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).