NCERT Solutions · Class 12 Chemistry Alcohols, Phenols and Ethers

Acid-catalysed hydration of an alkene is electrophilic addition — the pi electrons of the double bond are the nucleophile, and \(\displaystyle H_3O^+ \) supplies the electrophile that starts the attack, in three steps: protonation, water addition, deprotonation.Ethene is \(\displaystyle CH_2=CH_2 \): two \(\displaystyle sp^2 \) carbons joined by a sigma bond and a pi bond, the pi electrons sitting above and below the C–C axis, exposed and electron-rich. In dilute acid (e.g. dilute \(\displaystyle H_2SO_4 \) in water) the acid is only a catalyst — it is used up in the first step and handed back in the last one.Step $\displaystyle 1$ — Protonation of ethene by hydronium ion. The electrophile here is \(\displaystyle H_3O^+ \) (a proton attached to water). The pi electron pair of \(\displaystyle CH_2=CH_2 \) attacks one of the acidic hydrogens of \(\displaystyle H_3O^+ \). A new C–H sigma bond forms using the pi electrons, and simultaneously the O–H bond of \(\displaystyle H_3O^+ \) breaks heterolytically, both electrons of that bond staying on oxygen. This does two things at once: it converts one alkene carbon into an \(\displaystyle sp^3\) carbon carrying the new H, and it leaves the other carbon short an electron pair, so it becomes a carbocation.\[CH_2=CH_2 + H_3O^+ \longrightarrow CH_3-\overset{+}{C}H_2 + H_2O \]The intermediate \(\displaystyle CH_3-\overset{+}{C}H_2 \) is the ethyl carbocation — a primary carbocation with an empty \(\displaystyle p\) orbital on the terminal carbon — and a neutral water molecule is released as the leaving group from the original hydronium ion.Step $\displaystyle 2$ — Nucleophilic attack of water on the carbocation. Water, \(\displaystyle H_2O \), is a nucleophile because oxygen carries two lone pairs. One lone pair on the oxygen of a fresh water molecule attacks the positively charged carbon of \(\displaystyle CH_3-\overset{+}{C}H_2 \), forming a new C–O sigma bond. The oxygen, having just used a lone pair to bond, now carries a formal positive charge and three bonds (two H, one C) plus one remaining lone pair.\[CH_3-\overset{+}{C}H_2 + H_2O \longrightarrow CH_3-CH_2-\overset{+}{O}H_2 \]The product of this step, \(\displaystyle CH_3-CH_2-\overset{+}{O}H_2 \), is protonated ethanol — an oxonium ion (ethyl analogue of \(\displaystyle H_3O^+ \)) — not yet the neutral alcohol.Step $\displaystyle 3$ — Deprotonation restores the neutral alcohol and regenerates the catalyst. A second molecule of water now acts as a base rather than a nucleophile at carbon: a lone pair on its oxygen pulls off one of the two acidic protons sitting on the oxonium oxygen of \(\displaystyle CH_3-CH_2-\overset{+}{O}H_2 \). That O–H bond breaks heterolytically, its electrons staying on the ethanol oxygen, while the abstracted proton ends up bonded to the attacking water, recreating \(\displaystyle H_3O^+ \).\[CH_3-CH_2-\overset{+}{O}H_2 + H_2O \longrightarrow CH_3-CH_2-OH + H_3O^+ \]Because the \(\displaystyle H_3O^+ \) consumed in step $\displaystyle 1$ reappears here in step $\displaystyle 3$, the acid is never used up overall — it is a true catalyst, and the sequence can run again on another ethene molecule.The species named at each stage, in order: electrophile \(\displaystyle H_3O^+ \) → alkene \(\displaystyle CH_2=CH_2 \) → primary carbocation intermediate \(\displaystyle CH_3-\overset{+}{C}H_2 \) → nucleophile \(\displaystyle H_2O \) → oxonium-ion intermediate \(\displaystyle CH_3CH_2\overset{+}{O}H_2 \) → base \(\displaystyle H_2O \) → final product ethanol \(\displaystyle CH_3CH_2OH \), with \(\displaystyle H_3O^+ \) regenerated.Adding the three steps together, the net transformation is simple addition of \(\displaystyle H \) and \(\displaystyle OH \) across the double bond, with the acid cancelling out:\[CH_2=CH_2 + H_2O \xrightarrow{\ H^+ (\text{catalyst})\ } CH_3-CH_2-OH \]**Answer: The hydration of ethene to ethanol proceeds by acid-catalysed electrophilic addition in three steps — ($\displaystyle 1$) \(\displaystyle H_3O^+ \) protonates the ethene pi bond to give the primary carbocation \(\displaystyle CH_3\overset{+}{C}H_2 \) and \(\displaystyle H_2O \); ($\displaystyle 2$) a water molecule's lone pair attacks this carbocation to give the oxonium ion \(\displaystyle CH_3CH_2\overset{+}{O}H_2 \); ($\displaystyle 3$) a second water molecule removes a proton from this oxonium ion, giving the product ethanol , \(\displaystyle CH_3CH_2OH \) (condensed formula \(\displaystyle CH_3-CH_2-OH \)), and regenerating \(\displaystyle H_3O^+ \) so the acid is not consumed overall.

Three separate reactions get you from benzene all the way to phenol: sulphonate the ring, fuse the sulphonic acid with alkali to swap the group for oxygen, then liberate the free phenol with acid.Benzene itself has no group that alkali can attack, so the first job is to put a leaving group on the ring. Conc. \(\displaystyle \mathrm{H_2SO_4} \) does that by electrophilic aromatic substitution.Step $\displaystyle 1$ — Sulphonation (benzene + conc. \(\displaystyle \mathrm{H_2SO_4}\)). Conc. sulphuric acid generates the electrophile \(\displaystyle \mathrm{SO_3H^+} \) (effectively \(\displaystyle \mathrm{SO_3} \) protonated), which attacks the benzene ring exactly as \(\displaystyle \mathrm{NO_2^+} \) does in nitration. A ring carbon bonds to sulphur, a proton is lost from that carbon to restore aromaticity, and the ring keeps its aromatic sextet throughout — this is substitution, not addition. \[\mathrm{C_6H_6 + HO\!-\!SO_3H \; \xrightarrow{\Delta} \; C_6H_5\!-\!SO_3H \; + \; H_2O} \] The product is benzenesulphonic acid, \(\displaystyle \mathrm{C_6H_5SO_3H} \): a benzene ring carrying one \(\displaystyle \mathrm{-SO_3H} \) (sulphonic acid) group in place of one ring hydrogen.Step $\displaystyle 2$ — Alkali fusion with NaOH. \(\displaystyle \mathrm{-SO_3H} \) is first neutralised by NaOH to the sodium salt, and then, when that salt is fused (heated solid, no solvent) with excess NaOH at high temperature (around $\displaystyle 573$ K), the \(\displaystyle \mathrm{-SO_3Na} \) group is displaced from the ring by hydroxide acting as a nucleophile. The aryl carbon that bore sulphur now bonds to oxygen instead — a nucleophilic aromatic substitution at that one carbon, with sulphite ion leaving. \[\mathrm{C_6H_5\!-\!SO_3H \; + \; NaOH \; \longrightarrow \; C_6H_5\!-\!SO_3Na \; + \; H_2O} \] \[\mathrm{C_6H_5\!-\!SO_3Na \; + \; 2NaOH \; \xrightarrow{\text{fuse, } \sim 573\,K} \; C_6H_5\!-\!ONa \; + \; Na_2SO_3 \; + \; H_2O} \] The product \(\displaystyle \mathrm{C_6H_5ONa} \) is sodium phenoxide: a benzene ring bonded directly to \(\displaystyle \mathrm{-O^-Na^+} \). Note the trap here — this fusion step needs solid NaOH at high temperature, not NaOH in dilute aqueous solution; a cold aqueous NaOH wash only re-forms the sodium sulphonate salt of Step $\displaystyle 2$ and never touches the ring carbon.Step $\displaystyle 3$ — Acidification. Sodium phenoxide is the sodium salt of phenol, not phenol itself, because the negative charge sits on oxygen. Treating it with a dilute mineral acid (dilute HCl, or simply dilute \(\displaystyle \mathrm{H_2SO_4} \) — using the same acid already on hand) protonates that oxygen and pushes the equilibrium to the neutral compound. \[\mathrm{C_6H_5\!-\!ONa \; + \; HCl \; \longrightarrow \; C_6H_5\!-\!OH \; + \; NaCl} \]Put together, the three-stage route is \[\mathrm{C_6H_6 \;\xrightarrow[\Delta]{\text{conc. }H_2SO_4}\; C_6H_5SO_3H \;\xrightarrow[\text{fuse}]{NaOH}\; C_6H_5ONa \;\xrightarrow{H^+/H_2O}\; C_6H_5OH} \] and the final product \(\displaystyle \mathrm{C_6H_5OH} \) is phenol: a hydroxyl group bonded directly to a benzene ring carbon.Answer: benzene is sulphonated by conc. \(\displaystyle \mathrm{H_2SO_4} \) to benzenesulphonic acid ( \(\displaystyle \mathrm{C_6H_5SO_3H} \) ), which is fused with NaOH to give sodium phenoxide ( \(\displaystyle \mathrm{C_6H_5ONa} \) ), and acidification of the phenoxide liberates phenol, \(\displaystyle \mathrm{C_6H_5OH} \).

Three different reactions, because three different bonds have to be made -- Markovnikov addition across a double bond, a single backside \(\displaystyle \mathrm{(SN_{2})}\) attack by hydroxide on a primary halide, and a one-carbon chain extension through a Grignard reagent.(i) $\displaystyle 1$-phenylethanol from a suitable alkeneThe target, $\displaystyle 1$-phenylethanol, is C6H5-CH(OH)-CH3: a secondary alcohol with the -OH sitting on the benzylic carbon (the carbon joined directly to the ring). Its two-carbon skeleton, CH(OH)-CH3, is exactly what you get by adding water across a C=C bond, so the alkene to start from is styrene (phenylethene), C6H5-CH=CH2.Reaction: C6H5-CH=\(\displaystyle \mathrm{CH_{2}}\) + \(\displaystyle \mathrm{H_{2}O}\), with dilute \(\displaystyle \mathrm{H_{2}SO_{4}}\) (H+) as catalyst, gives C6H5-CH(OH)-CH3.Mechanism (acid-catalysed electrophilic addition): Step $\displaystyle 1$ -- A proton from the acid is the electrophile. The pi electrons of the C=C bond attack it, and the new C-H bond forms at the terminal, unsubstituted carbon (the \(\displaystyle \mathrm{CH_{2}}\) end). This leaves the positive charge on the other carbon -- the one bonded to the phenyl ring. Step $\displaystyle 2$ -- That carbocation is benzylic: its empty p orbital overlaps with the ring's pi system, so the positive charge is delocalised into the ring by resonance. This resonance stabilisation makes the benzylic cation far more stable than the alternative (a primary cation with no ring attached), which is exactly why the proton adds to the terminal carbon rather than the substituted one -- Markovnikov's rule, explained by cation stability rather than just memorised. Step $\displaystyle 3$ -- A water molecule, the nucleophile, uses a lone pair on oxygen to attack this electrophilic benzylic carbon, forming a new C-O bond and giving a protonated alcohol, C6H5-CH(OH2+)-CH3. Step $\displaystyle 4$ -- A second water molecule removes the extra proton from that oxonium ion, regenerating the \(\displaystyle \mathrm{H^{+}}\) catalyst and releasing the neutral alcohol.Product: $\displaystyle 1$-phenylethanol, C6H5-CH(OH)-CH3.(ii) cyclohexylmethanol using an alkyl halide by an \(\displaystyle \mathrm{SN_{2}}\) reactionThe target, cyclohexylmethanol , is a cyclohexane ring carrying a -CH2OH group: C6H11-CH2-OH. The carbon bearing the -OH is primary (bonded to just one other carbon, the ring carbon, plus two hydrogens) -- unhindered, which is exactly the pattern \(\displaystyle \mathrm{SN_{2}}\) needs.Starting halide: (bromomethyl)cyclohexane, C6H11-CH2-Br, a primary bromide.Reaction: C6H11-CH2-Br + aqueous KOH gives C6H11-CH2-OH + KBr.Mechanism (bimolecular nucleophilic substitution, SN2): Step $\displaystyle 1$ -- Hydroxide ion, OH-, is the nucleophile. Since the carbon carrying the bromine is primary and open, OH- approaches it directly from the side opposite the C-Br bond (backside attack). Step $\displaystyle 2$ -- In one concerted step the incoming C-O bond forms as the C-Br bond breaks and Br- departs. The transition state has the entering OH and the leaving Br both partially bonded to the central carbon, $\displaystyle 180$ degrees apart (a trigonal-bipyramidal arrangement) -- there is no separate carbocation stage. Step $\displaystyle 3$ -- Because everything happens in this single step, the rate depends on the concentrations of both the halide and the hydroxide -- which is the "bimolecular" in SN2.Product: cyclohexylmethanol, C6H11-CH2-OH.(iii) pentan-$\displaystyle 1$-ol using a suitable alkyl halideThe target, pentan-$\displaystyle 1$-ol, is a straight-chain, five-carbon primary alcohol: CH3-CH2-CH2-CH2-CH2-OH. Simply hydrolysing a five-carbon halide would just repeat method (ii), so "a suitable alkyl halide" here points to a route that builds the chain up by one carbon -- a Grignard reagent reacting with formaldehyde.Starting halide: $\displaystyle 1$-bromobutane, CH3-CH2-CH2-CH2-Br, a four-carbon primary halide.Step $\displaystyle 1$ -- forming the Grignard reagent: $\displaystyle 1$-bromobutane is treated with magnesium turnings in dry ether (moisture has to be kept out completely, since a Grignard reagent is destroyed instantly by water or any -OH group). Magnesium inserts into the C-Br bond: CH3CH2CH2CH2-Br + Mg, in dry ether, gives CH3CH2CH2CH2-MgBr (butylmagnesium bromide).Step $\displaystyle 2$ -- addition to formaldehyde: the C-Mg bond is strongly polarised, so the butyl carbon behaves as a carbanion (a carbon nucleophile). It attacks the electrophilic carbonyl carbon of formaldehyde, H-CHO -- the smallest possible carbonyl compound, contributing exactly one carbon to the product: CH3CH2CH2CH2-MgBr + H-CHO gives CH3CH2CH2CH2-CH2-O-MgBr.Step $\displaystyle 3$ -- hydrolysis: dilute acid (H3O+) protonates the alkoxide oxygen, releasing the free alcohol: CH3CH2CH2CH2-CH2-O-MgBr + \(\displaystyle \mathrm{H_{3}O^{+}}\) gives CH3CH2CH2CH2-CH2-OH + a basic magnesium salt.Product: pentan-$\displaystyle 1$-ol, \(\displaystyle \mathrm{CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH}\) -- five carbons in all, because the fifth carbon (originally formaldehyde's carbonyl carbon) has been bonded onto the four-carbon butyl group.Answer: (i) $\displaystyle 1$-phenylethanol, C6H5-CH(OH)-CH3, from acid-catalysed (Markovnikov) hydration of styrene: C6H5-CH=\(\displaystyle \mathrm{CH_{2}}\) + \(\displaystyle \mathrm{H_{2}O}\), \(\displaystyle \mathrm{H^{+}}\)/H2SO4. (ii) Cyclohexylmethanol, C6H11-CH2-OH, from \(\displaystyle \mathrm{SN_{2}}\) hydrolysis of (bromomethyl)cyclohexane: C6H11-CH2-Br + aq. KOH. (iii) Pentan-$\displaystyle 1$-ol, \(\displaystyle \mathrm{CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH}\), from $\displaystyle 1$-bromobutane via its Grignard reagent \(\displaystyle \mathrm{(CH_{3}CH_{2}CH_{2}CH_{2}MgBr)}\) reacting with formaldehyde, then hydrolysis with dilute acid.

Phenol reacts with a base (and even with reactive metals) the way a real acid does; ethanol does neither — because the phenoxide ion formed from phenol is resonance-stabilized and the ethoxide ion from ethanol is not.Reaction $\displaystyle 1$ — with a base (aqueous sodium hydroxide).Phenol , \(\displaystyle \text{C}_6\text{H}_5\text{-OH} \), is acidic enough to be neutralized by \(\displaystyle \text{NaOH} \), giving the water-soluble salt sodium phenoxide and water:\[\text{C}_6\text{H}_5\text{-OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{-O}^-\text{Na}^+ + \text{H}_2\text{O} \]Here the base attacks the acidic O–H proton of phenol directly; no ring chemistry is involved. Ethanol , \(\displaystyle \text{CH}_3\text{CH}_2\text{-OH} \), is placed in the same NaOH solution and nothing happens — its O–H is not acidic enough for \(\displaystyle \text{OH}^- \) to remove the proton, so ethanol is recovered unchanged. This single test — solubility/reaction with NaOH — is the standard way phenols are told apart from alcohols.Reaction $\displaystyle 2$ — with reactive metal sodium.Both phenol and ethanol will give up the O–H proton to sodium metal, but phenol does it far more vigorously, again because the resulting anion is more stable:\[2\,\text{C}_6\text{H}_5\text{-OH} + 2\,\text{Na} \rightarrow 2\,\text{C}_6\text{H}_5\text{-O}^-\text{Na}^+ + \text{H}_2\uparrow \]\[2\,\text{CH}_3\text{CH}_2\text{-OH} + 2\,\text{Na} \rightarrow 2\,\text{CH}_3\text{CH}_2\text{-O}^-\text{Na}^+ + \text{H}_2\uparrow \]Both liberate hydrogen gas and both give a metal alkoxide/phenoxide, but the phenol reaction is faster and goes further, because sodium phenoxide is a more stable (lower-energy) species to form than sodium ethoxide — again pointing to the greater acidity of the O–H bond in phenol.Why phenol is the stronger acid — comparing phenol and ethanol quantitatively and structurally.Acid strength is set by how stable the conjugate base is once the proton has left. Write the two conjugate bases and compare:Phenoxide ion, \(\displaystyle \text{C}_6\text{H}_5\text{-O}^- \): the negative charge on oxygen is not stuck on that one atom. The oxygen's lone pair can conjugate with the benzene \(\displaystyle \pi \) system, so the negative charge is delocalized onto the ortho and para carbons of the ring as well as staying on oxygen — five resonance structures altogether (one with charge on O, three with charge on ring carbons that are stabilized by the ring's own conjugation, and the underlying benzene resonance). Spreading a charge over several atoms always lowers the ion's energy, so the phenoxide ion is considerably more stable than a simple, charge-localized alkoxide.Ethoxide ion, \(\displaystyle \text{CH}_3\text{CH}_2\text{-O}^- \): there is no adjacent \(\displaystyle \pi \) system for the oxygen's negative charge to delocalize into. The charge sits entirely on the one oxygen atom (only inductive donation from the two alkyl C–H/C–C bonds, which if anything pushes electron density toward the already-negative oxygen and destabilizes it slightly). This ion is far less stable than phenoxide.There is a second, reinforcing effect on the neutral molecules themselves. In phenol the oxygen lone pair is partly delocalized into the ring even before ionization (this is also why the C–OH bond in phenol is shorter than a normal C–O single bond, carrying some double-bond character), which withdraws electron density from oxygen and makes the O–H bond more polarized, i.e. the proton is already held more loosely. In ethanol the sp3 carbon attached to O has no such conjugation, and the alkyl group's electron-donating (+I) effect pushes electron density onto oxygen, making the O–H bond less polarized and the proton harder to remove.Because the anion phenol produces on losing a proton is so much better stabilized than the anion ethanol produces, phenol loses its proton far more readily: \[K_a(\text{phenol}) \approx 1.0\times10^{-10}\ (\text{p}K_a \approx 10),\qquad K_a(\text{ethanol}) \approx 10^{-16}\ (\text{p}K_a \approx 16) \] so phenol is roughly a million times (\(\displaystyle 10^6\)) more acidic than ethanol — strong enough to react with aqueous NaOH, which ethanol cannot do at all.Answer: Two reactions showing phenol's acidity are (i) with aqueous NaOH, \(\displaystyle \text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O} \) (ethanol does not react), and (ii) with sodium metal, \(\displaystyle 2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2\uparrow \) (more vigorous than the analogous, slower reaction of ethanol). Phenol (\(\displaystyle \text{p}K_a \approx 10 \)) is much more acidic than ethanol (\(\displaystyle \text{p}K_a \approx 16 \)) because the phenoxide ion is resonance-stabilized by delocalization of the negative charge into the benzene ring, whereas the ethoxide ion has no such stabilization and is destabilized further by the electron-donating alkyl group.

The strength of a substituted phenol as an acid is set by how well its conjugate base -- the phenoxide ion -- can spread out the negative charge it carries; a group that pulls charge away and shares it onto an electronegative atom stabilises that anion and raises acidity, while a group that pushes extra electron density back onto the already-charged oxygen destabilises it and lowers acidity.Both compounds ionise the same way: loss of the phenolic proton gives a phenoxide ion, \[\text{Ar-OH} \rightleftharpoons \text{Ar-O}^- + \text{H}^+ \] where \(\displaystyle \text{Ar} \) is the substituted benzene ring. Which side of this equilibrium is favoured depends entirely on how stable \(\displaystyle \text{Ar-O}^- \) is, since the free proton \(\displaystyle \text{H}^+ \) is identical in both cases.Step $\displaystyle 1$ -- what the nitro group does in $\displaystyle 2$-nitrophenol (ortho-nitrophenol). In an unsubstituted phenoxide ion the negative charge on oxygen is already delocalised into the ring by resonance, onto the carbons ortho and para to it (\(\displaystyle \text{C-2} \), \(\displaystyle \text{C-4} \), \(\displaystyle \text{C-6} \) if the oxygen sits on \(\displaystyle \text{C-1} \)). In $\displaystyle 2$-nitrophenol the \(\displaystyle -\text{NO}_2 \) group sits at \(\displaystyle \text{C-2} \), exactly one of the ring positions that carries negative charge in a resonance form of the phenoxide. Because the nitrogen of \(\displaystyle -\text{NO}_2 \) is itself part of an \(\displaystyle \text{N=O} \) \(\displaystyle \pi \) system, the electron pair that resonance places on \(\displaystyle \text{C-2} \) does not have to stop there: it can conjugate one step further into the nitro group, pushing the negative charge out onto one of its oxygen atoms. This gives extra canonical (resonance) structures for the anion in which the negative charge sits on a nitro-group oxygen rather than on the ring carbon or the phenolic oxygen. Oxygen is far more electronegative than carbon, so it can carry a negative charge at much lower energy cost; spreading the charge this far -- onto two additional electronegative atoms -- stabilises the anion substantially. On top of this resonance (mesomeric, \(\displaystyle -\text{M} \)) effect, \(\displaystyle -\text{NO}_2 \) is also a strong inductively electron-withdrawing (\(\displaystyle -\text{I} \)) group, pulling electron density away from the ring and the phenoxide oxygen through the \(\displaystyle \sigma \)-bond framework as well. Both effects act in the same direction -- they withdraw electron density from the site of the negative charge and disperse it -- so the $\displaystyle 2$-nitrophenoxide ion is markedly more stable than an unsubstituted phenoxide ion.Step $\displaystyle 2$ -- what the methoxy group does in $\displaystyle 2$-methoxyphenol (ortho-methoxyphenol, guaiacol). The \(\displaystyle -\text{OCH}_3 \) group is attached through an oxygen atom that carries two lone pairs, one of which conjugates with the ring \(\displaystyle \pi \) system -- this makes \(\displaystyle -\text{OCH}_3 \) a resonance-donating (\(\displaystyle +\text{M} \)) group, not a withdrawing one. Resonance donation from \(\displaystyle -\text{OCH}_3 \) pushes electron density into the ring at the positions ortho and para to itself; since \(\displaystyle -\text{OCH}_3 \) sits on \(\displaystyle \text{C-2} \), one of those positions is \(\displaystyle \text{C-1} \) -- exactly where the phenolic oxygen (already bearing the negative charge in the phenoxide ion) is attached. So instead of drawing charge away from the anion and spreading it out, the methoxy group feeds extra electron density back onto the very oxygen that is already negatively charged. Concentrating negative charge on one atom rather than delocalising it raises the energy of the ion -- it is destabilising. \(\displaystyle -\text{OCH}_3 \) does have a weak inductive (\(\displaystyle -\text{I} \)) pull, since oxygen is electronegative, but this is far outweighed by its \(\displaystyle +\text{M} \) donation, so the net electronic effect of the group is electron-donating. The $\displaystyle 2$-methoxyphenoxide ion is therefore less stable than an unsubstituted phenoxide ion -- consistent with $\displaystyle 2$-methoxyphenol being reported as a slightly weaker acid than phenol itself (\(\displaystyle \text{p}K_a \) of phenol \(\displaystyle \approx 10.0 \); of $\displaystyle 2$-methoxyphenol \(\displaystyle \approx 9.9\text{-}10.0 \)), while $\displaystyle 2$-nitrophenol is a much stronger acid than phenol (\(\displaystyle \text{p}K_a \approx 7.2 \)).Step $\displaystyle 3$ -- put the two together. Acid strength tracks the stability of the conjugate base. The $\displaystyle 2$-nitrophenoxide ion is stabilised because \(\displaystyle -\text{NO}_2 \) withdraws the negative charge and disperses it onto its own, highly electronegative oxygen atoms. The $\displaystyle 2$-methoxyphenoxide ion is destabilised because \(\displaystyle -\text{OCH}_3 \) donates electron density and reinforces the negative charge already sitting on the phenolic oxygen. So the equilibrium \[\text{Ar-OH} \rightleftharpoons \text{Ar-O}^- + \text{H}^+ \] lies further to the right when \(\displaystyle \text{Ar} \) is the $\displaystyle 2$-nitrophenyl ring than when it is the $\displaystyle 2$-methoxyphenyl ring, meaning ortho-nitrophenol loses its proton more readily.Answer: Ortho-nitrophenol is more acidic than ortho-methoxyphenol because \(\displaystyle -\text{NO}_2 \) is a strong electron-withdrawing group (by both resonance and induction) that stabilises the phenoxide ion by delocalising its negative charge onto the nitro group's own oxygen atoms, whereas \(\displaystyle -\text{OCH}_3 \) is a net electron-donating group (its \(\displaystyle +\text{M} \) resonance dominates its weak \(\displaystyle -\text{I} \)) that destabilises the phenoxide ion by pushing extra electron density back onto the already-negative phenolic oxygen; the more stable conjugate base makes ortho-nitrophenol the stronger acid.

A lone pair on the oxygen of –OH is delocalised into the benzene ring by resonance, so the ring becomes electron-rich specifically at the ortho and para carbons – this +R (mesomeric) donation is stronger than oxygen's –I (inductive) pull, so the net effect is activation, not deactivation.Phenol is \(\displaystyle \text{C}_6\text{H}_5\text{-OH} \): a benzene ring with the –OH group bonded directly to one ring carbon — call it \(\displaystyle \mathrm{C_{1}}\), the carbon bearing the substituent. Oxygen carries two lone pairs. Because \(\displaystyle \mathrm{C_{1}}\) is sp2 and part of the ring's conjugated system, oxygen itself behaves as sp2-hybridised here, and one of its lone pairs sits in a p-orbital that is parallel to, and overlaps with, the p-orbitals making up the ring's \(\displaystyle \pi \) system. That parallel alignment is what lets the lone pair conjugate with the ring instead of staying localised on oxygen — exactly the geometry needed for resonance donation.Because that lone pair can delocalise, phenol is a resonance hybrid of five contributing structures, not one fixed Kekulé picture:Structure I — the ordinary no-charge Kekulé form: alternating ring double bonds, a single C1–O bond, oxygen neutral with two lone pairs.Structure II — one oxygen lone pair moves in to form a \(\displaystyle \mathrm{C_{1}}\)=O double bond; to keep every carbon tetravalent, the adjacent \(\displaystyle \mathrm{C_{1}}\)=\(\displaystyle \mathrm{C_{2}}\) ring double bond breaks and that electron pair shifts onto C2. Oxygen, now with only one lone pair and three bonds, carries a formal positive charge; \(\displaystyle \mathrm{C_{2}}\) carries the extra pair and a formal negative charge.Structure III — the mirror shift, using the ring double bond on the other side of \(\displaystyle \mathrm{C_{1}}\), moves the negative charge onto \(\displaystyle \mathrm{C_{6}}\) instead of \(\displaystyle \mathrm{C_{2}}\) (C2 and \(\displaystyle \mathrm{C_{6}}\) are the two, symmetry-equivalent ortho carbons).Structure IV — starting from structure II, the negative charge is pushed one bond further round the ring (through the \(\displaystyle \mathrm{C_{2}}\)=\(\displaystyle \mathrm{C_{3}}\) double bond) so it ends up on \(\displaystyle \mathrm{C_{4}}\), the para carbon, while \(\displaystyle \mathrm{C_{1}}\)=O stays a double bond and oxygen stays positive.Structure V — the same relay starting from structure III (through the \(\displaystyle \mathrm{C_{6}}\)=\(\displaystyle \mathrm{C_{5}}\) double bond) also lands the negative charge on \(\displaystyle \mathrm{C_{4}}\), again with oxygen positive.Across the whole set, the formal negative charge only ever appears at \(\displaystyle \mathrm{C_{2}}\), \(\displaystyle \mathrm{C_{6}}\) (both ortho) and \(\displaystyle \mathrm{C_{4}}\) (para) — never at \(\displaystyle \mathrm{C_{3}}\) or \(\displaystyle \mathrm{C_{5}}\) (the meta carbons), because there is no bonding pathway that carries the oxygen's donated pair to a meta position in an alternating (conjugated) system. So the electron density oxygen pushes into the ring lands exactly at the ortho and para carbons, making the ring as a whole more electron-rich than benzene, which has no such donor and no such resonance structures.This picture is not just formal bookkeeping — it shows up in a measurable way: the C1–O bond in phenol is shorter (about $\displaystyle 136$ pm) than a typical C–O single bond in a saturated alcohol (about $\displaystyle 143$ pm), because resonance structures II–V all give that bond partial double-bond character. The C–OH bond behaves as something between a single and a double bond, which is exactly what conjugation with the ring predicts.Two consequences follow directly:Rate of reaction — an electrophile is electron-seeking, so a ring carrying extra electron density reacts with it faster than benzene does. Phenol undergoes nitration, sulfonation, halogenation and related electrophilic aromatic substitutions much more readily than benzene. For example, phenol reacts with bromine water alone, at room temperature, with no \(\displaystyle \text{FeBr}_3 \) or other Lewis-acid catalyst needed to generate the electrophile — unlike benzene, which requires that catalyst. The product is $\displaystyle 2,4,6$-tribromophenol , \(\displaystyle \text{C}_6\text{H}_2\text{Br}_3\text{OH} \), because substitution occurs at all three ring positions (both ortho carbons and the one para carbon) that carry the extra electron density.Orientation of attack — since the extra electron density sits only at the ortho and para carbons, the electrophile bonds preferentially there, generating an arenium-ion (Wheland) intermediate in which the positive charge can be delocalised onto the very carbon that bears the oxygen — the most stabilised of the possible intermediates. Attack at a meta carbon gives an intermediate that cannot draw on that extra stabilisation, so meta product is minor. This dual role — speeding the reaction up and steering it to specific ring positions — is why –OH is classified as an activating, ortho/para-directing group.(Oxygen is also electronegative, so it does withdraw some electron density from the ring inductively through the sigma bonds, an effect that on its own would be deactivating. But inductive effects fall off quickly with distance, while the resonance donation described above reaches all the way round the conjugated ring; the resonance effect dominates, so the observed, net behaviour of the –OH group is activation, not deactivation.)**Answer: The lone pair on the –OH oxygen conjugates with the benzene ring's π system (a +R/mesomeric effect), giving five resonance structures in which extra negative charge appears only at the two ortho carbons and the para carbon (never at the meta carbons); this is confirmed by the shortened, partial-double-bond C–OH bond length in phenol. This resonance donation outweighs oxygen's inductive (–I) withdrawal, so the ring is net electron-rich: it reacts with electrophiles faster than benzene (e.g., phenol plus bromine water gives $\displaystyle 2,4,6$-tribromophenol with no catalyst needed) and directs the incoming electrophile specifically to the ortho and para positions.

Propan-$\displaystyle 1$-ol is a primary alcohol, and alkaline \(\displaystyle KMnO_4\) is a strong enough oxidant to push it all the way past the aldehyde stage to the acid; phenol's electron-rich ring, on the other hand, undergoes electrophilic substitution whose position and count depend entirely on the medium the reagent is dissolved in.(i) Oxidation of propan-$\displaystyle 1$-ol with alkaline \(\displaystyle KMnO_4\)Propan-$\displaystyle 1$-ol , \(\displaystyle CH_3-CH_2-CH_2-OH\), carries its \(\displaystyle -OH\) on a carbon attached to only one other carbon, so it is a primary alcohol. A primary alcohol is oxidised by removing the two hydrogens on the carbinol carbon (the C–H bonds on the carbon bearing \(\displaystyle -OH\)) to first give an aldehyde, \(\displaystyle CH_3-CH_2-CHO\) (propanal) . Alkaline \(\displaystyle KMnO_4\) does not stop there — unlike a mild, controlled oxidant, it is strong enough to abstract a further hydrogen (this time using the aldehyde's own C–H) and add oxygen, carrying the aldehyde on to the carboxylic acid. Because the medium is alkaline, the acid is trapped as its potassium salt; a later acidification step liberates the free acid.\[3\,CH_3CH_2CH_2OH + 4\,KMnO_4 \xrightarrow{\Delta} 3\,CH_3CH_2COOK + 4\,MnO_2\!\downarrow + KOH + 4\,H_2O \] \[CH_3CH_2COOK \xrightarrow{H_3O^+} CH_3CH_2COOH \]The product is propanoic acid , \(\displaystyle CH_3CH_2COOH\). (This is the step people get wrong: mild oxidants like \(\displaystyle PCC\) stop cleanly at the aldehyde, but alkaline \(\displaystyle KMnO_4\) — like acidic \(\displaystyle KMnO_4\) or \(\displaystyle K_2Cr_2O_7\) — is strong enough to go all the way to the acid; it never leaves the aldehyde as the isolated product.)(ii) Bromine in \(\displaystyle CS_2\) with phenolPhenol , \(\displaystyle C_6H_5OH\), is far more reactive toward electrophilic substitution than benzene because a lone pair on the oxygen delocalises into the ring, piling up electron density at the ortho and para carbons. In water, this reactivity is so high that bromine reacts at all three activated positions at once, giving $\displaystyle 2,4,6$-tribromophenol as a precipitate. To stop the reaction after a single substitution, the reaction is instead run in carbon disulfide, \(\displaystyle CS_2\) — a non-polar solvent — at low temperature ($\displaystyle 273$ K). The non-polar, low-temperature medium does not help ionise the \(\displaystyle Br-Br\) bond or stabilise the charged intermediate at every ring position the way water does, so bromination happens only once, at the ortho or para carbon:\[C_6H_5OH + Br_2 \xrightarrow[273\ K]{CS_2} \underset{\text{(2-bromophenol)}}{o\text{-}BrC_6H_4OH} \;+\; \underset{\text{(4-bromophenol)}}{p\text{-}BrC_6H_4OH} \]The two isomers are obtained as a mixture and separated afterward by exploiting their very different boiling points and solubilities (the ortho isomer's intramolecular hydrogen bonding to \(\displaystyle -OH\) makes it more volatile than the intermolecularly hydrogen-bonded para isomer). The products are ortho-bromophenol and para-bromophenol .(iii) Dilute \(\displaystyle HNO_3\) with phenolThe same activation of the ring lets phenol undergo nitration even with dilute nitric acid at room temperature, without needing the concentrated acid/heat that benzene requires. The nitronium-type electrophile substitutes a ring hydrogen at the ortho and para positions:\[C_6H_5OH + dil.\,HNO_3 \xrightarrow{298\ K} \underset{\text{(2-nitrophenol)}}{o\text{-}O_2NC_6H_4OH} \;+\; \underset{\text{(4-nitrophenol)}}{p\text{-}O_2NC_6H_4OH} \;+\; H_2O \]The mixture is separated by steam distillation: intramolecular hydrogen bonding between the \(\displaystyle -OH\) and the adjacent \(\displaystyle -NO_2\) group in the ortho isomer (chelation) leaves no \(\displaystyle -OH\) free to hydrogen-bond to a neighbouring molecule, so it is compact and steam-volatile; the para isomer instead hydrogen-bonds intermolecularly, is associated into a higher-boiling network, and is left behind. The products are ortho-nitrophenol and para-nitrophenol .(iv) Phenol with chloroform in the presence of aqueous \(\displaystyle NaOH\) — the Reimer–Tiemann reactionThis is a carbene insertion, not a simple substitution, so it has to be followed step by step.Step $\displaystyle 1$ — Aqueous \(\displaystyle NaOH\) deprotonates phenol's \(\displaystyle -OH\): \[C_6H_5OH + NaOH \rightarrow C_6H_5O^-\,Na^+ + H_2O \] The resulting phenoxide ion, \(\displaystyle C_6H_5O^-\), is even more electron-rich at its ortho and para carbons than neutral phenol, because the negative charge itself is now delocalised onto the ring in the resonance structures.Step $\displaystyle 2$ — \(\displaystyle NaOH\) also attacks the chloroform molecule, \(\displaystyle CHCl_3\), pulling off its one acidic hydrogen (acidic because it sits on a carbon holding three electron-withdrawing chlorines) to give the trichloromethyl carbanion, \(\displaystyle CCl_3^-\). This carbanion is unstable and immediately expels a chloride ion (alpha-elimination), generating a neutral, highly electrophilic species with only six electrons on carbon: dichlorocarbene, \(\displaystyle :CCl_2\).Step $\displaystyle 3$ — The electron-rich ortho carbon of the phenoxide ion attacks the electron-deficient carbene carbon of \(\displaystyle :CCl_2\), forming a new carbon–carbon bond. Loss of a proton restores aromaticity, giving the sodium salt of an ortho-substituted phenol carrying a \(\displaystyle -CHCl_2\) (dichloromethyl) group: sodium $\displaystyle 2$-(dichloromethyl)phenoxide.Step $\displaystyle 4$ — The excess aqueous alkali then hydrolyses this gem-dichloride: each \(\displaystyle C-Cl\) bond is replaced by \(\displaystyle C-OH\) in turn, passing through an unstable gem-diol, \(\displaystyle -CH(OH)_2\), which collapses by losing water to give the aldehyde group, \(\displaystyle -CHO\).Step $\displaystyle 5$ — Acidic work-up (dilute acid) protonates the phenoxide oxygen back to neutral \(\displaystyle -OH\), releasing the final product.\[C_6H_5OH + CHCl_3 + 4\,NaOH \xrightarrow{\Delta} \text{sodium salicylaldehyde (2-\(\displaystyle CHO\)-phenoxide)} + 3\,NaCl + 2\,H_2O \] \[\xrightarrow{H_3O^+} \underset{\text{salicylaldehyde}}{HO-C_6H_4-CHO\ (ortho)} \]The product is $\displaystyle 2$-hydroxybenzaldehyde, commonly called salicylaldehyde — the new \(\displaystyle -CHO\) group lands specifically ortho to the \(\displaystyle -OH\) because that is the carbon of the phenoxide ion carrying the greatest negative charge, the one that attacked the carbene.Answer: (i) propanoic acid , \(\displaystyle CH_3CH_2COOH\); (ii) a mixture of o-bromophenol and p-bromophenol ; (iii) a mixture of o-nitrophenol and p-nitrophenol ; (iv) salicylaldehyde ($\displaystyle 2$-hydroxybenzaldehyde), \(\displaystyle HOC_6H_4CHO\).

Each of these four names describes how a nucleophile — a phenoxide ion, an alkoxide ion, or the electron-rich phenoxide ring — attacks an electrophilic carbon to build a new C–C, C–H, or C–O bond. Work through them one at a time.(i) Kolbe's reactionPhenol is first converted to sodium phenoxide with NaOH: \[C_6H_5-OH + NaOH \rightarrow C_6H_5-O^-Na^+ + H_2O \] The phenoxide ion is a far better nucleophile than phenol because the negative charge on oxygen is delocalized by resonance onto the ring carbons — mainly the ortho and para positions — so those carbons carry extra electron density.Dry \(\displaystyle CO_2\) is passed into sodium phenoxide at $\displaystyle 400$ K under $\displaystyle 4$–$\displaystyle 7$ atm pressure. The carbon of \(\displaystyle CO_2\) is electrophilic (it sits between two electronegative oxygens), and the electron-rich ortho carbon of the phenoxide ring attacks it. This C–C bond-forming step gives a cyclohexadienone-carboxylate intermediate, which then loses a proton (tautomerises) to restore the aromatic ring, giving sodium salicylate. The ortho product dominates because the carboxylate that forms there is stabilised by an intramolecular hydrogen bond to the adjacent \(\displaystyle O^-\)/OH — a stabilisation the para isomer cannot get.Acidifying with dilute HCl replaces \(\displaystyle Na^+\) with \(\displaystyle H^+\): \[C_6H_5O^-Na^+ + CO_2 \xrightarrow[\ 4\text{–}7\ \text{atm}\ ]{400\ K} \; o\text{-}C_6H_4(OH)(COO^-Na^+) \xrightarrow{\ H_3O^+\ } \; o\text{-}C_6H_4(OH)(COOH) \] The product is salicylic acid, $\displaystyle 2$-hydroxybenzoic acid, condensed formula \(\displaystyle o\text{-}HOC_6H_4COOH\).(ii) Reimer–Tiemann reactionPhenol is treated with chloroform (\(\displaystyle CHCl_3\)) and aqueous NaOH, warmed to about $\displaystyle 340$ K, then acidified.The mechanism runs in two nucleophile-generating steps and one attack: 1. NaOH deprotonates \(\displaystyle CHCl_3\) — the three chlorines pull enough electron density off the C–H that the proton is (weakly) acidic — giving the trichloromethyl carbanion \(\displaystyle CCl_3^-\). 2. This carbanion undergoes α-elimination: it kicks out a chloride ion \(\displaystyle Cl^-\) from its own carbon, generating dichlorocarbene, \(\displaystyle :CCl_2\), a highly electron-deficient, highly reactive species with only six electrons on carbon. 3. Meanwhile phenol has also been converted to phenoxide by the NaOH. The electron-rich ortho carbon of the phenoxide ring attacks the empty orbital on the dichlorocarbene carbon, forming a new C–C bond. Re-aromatisation (a proton shift) then gives ortho-(dichloromethyl)phenol, \(\displaystyle o\text{-}HOC_6H_4CHCl_2\).The \(\displaystyle -CHCl_2\) group is a geminal dihalide attached to an activated (electron-rich) ring, so aqueous alkali hydrolyses it readily: both chlorines are displaced by \(\displaystyle OH^-\)/water and the resulting geminal diol loses water to close down to a carbonyl. On acidification this delivers the aldehyde: \[C_6H_5OH \xrightarrow[\ 340\ K\ ]{CHCl_3,\ NaOH} \; o\text{-}HOC_6H_4CHCl_2 \xrightarrow{\ NaOH,\ then\ H_3O^+\ } \; o\text{-}HOC_6H_4CHO \] The product is salicylaldehyde, $\displaystyle 2$-hydroxybenzaldehyde, as the major (ortho) isomer.(iii) Williamson ether synthesisAn alkoxide ion (made from an alcohol plus sodium metal or NaOH) reacts with a primary alkyl halide (or alkyl sulfonate) to give an ether. Example — sodium ethoxide with methyl iodide: \[C_2H_5-O^-Na^+ + CH_3-I \rightarrow C_2H_5-O-CH_3 + NaI \] Here \(\displaystyle C_2H_5O^-Na^+\) is sodium ethoxide (the nucleophile), \(\displaystyle CH_3I\) is methyl iodide (the electrophile, iodide is the leaving group), and the product \(\displaystyle C_2H_5-O-CH_3\) is ethyl methyl ether (methoxyethane).This is a bimolecular nucleophilic substitution, \(\displaystyle S_N2\): the lone pair on the alkoxide oxygen attacks the electrophilic methyl carbon from the side directly opposite the C–I bond (backside attack). As the new C–O bond forms, the C–I bond breaks in the same concerted step, and \(\displaystyle I^-\) departs. Because backside attack needs open space behind the carbon bearing the leaving group, this route works cleanly only with methyl and primary halides. With secondary or tertiary halides, the alkoxide instead acts as a base and pulls off a β-hydrogen, giving an alkene by elimination rather than the ether — the bulky alkyl group blocks the nucleophile's path to the back of the carbon.(iv) Unsymmetrical etherAn ether has the general formula \(\displaystyle R-O-R'\): two carbon groups joined through a single oxygen atom. When \(\displaystyle R\) and \(\displaystyle R'\) are the same group, the ether is symmetrical (simple) — for instance diethyl ether, \(\displaystyle C_2H_5-O-C_2H_5\), where both groups are ethyl.An unsymmetrical (mixed) ether is one where \(\displaystyle R\) and \(\displaystyle R'\) are different — different alkyl groups, or one alkyl and one aryl group. Example: ethyl methyl ether, \(\displaystyle CH_3-O-C_2H_5\) (systematic name methoxyethane), which has a methyl group on one side of the oxygen and an ethyl group on the other. Anisole, \(\displaystyle C_6H_5-O-CH_3\) (methoxybenzene), is also unsymmetrical since it joins a phenyl group and a methyl group through the same oxygen.Answer: (i) Kolbe's reaction: sodium phenoxide + \(\displaystyle CO_2\) ($\displaystyle 400$ K, $\displaystyle 4$–$\displaystyle 7$ atm) via ortho attack on the phenoxide ring, then acidification, gives salicylic acid, \(\displaystyle o\text{-}HOC_6H_4COOH\). (ii) Reimer–Tiemann reaction: phenol + \(\displaystyle CHCl_3\) + NaOH ($\displaystyle 340$ K) generates dichlorocarbene, which the phenoxide ring attacks at the ortho carbon; hydrolysis of the resulting \(\displaystyle -CHCl_2\) group gives salicylaldehyde, \(\displaystyle o\text{-}HOC_6H_4CHO\). (iii) Williamson ether synthesis: \(\displaystyle R'ONa + RX \xrightarrow{S_N2} R-O-R' + NaX\), e.g. \(\displaystyle C_2H_5ONa + CH_3I \rightarrow C_2H_5-O-CH_3 + NaI\) (ethyl methyl ether); works best with primary \(\displaystyle RX\) because backside attack needs an unhindered carbon. (iv) An unsymmetrical ether is \(\displaystyle R-O-R'\) with \(\displaystyle R \neq R'\), e.g. \(\displaystyle CH_3-O-C_2H_5\) (ethyl methyl ether) or \(\displaystyle C_6H_5-O-CH_3\) (anisole).

Acid dehydration is an elimination reaction that removes \(\displaystyle -OH\) from one carbon and \(\displaystyle -H\) from the next carbon, and it goes through a carbocation, not in one single step.The substrate is ethanol , \(\displaystyle \text{CH}_3-\text{CH}_2-\text{OH}\): a two-carbon chain with the hydroxyl group on the terminal carbon. Heating it with concentrated sulphuric acid at about $\displaystyle 443$ K removes water and leaves a carbon-carbon double bond. The mechanism runs in three steps.Step $\displaystyle 1$ — Protonation of the hydroxyl oxygen. Concentrated \(\displaystyle \text{H}_2\text{SO}_4\) is a strong proton donor. The oxygen of the \(\displaystyle -OH\) group has two lone pairs; one of these lone pairs attacks the proton \(\displaystyle \text{H}^+\) supplied by the acid. A new O–H bond forms, and the oxygen — now bonded to two hydrogens and the ethyl carbon — carries a formal positive charge.\[\text{CH}_3-\text{CH}_2-\ddot{\text{O}}\text{H} + \text{H}^+ \;\longrightarrow\; \text{CH}_3-\text{CH}_2-\overset{+}{\text{O}}\text{H}_2 \]The product of this step is protonated ethanol (an ethyloxonium ion). This step is fast and reversible — it is only an acid-base proton transfer, no bond to carbon has broken yet.Step $\displaystyle 2$ — Loss of water to form a carbocation (the slow, rate-determining step). Protonation has converted the poor leaving group \(\displaystyle -OH^-\) into the good leaving group \(\displaystyle -OH_2^+\) (water is far more stable, and hence a far better leaving group, than hydroxide). The carbon–oxygen bond now breaks heterolytically: both electrons of that bond leave with the departing water molecule, and the carbon is left electron-deficient.\[\text{CH}_3-\text{CH}_2-\overset{+}{\text{O}}\text{H}_2 \;\longrightarrow\; \text{CH}_3-\overset{+}{\text{C}}\text{H}_2 \;+\; \text{H}_2\text{O} \]This generates the ethyl carbocation, \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}_2\), a primary carbocation with an empty p-orbital on the carbon that used to bear the \(\displaystyle -OH\) group. Because carbon–oxygen bond cleavage (breaking a real covalent bond) is much harder than the proton transfer in Step $\displaystyle 1$, this step is slow — it is the rate-determining step of the whole sequence.Step $\displaystyle 3$ — Loss of a \(\displaystyle \beta\)-hydrogen to give the alkene. The carbocation is unstable and is trapped by a base already present in the mixture — the hydrogensulphate ion \(\displaystyle \text{HSO}_4^-\) (or a second molecule of water) — which removes a proton from the carbon adjacent to the positive centre (the \(\displaystyle \beta\)-carbon, here the \(\displaystyle \text{CH}_3\) group). As that C–H bond breaks, its bonding electron pair does not leave with the proton; instead it moves in to form a new \(\displaystyle \pi\) bond between the \(\displaystyle \alpha\)- and \(\displaystyle \beta\)-carbons, neutralising the positive charge.\[\text{CH}_3-\overset{+}{\text{C}}\text{H}_2 + \text{HSO}_4^- \;\longrightarrow\; \text{CH}_2=\text{CH}_2 + \text{H}_2\text{SO}_4 \]The base \(\displaystyle \text{HSO}_4^-\) picks up the proton and regenerates \(\displaystyle \text{H}_2\text{SO}_4\), which is why the acid is called a catalyst here — it is used up in Step $\displaystyle 1$ and returned in Step $\displaystyle 3$, so the net amount of sulphuric acid is unchanged at the end of the reaction.Overall transformation. Adding the three steps together, one molecule of water is eliminated across the two carbons of ethanol, and the concentrated acid comes back out unconsumed:\[\text{CH}_3-\text{CH}_2-\text{OH} \xrightarrow[443\ \text{K}]{\text{conc. } \text{H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]The product is ethene (ethylene) , \(\displaystyle \text{CH}_2=\text{CH}_2\) — the smallest alkene, with a carbon-carbon double bond in place of the original \(\displaystyle \text{C}-\text{OH}\) and \(\displaystyle \text{C}-\text{H}\) bonds that were broken in Steps $\displaystyle 2$ and 3.A short aside on why this mechanism matters here: this is the same three-step pattern (protonate the leaving group, lose it to form a carbocation, lose a \(\displaystyle \beta\)-proton to form the double bond) used for acid-catalysed dehydration of any alcohol; only the identity of the carbocation formed in Step $\displaystyle 2$ changes with the substrate.**Answer: The dehydration proceeds by protonation of the \(\displaystyle -OH\) oxygen by \(\displaystyle \text{H}_2\text{SO}_4\) to give \(\displaystyle \text{CH}_3-\text{CH}_2-\overset{+}{\text{O}}\text{H}_2\), loss of water in the rate-determining step to give the ethyl carbocation \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}_2\), and removal of a \(\displaystyle \beta\)-hydrogen by \(\displaystyle \text{HSO}_4^-\) to form the double bond, giving ethene: \(\displaystyle \text{CH}_3\text{CH}_2\text{OH} \xrightarrow[443\text{ K}]{\text{conc. }\text{H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}\).

(i) Propene → Propan-$\displaystyle 2$-ol: Markovnikov's rule decides which carbon gets the \(\displaystyle \mathrm{OH} \) — it lands on the carbon that can support the more stable carbocation, not on the terminal carbon.Propene, \(\displaystyle \mathrm{CH_3-CH=CH_2} \), is shaken with cold concentrated sulphuric acid, \(\displaystyle \mathrm{H_2SO_4} \). The alkene's \(\displaystyle \pi \) electrons attack a proton donated by the acid. Protonation occurs at the terminal \(\displaystyle \mathrm{=CH_2} \) carbon \(\displaystyle \mathrm{(C_{1})}\) because that leaves the positive charge on \(\displaystyle \mathrm{C_{2}}\), which is flanked by two methyl-type carbons and is therefore a secondary carbocation — more stable than the primary carbocation that would form if the proton went to \(\displaystyle \mathrm{C_{2}}\) instead. The conjugate base, hydrogensulphate ion \(\displaystyle \mathrm{HSO_4^-} \), then bonds to this carbocation at \(\displaystyle \mathrm{C_{2}}\), giving propan-$\displaystyle 2$-yl hydrogen sulphate: \[\mathrm{CH_3-CH=CH_2 + H_2SO_4 \rightarrow CH_3-CH(OSO_3H)-CH_3} \] This ester is boiled with water; water attacks the carbon–oxygen bond and hydrolyses it, releasing the alcohol and regenerating sulphuric acid: \[\mathrm{CH_3-CH(OSO_3H)-CH_3 + H_2O \xrightarrow{\Delta} CH_3-CH(OH)-CH_3 + H_2SO_4} \] Product: propan-$\displaystyle 2$-ol , \(\displaystyle \mathrm{CH_3-CH(OH)-CH_3} \) — the \(\displaystyle \mathrm{OH} \) sits on the middle carbon, exactly as Markovnikov's rule predicts.(ii) Benzyl chloride → Benzyl alcohol: this is a plain \(\displaystyle \mathrm{S_N2} \) displacement — a primary, non-hindered carbon lets hydroxide simply push chloride out from the back side.Benzyl chloride , \(\displaystyle \mathrm{C_6H_5-CH_2-Cl} \) (chloromethylbenzene), is warmed with aqueous sodium hydroxide. Hydroxide ion, \(\displaystyle \mathrm{OH^-} \), is a strong nucleophile; because the benzylic carbon is primary and unhindered, \(\displaystyle \mathrm{OH^-} \) approaches it from the side directly opposite the \(\displaystyle \mathrm{C-Cl} \) bond. In a single step, the new \(\displaystyle \mathrm{C-OH} \) bond forms as the \(\displaystyle \mathrm{C-Cl} \) bond breaks, and chloride ion leaves: \[\mathrm{C_6H_5CH_2Cl + NaOH \xrightarrow{H_2O} C_6H_5CH_2OH + NaCl} \] Product: benzyl alcohol , \(\displaystyle \mathrm{C_6H_5-CH_2-OH} \).(iii) Ethylmagnesium chloride → Propan-$\displaystyle 1$-ol: reacting a Grignard reagent with methanal (the smallest aldehyde) always adds exactly one carbon, so the $\displaystyle 2$-carbon ethyl group becomes the $\displaystyle 3$-carbon chain of propan-$\displaystyle 1$-ol.Ethylmagnesium chloride, \(\displaystyle \mathrm{CH_3CH_2-MgCl} \), is treated with methanal (formaldehyde), \(\displaystyle \mathrm{HCHO} \), in dry ether. The \(\displaystyle \mathrm{C-Mg} \) bond is strongly polarised (carbon \(\displaystyle \delta^- \), magnesium \(\displaystyle \delta^+ \)), so the ethyl carbon acts as a carbanion-like nucleophile. It attacks the electrophilic carbonyl carbon of \(\displaystyle \mathrm{HCHO} \); as the new \(\displaystyle \mathrm{C-C} \) bond forms, the \(\displaystyle \pi \) electrons of the \(\displaystyle \mathrm{C=O} \) bond move fully onto oxygen, giving a magnesium alkoxide: \[\mathrm{CH_3CH_2-MgCl + HCHO \rightarrow CH_3CH_2CH_2-O-MgCl} \] This addition product is hydrolysed with dilute acid, \(\displaystyle \mathrm{H_3O^+} \), which protonates the alkoxide oxygen: \[\mathrm{CH_3CH_2CH_2-O-MgCl + H_2O \rightarrow CH_3CH_2CH_2-OH + Mg(OH)Cl} \] Product: propan-$\displaystyle 1$-ol , \(\displaystyle \mathrm{CH_3-CH_2-CH_2-OH} \) — a primary alcohol, because the carbonyl partner \(\displaystyle \mathrm{HCHO} \) had no alkyl group of its own to leave behind on the carbinol carbon.(iv) Methylmagnesium bromide → $\displaystyle 2$-Methylpropan-$\displaystyle 2$-ol: adding a Grignard reagent to a ketone rather than an aldehyde is what produces a tertiary alcohol, because the carbonyl carbon already carries two alkyl groups before the Grignard's carbon is added as the third.Methylmagnesium bromide, \(\displaystyle \mathrm{CH_3-MgBr} \), is reacted with propan-$\displaystyle 2$-one (acetone) , \(\displaystyle \mathrm{CH_3-CO-CH_3} \), in dry ether. The nucleophilic methyl carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of acetone; the \(\displaystyle \mathrm{C=O} \) \(\displaystyle \pi \) electrons shift onto oxygen as the new \(\displaystyle \mathrm{C-C} \) bond forms, giving a magnesium alkoxide in which the carbinol carbon now bears three methyl groups: \[\mathrm{CH_3MgBr + CH_3COCH_3 \rightarrow (CH_3)_3C-OMgBr} \] Hydrolysis with dilute acid protonates the alkoxide oxygen and releases the alcohol: \[\mathrm{(CH_3)_3C-OMgBr + H_2O \rightarrow (CH_3)_3C-OH + Mg(OH)Br} \] Product: $\displaystyle 2$-methylpropan-$\displaystyle 2$-ol (tert-butyl alcohol) , \(\displaystyle \mathrm{(CH_3)_3C-OH} \) — a tertiary alcohol, since acetone contributed two methyl groups to the carbinol carbon and the Grignard reagent contributed the third.Answer: (i) \(\displaystyle \mathrm{CH_3CH{=}CH_2} \xrightarrow{\text{conc. } H_2SO_4} \mathrm{CH_3CH(OSO_3H)CH_3} \xrightarrow{H_2O,\ \Delta} \mathrm{CH_3CH(OH)CH_3}\) (propan-$\displaystyle 2$-ol). (ii) \(\displaystyle \mathrm{C_6H_5CH_2Cl} \xrightarrow{aq.\ NaOH} \mathrm{C_6H_5CH_2OH}\) (benzyl alcohol) via \(\displaystyle \mathrm{S_N2} \). (iii) \(\displaystyle \mathrm{CH_3CH_2MgCl} \xrightarrow{(i)\ HCHO\ \ (ii)\ H_3O^+} \mathrm{CH_3CH_2CH_2OH}\) (propan-$\displaystyle 1$-ol). (iv) \(\displaystyle \mathrm{CH_3MgBr} \xrightarrow{(i)\ CH_3COCH_3\ \ (ii)\ H_3O^+} \mathrm{(CH_3)_3COH}\) ($\displaystyle 2$-methylpropan-$\displaystyle 2$-ol).

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).