you that.The reaction used throughout is acid-catalysed hydration of an alkene — dilute \(\displaystyle H_2SO_4\) in water (or, to avoid any risk of a carbocation rearranging along the way, oxymercuration–demercuration: \(\displaystyle Hg(OAc)_2\), \(\displaystyle H_2O\)/THF, then \(\displaystyle NaBH_4\)). Both add \(\displaystyle H\) and \(\displaystyle OH\) across the double bond with the \(\displaystyle -OH\) landing on the more substituted (more highly alkylated) carbon, because that is the carbon that becomes the more stable carbocation.(i) $\displaystyle 1$-Methylcyclohexanol — a cyclohexane ring in which one ring carbon (call it C1) carries both a methyl group and the \(\displaystyle -OH\): condensed as cyclo-\(\displaystyle C_6H_{10}(CH_3)(OH)\).Work backward: remove the \(\displaystyle -OH\) from \(\displaystyle \mathrm{C_{1}}\) and an \(\displaystyle H\) from the neighbouring ring carbon \(\displaystyle \mathrm{C_{2}}\), and a double bond reappears between \(\displaystyle \mathrm{C_{1}}\) and C2. The alkene is $\displaystyle 1$-methylcyclohexene (double bond \(\displaystyle \mathrm{C_{1}}\)=\(\displaystyle \mathrm{C_{2}}\), methyl on C1).Mechanism: the pi electrons of the \(\displaystyle \mathrm{C_{1}}\)=\(\displaystyle \mathrm{C_{2}}\) double bond attack a proton of \(\displaystyle H_3O^+\); the new C–H bond forms at \(\displaystyle \mathrm{C_{2}}\) (which already carries only a ring-\(\displaystyle CH_2\) and one H), so the positive charge is left on C1. \(\displaystyle \mathrm{C_{1}}\) is now bonded to three carbons — the methyl group and two ring carbons (C6 and C2) — so this is a tertiary carbocation, the most stable option, which is why protonation happens at \(\displaystyle \mathrm{C_{2}}\) and not C1. A lone pair on the oxygen of a water molecule then attacks this electron-poor \(\displaystyle \mathrm{C_{1}}\), forming the new C–O bond and giving a protonated alcohol (an oxonium ion, \(\displaystyle R\text{-}OH_2^+\)). A second water molecule pulls the extra proton off that oxygen, regenerating \(\displaystyle H_3O^+\) and releasing neutral $\displaystyle 1$-methylcyclohexanol.Formula check: $\displaystyle 1$-methylcyclohexene is \(\displaystyle C_7H_{12}\); adding \(\displaystyle H_2O\) ($\displaystyle 2$ H and $\displaystyle 1$ O) gives \(\displaystyle C_7H_{14}O\), which is exactly $\displaystyle 1$-methylcyclohexanol.(ii) $\displaystyle 4$-Methylheptan-$\displaystyle 4$-ol — condensed as \(\displaystyle CH_3\text{-}CH_2\text{-}CH_2\text{-}C(CH_3)(OH)\text{-}CH_2\text{-}CH_2\text{-}CH_3\): a seven-carbon chain (heptane, numbered \(\displaystyle \mathrm{C_{1}}\) to C7) with both the \(\displaystyle -OH\) and a methyl branch on the middle carbon, C4.Removing the \(\displaystyle -OH\) from \(\displaystyle \mathrm{C_{4}}\) and an \(\displaystyle H\) from \(\displaystyle \mathrm{C_{3}}\) puts a double bond back between \(\displaystyle \mathrm{C_{3}}\) and C4. The alkene is $\displaystyle 4$-methylhept-$\displaystyle 3$-ene: \(\displaystyle CH_3\text{-}CH_2\text{-}CH=C(CH_3)\text{-}CH_2\text{-}CH_2\text{-}CH_3\).Mechanism: \(\displaystyle H_3O^+\) protonates \(\displaystyle \mathrm{C_{3}}\) (which, as an alkene carbon, carries only an ethyl chain and one H), so the double-bond electrons swing over and the positive charge sits on C4. \(\displaystyle \mathrm{C_{4}}\) is bonded to three carbon groups — the methyl branch, the propyl arm C5–C6–C7, and the (now single-bonded) \(\displaystyle \mathrm{C_{3}}\), which itself leads on to C2–C1 — so \(\displaystyle \mathrm{C_{4}}\) is a tertiary carbocation. Protonating \(\displaystyle \mathrm{C_{4}}\) instead would leave the cation on \(\displaystyle \mathrm{C_{3}}\) with only one carbon substituent (a primary cation), far less stable, so \(\displaystyle \mathrm{C_{3}}\) is where protonation actually occurs. Water's oxygen lone pair then attacks \(\displaystyle \mathrm{C_{4}}\), giving the oxonium ion, and a second water molecule removes the extra proton to release neutral $\displaystyle 4$-methylheptan-$\displaystyle 4$-ol.Formula check: $\displaystyle 4$-methylhept-$\displaystyle 3$-ene is \(\displaystyle C_8H_{16}\); adding \(\displaystyle H_2O\) gives \(\displaystyle C_8H_{18}O\), matching $\displaystyle 4$-methylheptan-$\displaystyle 4$-ol.(iii) Pentan-$\displaystyle 2$-ol — condensed as \(\displaystyle CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_2\text{-}CH_3\): the \(\displaystyle -OH\) sits on \(\displaystyle \mathrm{C_{2}}\) of a five-carbon chain, one carbon in from the end.Removing the \(\displaystyle -OH\) from \(\displaystyle \mathrm{C_{2}}\) and an \(\displaystyle H\) from \(\displaystyle \mathrm{C_{1}}\) restores a double bond between \(\displaystyle \mathrm{C_{1}}\) and C2. The alkene is pent-$\displaystyle 1$-ene: \(\displaystyle CH_2=CH\text{-}CH_2\text{-}CH_2\text{-}CH_3\).Mechanism: \(\displaystyle H_3O^+\) protonates the terminal carbon \(\displaystyle \mathrm{C_{1}}\) (which has two H's and is the less substituted alkene carbon), so the positive charge ends up on C2. \(\displaystyle \mathrm{C_{2}}\) is bonded to one carbon (the propyl chain C3–C4–C5) plus the now-\(\displaystyle CH_3\) group at \(\displaystyle \mathrm{C_{1}}\) — a secondary carbocation — which is more stable than the alternative (a primary cation at \(\displaystyle \mathrm{C_{1}}\) if protonation had instead occurred at C2), so this is the pathway that dominates. Water attacks \(\displaystyle \mathrm{C_{2}}\), and loss of a proton from the resulting oxonium ion gives neutral pentan-$\displaystyle 2$-ol as the major product (Markovnikov addition of water to a terminal alkene).Formula check: pent-$\displaystyle 1$-ene is \(\displaystyle C_5H_{10}\); adding \(\displaystyle H_2O\) gives \(\displaystyle C_5H_{12}O\), matching pentan-$\displaystyle 2$-ol.(iv) $\displaystyle 2$-Cyclohexylbutan-$\displaystyle 2$-ol — condensed as \(\displaystyle CH_3\text{-}C(OH)(C_6H_{11})\text{-}CH_2\text{-}CH_3\), where \(\displaystyle C_6H_{11}\) is the cyclohexyl group: a four-carbon (butane) chain whose \(\displaystyle \mathrm{C_{2}}\) carries the \(\displaystyle -OH\) and, in place of a hydrogen, a cyclohexyl ring.Removing the \(\displaystyle -OH\) from \(\displaystyle \mathrm{C_{2}}\) and an \(\displaystyle H\) from \(\displaystyle \mathrm{C_{3}}\) restores a double bond between \(\displaystyle \mathrm{C_{2}}\) and C3. The alkene is $\displaystyle 2$-cyclohexylbut-$\displaystyle 2$-ene: \(\displaystyle CH_3\text{-}C(C_6H_{11})=CH\text{-}CH_3\).Mechanism: \(\displaystyle H_3O^+\) protonates \(\displaystyle \mathrm{C_{3}}\) (which carries only a methyl group, \(\displaystyle \mathrm{C_{4}}\), and one H), so the positive charge is left on C2. \(\displaystyle \mathrm{C_{2}}\) is bonded to three carbon groups — the methyl at \(\displaystyle \mathrm{C_{1}}\), the cyclohexyl ring, and the now-ethyl group formed from C3–C4 — a tertiary carbocation, again the more stable option (protonating \(\displaystyle \mathrm{C_{2}}\) instead would strand the cation on \(\displaystyle \mathrm{C_{3}}\) with only one carbon substituent, a primary cation). Water's oxygen attacks \(\displaystyle \mathrm{C_{2}}\), forming the oxonium ion, and loss of a proton gives neutral $\displaystyle 2$-cyclohexylbutan-$\displaystyle 2$-ol.Formula check: $\displaystyle 2$-cyclohexylbut-$\displaystyle 2$-ene is \(\displaystyle \mathrm{C_{10}H_{18}}\); adding \(\displaystyle H_2O\) gives \(\displaystyle \mathrm{C_{10}H_{20}O}\), matching $\displaystyle 2$-cyclohexylbutan-$\displaystyle 2$-ol.In every case the same rule decides where the water adds: protonation occurs at whichever alkene carbon leaves the positive charge on the carbon that can best spread it out over the most alkyl groups (tertiary > secondary > primary), and that carbon is exactly where the \(\displaystyle -OH\) of the target alcohol needs to be.Answer: (i) $\displaystyle 1$-methylcyclohexanol from $\displaystyle 1$-methylcyclohexene; (ii) $\displaystyle 4$-methylheptan-$\displaystyle 4$-ol from $\displaystyle 4$-methylhept-$\displaystyle 3$-ene; (iii) pentan-$\displaystyle 2$-ol from pent-$\displaystyle 1$-ene; (iv) $\displaystyle 2$-cyclohexylbutan-$\displaystyle 2$-ol from $\displaystyle 2$-cyclohexylbut-$\displaystyle 2$-ene — each made by acid-catalysed (Markovnikov) hydration of the alkene, dil. \(\displaystyle H_2SO_4\)/\(\displaystyle H_2O\) (or oxymercuration–demercuration), with \(\displaystyle -OH\) delivered to the carbon that forms the more stable carbocation.

A secondary carbocation with a fully-substituted (tertiary) carbon sitting right next to it never survives as drawn — a hydride ion hops across from that neighbouring carbon so the positive charge ends up on the carbon that can spread it over three alkyl groups instead of two.Number the parent chain of $\displaystyle 3$-methylbutan-$\displaystyle 2$-ol so that C-$\displaystyle 1$ and C-$\displaystyle 4$ are the terminal methyls, C-$\displaystyle 2$ carries the \(\displaystyle -\text{OH}\), and C-$\displaystyle 3$ carries the methyl branch:\[\text{C-1(CH}_3\text{)} - \text{C-2(CH-OH)} - \text{C-3(CH-CH}_3\text{)} - \text{C-4(CH}_3\text{)} \]i.e., condensed, \(\displaystyle \text{CH}_3-\text{CH(OH)}-\text{CH(CH}_3\text{)}-\text{CH}_3\). C-$\displaystyle 2$ (the carbinol carbon) is attached to only two other carbons (C-$\displaystyle 1$ and C-$\displaystyle 3$), so it is a secondary centre, while C-$\displaystyle 3$ is already attached to three carbons (C-$\displaystyle 2$, C-$\displaystyle 4$, and the branch methyl) — it needs to lose only one more group to become a tertiary carbon.Step I — protonation. \(\displaystyle \text{HBr}\) first hands over a proton, not a bromide ion, to the most basic site in the molecule: the lone pair on the oxygen of \(\displaystyle -\text{OH}\). That lone pair attacks \(\displaystyle \text{H}^+\), turning the poor leaving group \(\displaystyle -\text{OH}\) into the good leaving group \(\displaystyle -\overset{+}{\text{O}}\text{H}_2\) (an oxonium ion):\[\text{CH}_3-\text{CH(OH)}-\text{CH(CH}_3\text{)}-\text{CH}_3 + \text{H}^+ \longrightarrow \text{CH}_3-\text{CH}(\overset{+}{\text{O}}\text{H}_2)-\text{CH(CH}_3\text{)}-\text{CH}_3 \]The bromide ion released alongside this proton, \(\displaystyle \text{Br}^-\), stays in solution and is not used until Step IV.Step II — ionisation to a secondary carbocation. The C–O bond of the oxonium ion breaks heterolytically: both bonding electrons leave with the oxygen, which departs as a neutral water molecule. C-$\displaystyle 2$ is left with only three bonds and an empty \(\displaystyle p\) orbital:\[\text{CH}_3-\text{CH}(\overset{+}{\text{O}}\text{H}_2)-\text{CH(CH}_3\text{)}-\text{CH}_3 \longrightarrow \text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH(CH}_3\text{)}-\text{CH}_3 \;+\; \text{H}_2\text{O} \]This cation at C-$\displaystyle 2$ is only secondary (flanked by C-$\displaystyle 1$ and C-$\displaystyle 3$), so the skeleton has not yet reached its most stable cationic form.Step III — the $\displaystyle 1,2$-hydride shift (the rearrangement the hint points to). C-$\displaystyle 3$, sitting immediately next to the electron-deficient C-$\displaystyle 2$, carries a hydrogen it can give up. That C-$\displaystyle 3$–H bonding pair migrates as a hydride ion, \(\displaystyle \text{H}:^-\), across the C-$\displaystyle 2$–C-$\displaystyle 3$ bond onto the empty orbital of C-2. C-$\displaystyle 2$ is filled and becomes an ordinary, neutral \(\displaystyle -\text{CH}_2-\) carbon; C-$\displaystyle 3$, having just lost that hydride, is left with only three bonds — to C-$\displaystyle 2$, to C-$\displaystyle 4$, and to the branch methyl — so the positive charge now sits on C-$\displaystyle 3$, which is tertiary:\[\text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH(CH}_3\text{)}-\text{CH}_3 \longrightarrow \text{CH}_3-\text{CH}_2-\overset{+}{\text{C}}(\text{CH}_3)-\text{CH}_3 \]The product of this step, \(\displaystyle \text{CH}_3-\text{CH}_2-\overset{+}{\text{C}}(\text{CH}_3)-\text{CH}_3\), is a tertiary carbocation: it carries three alkyl groups (one ethyl, two methyls) that donate electron density into the empty \(\displaystyle p\) orbital by hyperconjugation and the inductive effect, so it lies lower in energy than the secondary cation formed in Step II. A hydride shift from C-$\displaystyle 1$ was geometrically just as possible, but it would have moved the charge onto a primary carbon — less stable than the starting secondary cation — so that path is not taken; the C-$\displaystyle 3$ shift is followed because it is the route to a more stable cation, not merely an available one.Step IV — capture by bromide. The tertiary carbocation is now attacked by the bromide ion set free back in Step I. The lone pair on \(\displaystyle \text{Br}^-\) attacks the empty \(\displaystyle p\) orbital of the cationic carbon, forming a new C–Br \(\displaystyle \sigma\) bond:\[\text{CH}_3-\text{CH}_2-\overset{+}{\text{C}}(\text{CH}_3)-\text{CH}_3 \;+\; \text{Br}^- \longrightarrow \text{CH}_3-\text{CH}_2-\text{C(CH}_3\text{)(Br)}-\text{CH}_3 \]Renumbering this product chain to give the substituents the lowest locants puts both the bromine and the methyl branch on C-$\displaystyle 2$ of the four-carbon chain, so the compound formed is $\displaystyle 2$-bromo-$\displaystyle 2$-methylbutane — the rearranged, tertiary bromide, not the "expected" $\displaystyle 2$-bromo-$\displaystyle 3$-methylbutane that a plain substitution at the original carbinol carbon (C-$\displaystyle 2$) would have given.Answer: Protonation of the \(\displaystyle -\text{OH}\) group (Step I) is followed by loss of water to give the secondary carbocation \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH(CH}_3\text{)}-\text{CH}_3\) at C-$\displaystyle 2$ (Step II). A hydride ion migrates from C-$\displaystyle 3$ to C-$\displaystyle 2$ (Step III), converting this into the more stable tertiary carbocation \(\displaystyle \text{CH}_3-\text{CH}_2-\overset{+}{\text{C}}(\text{CH}_3)-\text{CH}_3\). Bromide ion then attacks this tertiary carbocation (Step IV) to give the rearranged product \(\displaystyle \text{CH}_3-\text{CH}_2-\text{C(CH}_3\text{)(Br)}-\text{CH}_3\), i.e., $\displaystyle 2$-bromo-$\displaystyle 2$-methylbutane (tert-amyl bromide).