NCERT Solutions for Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers

Concept and approach. An alcohol is classified by looking only at the carbinol carbon — the sp³ carbon that carries the –OH group — and counting how many other carbon atoms are directly bonded to it. If the carbinol carbon is attached to one carbon atom (or to none, as in methanol) the alcohol is primary (1°); if it is attached to two carbon atoms it is secondary (2°); if to three it is tertiary (3°). Equivalently, a primary alcohol has the group –CH₂OH, a secondary alcohol has >CH–OH and a tertiary alcohol has ≥C–OH with no hydrogen on the carbinol carbon. What lies further along the chain (a double bond, a benzene ring, branching) does not change the class — it only adds a second label such as allylic or benzylic.

Working for each case. In (i) the carbinol carbon is the CH₂ of –CH₂OH; its only carbon neighbour is the quaternary carbon of the tert-butyl group, so the alcohol is primary even though the molecule looks heavily branched (the branching is on the neighbouring carbon and makes the alcohol sterically hindered towards SN2, but it does not change the class). In (ii) and (iii) the carbinol carbon is again a –CH₂OH group. In (iv) the carbinol carbon carries H, OH, CH₃ and the phenyl ring, i.e. two carbon neighbours: secondary. In (v) the carbinol carbon carries H, OH, CH₃ and –CH₂C₆H₅: also two carbon neighbours: secondary. In (vi) the carbinol carbon carries OH, two CH₃ groups and the –CH=CH–C₆H₅ group — three carbon neighbours and no H: tertiary.

Why the classification matters. The class decides almost everything an alcohol does: (a) oxidation — a primary alcohol gives an aldehyde and then a carboxylic acid, a secondary alcohol gives a ketone, a tertiary alcohol resists oxidation because it has no hydrogen on the carbinol carbon; (b) Lucas test — a tertiary alcohol turns Lucas reagent cloudy at once, a secondary one in about five minutes, a primary one only on heating; (c) dehydration — the ease of loss of water is 3° > 2° > 1°; (d) acidity — 1° > 2° > 3° (alkyl groups raise the electron density on oxygen); (e) the carbocation formed in SN1/E1 reactions is more stable when the alcohol is tertiary (or allylic/benzylic).

Conclusion: Primary alcohols: (i), (ii), (iii); secondary alcohols: (iv) and (v); tertiary alcohol: (vi).

Concept. An allylic alcohol is one in which the –OH group is attached to an sp³ carbon that is directly bonded to a carbon–carbon double bond (an allylic carbon), i.e. the arrangement C=C–C–OH. The name comes from the allyl group CH₂=CH–CH₂–. Two things must both be true: (1) the carbon bearing –OH must be sp³ (otherwise the compound would be a vinylic alcohol or a phenol), and (2) one of its neighbours must be an sp² carbon of a C=C. A benzene ring does not count as a C=C for this purpose — a carbinol carbon next to a benzene ring makes the alcohol benzylic, not allylic.

Extra depth. Allylic alcohols are special because on ionisation (SN1) or in acid-catalysed substitution they give a resonance-stabilised allyl carbocation (CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂), so they react faster than ordinary alcohols and may give rearranged products (the nucleophile can attack either end of the allyl system). In (vi) the carbocation would be both tertiary and conjugated with C=C and the benzene ring, which makes it exceptionally stable. Allylic alcohols are also easily oxidised, with MnO₂ selectively oxidising allylic –OH to the carbonyl without touching the C=C.

Conclusion: The allylic alcohols are (ii) CH₂=CH–CH₂OH (prop-2-en-1-ol) and (vi) C₆H₅CH=CH–C(CH₃)₂OH.

Concept and rules used. The IUPAC name of an alcohol is built in five steps: (1) pick the longest continuous carbon chain that contains the carbon bearing –OH (the –OH group, as the principal functional group, must be on the parent chain); (2) number the chain from the end that gives the lowest locant to the carbon carrying –OH (this rule outranks the numbering of C=C bonds and of substituents); (3) replace the final “e” of the alkane by “ol” (“-ene” + “-ol” → “-en-ol”); (4) name every other group as a prefix with its locant, in alphabetical order (ignoring di-, tri-, but not iso/sec/tert as written in the full name); (5) write the name as one word, with the locant of –OH placed just before “ol”. For a ring, the carbon with –OH is always C-1 and the ring is then numbered in the direction that gives the substituents the lowest locants.

Points worth noticing. (a) In (i) the longest chain through the CH₂OH carbon is only four carbons even though the molecule has more: a chain that does not include the carbinol carbon is not allowed to be the parent. The isopropyl group is named (1-methylethyl) in the older NCERT scheme and (propan-2-yl) in the current IUPAC recommendations; both are acceptable. Alphabetical order of prefixes: chloro, ethyl, methylethyl. (b) In (ii) the –OH groups are two, so the suffix is “-diol” and the “e” of hexane is retained (hexane-1,3-diol, not hexan-1,3-diol); the numbering gives the two –OH groups the lowest set of locants {1,3}. (c) In (iv) the double bond is in the chain too, so the suffix reads “-en-3-ol”; the –OH has priority for the lowest locant: numbering from the C=C end gives –OH the locant 3, whereas numbering from the other end would give –OH locant 4 (and C=C locant 5), which is worse. (d) In (v) the –OH carbon is C1 and then the C=C gets the lowest locant (2), giving but-2-en-1-ol.

Conclusion: (i) 4-chloro-3-ethyl-2-(1-methylethyl)butan-1-ol; (ii) 2,5-dimethylhexane-1,3-diol; (iii) 3-bromocyclohexanol; (iv) hex-1-en-3-ol; (v) 2-bromo-3-methylbut-2-en-1-ol.

Concept. A Grignard reagent R–MgX has a polar C–Mg bond (Cᵟ⁻–Mgᵟ⁺), so the carbon bound to magnesium acts as a powerful carbon nucleophile. Methanal (formaldehyde, HCHO) has an electrophilic carbonyl carbon; the R group adds to it to give an alkoxide–magnesium halide adduct, R–CH₂–O⁻MgX⁺, which on hydrolysis with dilute acid or aqueous NH₄Cl gives the primary alcohol R–CH₂OH. Methanal is the only carbonyl compound that gives a primary alcohol with a Grignard reagent (other aldehydes give secondary alcohols and ketones give tertiary alcohols). The key to a retrosynthesis is therefore: the alcohol R–CH₂OH = the group R of the Grignard reagent + the CH₂OH carbon contributed by HCHO. So remove the –CH₂OH from the target; what remains is R.

Reaction sequences.

Why each condition is needed. The reagent must be prepared in dry (anhydrous) ether because the Grignard reagent is a very strong base: any water, alcohol or other proton source destroys it (R–MgX + H₂O → R–H + Mg(OH)X). Ether is the solvent of choice because its oxygen lone pairs coordinate to magnesium and stabilise the reagent. Methanal is a gas, so it is usually generated by heating paraformaldehyde and passing the vapour into the Grignard solution. The final hydrolysis step is done after the addition is complete; adding water earlier would simply convert the Grignard reagent to the alkane.

Conclusion: (i) (CH₃)₂CH–MgBr + HCHO, then H₃O⁺, gives (CH₃)₂CHCH₂OH; (ii) cyclohexylmagnesium bromide + HCHO, then H₃O⁺, gives cyclohexylmethanol.

Concept. The three parts test two different routes to alcohols. (i) Acid-catalysed hydration of an alkene adds H and OH across the C=C double bond according to Markovnikov’s rule — the –OH goes to the carbon that carries fewer hydrogens (the one that gives the more stable carbocation). (ii) and (iii) Reduction of carbonyl compounds by sodium borohydride. NaBH₄ is a mild, selective source of hydride ion (H⁻): it reduces aldehydes and ketones (C=O → CH–OH) to alcohols after protonation by the solvent, but it is not strong enough to reduce esters, carboxylic acids or C=C bonds; those need the much stronger LiAlH₄ (acids, esters) or catalytic hydrogenation (C=C).

Mechanism of the borohydride reduction (parts ii and iii). The B–H bond of BH₄⁻ delivers H⁻ to the electrophilic carbonyl carbon; the carbonyl oxygen becomes an alkoxide (R–CH₂–O⁻ attached to boron), which picks up a proton from the protic solvent (ethanol/water) or during aqueous work-up to give the alcohol. Each BH₄⁻ can deliver four hydrides in turn. Why the ester survives in (ii): the carbonyl carbon of an ester is less electrophilic than that of a ketone because the lone pair of the –OR oxygen is donated into the C=O (resonance), so NaBH₄ reacts with the ketone far faster than with the ester; the reaction is therefore chemoselective. With LiAlH₄ both groups would be reduced (to a diol).

Extra depth. The three products are a secondary alcohol (i), a secondary alcohol that retains an ester (ii) and a primary alcohol (iii) — in line with the rule “aldehyde → primary alcohol, ketone → secondary alcohol”. In (iii) the product has a chiral carbon (C2), but it forms as a racemic mixture because hydride can attack the planar C=O from either face of the achiral/racemic aldehyde.

Conclusion: (i) propan-2-ol; (ii) methyl (2-hydroxycyclohexyl)ethanoate (only the ketone is reduced); (iii) 2-methylbutan-1-ol.

Concept. In all three reagents the net change is the replacement of the –OH group of the alcohol by a halogen atom: R–OH → R–X. Because –OH⁻ is a poor leaving group, it must first be converted into a better one — protonation by the acid (HCl/ZnCl₂, HBr) gives R–OH₂⁺ which loses water, while SOCl₂ converts –OH into the chlorosulphite ester R–O–S(=O)Cl whose leaving group (SO₂ + Cl⁻) departs readily. The mechanism then depends on the class of the alcohol: primary alcohols react by SN2 (halide attacks the protonated alcohol from the back side — slow, needs heat), while tertiary alcohols react by SN1 (the C–O bond ionises to a stable carbocation, then X⁻ adds — fast, at room temperature).

Reactivity facts to remember. (1) The reactivity of alcohols towards HX is 3° > 2° > 1° (the stability of the carbocation or the ease of SN1) and for the acids HI > HBr > HCl (nucleophilicity/acidity of the hydrogen halide). (2) Anhydrous ZnCl₂, a Lewis acid, coordinates to the oxygen of the alcohol and makes the C–O bond easier to break; this mixture is the Lucas reagent and the difference in the speed of the reaction (immediate, five minutes, none) is the basis of the Lucas test for distinguishing 3°, 2° and 1° alcohols with up to six carbons. (3) Thionyl chloride is the reagent of choice for preparing alkyl chlorides from 1° and 2° alcohols because both by-products (SO₂ and HCl) are gases and escape, leaving a pure chloroalkane without any need for a tedious separation. (4) Because (ii) goes through a tertiary carbocation, the chlorine ends up on the same carbon that carried the –OH — no rearrangement is possible here.

Conclusion: (i) Butan-1-ol gives 1-chlorobutane (a, c) and 1-bromobutane (b); (ii) 2-methylbutan-2-ol gives 2-chloro-2-methylbutane (a, c) and 2-bromo-2-methylbutane (b).

Concept. When an alcohol is heated with a strong protic acid (conc. H₂SO₄ or H₃PO₄) water is eliminated and an alkene is formed. The reaction is an E1 elimination: (1) the –OH is protonated to –OH₂⁺; (2) water leaves slowly to give a carbocation (the rate-determining step); (3) a base (H₂O or HSO₄⁻) removes a proton from a carbon adjacent to the carbocation, and the electrons form the C=C. Two rules govern the major product: Saytzeff (Zaitsev) rule — the more highly substituted (more stable) alkene predominates; and carbocation rearrangement — if a 1,2-hydride or alkyl shift can give a more stable carbocation, it will happen before the proton is lost. The ease of dehydration is 3° > 2° > 1° (the stability of the carbocation).

Explanation of (ii). A primary carbocation is so unstable that it is not formed as a free intermediate; the departure of water from the protonated butan-1-ol is assisted by migration of a hydrogen with its bonding electron pair from C2 to C1 (a 1,2-hydride shift), which directly gives the more stable secondary carbocation. From it, loss of a proton from C3 gives but-2-ene (a disubstituted alkene — more stable by hyperconjugation, hence major) while loss of a proton from C1 gives but-1-ene (monosubstituted — minor). Trans-but-2-ene is favoured over the cis isomer because of lower steric strain. In practice the dehydration of primary alcohols is normally carried out under conditions where an E2-like pathway (and not a free carbocation) also contributes, but NCERT’s expected answer is the mixture with but-2-ene as the major product.

Conclusion: (i) 1-methylcyclohexene; (ii) a mixture of but-1-ene and but-2-ene in which but-2-ene is the major product (formed after rearrangement to the secondary carbocation).

Concept. The acidity of a phenol depends on how well the phenoxide ion (the conjugate base) is stabilised: the more stable the anion, the more readily the phenol gives up its proton and the stronger the acid (lower pKₐ). The phenoxide ion is stabilised by resonance because the negative charge on oxygen is delocalised into the ring at the ortho and para carbons. A nitro group at an ortho or para position joins the delocalisation: the negative charge from the ring carbon bearing the –NO₂ can be pushed onto the oxygen atoms of the nitro group (a –M or –R effect), giving two extra resonance structures in which the charge sits on the electronegative oxygen of NO₂. The ion is stabilised further by the electron-withdrawing –I effect of –NO₂. At the meta position the nitro group cannot take part in the resonance (no structure puts a charge on the carbon bearing –NO₂), so only the –I effect operates, and m-nitrophenol is less acidic than the ortho and para isomers.

p-Nitrophenoxide ion — five contributing structures (I: charge on O; II, III: charge on the two ortho carbons; IV: charge on the para carbon that bears NO₂; V: charge delocalised onto the O atoms of the nitro group, the “aci-nitro” form).

o-Nitrophenoxide ion — five contributing structures (I: charge on O; II: charge on the ortho carbon bearing NO₂; III: charge on the para carbon; IV: charge on the other ortho carbon; V: charge on the nitro group oxygen).

Extra depth. The pKₐ values fall in the order phenol (10.0) > m– (8.3) > o– (7.2) > p-nitrophenol (7.1) > 2,4-dinitrophenol (4.1) > 2,4,6-trinitrophenol or picric acid (0.25 — as strong as a mineral acid). Every additional nitro group at ortho/para positions increases the acidity by providing more delocalisation of the negative charge. o-Nitrophenol is slightly weaker than the para isomer even though the –I effect is stronger at the ortho position, because the O–H of the neutral ortho isomer is already held by an intramolecular hydrogen bond with the adjacent nitro group, which makes it harder to remove the proton.

Conclusion: The o- and p-nitrophenoxide ions each have five resonance structures, two of which place the negative charge on the nitro-group oxygens; this extra delocalisation (plus the –I effect of NO₂) stabilises the anion and makes o- and p-nitrophenol more acidic than phenol.

Concept. Both reactions are electrophilic aromatic substitutions of phenol carried out in alkaline medium, and both introduce a new group at the ortho position (next to –OH). In alkali phenol exists as the phenoxide ion (C₆H₅O⁻), in which the oxygen carries a full negative charge that is delocalised into the ring at the ortho and para carbons. The phenoxide ring is therefore far more electron-rich (more reactive) than phenol itself, and it can attack even weak electrophiles — dichlorocarbene in the Reimer–Tiemann reaction and carbon dioxide in the Kolbe reaction. The ortho product predominates because the sodium ion of the phenoxide is held close to the ortho position (chelation) and directs the electrophile there.

(i) Reimer–Tiemann reaction — phenol is treated with chloroform in the presence of aqueous sodium hydroxide at about 340 K; a –CHO group enters the ring at the ortho position, giving salicylaldehyde (2-hydroxybenzaldehyde) after acidification.

Steps of the mechanism: (1) NaOH removes HCl from chloroform to give the carbanion ⁻CCl₃, which loses Cl⁻ to give the electron-deficient species dichlorocarbene (:CCl₂); (2) the phenoxide ring attacks :CCl₂ at the ortho carbon, giving an intermediate that regains aromaticity (by proton shift) as o-(dichloromethyl)phenoxide, the “substituted benzal chloride”; (3) alkali hydrolyses the –CHCl₂ group (via –CH(OH)₂, which loses water) to –CHO; (4) acidification of the sodium salt of the product liberates salicylaldehyde. A minor amount of the para isomer (4-hydroxybenzaldehyde) is also formed.

(ii) Kolbe’s reaction — phenol is first converted into sodium phenoxide with NaOH; the phenoxide is then heated with carbon dioxide (400 K, 4–7 atmospheres pressure), and the resulting sodium salicylate is acidified to give salicylic acid (2-hydroxybenzoic acid).

Here the electrophile is carbon dioxide, whose carbon atom is electron-deficient (δ⁺); the ortho carbon of the phenoxide ion attacks it to give a carboxylate group, and the proton lost from the ring carbon ends up on oxygen. Salicylic acid is the starting material for aspirin (acetylsalicylic acid, prepared by acetylation with acetic anhydride) and for methyl salicylate (oil of wintergreen), a topical analgesic.

Conclusion: (i) Phenol + CHCl₃ + aq. NaOH (340 K), then H⁺ → salicylaldehyde; (ii) sodium phenoxide + CO₂ (400 K, 4–7 atm), then H⁺ → salicylic acid.

Concept. The Williamson synthesis prepares an ether R–O–R′ by the reaction of an alkoxide ion (R′O⁻) with an alkyl halide (R–X). It is an SN2 reaction: the alkoxide attacks the carbon bearing the halogen from the back side, and X⁻ leaves in the same step. Because an alkoxide is also a strong base, it can bring about E2 elimination instead; the substitution is favoured only when the halide is primary (and, to a lesser extent, secondary under mild conditions), while tertiary halides give almost entirely alkenes. So for an unsymmetrical ether the golden rule is: the alkyl halide should be the less hindered (primary) component and the alkoxide the more hindered one.

Choosing the pair. The target 2-ethoxy-3-methylpentane is CH₃CH₂–O–CH(CH₃)–CH(CH₃)–CH₂CH₃. It can be disconnected in two ways: (A) ethyl bromide (primary) + the sodium alkoxide of 3-methylpentan-2-ol (secondary, bulky), or (B) 2-bromo-3-methylpentane (secondary, hindered) + sodium ethoxide. Route B would give mainly alkenes by E2, so the correct choice is route A.

Steps. Both the question’s starting materials are alcohols, so each must be converted into the appropriate partner:

Extra depth. If the alternative route B were tried, the strongly basic ethoxide would abstract a β-hydrogen from 2-bromo-3-methylpentane and the product would be mostly 3-methylpent-2-ene (the more substituted alkene) rather than the ether; this is exactly the “limitation of Williamson synthesis” discussed in exercise 7.25. The SN2 step proceeds with inversion of configuration at the electrophilic carbon, but ethyl bromide has no stereocentre, so the stereochemistry of the 3-methylpentan-2-ol part is retained unchanged in the ether (the C–O bond of the alcohol is never broken).

Conclusion: C₂H₅OH + HBr → C₂H₅Br; 3-methylpentan-2-ol + Na → its sodium alkoxide; alkoxide + C₂H₅Br (SN2) → 2-ethoxy-3-methylpentane + NaBr.

Concept. 1-Methoxy-4-nitrobenzene (p-nitroanisole) is an alkyl aryl ether, and alkyl aryl ethers are made by the Williamson synthesis from a phenoxide ion + an alkyl halide: Ar–O⁻ + R–X → Ar–O–R + X⁻. The reaction relies on an SN2 attack by the nucleophile on an sp³-hybridised carbon that carries a good leaving group. So the question is whether the C–X bond that has to be broken belongs to an alkyl group (easy SN2) or to an aryl ring (very difficult).

Why an aryl halide is a poor substrate for the SN2 route. (1) The C–Br bond is attached to an sp² carbon; the carbon has more s-character and the C–Br bond is shorter and stronger. (2) The lone pair of bromine is delocalised into the ring (resonance), giving the C–Br bond partial double-bond character. (3) The nucleophile would have to approach from the back side of the C–Br bond, exactly where the ring itself stands in the way — back-side attack is geometrically impossible. (4) The repulsion between the electron-rich π cloud of the ring and the incoming nucleophile also hinders attack. Although the para –NO₂ group activates the ring for nucleophilic aromatic substitution (SNAr) by stabilising the negatively charged intermediate, this requires forcing conditions and is not the Williamson reaction; the textbook answer is therefore set (ii), which uses a simple SN2 step with a methyl halide.

Extra depth. The nitro group makes the phenol more acidic (pKₐ 7.1), so sodium 4-nitrophenoxide is readily formed even with a weak base like Na₂CO₃ or NaOH; the anion is a somewhat weaker nucleophile than ordinary phenoxide (its charge is delocalised onto the nitro group) but it is still adequate to attack CH₃Br, which is an excellent SN2 substrate. The product, p-nitroanisole, is used in dye intermediates; its reduction gives p-anisidine.

Conclusion: Set (ii), sodium 4-nitrophenoxide + CH₃Br, is appropriate because the SN2 attack takes place on a primary alkyl halide; in set (i) the halogen is on an sp² carbon of the ring, so it cannot be displaced by methoxide under Williamson conditions.

Concept. This question covers the two types of reaction of ethers: (1) cleavage of the C–O bond by hydrogen halides (parts i, ii and iv) and (2) electrophilic aromatic substitution of an aryl alkyl ether (part iii). In cleavage, the ether oxygen is first protonated to an oxonium ion R–O⁺(H)–R′; the halide ion then attacks one of the two carbon atoms and the other group leaves as an alcohol: R–O–R′ + HX → R–X + R′–OH. Which C–O bond breaks depends on the mechanism: (a) if both groups are primary or secondary alkyl, the halide attacks the smaller, less hindered carbon by SN2; (b) if one group is tertiary (or benzylic/allylic), the C–O bond to that group breaks by SN1 through the stable carbocation; (c) in an aryl alkyl ether only the alkyl–O bond breaks, because the aryl–O bond has partial double-bond character and the C(sp²)–O carbon cannot be attacked by X⁻ — so the products are a phenol and an alkyl halide, never an aryl halide.

Extra depth. The order of reactivity of the hydrogen halides is HI > HBr > HCl, since I⁻ is the best nucleophile and HI is the strongest acid; HCl does not cleave simple ethers readily. In (iii) the nitration of an alkoxybenzene is faster than that of benzene because the oxygen lone pair stabilises the Wheland (arenium) intermediate for ortho and para attack; the para isomer predominates for steric reasons. With a large excess of HBr, the phenol of part (ii) does not react further (it does not give bromobenzene), and the ethanol of part (iv) would eventually be converted to ethyl iodide with excess hot HI.

Conclusion: (i) CH₃Br + CH₃CH₂CH₂OH; (ii) C₆H₅OH + C₂H₅Br; (iii) 4-nitroethoxybenzene (major) + 2-nitroethoxybenzene (minor); (iv) (CH₃)₃C–I + C₂H₅OH.

Concept and rules. Three families are mixed in this question, and each has its own naming rule. Alcohols (i–iv): choose the longest chain that contains the carbon bearing –OH, number so that the –OH gets the lowest locant (then the substituents get the lowest set), change “-e” to “-ol” (keep the “e” before di-, tri-: -diol, -triol) and cite substituents alphabetically. Phenols (v–viii): phenol is the parent; the carbon bearing –OH is C-1, and the substituents (CH₃) are numbered to give the lowest locants. Ethers (ix–xii): named as alkoxyalkanes/aryloxy-alkanes — the larger group is the parent, the smaller one becomes an alkoxy prefix (methoxy-, ethoxy-, phenoxy-). For each compound I first identify the functional group, then the parent chain, then the locants.

Points that cause mistakes. (a) In (i) there are two ways to number the five-carbon chain and the –OH is at C3 both ways; the tie is broken by the substituent locants — the set {2,2,4} is lower than {2,4,4} at the first point of difference. (b) In (ii) the longest chain is seven carbons and includes the ethyl-bearing carbon, which is why the ethyl is quoted as a substituent on C5; the diol locants {2,4} are chosen over {4,6}. (c) In the diols/triol, the “e” of the alkane is kept: butane-2,3-diol, propane-1,2,3-triol. (d) In (vii) both numbering directions are tried: starting at the carbon bearing –OH, one direction gives methyls at 2 and 5, the other at 3 and 6, so {2,5} is correct. (e) In the ethers the choice of the parent is by size, not alphabet: the C₇ chain of (xi) is the parent and the phenyl–O group is phenoxy; in (ix) the four-carbon branched chain is the parent.

Conclusion: (i) 2,2,4-trimethylpentan-3-ol; (ii) 5-ethylheptane-2,4-diol; (iii) butane-2,3-diol; (iv) propane-1,2,3-triol; (v) 2-methylphenol; (vi) 4-methylphenol; (vii) 2,5-dimethylphenol; (viii) 2,6-dimethylphenol; (ix) 1-methoxy-2-methylpropane; (x) ethoxybenzene; (xi) 1-phenoxyheptane; (xii) 2-ethoxybutane.

Concept. To convert an IUPAC name into a structure, read it from right to left: the suffix tells the functional group (-ol, -diol, -triol, -en-ol), the root gives the number of carbon atoms of the parent chain or ring (but, pent, hex …), and the prefixes with locants say what is attached where. Draw the parent skeleton, number it, add the –OH groups at the stated carbons, add the substituents, and finally complete each carbon to four bonds with hydrogens. As a check, count the carbons in the name and in the drawing — they must match. Ethers named as alkoxy-alkanes are drawn as R–O–R′ with the alkoxy group on the numbered carbon of the parent alkane.

Notes on the drawing. In (iii) the three –OH groups are on C1 (primary), C3 and C5 (both tertiary, each also carrying a methyl group); the compound therefore has one primary and two tertiary alcohol groups. In (viii) the carbinol carbon carries four different atoms/groups only if the two ethyl groups were different — here two are identical, so the molecule is achiral. In (ix) “cyclopent-3-en-1-ol” — the ring is numbered from the carbon bearing –OH as C1, and the double bond gets the lowest possible locants (here 3,4, which lie symmetrically on either side of the carbinol carbon). In (iv) the two ethyl groups sit on adjacent ring carbons next to the –OH carbon, positions 2 and 3.

Conclusion: The ten structures are: (i) CH₃C(OH)(CH₃)CH₂CH₃; (ii) C₆H₅CH₂CH(OH)CH₃; (iii) HOCH₂CH₂C(OH)(CH₃)CH₂C(OH)(CH₃)₂; (iv) 2,3-diethylphenol; (v) CH₃CH₂CH₂OCH₂CH₃; (vi) CH₃CH(OC₂H₅)CH(CH₃)CH₂CH₃; (vii) C₆H₁₁CH₂OH; (viii) (C₂H₅)₂C(OH)C₆H₁₁; (ix) cyclopent-3-en-1-ol; (x) HOCH₂CH₂CH(C₂H₅)CH₂Cl.

Concept. C₅H₁₂O is a saturated formula (degree of unsaturation 0), so every isomer is either an alcohol (C₅H₁₁–OH) or an ether (R–O–R′). Only the alcohols are asked for here, and they are generated systematically: the C₅H₁₁– group is a pentyl group, so the number of alcohols equals the number of structurally different pentyl groups, i.e. the number of different places in which –OH can replace a hydrogen in the three isomeric pentanes (n-pentane, 2-methylbutane, 2,2-dimethylpropane). n-Pentane gives three (C1, C2, C3 — the other two are repeats by symmetry), 2-methylbutane gives four (the CH₃ groups on C2, the CH, the CH₂, the terminal CH₃), and 2,2-dimethylpropane gives one. Total: 3 + 4 + 1 = 8 isomeric alcohols (four primary, three secondary and one tertiary — verified in the table below).

Part (ii) — classification. The carbinol carbon (carrying –OH) is bonded to one carbon in pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol and 2,2-dimethylpropan-1-ol — these are primary alcohols (four). It is bonded to two carbons in pentan-2-ol, pentan-3-ol and 3-methylbutan-2-ol — secondary alcohols (three). It is bonded to three carbons in 2-methylbutan-2-ol — the only tertiary alcohol (one). Check: 4 + 3 + 1 = 8.

Extra depth. Three of the eight alcohols — pentan-2-ol, 2-methylbutan-1-ol and 3-methylbutan-2-ol — contain a carbon attached to four different groups, so each exists as a pair of enantiomers; including these, the total number of stereoisomeric C₅H₁₁OH alcohols is 11. The molecular formula C₅H₁₂O also has six constitutional isomers that are ethers (e.g. 1-methoxybutane, 2-methoxybutane, 1-methoxy-2-methylpropane, 2-methoxy-2-methylpropane, 1-ethoxypropane, 2-ethoxypropane) — these are functional isomers of the alcohols. Boiling points of the alcohols show how branching lowers the boiling point: pentan-1-ol 411 K, 3-methylbutan-1-ol 405 K, pentan-2-ol 392 K, 2-methylbutan-2-ol 375 K.

Conclusion: There are eight alcohols of formula C₅H₁₂O: primary — pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol; secondary — pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol; tertiary — 2-methylbutan-2-ol.

Concept. The boiling point of a liquid is the temperature at which the vapour pressure equals the external pressure; to boil, the molecules must overcome the intermolecular forces that hold them in the liquid. The stronger these forces, the more energy (higher temperature) is needed. Propan-1-ol (CH₃CH₂CH₂OH, M = 60 g mol⁻¹, b.p. ≈ 370 K) and butane (CH₃CH₂CH₂CH₃, M = 58 g mol⁻¹, b.p. ≈ 273 K) have almost the same molar mass and the same number of carbon atoms in their skeletons — so their van der Waals (London dispersion) forces are comparable. Yet propanol boils about 100 K higher; the explanation must therefore lie in a different, stronger kind of intermolecular attraction.

Explanation. The molecules of butane are non-polar hydrocarbons; the only forces between them are weak van der Waals (dispersion) forces. In propan-1-ol the –OH group is strongly polarised (O is highly electronegative: Oᵟ⁻–Hᵟ⁺). The partially positive hydrogen of one molecule is attracted to a lone pair on the oxygen of a neighbouring molecule, forming an intermolecular hydrogen bond (O–H···O, typically 20 kJ mol⁻¹ or more per bond, compared with 2–5 kJ mol⁻¹ for dispersion forces between molecules of this size). Each alcohol molecule can both donate and accept hydrogen bonds, so the liquid is held together as a network of associated molecules (chains and rings). A great deal of extra thermal energy is needed to break these bonds before the molecules can escape into the vapour, which is why propanol has a much higher boiling point than butane.

Extra depth. The comparison between propanol and its isomer methoxyethane (an ether of exactly the same formula) proves that the polarity of the C–O bond alone is not responsible: the ether has no O–H bond, so it cannot form hydrogen bonds with itself and boils only a few degrees above butane. The same logic explains why ethanol (351 K) boils so much higher than propane (231 K) and methoxymethane (249 K), and why boiling points of alcohols rise along the series methanol (338 K) → ethanol (351 K) → propan-1-ol (370 K) → butan-1-ol (391 K) as the van der Waals forces add to the hydrogen bonding.

Conclusion: Propanol molecules are associated by intermolecular O–H···O hydrogen bonds, which butane molecules cannot form; breaking these bonds needs extra energy, so propanol has a much higher boiling point than butane.

Concept. Dissolving a substance in water requires that the new interactions between solute and water molecules compensate for the loss of the solute–solute interactions and for breaking some of the strong hydrogen bonds between water molecules (“like dissolves like”). Water is a highly polar solvent whose molecules are linked by an extensive network of hydrogen bonds. A solute dissolves well in water only if it can take part in hydrogen bonding with water or is ionic.

Explanation. An alcohol R–OH has a polar –OH group, in which oxygen carries a partial negative charge with two lone pairs and hydrogen carries a partial positive charge. The oxygen of the alcohol can accept a hydrogen bond from water (O–H···O), and the hydrogen of the alcohol can donate a hydrogen bond to a water oxygen. The energy released by forming these alcohol–water hydrogen bonds compensates for the energy needed to break water–water and alcohol–alcohol hydrogen bonds, so the alcohol dissolves. A hydrocarbon (alkane) of comparable molar mass, however, is non-polar and has no group that can form hydrogen bonds with water; the weak water–hydrocarbon attraction cannot make up for the energy required to break the water–water hydrogen bonds, so the hydrocarbon is virtually insoluble and forms a separate layer.

Effect of the size of the alkyl group. The –OH group is the hydrophilic (water-loving) part and the alkyl chain the hydrophobic (water-repelling) part. As the carbon chain lengthens, the hydrophobic part dominates and solubility decreases: methanol, ethanol and propanols are miscible with water in all proportions, butanols are only partly soluble, and alcohols with more than about five carbons are almost insoluble. Branching increases solubility among isomers (the molecule is more compact, with a smaller hydrophobic surface). Phenols also dissolve to some extent in water (phenol ≈ 8 g per 100 mL at 293 K) for the same reason, while polyhydric alcohols such as glycerol and ethylene glycol are completely miscible because they have several –OH groups.

Conclusion: Alcohols dissolve in water because their –OH group forms hydrogen bonds with water molecules, whereas hydrocarbons of similar mass are non-polar and cannot hydrogen-bond with water, so they are practically insoluble.

Meaning. Hydroboration–oxidation (developed by H. C. Brown, Nobel Prize 1979) is a two-stage method for converting an alkene into an alcohol. In the first stage (hydroboration) diborane, (BH₃)₂ — or its more convenient complex BH₃·THF — adds across the C=C bond of the alkene to form a trialkylborane, R₃B. In the second stage (oxidation) the trialkylborane is oxidised by hydrogen peroxide in the presence of aqueous sodium hydroxide, which replaces the carbon–boron bond by a carbon–oxygen bond and gives the alcohol (and sodium borate). The overall result is the addition of H and OH across the double bond with anti-Markovnikov orientation: the –OH becomes attached to the carbon of the double bond that carries more hydrogen atoms (the less substituted carbon) — the opposite regiochemistry to acid-catalysed hydration.

Example: propene → propan-1-ol.

Why anti-Markovnikov? The boron atom of BH₃ is electron-deficient (an electrophile) while the hydrogen carries a partial negative charge (B is less electronegative than H, so B–H bond: Bᵟ⁺–Hᵟ⁻). In the addition step the boron attaches to the less hindered, less substituted carbon (steric reasons, and because the partial positive charge in the four-centre transition state develops better on the more substituted carbon), and the hydride is transferred to the more substituted carbon, in a single concerted step with syn addition (H and B add to the same face). Repeating this with the three B–H bonds gives R₃B. In the oxidation stage, HOO⁻ (from H₂O₂ + OH⁻) attacks boron, an alkyl group migrates from boron to oxygen with retention of configuration, and hydrolysis releases the alcohol; so the –OH ends up exactly where boron was.

Another example. 1-Methylcyclopentene on hydroboration–oxidation gives trans-2-methylcyclopentanol: the boron enters at the less substituted ring carbon (and the hydride at the tertiary carbon, from the same face), and the –OH finally replaces boron with retention — a result that cannot be obtained by acid hydration, which would give the tertiary 1-methylcyclopentanol.

Conclusion: Hydroboration–oxidation converts an alkene into an alcohol by adding diborane to give a trialkylborane and then oxidising it with H₂O₂/OH⁻; the net addition of H₂O is anti-Markovnikov, e.g. propene → propan-1-ol.

Concept. A monohydric phenol has one –OH group attached directly to a carbon of a benzene ring. The formula C₇H₈O has 4 degrees of unsaturation, exactly what is needed for a benzene ring (one ring + three C=C), so the seven carbons are: six in the ring and one extra carbon that must be a methyl group. The phenol skeleton C₆H₅OH accounts for C₆H₆O; replacing one ring hydrogen by –CH₃ gives C₇H₈O. The isomers differ only in the position of the methyl group relative to the –OH group: ortho (1,2), meta (1,3) and para (1,4), so there are three monohydric phenols of this formula — the three cresols (methylphenols).

Why not benzyl alcohol or anisole? Two more compounds have the same molecular formula C₇H₈O — benzyl alcohol (C₆H₅CH₂OH) and methoxybenzene or anisole (C₆H₅OCH₃) — but they are not phenols: in benzyl alcohol the –OH is on an sp³ carbon (a benzylic alcohol) and in anisole there is no –OH group at all (it is an ether). They are functional isomers of the cresols; the question restricts the answer to monohydric phenols, so only the three cresols are correct. So the five isomers of C₇H₈O containing a benzene ring are: three cresols + benzyl alcohol + anisole.

Extra depth — properties. The cresols are colourless crystalline solids or liquids (m.p. o 304 K, m 285 K, p 308 K) found in coal tar and used in disinfectants (Lysol is a solution of cresols in soap); like phenol they are weak acids (pKₐ ≈ 10.1–10.3, slightly weaker than phenol because the methyl group is electron-releasing) and they undergo electrophilic substitution at the ring positions activated by –OH. Their ring positions are not equivalent, so each cresol gives a different set of nitration and bromination products; p-cresol, with the para position blocked, gives only ortho substitution products.

Conclusion: The monohydric phenols of formula C₇H₈O are 2-methylphenol (o-cresol), 3-methylphenol (m-cresol) and 4-methylphenol (p-cresol).

Concept. Steam distillation separates a substance that is volatile in steam, immiscible with water and unreactive with it, from non-volatile impurities. A compound is carried over with steam if its molecules are only weakly attached to one another; a compound whose molecules are strongly associated stays behind. Since o- and p-nitrophenol have the same molecular formula and molar mass (139 g mol⁻¹), the difference in volatility must come from the nature of the hydrogen bonding in the two isomers.

Answer: o-nitrophenol is steam volatile. In o-nitrophenol the –OH group and the –NO₂ group are on adjacent carbons, close enough for the hydrogen of –OH to form a hydrogen bond with an oxygen atom of the nitro group in the same molecule — a six-membered chelate ring (intramolecular hydrogen bonding). This internal bond uses up the hydrogen-bonding ability of the molecule, so o-nitrophenol molecules are not strongly attracted to one another, the substance is less soluble in water, has a lower melting point (318 K) and boiling point and a higher vapour pressure, and it distils over with steam. In p-nitrophenol the –OH and –NO₂ groups are at opposite ends of the ring and cannot reach each other; each –OH hydrogen-bonds to a neighbouring molecule (intermolecular hydrogen bonding) and also to water. The molecules therefore are associated into long chains, which makes p-nitrophenol less volatile (higher m.p. 387 K and b.p.), more soluble in water, and it is left behind in the distillation flask.

Practical application. Nitration of phenol with dilute HNO₃ at 298 K gives a mixture of o- and p-nitrophenols; on steam distillation the ortho isomer (yellow oil/crystals) collects in the receiver with water, while the para isomer remains as a residue in the flask and can be recovered by crystallisation. This is the standard laboratory method of separating the two isomers. The same reasoning explains why salicylaldehyde (o-hydroxybenzaldehyde) is a liquid and steam volatile, while its p-isomer is a solid, and why o-hydroxybenzoic acid has a lower solubility than the para isomer.

Conclusion: o-Nitrophenol is steam volatile because it has intramolecular hydrogen bonding, whereas p-nitrophenol is associated through intermolecular hydrogen bonding and is not volatile with steam.

Concept. The cumene process (also called the Hock process) is the main industrial route to phenol and gives acetone as a valuable by-product. Cumene is isopropylbenzene, C₆H₅CH(CH₃)₂, itself made by the Friedel–Crafts alkylation of benzene with propene (in the presence of H₃PO₄ or AlCl₃). The cheap raw materials (benzene, propene, air) and the commercially useful co-product explain why most of the world’s phenol (and acetone) is produced this way. The key point is that the benzylic C–H bond of cumene is very easily oxidised by air through a stable benzylic free radical.

Equations.

(1) Oxidation: C₆H₅–CH(CH₃)₂ + O₂ → C₆H₅–C(CH₃)₂–O–OH (cumene hydroperoxide). Free radicals start a chain reaction in which the tertiary benzylic radical C₆H₅C·(CH₃)₂ (stabilised by the ring) combines with O₂ to give a peroxyl radical, which abstracts H from another cumene molecule.

(2) Acid-catalysed rearrangement and hydrolysis: C₆H₅–C(CH₃)₂–O–OH + H₂O (dil. H⁺) → C₆H₅OH + CH₃COCH₃ (phenol + propanone). In the mechanism (Hock rearrangement): the terminal –OH of the hydroperoxide is protonated and leaves as water as the phenyl group migrates from carbon to oxygen, giving an oxocarbenium ion (C₆H₅–O–C⁺(CH₃)₂); water adds to the carbocation, and the resulting hemiketal breaks down to phenol and acetone.

Advantages. No chlorine or sulphonic acid wastes (unlike the older chlorobenzene and benzenesulphonic acid routes); both products are in demand — phenol for Bakelite and other resins, bisphenol A, caprolactam (nylon-6), aspirin and dyes; acetone as a solvent and for making methyl methacrylate (acrylic glass). The two products are separated by fractional distillation (phenol b.p. 455 K; acetone b.p. 329 K).

Conclusion: Cumene is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid gives phenol and acetone: C₆H₅CH(CH₃)₂ → (O₂) C₆H₅C(CH₃)₂OOH → (H₃O⁺) C₆H₅OH + CH₃COCH₃.

Concept. Haloarenes are unreactive towards nucleophiles because the C–Cl bond is on an sp² carbon, is shortened by resonance with the ring (partial double-bond character) and is shielded by the π electron cloud. Replacing the halogen of chlorobenzene by –OH therefore needs drastic conditions: a hydroxide nucleophile (aqueous NaOH), high temperature and very high pressure. This is the Dow process. The product at the end of the reaction is not phenol but its sodium salt, because phenol (pKₐ ≈ 10) is acidic enough to react with excess NaOH; therefore the mixture has to be acidified in a second step to liberate the free phenol.

Equations. (1) C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O (623 K, 300 atm) — nucleophilic substitution of Cl by OH⁻, followed at once by neutralisation of the phenol formed by the second mole of NaOH. (2) C₆H₅ONa + HCl → C₆H₅OH + NaCl — the stronger acid HCl displaces the weak acid phenol from its salt. Carbon dioxide can also be used to liberate phenol, since carbonic acid is stronger than phenol: C₆H₅ONa + CO₂ + H₂O → C₆H₅OH + NaHCO₃.

Extra depth — why the conditions are so severe, and how substituents help. The nucleophilic substitution of chlorobenzene proceeds by an addition–elimination (or partly an elimination–addition, benzyne) pathway in which the incoming nucleophile must overcome the electron-rich ring. If strong electron-withdrawing groups are present at the ortho and/or para positions the reaction becomes much easier: p-nitrochlorobenzene reacts with aqueous NaOH at 443 K (170 °C); 2,4-dinitrochlorobenzene reacts at 368 K; and 2,4,6-trinitrochlorobenzene (picryl chloride) reacts with water at room temperature. The –NO₂ groups stabilise the negatively charged intermediate (the Meisenheimer complex) by delocalising the charge onto the nitro oxygens at the ortho/para positions. A meta –NO₂ group gives no such stabilisation and has hardly any effect.

Conclusion: Chlorobenzene is heated with aqueous NaOH at 623 K and 300 atm to give sodium phenoxide, which on acidification with HCl gives phenol: C₆H₅Cl → (NaOH) C₆H₅ONa → (H⁺) C₆H₅OH.

Concept. The acid-catalysed hydration of an alkene is an electrophilic addition in which water is added across the C=C bond in the presence of a strong acid catalyst (dilute H₂SO₄ or H₃PO₄ at high temperature and pressure). The π bond of the alkene is electron-rich and attacks the electrophile H⁺ (supplied as H₃O⁺ in the aqueous acid). For ethene the carbocation formed is a primary ethyl cation, which is high in energy but is the only possible one; in practice the protonation and the attack of water occur in an essentially concerted fashion. The catalyst H⁺ is regenerated at the end, so the overall reaction is CH₂=CH₂ + H₂O → CH₃CH₂OH.

Mechanism (three steps).

Step 1 — Protonation of ethene (rate-determining). H₃O⁺ is generated from the acid and water (H₂O + H⁺ → H₃O⁺). The π electrons of ethene attack the proton of H₃O⁺; a new C–H σ bond forms and a carbocation results: CH₂=CH₂ + H₃O⁺ → CH₃–C⁺H₂ + H₂O.

Step 2 — Nucleophilic attack of water on the carbocation. A water molecule donates a lone pair to the positively charged carbon: CH₃–C⁺H₂ + H₂O → CH₃CH₂–O⁺H₂ (protonated ethanol, an ethyl-oxonium ion).

Step 3 — Deprotonation. A water molecule removes a proton from the oxonium ion, giving ethanol and regenerating the hydronium ion catalyst: CH₃CH₂–O⁺H₂ + H₂O → CH₃CH₂OH + H₃O⁺.

Extra depth. The industrial hydration of ethene (about 570 K, 60–70 atm, H₃PO₄ on silica gel) is the major commercial route to synthetic ethanol. For unsymmetrical alkenes such as propene the proton adds so as to give the more stable carbocation (the secondary isopropyl cation), which explains Markovnikov’s rule: propene gives propan-2-ol, not propan-1-ol. For alkenes whose carbocation can rearrange (3-methylbut-1-ene, for example) the product may be a rearranged alcohol. Since all three steps are reversible, the same mechanism run backwards explains the acid-catalysed dehydration of ethanol to ethene (compare exercise 7.19).

Conclusion: Ethene is protonated by H₃O⁺ to the ethyl carbocation, water attacks the carbocation to give protonated ethanol, and loss of a proton gives ethanol and regenerates the acid catalyst.

Concept. This is the older sulphonation (benzenesulphonic acid) route to phenol. Benzene undergoes electrophilic substitution with concentrated sulphuric acid (in practice oleum, H₂SO₄·SO₃) to give benzenesulphonic acid. The –SO₃H group can then be replaced by –OH when the sodium salt of the sulphonic acid is fused with solid NaOH at a high temperature (≈ 573 K): a nucleophilic substitution under drastic conditions, giving sodium phenoxide. Acidification then liberates phenol. The sequence uses only the three reagents named in the question (the acid for the last step is dilute H₂SO₄ or any strong acid).

Equations.

(1) Sulphonation: C₆H₆ + H₂SO₄ (conc., or oleum) → C₆H₅SO₃H + H₂O. (The electrophile is SO₃, present in oleum and formed in conc. H₂SO₄; the reaction is reversible, so excess acid is used.)

(2) Neutralisation: C₆H₅SO₃H + NaOH → C₆H₅SO₃Na + H₂O (sodium benzenesulphonate).

(3) Alkali fusion: C₆H₅SO₃Na + 2NaOH → C₆H₅ONa + Na₂SO₃ + H₂O (solid NaOH, ≈ 573 K).

(4) Acidification: 2C₆H₅ONa + H₂SO₄ (dilute) → 2C₆H₅OH + Na₂SO₄ (or C₆H₅ONa + HCl → C₆H₅OH + NaCl).

Extra depth. The method was the first industrial synthesis of phenol (Wurtz and Kekulé, 1867) but is now largely replaced by the cumene process because it consumes large quantities of acid and alkali and produces sodium sulphite waste. The alkali fusion of aryl sulphonates is also the traditional route to naphthols and other phenolic dye intermediates. The –SO₃H group is a good leaving group in this reaction because the sulphite ion is stable; the reaction is believed to occur by addition of hydroxide to the ring carbon bearing sulphur and loss of sulphite.

Conclusion: C₆H₆ + H₂SO₄ → C₆H₅SO₃H; C₆H₅SO₃H + NaOH → C₆H₅SO₃Na; C₆H₅SO₃Na + 2NaOH (fusion) → C₆H₅ONa + Na₂SO₃ + H₂O; C₆H₅ONa + H⁺ → C₆H₅OH.

Concept. Three different bond-forming strategies are asked for, and the first step in each case is to work backwards from the target (retrosynthesis). (i) An alcohol can be made from an alkene by adding H₂O across the double bond; the regiochemistry is controlled by Markovnikov’s rule (acid-catalysed hydration) or the anti-Markovnikov rule (hydroboration–oxidation). (ii) and (iii) An alcohol can be made from an alkyl halide by nucleophilic substitution of the halogen by a hydroxide ion: R–X + OH⁻ → R–OH + X⁻. For primary halides this is a clean SN2 reaction (the nucleophile attacks the carbon from the back side while the halide leaves); the solvent is aqueous (or aqueous–alcoholic, at low temperature) to avoid elimination.

(i) 1-Phenylethanol from an alkene. The target C₆H₅–CH(OH)–CH₃ has the –OH on the carbon attached to the ring (a secondary, benzylic alcohol). Removing H₂O gives the alkene styrene (phenylethene), C₆H₅–CH=CH₂. Acid-catalysed hydration places H on the terminal CH₂ and –OH on the carbon next to the ring, because the benzylic carbocation C₆H₅–C⁺H–CH₃ (resonance-stabilised by the ring) is far more stable than the primary carbocation C₆H₅CH₂CH₂⁺. Hence Markovnikov addition gives exactly 1-phenylethanol.

(ii) Cyclohexylmethanol from an alkyl halide by SN2. The target is C₆H₁₁–CH₂–OH, a primary alcohol. Replacing –OH by a halogen gives the alkyl halide C₆H₁₁–CH₂–X (chloromethyl- or bromomethylcyclohexane), a primary halide in which the leaving group sits on an unhindered CH₂ — ideal for SN2. Treatment with aqueous NaOH (or KOH) displaces the halide.

(iii) Pentan-1-ol from an alkyl halide. The target CH₃CH₂CH₂CH₂CH₂OH is a primary alcohol; the matching alkyl halide is the 1-halopentane (1-chloropentane or 1-bromopentane). Aqueous NaOH again effects the SN2 substitution.

Extra depth and alternatives. For (iii) it would be wrong to start with 2-bromopentane (that would give pentan-2-ol) or with 1-pentene and acid (that would give pentan-2-ol by Markovnikov addition); to get the primary alcohol from the alkene one would use hydroboration–oxidation of pent-1-ene. The substitution works with primary halides only: with a secondary or tertiary halide the strongly basic hydroxide would give an alkene by elimination and, for tertiary, an SN1 hydrolysis; this is why only primary halides are specified in (ii) and (iii). The reactions of (ii) and (iii) can also be done through the Grignard route (e.g. C₄H₉MgBr + HCHO → pentan-1-ol).

Conclusion: (i) C₆H₅CH=CH₂ + H₂O/H⁺ → 1-phenylethanol; (ii) C₆H₁₁CH₂Br + aq. NaOH → cyclohexylmethanol; (iii) CH₃(CH₂)₄Br + aq. NaOH → pentan-1-ol.

Concept. Phenol is a weak Brønsted acid (pKₐ ≈ 10, Kₐ ≈ 1.3 × 10⁻¹⁰): it can donate the proton of its –OH group to a base to form the phenoxide ion, C₆H₅O⁻. The acidity of the O–H bond of phenol is much greater than that of alcohols because the phenoxide ion is stabilised by resonance, whereas an alkoxide ion is not. The acidic character is demonstrated by two reactions of the O–H hydrogen: reaction with an active metal (like an acid) and reaction with an alkali (like an acid with a base).

Reaction 1 — with sodium metal (hydrogen is evolved):

Reaction 2 — with aqueous sodium hydroxide (a salt and water — neutralisation):

Phenol is, however, a much weaker acid than carboxylic acids: it does not react with sodium carbonate or sodium hydrogencarbonate (it is weaker than carbonic acid, pKₐ 6.4), so no CO₂ is liberated — this is used to distinguish phenols from carboxylic acids. Phenol can be recovered from its salt by CO₂ or any stronger acid: C₆H₅ONa + CO₂ + H₂O → C₆H₅OH + NaHCO₃.

Comparison with ethanol. Ethanol also reacts with sodium (2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂), but it does not react with aqueous NaOH: the equilibrium C₂H₅OH + OH⁻ ⇌ C₂H₅O⁻ + H₂O lies far to the left, because ethanol (pKₐ 15.9) is a weaker acid than water (pKₐ 15.7). Phenol (pKₐ ≈ 10) is about a million times (10⁵·⁹) more acidic than ethanol, and reacts quantitatively with NaOH. The reasons are: (1) in the phenoxide ion the negative charge on oxygen is delocalised over the ring (to the ortho and para carbons) through resonance, which stabilises the conjugate base and favours ionisation; in the ethoxide ion the charge stays localised on oxygen and is in fact intensified by the electron-releasing (+I) ethyl group, destabilising the ion; (2) the oxygen of phenol is attached to an sp² carbon, which is more electronegative than the sp³ carbon of ethanol, so the O–H bond of phenol is more polarised.

Conclusion: Phenol reacts with Na (giving H₂ and sodium phenoxide) and with NaOH (giving sodium phenoxide and water), showing its acidic nature; it is far more acidic than ethanol (pKₐ 10 against 15.9) because the phenoxide ion is stabilised by resonance, while ethoxide is destabilised by the +I effect of the alkyl group.

Concept. The strength of a phenol as an acid is decided by the stability of its conjugate base, the phenoxide ion, relative to the un-ionised phenol. A substituent that stabilises the phenoxide ion increases acidity; one that destabilises it decreases acidity. Electron-withdrawing groups (–NO₂, –CN, –CHO, –COOH, halogens) pull electron density away from the ring and from the negatively charged oxygen, spreading the negative charge and stabilising the anion; electron-donating groups (–CH₃, –OCH₃, –NH₂, –OH) push electrons towards the oxygen and into the ring, intensifying the negative charge and destabilising the anion.

o-Nitrophenol. The nitro group is a strong electron-withdrawing group through both the –I (inductive) and the –R/–M (resonance) effects. In the o-nitrophenoxide ion the negative charge, delocalised from oxygen to the ortho and para carbons of the ring, reaches the ring carbon that bears –NO₂ and can then be accommodated on the electronegative oxygen atoms of the nitro group (two extra resonance structures; see intext 7.8). The resulting ion is considerably stabilised and the phenol ionises more readily: pKₐ = 7.2, compared with 10.0 for phenol.

o-Methoxyphenol (guaiacol). The methoxy group (–OCH₃) is electron-donating by resonance (+R: a lone pair of oxygen delocalises into the ring) even though it is mildly electron-withdrawing by the inductive effect (–I). At the ortho position the +R effect and the –I effect nearly cancel, and there is also a stabilising intramolecular hydrogen bond in the neutral molecule between the –OH hydrogen and the methoxy oxygen, which makes the proton harder to remove. As a result, the phenoxide ion is not stabilised (its charge is slightly intensified in the ring), and o-methoxyphenol is about as weak an acid as phenol or even weaker: pKₐ ≈ 10.0–10.2.

Extra depth. A useful way to remember the substituent effect on acidity is the Hammett-type trend: acid strength increases with the electron-withdrawing power of the substituent (2,4,6-trinitrophenol, picric acid, pKₐ = 0.25 is as strong as a mineral acid) and decreases with electron-releasing power (the cresols, pKₐ 10.1–10.3, are slightly weaker acids than phenol). The same principle explains the acid strengths of substituted benzoic acids and of the aliphatic acids ClCH₂COOH > CH₃COOH.

Conclusion: The electron-withdrawing –NO₂ group stabilises the o-nitrophenoxide ion by –I and –R effects, whereas the electron-releasing –OCH₃ group destabilises the o-methoxyphenoxide ion; hence o-nitrophenol (pKₐ 7.2) is much more acidic than o-methoxyphenol.

Concept. In electrophilic aromatic substitution the benzene ring acts as the nucleophile: its π electrons attack the electrophile E⁺ to give a positively charged carbocation intermediate (the arenium or sigma-complex), and loss of H⁺ then restores aromaticity. The rate depends on how readily the ring donates electrons and how stable the arenium ion is. A substituent that donates electron density to the ring activates it (faster reaction than benzene) and stabilises the arenium ion; one that withdraws electrons deactivates it.

How –OH activates the ring. The oxygen atom of the –OH group has two lone pairs. Two electronic effects operate: (1) the –I (inductive) effect — oxygen is more electronegative than carbon and withdraws σ-electrons slightly; (2) the +R (resonance or mesomeric) effect — a lone pair on oxygen is in a p orbital that overlaps with the π system of the ring and is delocalised into it. For the –OH group, the +R effect is much stronger than the –I effect, so the net result is a donation of electron density into the ring. The resonance structures of phenol show negative charge (increased electron density) at the ortho and para positions, not at the meta positions:

Because of this, (a) the ring of phenol is richer in electrons than benzene, so it reacts faster with electrophiles (bromine water reacts with phenol at room temperature, without a catalyst, to give 2,4,6-tribromophenol at once, whereas benzene needs FeBr₃ and gives only bromobenzene); (b) the arenium ion formed by attack at the ortho or para position has an extra resonance structure in which the oxygen bears the positive charge (O⁺=C), with every atom having a complete octet — a particularly stable structure; attack at the meta position gives no such structure. Hence the –OH group is an activating group and an ortho, para-director.

Extra depth. In the phenoxide ion (in alkaline solution) the oxygen carries a full negative charge, so its +R effect is even stronger and the ring is still more reactive; this is why phenoxide ions undergo the Kolbe and Reimer–Tiemann reactions with weak electrophiles (CO₂, :CCl₂). The ortho/para-directing and activating effect of –OH is shared by –OR, –NH₂ and –NHR groups (with lone pairs on the atom attached to the ring), whereas groups such as –NO₂, –COOH and –CN, which withdraw electrons by –I and –R, deactivate the ring and direct incoming groups to the meta position.

Conclusion: The lone pair on the oxygen of –OH is delocalised into the ring (+R effect, stronger than –I), raising the electron density at the ortho and para positions and stabilising the arenium ion; hence –OH activates the ring and directs the electrophile to ortho and para positions.

Concept. The four reactions cover the two main kinds of chemistry of the hydroxy compounds: oxidation of the –CH₂OH group of a primary alcohol (part i), and electrophilic substitution in the ring of phenol (parts ii–iv), where the strongly activating –OH group directs the electrophile to the ortho and para positions and allows the use of mild conditions.

(i) Oxidation of propan-1-ol with alkaline KMnO₄. A primary alcohol is oxidised first to the aldehyde and then, because the aldehyde is oxidised more easily than the alcohol, all the way to the carboxylic acid with the same number of carbon atoms. In alkaline permanganate the acid is obtained as its potassium salt, and the purple MnO₄⁻ is reduced to brown MnO₂; the acid is liberated by acidification.

Balanced: 3CH₃CH₂CH₂OH + 4MnO₄⁻ → 3CH₃CH₂COO⁻ + 4MnO₂ + 4H₂O + OH⁻.

(ii) Phenol with bromine in CS₂ (non-polar solvent, low temperature). In a solvent of low polarity the ionisation of phenol (and the activity of Br₂) is suppressed, so only monobromination occurs; a mixture of o– and p-bromophenol results, with the para isomer the major product (less steric hindrance).

Contrast: with bromine water (a polar solvent in which phenol ionises to phenoxide) phenol gives a white precipitate of 2,4,6-tribromophenol: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr.

(iii) Phenol with dilute HNO₃. At low temperature (298 K) dilute nitric acid nitrates phenol to a mixture of o– and p-nitrophenol (the ortho isomer is steam volatile and can be separated from the para isomer by steam distillation).

With concentrated nitric acid, 2,4,6-trinitrophenol (picric acid) is formed, but the yield is poor because of oxidation; picric acid is therefore prepared by sulphonating phenol first and then treating the disulphonic acid with conc. HNO₃.

(iv) Phenol with chloroform and aqueous NaOH (Reimer–Tiemann reaction). Phenol is heated with CHCl₃ and aqueous NaOH at 340 K, and the product is acidified: salicylaldehyde (2-hydroxybenzaldehyde) is formed.

Conclusion: (i) CH₃CH₂CH₂OH → (alk. KMnO₄) CH₃CH₂COOH; (ii) C₆H₅OH + Br₂ (CS₂) → p- and o-bromophenol; (iii) C₆H₅OH + dil. HNO₃ → o- and p-nitrophenol; (iv) C₆H₅OH + CHCl₃/aq. NaOH → salicylaldehyde.

Concept. The four terms are all named reactions or definitions from this chapter. The first two are ortho-substitutions of phenol through the phenoxide ion, the third is the general method for ethers, and the fourth is a class of ethers that Williamson’s method can prepare.

(i) Kolbe’s reaction. Phenol is converted into sodium phenoxide by sodium hydroxide; the phenoxide, which is much more reactive than phenol towards electrophilic substitution, is heated with carbon dioxide (400 K, 4–7 atm). The weakly electrophilic CO₂ attacks the ortho position and sodium salicylate is formed; acidification gives salicylic acid (2-hydroxybenzoic acid), the raw material for aspirin.

(ii) Reimer–Tiemann reaction. On treating phenol with chloroform in the presence of aqueous sodium hydroxide at about 340 K, a –CHO group is introduced at the ortho position of the benzene ring (the intermediate substituted benzal chloride is hydrolysed by the alkali), and acidification gives salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene, :CCl₂, produced from CHCl₃ and OH⁻.

(iii) Williamson ether synthesis. This is the most important laboratory method for preparing ethers. An alkyl halide is treated with sodium alkoxide (or a phenoxide); the alkoxide ion, a strong nucleophile, displaces the halide by an SN2 mechanism: R–X + R′–O⁻Na⁺ → R–O–R′ + NaX. Both symmetrical and unsymmetrical ethers can be made: e.g. CH₃CH₂ONa + CH₃CH₂Br → CH₃CH₂OCH₂CH₃ (diethyl ether) or C₆H₅ONa + CH₃I → C₆H₅OCH₃ (anisole). The halide should be primary (secondary and tertiary halides give alkenes through elimination, since an alkoxide is also a strong base).

(iv) Unsymmetrical (mixed) ether. An ether in which the two groups attached to the oxygen atom are different is called an unsymmetrical or mixed ether, R–O–R′ (R ≠ R′). Examples: ethyl methyl ether (CH₃OC₂H₅, methoxyethane), methyl phenyl ether (anisole, C₆H₅OCH₃) and ethyl phenyl ether (C₆H₅OC₂H₅, phenetole). They are best prepared by Williamson synthesis (the dehydration of a mixture of two alcohols gives a mixture of three ethers and is therefore unsuitable). In contrast, a symmetrical or simple ether has identical groups (C₂H₅OC₂H₅, diethyl ether).

Conclusion: Kolbe’s reaction gives salicylic acid from sodium phenoxide and CO₂; Reimer–Tiemann gives salicylaldehyde from phenol, CHCl₃ and aq. NaOH; Williamson synthesis forms ethers from an alkyl halide and an alkoxide by SN2; an unsymmetrical ether has two different groups on oxygen (e.g. CH₃OC₂H₅).

Concept. When ethanol is heated with excess concentrated sulphuric acid (or H₃PO₄) at 443 K (170 °C), the elements of water are removed from adjacent carbons (β-elimination) and ethene is obtained: CH₃CH₂OH → CH₂=CH₂ + H₂O. The –OH group is a poor leaving group (OH⁻ is a strong base), so the acid first converts it into a good leaving group, H₂O. The mechanism is E1 (unimolecular elimination) and proceeds through three steps; the first and last steps are fast acid–base steps while the second step, in which the carbocation is formed, is slow and rate-determining. The conditions are crucial: at the lower temperature of 413 K with excess alcohol, the same reagents give diethyl ether by substitution (SN2) instead.

Step 1 — Protonation of the alcohol (fast, reversible). The oxygen of ethanol, which has lone pairs, accepts a proton from the acid to form ethyl-oxonium ion (protonated ethanol): CH₃CH₂–OH + H⁺ → CH₃CH₂–O⁺H₂.

Step 2 — Formation of the carbocation (slow, rate-determining). The protonated alcohol loses a water molecule (a good leaving group) to give the ethyl carbocation: CH₃CH₂–O⁺H₂ → CH₃–C⁺H₂ + H₂O.

Step 3 — Loss of a proton to form the alkene (fast). A base (H₂O or HSO₄⁻) removes a proton from the carbon next to the carbocation; the electrons of the C–H bond form the new π bond, giving ethene and regenerating the acid catalyst: CH₃–C⁺H₂ + H₂O → CH₂=CH₂ + H₃O⁺ (H⁺ regenerated).

Extra depth. The ease of dehydration of alcohols follows the order tertiary > secondary > primary, which parallels the stability of the carbocation formed in step 2: tertiary alcohols dehydrate at about 358 K (even with dilute acid), secondary at 440 K, primary at 443 K. In ethanol the intermediate is a (relatively unstable) primary carbocation, and that is why a high temperature and concentrated acid are needed. When a β-hydrogen is available on more than one carbon, Saytzeff’s rule predicts that the more substituted alkene is the major product (e.g. butan-2-ol → but-2-ene rather than but-1-ene). At 413 K the same acid gives ethoxyethane through a competing SN2 reaction between an alcohol molecule and the protonated alcohol: this temperature control is the basis of the preparation of both ethene and diethyl ether from ethanol.

Conclusion: Ethanol is protonated by the acid, the protonated alcohol loses water to give the ethyl carbocation (slow step), and loss of a β-proton gives ethene, regenerating the acid catalyst.

Concept. The four conversions use three of the standard preparations of alcohols: hydration of an alkene (i), nucleophilic substitution of an alkyl halide by hydroxide (ii), and reaction of a Grignard reagent with a carbonyl compound (iii and iv). In the Grignard reactions the carbonyl compound that must be chosen is the one that supplies the missing carbon atoms and gives the correct class of alcohol: HCHO → primary alcohol, other aldehydes → secondary, ketones → tertiary alcohol. A useful method is to count the carbons added and to look at the class of the target alcohol.

(i) Propene → propan-2-ol — acid-catalysed hydration (Markovnikov addition): the proton adds to the terminal CH₂ so that the more stable secondary carbocation CH₃–C⁺H–CH₃ forms, then water adds.

(In the older industrial process propene is absorbed in conc. H₂SO₄ to form isopropyl hydrogen sulphate, which is hydrolysed by boiling with water.)

(ii) Benzyl chloride → benzyl alcohol — nucleophilic substitution (hydrolysis) of the benzylic C–Cl bond by hydroxide; benzylic halides are very reactive (both SN1 and SN2), so the reaction takes place readily with aqueous NaOH or KOH.

(iii) Ethylmagnesium chloride → propan-1-ol — the target has one more carbon than the ethyl group and is primary, so the carbonyl partner must be methanal (HCHO). The ethyl carbanion of the Grignard reagent adds to the carbonyl carbon, and hydrolysis gives the alcohol.

(iv) Methylmagnesium bromide → 2-methylpropan-2-ol — the target (CH₃)₃C–OH is a tertiary alcohol with the carbinol carbon carrying three methyl groups; one methyl comes from the Grignard reagent, so the carbonyl compound must be a ketone with two methyl groups — propanone (acetone), CH₃COCH₃.

Conclusion: (i) propene + H₂O/H⁺; (ii) benzyl chloride + aq. NaOH; (iii) C₂H₅MgCl + HCHO, then H₂O/H⁺; (iv) CH₃MgBr + CH₃COCH₃, then H₂O/H⁺.

Concept. The reagent decides the product, so the first step is to decide what kind of change is wanted: an oxidation that stops at the aldehyde needs a mild, anhydrous oxidant; one that proceeds to the acid needs a strong, aqueous oxidant; a ring bromination that goes to all three activated positions needs a polar solvent (water); dehydration needs a strong dehydrating acid and heat; and the reduction of a ketone to an alcohol needs a hydride or catalytic hydrogenation. Each choice below is therefore justified.

Extra depth. Distinguish the oxidation levels: primary alcohol → aldehyde → carboxylic acid (same number of carbons); secondary alcohol → ketone (CrO₃, Jones reagent, or KMnO₄); tertiary alcohol → no oxidation under mild conditions (no hydrogen on the carbinol carbon) — under drastic conditions C–C bond cleavage gives a mixture of acids and ketones with fewer carbons. Strong oxidants such as KMnO₄ also oxidise the C=C bond and alkyl side chains of arenes, so PCC or Collins/Swern-type reagents are used when other oxidisable groups are present. In (iii), a very similar-looking reagent (Br₂ in CS₂ at low temperature) would stop at the mono-bromo stage (o- and p-bromophenol).

Conclusion: (i) acidified KMnO₄ or K₂Cr₂O₇; (ii) PCC (or Cu, 573 K); (iii) bromine water; (iv) alkaline KMnO₄ (then H⁺); (v) conc. H₂SO₄ at 443 K; (vi) NaBH₄ (or LiAlH₄, or H₂/Pd).

Concept. Ethanol (CH₃CH₂OH) and methoxymethane (dimethyl ether, CH₃OCH₃) are functional isomers — both have the molecular formula C₂H₆O and the same molar mass (46 g mol⁻¹). Because the molar masses are identical, the van der Waals (dispersion) forces are about the same, so any difference in boiling point must arise from the specific intermolecular forces due to the functional group. Ethanol boils at 351 K while methoxymethane boils at only 248 K — a difference of more than 100 K.

Explanation. Ethanol contains an –OH group in which a hydrogen is directly bonded to the very electronegative oxygen. The O–H bond is highly polar, so the hydrogen carries a significant partial positive charge and can form a hydrogen bond with the lone pair of oxygen on another ethanol molecule (O–H···O). Each ethanol molecule can act as both donor and acceptor, so the molecules are linked into an extensively associated liquid; a considerable amount of energy is required to break these intermolecular hydrogen bonds before the molecules can vaporise. In methoxymethane the oxygen is bonded to two carbon atoms and there is no hydrogen attached to oxygen; the molecules therefore cannot hydrogen-bond with one another. They are held only by weak dipole–dipole attractions (the C–O–C dipole, μ ≈ 1.3 D) and dispersion forces, so little energy is needed to separate them and the boiling point is low (the compound is a gas at room temperature).

Extra depth. The same factor accounts for the whole pattern of boiling points of comparable-mass compounds: alcohols (strong H-bonding) > aldehydes/ketones, ethers, haloalkanes (dipole–dipole) > alkanes (dispersion only). For C₂–C₄ compounds: propane 231 K < methoxymethane 248 K < ethanol 351 K. Note also that although an ether molecule cannot hydrogen-bond to another ether molecule, its oxygen can accept hydrogen bonds from water or alcohols, which is why ethers dissolve in water to about the same extent as alcohols of similar molar mass.

Conclusion: Ethanol molecules are associated through intermolecular hydrogen bonds (O–H···O), which are absent in methoxymethane; therefore ethanol (351 K) boils at a much higher temperature than methoxymethane (248 K).

Concept. In the IUPAC system an ether R–O–R′ is named as an alkoxy-alkane (or alkoxy-arene): the larger (longer or more complex) hydrocarbon group is chosen as the parent, and the smaller group together with the oxygen is named as an alkoxy substituent (methoxy, ethoxy, propoxy, phenoxy …). The parent chain is numbered so that the alkoxy group (and other substituents) get the lowest locants; the prefixes are written in alphabetical order. For a ring, the carbon bearing the –OR group is generally C-1 unless a higher-priority group is present.

Points of care. (a) In (i) the alkyl group on the left is ethyl (two carbons) while the right side is the four-carbon 2-methylpropyl group; the larger group is the parent, so the name is 1-ethoxy-2-methylpropane, not “isobutyl ethyl ether” (that is the common name). (b) In (ii) the substituent prefixes are written alphabetically: chloro before methoxy; the locants are {1,2} either way, so alphabetical order decides that Cl gets 1. (c) In (iii) the two substituents are at the 1,4-positions of benzene; alphabetical order gives methoxy locant 1, nitro locant 4. (d) For ring (v) the –OR carbon is not automatically C-1: the lowest set of locants for all substituents is chosen — {1,1,4}, which means the carbon bearing the two methyl groups is C-1 and the ethoxy-bearing carbon is C-4.

Conclusion: (i) 1-ethoxy-2-methylpropane; (ii) 1-chloro-2-methoxyethane; (iii) 1-methoxy-4-nitrobenzene; (iv) 1-methoxypropane; (v) 4-ethoxy-1,1-dimethylcyclohexane; (vi) ethoxybenzene.

Concept. In Williamson’s synthesis an alkoxide (or phenoxide) ion displaces the halide from a primary alkyl halide by an SN2 mechanism: R–O⁻ + R′–X → R–O–R′ + X⁻. For an unsymmetrical ether there are two ways to disconnect R–O–R′, and the correct choice is the one in which the halide is primary (or methyl) and the alkoxide carries the more hindered or tertiary group. A tertiary alkyl halide must never be used (it eliminates to an alkene) and an aryl halide must never be used as the substrate (it does not undergo SN2). For alkyl aryl ethers the phenoxide ion is therefore the nucleophile and the alkyl halide is the electrophile.

Why these choices? (i) A symmetrical ether: the same propyl group appears on both sides, so propoxide and a propyl halide (primary — good). (ii) The aryl–O bond cannot be formed by SN2 on bromobenzene, so sodium phenoxide supplies the oxygen and the primary ethyl halide supplies the ethyl group; the reverse combination (C₆H₅Br + C₂H₅ONa) gives no reaction under these conditions. (iii) A tertiary alkyl group is introduced as the alkoxide (CH₃)₃C–O⁻, while the methyl group comes from CH₃Br; the opposite combination ((CH₃)₃C–Br + CH₃ONa) would give 2-methylpropene by E2 elimination — the point examined in exercise 7.25. (iv) Both combinations (CH₃ONa + C₂H₅Br, or C₂H₅ONa + CH₃Br) work because both halides are primary or methyl; either is accepted.

Preparing the alkoxides. The alkoxide is made from the alcohol with sodium (2ROH + 2Na → 2RONa + H₂) or sodium hydride; the phenoxide is made with NaOH (C₆H₅OH + NaOH → C₆H₅ONa + H₂O) because phenol is acidic enough.

Conclusion: (i) CH₃CH₂CH₂ONa + CH₃CH₂CH₂Br; (ii) C₆H₅ONa + C₂H₅Br; (iii) (CH₃)₃CONa + CH₃Br; (iv) CH₃ONa + C₂H₅Br — each gives the ether and NaBr by an SN2 reaction.

Concept. The Williamson synthesis, R–X + R′O⁻ → R–O–R′ + X⁻, is a bimolecular nucleophilic substitution (SN2). The alkoxide ion is both a good nucleophile and a strong base. Which of these two properties wins depends on the structure of the alkyl halide: if the carbon bearing the halogen is easily approached from the back side (a methyl or primary halide) substitution takes place and an ether forms; if the carbon is crowded (secondary, tertiary or hindered primary halides) the alkoxide cannot reach it and instead abstracts a β-hydrogen, giving an alkene by E2 elimination. In addition, aryl and vinyl halides, whose C–X carbon is sp², cannot undergo SN2 at all. These three facts are the limitations of the method.

The golden rule. For an unsymmetrical ether R–O–R′ always disconnect so that the halide is the less hindered (methyl or primary) component and the alkoxide carries the more hindered (secondary or tertiary) group. Thus 2-methoxy-2-methylpropane is made from sodium tert-butoxide and methyl bromide (and never from tert-butyl bromide and sodium methoxide). If both groups are tertiary (di-tert-butyl ether) the Williamson reaction cannot be used at all; such ethers can be made only by special methods (e.g. acid-catalysed addition of an alcohol to an alkene).

Extra depth. Higher temperature favours elimination over substitution (elimination has the higher activation entropy), so mild conditions are used; solvents of high polarity but not hydrogen-bond donating (DMF, DMSO) increase the nucleophilicity of the alkoxide and favour SN2. Alkyl tosylates and sulphates may be used instead of halides. The intramolecular version of the Williamson reaction, a halohydrin cyclising with base, is the standard route to epoxides (oxiranes).

Conclusion: Williamson synthesis works well only with methyl and primary halides; secondary and tertiary halides give alkenes by elimination (e.g. (CH₃)₃CBr + CH₃ONa → 2-methylpropene), and aryl halides do not react by SN2, so the hindered group must be placed on the alkoxide, not on the halide.

Concept. 1-Propoxypropane (dipropyl ether, CH₃CH₂CH₂–O–CH₂CH₂CH₃) is a symmetrical ether, which can be made from a primary alcohol by acid-catalysed intermolecular dehydration: two molecules of the alcohol lose one molecule of water. With excess propan-1-ol and concentrated sulphuric acid at 413 K (140 °C) the product is the ether; at the higher temperature of 443 K the same reagents would give propene by elimination. Overall: 2CH₃CH₂CH₂OH → (conc. H₂SO₄, 413 K) CH₃CH₂CH₂OCH₂CH₂CH₃ + H₂O. (The alternative is Williamson synthesis: propan-1-ol + Na → CH₃CH₂CH₂ONa, then + CH₃CH₂CH₂Br.)

Mechanism (SN2 — for primary alcohols).

Step 1 — Protonation of the alcohol. The oxygen of a propan-1-ol molecule accepts a proton from H₂SO₄ and forms the protonated alcohol, an alkyl-oxonium ion (a species with an excellent leaving group): CH₃CH₂CH₂–OH + H⁺ → CH₃CH₂CH₂–O⁺H₂.

Step 2 — Nucleophilic attack (SN2). A second, unprotonated alcohol molecule, using a lone pair on oxygen, attacks the α-carbon of the protonated alcohol from the back side and displaces a molecule of water; this gives a protonated ether: CH₃CH₂CH₂–O⁺H₂ + HO–CH₂CH₂CH₃ → CH₃CH₂CH₂–O⁺(H)–CH₂CH₂CH₃ + H₂O.

Step 3 — Deprotonation. The protonated ether loses a proton to a base (water or HSO₄⁻), giving the ether and regenerating the acid catalyst: CH₃CH₂CH₂–O⁺(H)–CH₂CH₂CH₃ → CH₃CH₂CH₂–O–CH₂CH₂CH₃ + H⁺.

Extra depth. The ether is removed by distillation as it forms (b.p. of dipropyl ether 363 K), so the reaction goes to completion; fresh alcohol is added continuously (the Williamson “continuous etherification” in the manufacture of diethyl ether). The method gives satisfactory yields only for primary alcohols; secondary and tertiary alcohols give alkenes (exercise 7.27). If two different primary alcohols are used, a mixture of three ethers (two symmetrical and one unsymmetrical) is obtained, so the method is practically restricted to symmetrical ethers.

Conclusion: Propan-1-ol is heated with conc. H₂SO₄ at 413 K: one molecule is protonated, a second molecule attacks it by SN2 with loss of water, and deprotonation gives 1-propoxypropane.

Concept. The acid-catalysed dehydration of alcohols to ethers works through an SN2 attack of one alcohol molecule on another, protonated one (see exercise 7.26). But the same reaction mixture — an alcohol, a strong acid and heat — can also lead to an alkene by elimination (E1/E2). Which of the two competing reactions prevails depends on the structure of the alcohol: SN2 needs an unhindered carbon at the back side, while E1 needs a stable carbocation and a β-hydrogen. The structure of the alcohol therefore decides the product.

Secondary and tertiary alcohols give alkenes, not ethers. (1) Steric hindrance: in a secondary or tertiary alkyl-oxonium ion the carbon bearing the leaving group (water) is surrounded by two or three alkyl groups, so a second alcohol molecule cannot approach it from the rear to give SN2 substitution. (2) Ease of carbocation formation: secondary and especially tertiary alcohols ionise readily in acid to give relatively stable 2° or 3° carbocations (R⁺), and the carbocations quickly lose a β-proton (E1) to give alkenes — a faster process than the attack by another weak nucleophile (alcohol). (3) Temperature: the ether-forming temperature of 413 K is already above the dehydration temperature of 3° alcohols (358 K) and close to that of 2° alcohols (440 K), so alkenes are formed under those conditions.

Extra depth. Di-tert-butyl ether cannot be made by this route (nor by Williamson’s). Tertiary ethers with a methyl or primary group are, however, made industrially by the addition of a primary alcohol to an alkene in the presence of an acid catalyst — for example methyl tert-butyl ether (MTBE) from methanol and 2-methylpropene — a reaction that proceeds through the stable tertiary carbocation. Unsymmetrical ethers from two different alcohols are not obtained cleanly either (a mixture of three ethers).

Conclusion: Secondary and tertiary alcohols give alkenes on acid-catalysed dehydration because their carbocations form easily and eliminate a proton, while steric hindrance blocks the SN2 attack required for ether formation.

Concept. Ethers are cleaved by hot concentrated hydrogen iodide (HI), the most reactive of the hydrogen halides, through protonation of the oxygen followed by attack of I⁻. The general equation is R–O–R′ + HI → R–I + R′–OH, and with excess HI the alcohol is further converted to R′–I. Which C–O bond is cleaved depends on the mechanism: (a) SN2 — I⁻ attacks the less hindered (smaller) alkyl carbon; (b) SN1 — if one group is tertiary, benzylic or allylic the C–O bond to that group breaks to give the stable carbocation; (c) in alkyl aryl ethers the O–aryl bond never breaks (partial double bond character, sp² carbon) so a phenol and an alkyl iodide are formed.

(i) 1-Propoxypropane (a symmetrical ether, CH₃CH₂CH₂–O–CH₂CH₂CH₃). Both groups are the same primary group, so the cleavage gives one mole of propyl iodide and one mole of propan-1-ol; with excess HI the alcohol is converted to a second mole of propyl iodide.

(ii) Methoxybenzene (anisole, C₆H₅–O–CH₃). The C(sp²)–O bond is stronger than the CH₃–O bond (partial double-bond character from conjugation of the oxygen lone pair with the ring), and a phenyl cation is unstable, so I⁻ attacks the methyl carbon (SN2) and phenol is released. Phenol does not react further with HI (it does not give iodobenzene).

(iii) Benzyl ethyl ether (C₆H₅CH₂–O–CH₂CH₃). Both carbons attached to oxygen are primary, but the benzylic carbon is much more reactive towards nucleophilic substitution than the ethyl carbon (the benzyl cation, or the SN2 transition state at a benzylic carbon, is stabilised by the ring). So the C₆H₅CH₂–O bond is cleaved: I⁻ ends up on the benzylic carbon, and ethanol is released; with excess HI the ethanol is converted into ethyl iodide.

Conclusion: (i) CH₃CH₂CH₂OCH₂CH₂CH₃ + HI → CH₃CH₂CH₂I + CH₃CH₂CH₂OH; (ii) C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I; (iii) C₆H₅CH₂OC₂H₅ + HI → C₆H₅CH₂I + C₂H₅OH (the benzylic C–O bond breaks).

Concept. In aromatic electrophilic substitution (nitration, halogenation, Friedel–Crafts) the electrophile E⁺ is attacked by the π electrons of the ring to give a positively charged arenium ion; the more the substituent on the ring can stabilise that ion, the faster the reaction (activation) and the more the attack is directed to the positions where the stabilisation is greatest (orientation). In an aryl alkyl ether such as anisole (methoxybenzene, C₆H₅–O–CH₃) the oxygen atom is attached to the ring, and its two lone pairs are available for interaction with the ring’s π system.

(i) Activation of the ring. The oxygen of the alkoxy group exerts two opposing effects: it withdraws electrons through the σ bond (–I effect, since O is more electronegative than C), but it also donates a lone pair into the π system of the ring (the +R or +M effect). For the –OR group the +R effect dominates, so there is a net increase of electron density in the ring. The ring of anisole is therefore more nucleophilic than benzene: nitration, halogenation and Friedel–Crafts reactions of anisole take place under milder conditions than for benzene (for example, bromination in ethanoic acid proceeds without any Lewis-acid catalyst). The group is a strong activating group.

(ii) Ortho–para orientation. The resonance structures of anisole show that the lone-pair donation places the negative charge (extra electron density) specifically on the ortho and para carbons:

Attack by E⁺ at the ortho or para carbon therefore leads to an arenium ion that has four resonance structures, including one in which the oxygen bears the positive charge and every atom has a complete octet (the oxonium-type structure R–O⁺=C) — a very stable contributor. Attack at the meta position gives only three resonance structures and none with the positive charge on the carbon bearing –OR, so the intermediate is less stable and the reaction is slower. The substitution therefore takes place at the ortho and para positions; the para product is usually the major one because the ortho position is hindered by the bulk of the alkoxy group.

Extra depth. The same explanation covers phenol (–OH) and the amino group (–NH₂): in each case an atom with a lone pair is directly attached to the ring (+R dominates –I). The alkoxy group is a slightly weaker activator than –OH or the phenoxide ion because the alkyl group on oxygen makes the lone pair a little less available. Contrast: groups such as –NO₂, –CN, –COOH which withdraw electrons by both –I and –R effects deactivate the ring and direct incoming groups to the meta position.

Conclusion: The alkoxy group donates a lone pair into the ring by the +R effect (which outweighs –I), activating the ring and increasing electron density at the ortho and para positions; hence it is an activating, ortho/para-directing group.

Concept. Methoxymethane (dimethyl ether, CH₃–O–CH₃) is a symmetrical ether with two primary (methyl) groups. Ethers are cleaved by hot concentrated hydrogen iodide, and because both alkyl groups are methyl — which cannot form a carbocation at all — the reaction must be an SN2 reaction. The ether oxygen is a weak base, so the acid first protonates it (turning the poor leaving group CH₃O⁻ into the good leaving group CH₃OH); the iodide ion, a very good nucleophile, then attacks one of the methyl carbons from the back side.

Step 1 — Protonation of the ether oxygen (fast, reversible). CH₃–O–CH₃ + H–I → CH₃–O⁺(H)–CH₃ + I⁻ (a dimethyl-oxonium ion).

Step 2 — SN2 attack of iodide (slow, rate-determining). I⁻ attacks a methyl carbon from the side opposite to the C–O bond; the C–O bond breaks as the C–I bond forms, and methanol leaves: I⁻ + CH₃–O⁺(H)–CH₃ → CH₃–I + CH₃–OH.

Step 3 — Conversion of methanol to methyl iodide (with excess HI). The methanol produced is itself protonated by HI and undergoes the same SN2 reaction: CH₃OH + HI → CH₃–O⁺H₂ + I⁻ → CH₃I + H₂O.

Overall: CH₃OCH₃ + 2HI → 2CH₃I + H₂O (with excess HI at high temperature); with only one mole of HI the products are CH₃I and CH₃OH.

Extra depth. The order of reactivity of the hydrogen halides is HI > HBr > HCl: HI is the strongest acid (the best proton donor) and I⁻ the best nucleophile and best leaving-group partner; HCl rarely cleaves simple ethers. Contrast the mechanism with that of an ether containing a tertiary group (e.g. CH₃–O–C(CH₃)₃), which cleaves by SN1 with the tertiary halide, (CH₃)₃CI, and CH₃OH as the products (intext 7.12 iv).

Conclusion: HI protonates the ether oxygen; iodide then attacks a methyl carbon by SN2 to give CH₃I and CH₃OH, and with excess HI the methanol is converted into a second molecule of CH₃I.

Concept. The four reactions are the standard electrophilic substitutions of anisole (methoxybenzene). The methoxy group is an activating, ortho/para-directing group (exercise 7.29), so every reaction gives a mixture of the ortho and para substituted anisoles in which the para isomer predominates (the ortho position is hindered by the –OCH₃ group). Because the ring is activated, milder conditions than for benzene are enough — in (iii) no catalyst is needed at all. In the Friedel–Crafts reactions the Lewis acid anhydrous AlCl₃ generates the electrophile (a carbocation or acylium ion) from the halide.

(i) Friedel–Crafts alkylation (CH₃Cl, anhydrous AlCl₃, CS₂ solvent): the electrophile is CH₃⁺ (from CH₃Cl–AlCl₃); products: 4-methoxytoluene (major) and 2-methoxytoluene (minor).

(ii) Nitration (conc. HNO₃ + conc. H₂SO₄): the electrophile is the nitronium ion, NO₂⁺; products: 4-nitroanisole (major) and 2-nitroanisole (minor).

(iii) Bromination in ethanoic acid (Br₂ in CH₃COOH, no Fe or FeBr₃ needed): the ring is so reactive that the polarised Br₂ is a sufficient electrophile; products: 4-bromoanisole (major, about 90 %) and 2-bromoanisole (minor).

(iv) Friedel–Crafts acylation (acetylation) (CH₃COCl, anhydrous AlCl₃): the electrophile is the acylium ion CH₃–C⁺=O; products: 4-methoxyacetophenone (major) and 2-methoxyacetophenone (minor).

Extra depth. The para isomer is usually obtained in greater amount because (a) the ortho positions are shielded by the methoxy group and (b) the para arenium ion has a slightly lower energy. In Friedel–Crafts alkylation polyalkylation is a problem because the alkyl-substituted product is even more reactive than anisole itself, whereas acylation stops at the monoketone since the acyl group deactivates the ring; also at least one equivalent of AlCl₃ is required for acylation because it complexes with the ketone product. In the Friedel–Crafts reaction with an ether there is also a risk of O-demethylation (cleavage of the aryl–OCH₃ ether by AlCl₃), so low temperatures are used.

Conclusion: Anisole gives (i) 4-methoxytoluene + 2-methoxytoluene, (ii) 4-nitro- + 2-nitroanisole, (iii) 4-bromo- + 2-bromoanisole and (iv) 4-methoxy- + 2-methoxyacetophenone, the para product being major in each case.

Concept. The method asked for is the acid-catalysed hydration of alkenes: an alkene + H₂O in the presence of dilute H₂SO₄ gives an alcohol, with the –OH adding to the carbon of the double bond that carries fewer hydrogens (Markovnikov addition, via the more stable carbocation). To plan the synthesis work backwards: in the target alcohol the carbinol carbon (C–OH) was one of the two carbons of the C=C; the other doubly bonded carbon is an adjacent carbon that now carries an extra hydrogen. Removing H₂O by taking H from a neighbouring carbon and OH from the carbinol carbon gives the required alkene(s). For tertiary alcohols several alkenes lead to the same product; for secondary alcohols the correct choice is an alkene for which Markovnikov addition gives the desired alcohol rather than an isomer.

(i) 1-Methylcyclohexan-1-ol (3°). The carbinol carbon (ring carbon with CH₃ and OH) was part of the double bond: the alkene is 1-methylcyclohexene (double bond inside the ring, at the carbon carrying CH₃) or methylenecyclohexane (double bond to the CH₂). H⁺ adds to the less substituted carbon, forming the tertiary 1-methylcyclohexyl cation, and water gives the alcohol.

(ii) 4-Methylheptan-4-ol (3°). The carbinol carbon (C4) carries CH₃ and two propyl groups. Removing H₂O gives 4-methylhept-3-ene (CH₃CH₂CH=C(CH₃)CH₂CH₂CH₃); since the two propyl chains are identical there is only one such alkene with the double bond in the chain (the other possibility is 2-propylpent-1-ene, CH₂=C(CH₂CH₂CH₃)₂). Protonation of the carbon C3 gives the tertiary carbocation at C4, and water adds there.

(iii) Pentan-2-ol (2°). The –OH is on C2; the alkene is pent-1-ene, CH₂=CH–CH₂CH₂CH₃: H⁺ adds to the terminal CH₂ (carbon with more hydrogens) to give the secondary carbocation at C2, and water adds there (Markovnikov). Using pent-2-ene would give a mixture of pentan-2-ol and pentan-3-ol, so pent-1-ene is the appropriate alkene.

(iv) 2-Cyclohexylbutan-2-ol (3°). The carbinol carbon (C2 of the butane chain) carries a cyclohexyl group, CH₃ and C₂H₅. The alkenes that give this alcohol are 2-cyclohexylbut-2-ene (CH₃–C(C₆H₁₁)=CH–CH₃) or 2-cyclohexylbut-1-ene (CH₂=C(C₆H₁₁)–CH₂CH₃); either one is protonated to give the same tertiary carbocation C₆H₁₁–C⁺(CH₃)–CH₂CH₃, which adds water.

Conclusion: By acid-catalysed hydration (H₂O/H⁺): (i) from 1-methylcyclohexene; (ii) from 4-methylhept-3-ene; (iii) from pent-1-ene; (iv) from 2-cyclohexylbut-2-ene (or 2-cyclohexylbut-1-ene).

Concept. The starting alcohol, 3-methylbutan-2-ol, is a secondary alcohol, but the product is a tertiary bromide, 2-bromo-2-methylbutane — the bromine is not on the carbon that carried the –OH. This is possible because the reaction goes by SN1: the C–O bond breaks first to give a carbocation, and a carbocation that can rearrange to a more stable one does so before the nucleophile attacks. The rearrangement takes place by a 1,2-hydride shift: a hydrogen atom, with its bonding pair of electrons, moves from the adjacent carbon to the positively charged carbon; the positive charge thus moves to the carbon that lost the hydrogen. The driving force is the greater stability of the tertiary carbocation (three alkyl groups, nine hyperconjugating C–H bonds) compared with the secondary one.

Step I — Protonation of the –OH group. The hydroxyl oxygen picks up a proton from HBr to form the protonated alcohol (an oxonium ion with a good leaving group, H₂O): (CH₃)₂CH–CH(OH)–CH₃ + H⁺ → (CH₃)₂CH–CH(O⁺H₂)–CH₃.

Step II — Loss of water: formation of a secondary carbocation (slow). The C–O bond breaks heterolytically and a water molecule leaves, forming the 2° carbocation: (CH₃)₂CH–CH(O⁺H₂)–CH₃ → (CH₃)₂CH–C⁺H–CH₃ + H₂O.

Step III — 1,2-Hydride shift: rearrangement to the tertiary carbocation (fast). The hydrogen on C3 (the tertiary C–H carbon next to the cationic C2) migrates with its electron pair to C2. C2 thereby becomes a CH₂ group, and the positive charge moves to C3, which now carries two methyl groups and the CH₂ group: (CH₃)₂CH–C⁺H–CH₃ becomes (CH₃)₂C⁺–CH₂–CH₃ (a tertiary carbocation, the 2-methylbutan-2-yl cation).

Step IV — Attack of bromide ion. The nucleophile Br⁻ attacks the positively charged tertiary carbon, giving the final product: (CH₃)₂C⁺–CH₂CH₃ + Br⁻ → (CH₃)₂C(Br)–CH₂CH₃ (2-bromo-2-methylbutane).

Extra depth. A small amount of the “unrearranged” product, 2-bromo-3-methylbutane, may form from the secondary carbocation before it rearranges; but the 3° cation is so much more stable (by about 12–16 kJ mol⁻¹) and the shift so fast (the energy barrier of a hydride shift is only a few kJ mol⁻¹) that the rearranged tertiary bromide is the major product. This is the same type of rearrangement that explains the dehydration of butan-1-ol to but-2-ene (intext 7.7) and the formation of 2-methylbutan-2-ol derivatives in many SN1/E1 reactions of branched alcohols. Rearrangements are the signature of carbocation intermediates; reactions that go by SN2 (primary alcohols with HBr) never rearrange.

Conclusion: The protonated alcohol loses water to give a secondary carbocation, which rearranges by a 1,2-hydride shift to the more stable tertiary carbocation; Br⁻ then attacks it to give 2-bromo-2-methylbutane.