Chapter 7: Alcohols, Phenols and Ethers

An alcohol is classified by counting how many carbon atoms are directly bonded to the carbon that carries the $$\mathrm{-OH}$$ group: primary (1°) — one carbon, secondary (2°) — two carbons, tertiary (3°) — three carbons.

In $$\mathrm{(CH_3)_3C-CH_2OH}$$, the $$\mathrm{-OH}$$ is on the $$\mathrm{-CH_2-}$$ carbon. This carbon is bonded to only one other carbon atom (the central quaternary carbon).

Hence it is a primary (1°) alcohol.

In $$\mathrm{H_2C=CH-CH_2OH}$$, the $$\mathrm{-OH}$$ is on the $$\mathrm{-CH_2-}$$ carbon, which is bonded to only one other carbon (the $$\mathrm{=CH-}$$ carbon).

Hence it is a primary (1°) alcohol. (It is also an allylic alcohol, since the $$\mathrm{-OH}$$ carbon is attached to a carbon of a $$\mathrm{C=C}$$ double bond.)

In $$\mathrm{CH_3-CH_2-CH_2-OH}$$, the $$\mathrm{-OH}$$ is on the terminal $$\mathrm{-CH_2-}$$ carbon, which is bonded to only one other carbon.

Hence it is a primary (1°) alcohol.

In $$\mathrm{C_6H_5-CH(OH)-CH_3}$$, the $$\mathrm{-OH}$$ carbon is bonded to two other carbons — one of the phenyl ring and one of the methyl group.

Hence it is a secondary (2°) alcohol.

In $$\mathrm{C_6H_5-CH_2-CH(OH)-CH_3}$$, the $$\mathrm{-OH}$$ carbon is bonded to two other carbons — the $$\mathrm{-CH_2-}$$ carbon and the $$\mathrm{-CH_3}$$ carbon.

Hence it is a secondary (2°) alcohol.

In $$\mathrm{C_6H_5-CH=CH-C(CH_3)_2-OH}$$, the $$\mathrm{-OH}$$ carbon is bonded to three other carbons — the two methyl carbons and the $$\mathrm{=CH-}$$ carbon.

Hence it is a tertiary (3°) alcohol. (It is also allylic, since the $$\mathrm{-OH}$$ carbon is attached to a $$\mathrm{C=C}$$ double bond.)

The principal characteristic group $$\mathrm{-OH}$$ must lie on the parent chain, so the parent chain must include the carbon of the $$\mathrm{-CH_2OH}$$ group. That carbon is terminal, so it becomes C-1.

The longest chain that starts at $$\mathrm{-CH_2OH}$$ runs:

$$\mathrm{\underset{1}{HOCH_2}-\underset{2}{CH}-\underset{3}{CH}-\underset{4}{CH_2}-\underset{5}{CH_3}}$$

This is a 5-carbon chain $$\Rightarrow$$ pentan-1-ol.

Substituents on this chain:

• C-2 carries a $$\mathrm{-CH(CH_3)_2}$$ group $$\Rightarrow$$ isopropyl (propan-2-yl).
• C-3 carries a $$\mathrm{-CH_2Cl}$$ group $$\Rightarrow$$ chloromethyl.

Citing substituents alphabetically (chloromethyl before isopropyl):

Name: 3-(Chloromethyl)-2-isopropylpentan-1-ol [3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol].

The molecule has two $$\mathrm{-OH}$$ groups (a diol): one on the chain and one in the $$\mathrm{-CH_2OH}$$ branch. The parent chain must contain both hydroxyl-bearing carbons.

The longest chain that includes both $$\mathrm{-OH}$$ carbons is:

$$\mathrm{\underset{1}{HOCH_2}-\underset{2}{CH}-\underset{3}{CH(OH)}-\underset{4}{CH_2}-\underset{5}{CH}-\underset{6}{CH_3}}$$

This is a 6-carbon chain $$\Rightarrow$$ hexane-diol. Numbering from the $$\mathrm{-CH_2OH}$$ end gives the $$\mathrm{-OH}$$ groups locants $$\{1,3\}$$; from the other end they would be $$\{4,6\}$$. Since $$\{1,3\}$$ is lower, we number from the $$\mathrm{-CH_2OH}$$ end.

$$\mathrm{-OH}$$ groups at C-1 and C-3 $$\Rightarrow$$ hexane-1,3-diol. Methyl substituents at C-2 and C-5.

Name: 2,5-Dimethylhexane-1,3-diol.

The ring is named as cyclohexanol because $$\mathrm{-OH}$$ is the principal characteristic group; its carbon is C-1.

The $$\mathrm{-Br}$$ atom occupies the 1,3-position relative to $$\mathrm{-OH}$$, i.e. C-3.

Name: 3-Bromocyclohexan-1-ol (3-bromocyclohexanol).

The compound is $$\mathrm{H_2C=CH-CH(OH)-CH_2-CH_2-CH_3}$$. The parent chain has 6 carbons and contains both the $$\mathrm{C=C}$$ double bond and the $$\mathrm{-OH}$$ group, so the parent is a hexenol.

The $$\mathrm{-OH}$$ group is the principal characteristic group. It must receive the lowest possible locant, and this requirement takes priority over the locant given to the double bond.

The $$\mathrm{-OH}$$ bearing carbon is the third carbon counting from the $$\mathrm{CH_2=}$$ end and the fourth counting from the $$\mathrm{CH_3}$$ end. Its locant can therefore only be 3 or 4 — it can never be 1, because that carbon is not at the end of the chain. The lower value, 3, is selected, so the chain is numbered from the $$\mathrm{CH_2=}$$ end:

$$\mathrm{\underset{1}{CH_2}=\underset{2}{CH}-\underset{3}{CH(OH)}-\underset{4}{CH_2}-\underset{5}{CH_2}-\underset{6}{CH_3}}$$

This gives $$\mathrm{-OH}$$ at C-3 and the double bond between C-1 and C-2. (Numbering from the other end would place $$\mathrm{-OH}$$ at the higher locant 4, which is not permitted.)

Name: Hex-1-en-3-ol.

The longest chain has 4 carbons. Numbering so that $$\mathrm{-OH}$$ gets the lowest locant (from the $$\mathrm{-CH_2OH}$$ end):

$$\mathrm{\underset{1}{HOCH_2}-\underset{2}{C(Br)}=\underset{3}{C(CH_3)}-\underset{4}{CH_3}}$$

$$\mathrm{-OH}$$ at C-1; double bond between C-2 and C-3; $$\mathrm{Br}$$ at C-2; $$\mathrm{CH_3}$$ at C-3.

Name: 2-Bromo-3-methylbut-2-en-1-ol.

Butanal is the aldehyde $$\mathrm{CH_3CH_2CH_2CHO}$$. Catalytic reduction adds $$\mathrm{H_2}$$ across the $$\mathrm{C=O}$$ bond in the presence of a metal catalyst, converting the aldehyde to a primary alcohol:

$$\mathrm{CH_3CH_2CH_2CHO \xrightarrow{H_2/\,Ni} CH_3CH_2CH_2CH_2OH}$$

The product is $$\mathrm{CH_3CH_2CH_2CH_2OH}$$.

Acid-catalysed hydration adds a molecule of water across the double bond. Addition follows Markovnikov's rule: the $$\mathrm{H}$$ adds to the double-bond carbon already bearing more hydrogens, and $$\mathrm{-OH}$$ to the other, because the more stable secondary carbocation is formed.

$$\mathrm{CH_3-CH=CH_2 \xrightarrow[\text{dil. }H_2SO_4]{H_2O} CH_3-CH(OH)-CH_3}$$

The product is propan-2-ol (a secondary alcohol).

Propanone is the ketone $$\mathrm{(CH_3)_2C=O}$$. The Grignard reagent $$\mathrm{CH_3MgBr}$$ adds its alkyl group to the carbonyl carbon, giving an alkoxide (magnesium salt). Subsequent hydrolysis gives the alcohol.

$$\mathrm{(CH_3)_2C=O + CH_3MgBr \rightarrow (CH_3)_3C-OMgBr}$$

$$\mathrm{(CH_3)_3C-OMgBr \xrightarrow{H_2O/H^+} (CH_3)_3C-OH + Mg(OH)Br}$$

A ketone with a Grignard reagent yields a tertiary alcohol. The product is $$\mathrm{(CH_3)_3C-OH}$$.

A Grignard reagent adds to methanal ($$\mathrm{HCHO}$$) and, on hydrolysis, gives a primary alcohol containing one carbon more than the Grignard alkyl group:

$$\mathrm{R-MgX + HCHO \rightarrow R-CH_2-OMgX \xrightarrow{H_2O/H^+} R-CH_2OH}$$

In the target $$\mathrm{(CH_3)_2CH-CH_2OH}$$, the $$\mathrm{-CH_2OH}$$ carbon comes from methanal, so $$\mathrm{R = (CH_3)_2CH-}$$ (isopropyl). The required Grignard reagent is isopropylmagnesium bromide.

$$\mathrm{(CH_3)_2CH-MgBr + HCHO \rightarrow (CH_3)_2CH-CH_2-OMgBr}$$

$$\mathrm{(CH_3)_2CH-CH_2-OMgBr \xrightarrow{H_2O/H^+} (CH_3)_2CH-CH_2OH}$$

In cyclohexylmethanol, $$\mathrm{C_6H_{11}-CH_2OH}$$, the $$\mathrm{-CH_2OH}$$ carbon is supplied by methanal. Hence $$\mathrm{R = C_6H_{11}-}$$ (cyclohexyl), and the required Grignard reagent is cyclohexylmagnesium bromide.

$$\mathrm{C_6H_{11}-MgBr + HCHO \rightarrow C_6H_{11}-CH_2-OMgBr}$$

$$\mathrm{C_6H_{11}-CH_2-OMgBr \xrightarrow{H_2O/H^+} C_6H_{11}-CH_2OH}$$

The product is cyclohexylmethanol.

This is acid-catalysed hydration of propene. Water adds across the double bond following Markovnikov's rule — $$\mathrm{-OH}$$ goes to the more substituted carbon (via the more stable secondary carbocation).

$$\mathrm{CH_3-CH=CH_2 \xrightarrow{H_2O/H^+} CH_3-CH(OH)-CH_3}$$

The product is propan-2-ol.

Sodium borohydride, $$\mathrm{NaBH_4}$$, is a mild reducing agent. It reduces aldehyde and ketone carbonyls to alcohols, but it does not reduce ester ($$\mathrm{-COOCH_3}$$) groups.

Therefore, only the ring ketone $$\mathrm{C=O}$$ is reduced to a secondary alcohol ($$\mathrm{>CH-OH}$$); the $$\mathrm{-CH_2COOCH_3}$$ ester group remains unchanged.

Product: the cyclohexanone ring becomes a cyclohexanol ring (i.e. the carbonyl carbon now bears $$\mathrm{H}$$ and $$\mathrm{OH}$$), still carrying the $$\mathrm{-CH_2-C(=O)-OCH_3}$$ group on the adjacent carbon — methyl 2-(2-hydroxycyclohexyl)acetate.

$$\mathrm{NaBH_4}$$ reduces the aldehyde group $$\mathrm{-CHO}$$ to a primary alcohol group $$\mathrm{-CH_2OH}$$.

$$\mathrm{CH_3-CH_2-CH(CH_3)-CHO \xrightarrow{NaBH_4} CH_3-CH_2-CH(CH_3)-CH_2OH}$$

The product is 2-methylbutan-1-ol.

All six compounds are alcohols, so all show intermolecular hydrogen bonding. Two factors decide the boiling points:

• Molecular mass: boiling point rises as the carbon chain lengthens (greater van der Waals forces and larger surface area).
• Branching: for isomers, a more branched/compact molecule has a smaller surface area, so weaker van der Waals forces and a lower boiling point. Butan-2-ol is more branched than butan-1-ol, so it boils lower.

Ordering by carbon number, and placing butan-2-ol below butan-1-ol:

methanol (1C) < ethanol (2C) < propan-1-ol (3C) < butan-2-ol < butan-1-ol (4C) < pentan-1-ol (5C).

Here the four compounds belong to different families, so the nature of intermolecular forces decides the order:

• n-Butane ($$\mathrm{C_4H_{10}}$$): a non-polar hydrocarbon — only weak van der Waals forces. Lowest boiling point.
• Ethoxyethane ($$\mathrm{C_2H_5OC_2H_5}$$): a polar ether — weak dipole-dipole forces but no intermolecular hydrogen bonding among ether molecules.
• Pentanal ($$\mathrm{C_4H_9CHO}$$): a polar aldehyde with a strongly polar $$\mathrm{C=O}$$ group — appreciable dipole-dipole attraction, but still no hydrogen bonding.
• Pentan-1-ol ($$\mathrm{C_5H_{11}OH}$$): an alcohol with strong intermolecular hydrogen bonding. Highest boiling point.

Increasing boiling point:

n-butane < ethoxyethane < pentanal < pentan-1-ol.

In 3-methylphenol the ring carries $$\mathrm{-OH}$$ at C-1 and $$\mathrm{-CH_3}$$ at C-3. Both groups are ortho/para-directing, and their directing effects reinforce each other.

Relative to $$\mathrm{-OH}$$ (C-1), the ortho/para positions are C-2, C-4, C-6. Relative to $$\mathrm{-CH_3}$$ (C-3), the ortho/para positions are C-2, C-4, C-6 as well. So $$\mathrm{-NO_2}$$ is directed to C-2, C-4 and C-6.

Position C-2 lies between the two substituents and is sterically crowded, so the major mononitration products carry $$\mathrm{-NO_2}$$ at C-4 (para to $$\mathrm{-OH}$$) and C-6 (ortho to $$\mathrm{-OH}$$).

Major products: 3-methyl-4-nitrophenol and 3-methyl-6-nitrophenol (a small amount of 3-methyl-2-nitrophenol also forms).

On dinitration, two $$\mathrm{-NO_2}$$ groups enter the positions activated by both $$\mathrm{-OH}$$ (C-1) and $$\mathrm{-CH_3}$$ (C-3) — i.e. positions that are ortho/para to $$\mathrm{-OH}$$.

The two $$\mathrm{-NO_2}$$ groups go to C-4 (para to $$\mathrm{-OH}$$) and C-6 (ortho to $$\mathrm{-OH}$$), both of which are also activated by $$\mathrm{-CH_3}$$.

Major product: 3-methyl-4,6-dinitrophenol.

Phenyl methanoate is the ester $$\mathrm{H-COO-C_6H_5}$$ (phenyl formate). The acyloxy group $$\mathrm{-O-CO-H}$$ is attached to the ring.

The oxygen attached to the ring still carries lone pairs that it can donate by resonance, so $$\mathrm{-OCOH}$$ is an ortho/para-directing group (though, because the lone pair is partly drawn towards the carbonyl, it is much less strongly activating than $$\mathrm{-OH}$$).

Mononitration therefore gives the ortho and para nitro products, the para isomer being major:

Major product: 4-nitrophenyl methanoate (with some 2-nitrophenyl methanoate).

Butan-1-ol, $$\mathrm{CH_3CH_2CH_2CH_2OH}$$, is a primary alcohol. With each reagent the $$\mathrm{-OH}$$ group is replaced by a halogen.

(a) With $$\mathrm{HCl}$$ in the presence of anhydrous $$\mathrm{ZnCl_2}$$ (Lucas reagent):

$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow[\;]{HCl,\,ZnCl_2} CH_3CH_2CH_2CH_2Cl}$$ (1-chlorobutane)

(b) With $$\mathrm{HBr}$$:

$$\mathrm{CH_3CH_2CH_2CH_2OH + HBr \rightarrow CH_3CH_2CH_2CH_2Br + H_2O}$$ (1-bromobutane)

(c) With $$\mathrm{SOCl_2}$$ (thionyl chloride):

$$\mathrm{CH_3CH_2CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2CH_2Cl + SO_2\uparrow + HCl\uparrow}$$ (1-chlorobutane)

2-Methylbutan-2-ol, $$\mathrm{CH_3-C(OH)(CH_3)-CH_2CH_3}$$, is a tertiary alcohol; the $$\mathrm{-OH}$$ group is replaced by halogen in each case.

(a) With $$\mathrm{HCl/ZnCl_2}$$ (a tertiary alcohol reacts at once, even without $$\mathrm{ZnCl_2}$$):

$$\mathrm{CH_3-C(OH)(CH_3)-CH_2CH_3 \xrightarrow{HCl} CH_3-C(Cl)(CH_3)-CH_2CH_3}$$ (2-chloro-2-methylbutane)

(b) With $$\mathrm{HBr}$$:

$$\mathrm{CH_3-C(OH)(CH_3)-CH_2CH_3 + HBr \rightarrow CH_3-C(Br)(CH_3)-CH_2CH_3 + H_2O}$$ (2-bromo-2-methylbutane)

(c) With $$\mathrm{SOCl_2}$$:

$$\mathrm{CH_3-C(OH)(CH_3)-CH_2CH_3 + SOCl_2 \rightarrow CH_3-C(Cl)(CH_3)-CH_2CH_3 + SO_2 + HCl}$$ (2-chloro-2-methylbutane)

Acid-catalysed dehydration follows Saytzeff's rule: the major alkene is the more highly substituted (more stable) one, formed by loss of a $$\beta$$-hydrogen from the carbon with fewer hydrogens.

In 1-methylcyclohexanol, $$\mathrm{-OH}$$ is on the ring carbon that also bears the $$\mathrm{-CH_3}$$. Loss of water gives a tertiary carbocation on that ring carbon. A $$\beta$$-hydrogen can be removed either from a ring $$\mathrm{-CH_2-}$$ or from the $$\mathrm{-CH_3}$$ group.

• Loss of a ring $$\beta$$-H gives 1-methylcyclohex-1-ene — a trisubstituted (more stable) double bond.
• Loss of an H from $$\mathrm{-CH_3}$$ gives methylenecyclohexane — a less substituted double bond.

Hence the major product is 1-methylcyclohex-1-ene (1-methylcyclohexene).

Butan-1-ol, $$\mathrm{CH_3CH_2CH_2CH_2OH}$$, on protonation and loss of water would give a primary carbocation, which is unstable. It rearranges by a 1,2-hydride shift to the more stable secondary carbocation $$\mathrm{CH_3CH_2\overset{+}{C}HCH_3}$$.

Loss of a $$\beta$$-hydrogen from this secondary carbocation, following Saytzeff's rule, gives the more substituted alkene:

$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3-CH=CH-CH_3}$$

Hence the major product is but-2-ene (a disubstituted alkene), rather than the less stable but-1-ene.

On treating phenol with chloroform in the presence of aqueous sodium hydroxide at 340 K, a $$\mathrm{-CHO}$$ group is introduced at the ortho position of the ring. The reactive electrophile is dichlorocarbene, $$\mathrm{:CCl_2}$$, generated from $$\mathrm{CHCl_3}$$ and $$\mathrm{NaOH}$$.

The intermediate is an o-substituted benzal chloride, which on alkaline hydrolysis gives the salt of salicylaldehyde; acidification then liberates salicylaldehyde.

$$\mathrm{C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\,K} \underset{\text{(sodium salt)}}{o\text{-}NaO\text{-}C_6H_4\text{-}CHO} + 3NaCl + 2H_2O}$$

$$\mathrm{o\text{-}NaO\text{-}C_6H_4\text{-}CHO \xrightarrow{H^+} \underset{\text{salicylaldehyde}}{o\text{-}HO\text{-}C_6H_4\text{-}CHO}}$$

The product is 2-hydroxybenzaldehyde (salicylaldehyde).

Phenol is first converted to sodium phenoxide by reaction with $$\mathrm{NaOH}$$. The phenoxide ion is even more reactive than phenol towards electrophilic substitution, so it reacts with the weak electrophile carbon dioxide.

$$\mathrm{C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O}$$

Sodium phenoxide is treated with $$\mathrm{CO_2}$$ under pressure (about 4-7 atm) at 400 K, giving sodium salicylate; acidification yields salicylic acid.

$$\mathrm{C_6H_5ONa + CO_2 \xrightarrow{400\,K,\,pressure} \underset{\text{sodium salicylate}}{o\text{-}HO\text{-}C_6H_4\text{-}COONa}}$$

$$\mathrm{o\text{-}HO\text{-}C_6H_4\text{-}COONa \xrightarrow{H^+} \underset{\text{salicylic acid}}{o\text{-}HO\text{-}C_6H_4\text{-}COOH}}$$

The product is 2-hydroxybenzoic acid (salicylic acid).

Williamson synthesis proceeds by an $$\mathrm{S_N2}$$ attack of the alkoxide ion on the alkyl halide. Here the alkyl halide is tert-butyl chloride, $$\mathrm{(CH_3)_3C-Cl}$$ — a tertiary halide.

A tertiary halide is too hindered for $$\mathrm{S_N2}$$ substitution. Instead, the ethoxide ion ($$\mathrm{C_2H_5O^-}$$) acts as a strong base and abstracts a $$\beta$$-hydrogen, causing elimination (dehydrohalogenation).

$$\mathrm{(CH_3)_3C-Cl + C_2H_5O^-Na^+ \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH + NaCl}$$

So the major product is the alkene 2-methylpropene (isobutylene), not the ether.

For Williamson synthesis to work well, the alkyl halide should be primary (best for $$\mathrm{S_N2}$$), while the alkoxide may be of any class.

So we must reverse the roles: take the tertiary group as the alkoxide and the ethyl group as the (primary) halide.

$$\mathrm{(CH_3)_3C-ONa + C_2H_5-Br \rightarrow (CH_3)_3C-O-C_2H_5 + NaBr}$$

Here sodium t-butoxide attacks the primary halide ethyl bromide by $$\mathrm{S_N2}$$, giving t-butyl ethyl ether smoothly.

When an ether is heated with HI, the $$\mathrm{C-O}$$ bond is cleaved. With one mole of HI, the alkyl iodide formed corresponds to the smaller, less hindered alkyl group (formed by $$\mathrm{S_N2}$$ attack of $$\mathrm{I^-}$$ on the protonated ether), while the bulkier group is released as the alcohol.

Both groups here are primary, so $$\mathrm{I^-}$$ attacks the smaller one — the ethyl group:

$$\mathrm{CH_3CH_2CH(CH_3)CH_2-O-CH_2CH_3 + HI \rightarrow CH_3CH_2CH(CH_3)CH_2OH + CH_3CH_2I}$$

Products: 2-methylbutan-1-ol and iodoethane (ethyl iodide).

This ether has one primary group (n-propyl) and one tertiary group ($$\mathrm{-C(CH_3)_2CH_2CH_3}$$).

When one alkyl group is tertiary, cleavage occurs by an $$\mathrm{S_N1}$$ path: the protonated ether breaks to give the stable tertiary carbocation, which then combines with $$\mathrm{I^-}$$. The other (primary) group leaves as the alcohol.

$$\mathrm{CH_3CH_2CH_2-O-C(CH_3)_2CH_2CH_3 + HI \rightarrow CH_3CH_2CH_2OH + (CH_3)_2C(I)CH_2CH_3}$$

Products: propan-1-ol and 2-iodo-2-methylbutane.

In an alkyl aryl ether, the $$\mathrm{O-C(aryl)}$$ bond is not cleaved by HI, because the aryl carbon is $$\mathrm{sp^2}$$ and cannot undergo nucleophilic substitution; a phenol is therefore one product.

Cleavage occurs at the $$\mathrm{O-CH_2(benzyl)}$$ bond. The benzyl group is released and combines with $$\mathrm{I^-}$$ to give benzyl iodide; the aryl part is set free as phenol.

$$\mathrm{C_6H_5-CH_2-O-C_6H_5 + HI \rightarrow C_6H_5-CH_2I + C_6H_5-OH}$$

Products: benzyl iodide and phenol.

In this set the halogen is attached to the benzene ring — it is an aryl halide (haloarene). The $$\mathrm{C-Br}$$ bond of a haloarene has partial double-bond character (due to resonance between the lone pairs of the halogen and the ring) and the carbon is $$\mathrm{sp^2}$$. Hence aryl halides do not undergo the easy $$\mathrm{S_N2}$$ substitution needed in Williamson synthesis.

Therefore this set, $$\mathrm{p\text{-}Br\text{-}C_6H_4\text{-}NO_2 + CH_3ONa}$$, is not appropriate — methoxide cannot readily displace the ring-bound bromine.

In this set the alkylating agent is $$\mathrm{CH_3Br}$$ — a primary alkyl halide, which readily undergoes $$\mathrm{S_N2}$$ substitution. The other reactant, sodium p-nitrophenoxide, supplies the aryloxide ion.

The phenoxide oxygen attacks the methyl carbon of $$\mathrm{CH_3Br}$$, displacing bromide:

$$\mathrm{p\text{-}NaO\text{-}C_6H_4\text{-}NO_2 + CH_3Br \rightarrow p\text{-}CH_3O\text{-}C_6H_4\text{-}NO_2 + NaBr}$$

This cleanly gives 1-methoxy-4-nitrobenzene. Hence set (ii) is the appropriate set of reactants.

Methyl propyl ether is protonated by HBr; then bromide ion attacks by $$\mathrm{S_N2}$$ at the smaller, less hindered alkyl group — the methyl group. The propyl group is released as the alcohol.

$$\mathrm{CH_3CH_2CH_2-O-CH_3 + HBr \rightarrow CH_3CH_2CH_2OH + CH_3Br}$$

Products: propan-1-ol and bromomethane (methyl bromide).

In an aryl alkyl ether the $$\mathrm{O-C(aryl)}$$ bond is not broken (the aryl carbon is $$\mathrm{sp^2}$$ and resists nucleophilic substitution), so a phenol is one product.

Cleavage occurs at the $$\mathrm{O-C_2H_5}$$ bond; bromide attacks the ethyl carbon:

$$\mathrm{C_6H_5-O-C_2H_5 + HBr \rightarrow C_6H_5-OH + C_2H_5Br}$$

Products: phenol and bromoethane (ethyl bromide).

A mixture of conc. $$\mathrm{HNO_3}$$ and conc. $$\mathrm{H_2SO_4}$$ nitrates the ring. The ethoxy group $$\mathrm{-OC_2H_5}$$ donates its lone pair into the ring, so it is activating and ortho/para-directing.

Nitration therefore occurs at the ortho and para positions, the para product being major:

$$\mathrm{C_6H_5-OC_2H_5 \xrightarrow[\text{conc. }H_2SO_4]{\text{conc. }HNO_3} o\text{- and } p\text{-}O_2N\text{-}C_6H_4\text{-}OC_2H_5}$$

Products: 2-nitrophenetole (1-ethoxy-2-nitrobenzene) and, mainly, 4-nitrophenetole (1-ethoxy-4-nitrobenzene).

This ether has a tertiary group, $$\mathrm{(CH_3)_3C-}$$, and a primary ethyl group. After protonation of the ether oxygen, cleavage follows an $$\mathrm{S_N1}$$ path: the bond breaks to give the stable tertiary carbocation $$\mathrm{(CH_3)_3C^+}$$, which then combines with $$\mathrm{I^-}$$.

The ethyl group leaves with the oxygen as ethanol.

$$\mathrm{(CH_3)_3C-OC_2H_5 + HI \rightarrow (CH_3)_3C-I + C_2H_5OH}$$

Products: 2-iodo-2-methylpropane (tert-butyl iodide) and ethanol.

The longest chain has 5 carbons (pentane) and bears $$\mathrm{-OH}$$ on the middle carbon.

$$\mathrm{-OH}$$ is the principal group. From either end its locant is 3, so we choose the numbering that gives the methyl substituents the lower set of locants. Numbering so that the $$\mathrm{C(CH_3)_2}$$ carbon is C-2:

$$\mathrm{\underset{1}{CH_3}-\underset{2}{C(CH_3)_2}-\underset{3}{CH(OH)}-\underset{4}{CH(CH_3)}-\underset{5}{CH_3}}$$

This gives methyl locants $$\{2,2,4\}$$ (lower than $$\{2,4,4\}$$ from the other end). $$\mathrm{-OH}$$ at C-3.

Name: 2,2,4-Trimethylpentan-3-ol.

The longest chain containing both $$\mathrm{-OH}$$ groups has 7 carbons (heptane); it is a diol with an ethyl substituent.

Numbering from the left end gives the $$\mathrm{-OH}$$ groups the locants $$\{2,4\}$$; from the right end they would be $$\{4,6\}$$. The lower set $$\{2,4\}$$ is chosen:

$$\mathrm{\underset{1}{CH_3}-\underset{2}{CH(OH)}-\underset{3}{CH_2}-\underset{4}{CH(OH)}-\underset{5}{CH(C_2H_5)}-\underset{6}{CH_2}-\underset{7}{CH_3}}$$

$$\mathrm{-OH}$$ at C-2 and C-4; ethyl substituent at C-5.

Name: 5-Ethylheptane-2,4-diol.

The chain has 4 carbons (butane) with two $$\mathrm{-OH}$$ groups — a diol.

$$\mathrm{\underset{1}{CH_3}-\underset{2}{CH(OH)}-\underset{3}{CH(OH)}-\underset{4}{CH_3}}$$

$$\mathrm{-OH}$$ groups are at C-2 and C-3.

Name: Butane-2,3-diol.

The chain has 3 carbons (propane), and every carbon carries an $$\mathrm{-OH}$$ group — it is a triol.

$$\mathrm{\underset{1}{CH_2(OH)}-\underset{2}{CH(OH)}-\underset{3}{CH_2(OH)}}$$

$$\mathrm{-OH}$$ groups at C-1, C-2 and C-3.

Name: Propane-1,2,3-triol (common name: glycerol).

The compound is a phenol, so the $$\mathrm{-OH}$$-bearing carbon is C-1. The methyl group is on the adjacent (ortho) carbon, C-2.

Name: 2-Methylphenol (common name: o-cresol).

The $$\mathrm{-OH}$$-bearing carbon is C-1. The methyl group is on the para carbon, C-4.

Name: 4-Methylphenol (common name: p-cresol).

The $$\mathrm{-OH}$$-bearing carbon is C-1; the two methyl groups are at C-3 and C-5.

Name: 3,5-Dimethylphenol.

The $$\mathrm{-OH}$$-bearing carbon is C-1; the two methyl groups occupy the two ortho positions, C-2 and C-6.

Name: 2,6-Dimethylphenol.

This is an ether. The larger alkyl group, $$\mathrm{-CH_2-CH(CH_3)-CH_3}$$, is taken as the parent (propane chain with a methyl branch); the smaller group $$\mathrm{CH_3-O-}$$ is named as a methoxy substituent.

$$\mathrm{\underset{1}{CH_2}-\underset{2}{CH(CH_3)}-\underset{3}{CH_3}}$$, with $$\mathrm{-OCH_3}$$ on C-1 and $$\mathrm{-CH_3}$$ on C-2.

Name: 1-Methoxy-2-methylpropane.

This is an aryl alkyl ether. The benzene ring is taken as the parent, and the $$\mathrm{-OC_2H_5}$$ group is named as an ethoxy substituent on it.

Name: Ethoxybenzene (common name: phenetole).

This ether joins a phenyl group and an n-heptyl group ($$\mathrm{C_7H_{15}}$$) through oxygen. The longer 7-carbon chain (heptane) is taken as the parent; the $$\mathrm{C_6H_5-O-}$$ group is named as a phenoxy substituent at C-1.

Name: 1-Phenoxyheptane.

This ether joins an ethyl group and a butan-2-yl group, $$\mathrm{-CH(CH_3)-CH_2-CH_3}$$. The larger group is a 4-carbon chain (butane); the $$\mathrm{C_2H_5-O-}$$ group is the ethoxy substituent.

$$\mathrm{\underset{1}{CH_3}-\underset{2}{CH(OC_2H_5)}-\underset{3}{CH_2}-\underset{4}{CH_3}}$$ — the ethoxy group is on C-2.

Name: 2-Ethoxybutane.

Parent: butane (4 carbons). $$\mathrm{-OH}$$ at C-2 and a methyl substituent also at C-2.

Structure: $$\mathrm{CH_3-C(CH_3)(OH)-CH_2-CH_3}$$

The $$\mathrm{-OH}$$ carbon is bonded to three other carbons, so it is a tertiary alcohol.

Parent: propane (3 carbons). A phenyl group is at C-1 and $$\mathrm{-OH}$$ at C-2.

Structure: $$\mathrm{C_6H_5-CH_2-CH(OH)-CH_3}$$

Parent: hexane (6 carbons). $$\mathrm{-OH}$$ groups at C-1, C-3, C-5; methyl groups at C-3 and C-5.

$$\mathrm{\underset{1}{CH_2(OH)}-\underset{2}{CH_2}-\underset{3}{C(CH_3)(OH)}-\underset{4}{CH_2}-\underset{5}{C(CH_3)(OH)}-\underset{6}{CH_3}}$$

Structure: $$\mathrm{HOCH_2-CH_2-C(CH_3)(OH)-CH_2-C(CH_3)(OH)-CH_3}$$

Parent: phenol (benzene ring with $$\mathrm{-OH}$$ at C-1). Two ethyl ($$\mathrm{-C_2H_5}$$) groups are present at C-2 and C-3.

Structure: a benzene ring bearing $$\mathrm{-OH}$$ at position 1 and $$\mathrm{-C_2H_5}$$ groups at positions 2 and 3 (the two carbons adjacent to, and next-but-one from, the $$\mathrm{-OH}$$).

Parent: propane (3 carbons), bearing an ethoxy ($$\mathrm{-OC_2H_5}$$) group at C-1.

Structure: $$\mathrm{CH_3-CH_2-CH_2-O-C_2H_5}$$ (ethyl propyl ether).

Parent: pentane (5 carbons). An ethoxy ($$\mathrm{-OC_2H_5}$$) group is at C-2 and a methyl group at C-3.

$$\mathrm{\underset{1}{CH_3}-\underset{2}{CH(OC_2H_5)}-\underset{3}{CH(CH_3)}-\underset{4}{CH_2}-\underset{5}{CH_3}}$$

Structure: $$\mathrm{CH_3-CH(OC_2H_5)-CH(CH_3)-CH_2-CH_3}$$

The parent is methanol ($$\mathrm{CH_3OH}$$); one hydrogen of the methyl is replaced by a cyclohexyl group.

Structure: a cyclohexane ring carrying a $$\mathrm{-CH_2OH}$$ group, i.e. $$\mathrm{C_6H_{11}-CH_2OH}$$.

Parent: pentane (5 carbons). $$\mathrm{-OH}$$ at C-3, and a cyclohexyl group also at C-3.

$$\mathrm{\underset{1}{CH_3}-\underset{2}{CH_2}-\underset{3}{C(OH)(C_6H_{11})}-\underset{4}{CH_2}-\underset{5}{CH_3}}$$

Structure: $$\mathrm{CH_3CH_2-C(OH)(C_6H_{11})-CH_2CH_3}$$ (C-3 bears $$\mathrm{-OH}$$ and a cyclohexyl ring).

The parent is a cyclopentane ring. $$\mathrm{-OH}$$ is at C-1 and a $$\mathrm{C=C}$$ double bond lies between C-3 and C-4.

Structure: a five-membered carbon ring; carbon 1 carries $$\mathrm{-OH}$$, and there is a double bond between carbons 3 and 4 (C-1 and the double bond are separated by one $$\mathrm{CH_2}$$ on each side).

Parent: butane (4 carbons). $$\mathrm{-OH}$$ at C-1, an ethyl group at C-3, and a chlorine at C-4.

$$\mathrm{\underset{1}{CH_2(OH)}-\underset{2}{CH_2}-\underset{3}{CH(C_2H_5)}-\underset{4}{CH_2Cl}}$$

Structure: $$\mathrm{HOCH_2-CH_2-CH(C_2H_5)-CH_2Cl}$$

$$\mathrm{C_5H_{12}O}$$ is a saturated compound. Considering all possible carbon skeletons (straight, one branch, two branches) and all positions of the $$\mathrm{-OH}$$ group, there are eight isomeric alcohols:

An alcohol is primary (1°), secondary (2°) or tertiary (3°) according to whether the carbon bearing $$\mathrm{-OH}$$ is attached to one, two or three other carbon atoms.

1-Phenylethanol is $$\mathrm{C_6H_5-CH(OH)-CH_3}$$ — the $$\mathrm{-OH}$$ is on the carbon attached to the ring (a secondary, benzylic position).

The suitable alkene is styrene (phenylethene), $$\mathrm{C_6H_5-CH=CH_2}$$. On acid-catalysed hydration, water adds following Markovnikov's rule: protonation gives the more stable benzylic secondary carbocation $$\mathrm{C_6H_5-\overset{+}{C}H-CH_3}$$, so $$\mathrm{-OH}$$ ends up on the carbon next to the ring.

$$\mathrm{C_6H_5-CH=CH_2 \xrightarrow[H^+]{H_2O} C_6H_5-CH(OH)-CH_3}$$

Cyclohexylmethanol is $$\mathrm{C_6H_{11}-CH_2OH}$$. The corresponding alkyl halide is cyclohexylmethyl chloride (chloromethylcyclohexane), $$\mathrm{C_6H_{11}-CH_2Cl}$$ — a primary halide, ideal for $$\mathrm{S_N2}$$.

Heating it with aqueous sodium hydroxide, the hydroxide ion attacks the primary carbon by $$\mathrm{S_N2}$$, displacing chloride:

$$\mathrm{C_6H_{11}-CH_2Cl + NaOH\,(aq) \xrightarrow{S_N2} C_6H_{11}-CH_2OH + NaCl}$$

Pentan-1-ol is $$\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$$. The suitable alkyl halide is 1-bromopentane (or 1-chloropentane), $$\mathrm{CH_3CH_2CH_2CH_2CH_2Br}$$ — a primary halide.

On heating with aqueous sodium hydroxide, hydroxide ion substitutes the halogen:

$$\mathrm{CH_3CH_2CH_2CH_2CH_2Br + NaOH\,(aq) \rightarrow CH_3CH_2CH_2CH_2CH_2OH + NaBr}$$

Alkaline $$\mathrm{KMnO_4}$$ is a strong oxidising agent. It oxidises a primary alcohol all the way to a carboxylic acid (through the aldehyde stage).

$$\mathrm{CH_3CH_2CH_2OH \xrightarrow[\Delta]{alk.\,KMnO_4} CH_3CH_2COOH}$$

Propan-1-ol is oxidised to propanoic acid.

When phenol is treated with bromine in a non-polar solvent such as carbon disulphide ($$\mathrm{CS_2}$$) at low temperature, the reaction is controlled and gives mainly the monobromo product, with the para isomer predominating.

$$\mathrm{C_6H_5OH + Br_2 \xrightarrow[273-278\,K]{CS_2} p\text{-}Br\text{-}C_6H_4\text{-}OH + HBr}$$

The major product is 4-bromophenol (a small amount of 2-bromophenol is also formed). The low temperature and non-polar solvent prevent polybromination.

With dilute nitric acid at low temperature, phenol undergoes mononitration. The $$\mathrm{-OH}$$ group is ortho/para-directing, so a mixture of o-nitrophenol and p-nitrophenol is obtained.

$$\mathrm{C_6H_5OH + HNO_3\,(dil.) \rightarrow o\text{-}O_2N\text{-}C_6H_4\text{-}OH + p\text{-}O_2N\text{-}C_6H_4\text{-}OH + H_2O}$$

The ortho and para nitrophenols formed are then separated by steam distillation (o-nitrophenol being steam volatile).

This is the Reimer-Tiemann reaction. Phenol, on treatment with chloroform and aqueous sodium hydroxide at 340 K, gets a $$\mathrm{-CHO}$$ group introduced at the ortho position (the reactive intermediate is dichlorocarbene, $$\mathrm{:CCl_2}$$).

$$\mathrm{C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\,K} o\text{-}NaO\text{-}C_6H_4\text{-}CHO + 3NaCl + 2H_2O}$$

On acidification, salicylaldehyde (2-hydroxybenzaldehyde) is obtained:

$$\mathrm{o\text{-}NaO\text{-}C_6H_4\text{-}CHO \xrightarrow{H^+} o\text{-}HO\text{-}C_6H_4\text{-}CHO}$$

Kolbe's reaction: When phenol is treated with sodium hydroxide, sodium phenoxide is formed. The phenoxide ion is even more reactive towards electrophilic substitution than phenol itself, so it can react with a weak electrophile such as carbon dioxide.

$$\mathrm{C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O}$$

Sodium phenoxide is heated with $$\mathrm{CO_2}$$ at about 400 K under a pressure of 4-7 atm. The $$\mathrm{CO_2}$$ molecule acts as the electrophile and a $$\mathrm{-COOH}$$ group is introduced at the ortho position; acidification then liberates salicylic acid.

$$\mathrm{C_6H_5ONa \xrightarrow[\text{4-7 atm}]{CO_2,\ 400\,K} \underset{\text{sodium salicylate}}{o\text{-}HO\text{-}C_6H_4\text{-}COONa} \xrightarrow{H^+} \underset{\text{salicylic acid}}{o\text{-}HO\text{-}C_6H_4\text{-}COOH}}$$

Why ortho substitution predominates: the negatively charged oxygen of the phenoxide ion activates the ring at the ortho and para positions. The ortho product is favoured because the $$\mathrm{Na^+}$$ ion coordinates the phenoxide oxygen and holds the $$\mathrm{CO_2}$$ molecule close to the adjacent (ortho) carbon; in addition, the ortho product (sodium salicylate) is stabilised by intramolecular hydrogen bonding between the $$\mathrm{-OH}$$ and $$\mathrm{-COO^-}$$ groups.

The product, 2-hydroxybenzoic acid (salicylic acid), is used in making aspirin.

Reimer-Tiemann reaction: When phenol is treated with chloroform in the presence of aqueous sodium hydroxide at 340 K, a $$\mathrm{-CHO}$$ group is introduced at the ortho position of the ring.

$$\mathrm{NaOH}$$ reacts with $$\mathrm{CHCl_3}$$ to generate dichlorocarbene ($$\mathrm{:CCl_2}$$), the electrophile. It attacks the (highly reactive) phenoxide ring ortho to oxygen, and the resulting o-substituted benzal chloride is hydrolysed by alkali.

$$\mathrm{C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{340\,K} o\text{-}NaO\text{-}C_6H_4\text{-}CHO + 3NaCl + 2H_2O}$$

$$\mathrm{o\text{-}NaO\text{-}C_6H_4\text{-}CHO \xrightarrow{H^+} o\text{-}HO\text{-}C_6H_4\text{-}CHO}$$

The product is salicylaldehyde (2-hydroxybenzaldehyde).

Williamson ether synthesis is an important laboratory method of preparing ethers (both symmetrical and unsymmetrical). An alkyl halide is allowed to react with a sodium alkoxide (or aryloxide).

The alkoxide ion is a strong nucleophile; it attacks the carbon of the alkyl halide in an $$\mathrm{S_N2}$$ reaction, displacing the halide ion and forming the ether.

$$\mathrm{R-O^-Na^+ + R'-X \rightarrow R-O-R' + NaX}$$

Example — preparation of ethoxyethane (diethyl ether):

$$\mathrm{C_2H_5ONa + C_2H_5Br \rightarrow C_2H_5-O-C_2H_5 + NaBr}$$

The method works best when the alkyl halide is primary; with secondary or tertiary halides, elimination competes.

An unsymmetrical (or mixed) ether is an ether in which the two groups attached to the oxygen atom are different from each other.

If both groups are the same, the ether is symmetrical (e.g. $$\mathrm{C_2H_5-O-C_2H_5}$$, ethoxyethane); if they differ, it is unsymmetrical.

Examples of unsymmetrical ethers:

• $$\mathrm{CH_3-O-C_2H_5}$$ — methoxyethane (ethyl methyl ether)
• $$\mathrm{C_6H_5-O-CH_3}$$ — methoxybenzene (anisole)

Propene is converted to propan-2-ol by acid-catalysed hydration. Water adds across the double bond following Markovnikov's rule, so $$\mathrm{-OH}$$ goes to the more substituted (central) carbon.

$$\mathrm{CH_3-CH=CH_2 \xrightarrow[dil.\,H_2SO_4]{H_2O} CH_3-CH(OH)-CH_3}$$

Benzyl chloride, $$\mathrm{C_6H_5CH_2Cl}$$, is a primary (benzylic) halide. Heating it with aqueous sodium hydroxide (or aqueous $$\mathrm{KOH}$$) replaces the chlorine by $$\mathrm{-OH}$$ (nucleophilic substitution), giving benzyl alcohol.

$$\mathrm{C_6H_5CH_2Cl + NaOH\,(aq) \rightarrow C_6H_5CH_2OH + NaCl}$$

A Grignard reagent reacts with methanal ($$\mathrm{HCHO}$$) to give, after hydrolysis, a primary alcohol with one extra carbon.

Ethylmagnesium chloride adds to methanal, and the adduct is hydrolysed:

$$\mathrm{C_2H_5MgCl + HCHO \rightarrow C_2H_5-CH_2-OMgCl}$$

$$\mathrm{C_2H_5-CH_2-OMgCl \xrightarrow{H_2O/H^+} C_2H_5-CH_2-OH}$$

i.e. $$\mathrm{CH_3CH_2CH_2OH}$$ — propan-1-ol.

2-Methylpropan-2-ol is the tertiary alcohol $$\mathrm{(CH_3)_3C-OH}$$. A tertiary alcohol is obtained when a Grignard reagent reacts with a ketone.

Methylmagnesium bromide is allowed to react with propanone (acetone, $$\mathrm{(CH_3)_2C=O}$$); the adduct is then hydrolysed.

$$\mathrm{(CH_3)_2C=O + CH_3MgBr \rightarrow (CH_3)_3C-OMgBr}$$

$$\mathrm{(CH_3)_3C-OMgBr \xrightarrow{H_2O/H^+} (CH_3)_3C-OH}$$

To oxidise a primary alcohol all the way to a carboxylic acid, a strong oxidising agent is needed.

Reagent: acidified potassium permanganate ($$\mathrm{KMnO_4/H^+}$$) — alkaline $$\mathrm{KMnO_4}$$ or acidified potassium dichromate ($$\mathrm{K_2Cr_2O_7/H^+}$$) may also be used.

$$\mathrm{RCH_2OH \xrightarrow{KMnO_4/H^+} RCOOH}$$

To stop the oxidation of a primary alcohol at the aldehyde stage (without further oxidation to the acid), a mild oxidising agent is required.

Reagent: pyridinium chlorochromate (PCC), used in dichloromethane.

$$\mathrm{RCH_2OH \xrightarrow[CH_2Cl_2]{PCC} RCHO}$$

To brominate all three activated positions of phenol at once, a polar medium is used so that the reaction is fast.

Reagent: bromine water (an aqueous solution of bromine, $$\mathrm{Br_2}$$ in water).

$$\mathrm{C_6H_5OH + 3Br_2\,(aq) \rightarrow 2,4,6\text{-tribromophenol} + 3HBr}$$

(In water, phenol is highly activated, so all the ortho and para positions are brominated, giving a white precipitate of 2,4,6-tribromophenol.)

Benzyl alcohol ($$\mathrm{C_6H_5CH_2OH}$$) is a primary alcohol; to oxidise it fully to benzoic acid ($$\mathrm{C_6H_5COOH}$$), a strong oxidising agent is used.

Reagent: acidified potassium permanganate ($$\mathrm{KMnO_4/H^+}$$) — acidified potassium dichromate ($$\mathrm{K_2Cr_2O_7/H^+}$$) also works.

$$\mathrm{C_6H_5CH_2OH \xrightarrow{KMnO_4/H^+} C_6H_5COOH}$$

Dehydration (removal of water) of an alcohol to an alkene requires a hot acid catalyst.

Reagent: concentrated sulphuric acid ($$\mathrm{conc.\ H_2SO_4}$$) on heating (about 440 K). Alternatively, 85% phosphoric acid.

$$\mathrm{CH_3CH(OH)CH_3 \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH=CH_2 + H_2O}$$

Converting a ketone (butan-2-one) to a secondary alcohol (butan-2-ol) requires reduction of the carbonyl group.

Reagent: sodium borohydride ($$\mathrm{NaBH_4}$$) — lithium aluminium hydride ($$\mathrm{LiAlH_4}$$), or $$\mathrm{H_2}$$ with a Ni/Pt/Pd catalyst, may also be used.

$$\mathrm{CH_3-CO-CH_2CH_3 \xrightarrow{NaBH_4} CH_3-CH(OH)-CH_2CH_3}$$

This ether joins an ethyl group and an isobutyl group, $$\mathrm{-CH_2-CH(CH_3)-CH_3}$$. The larger group is taken as the parent: a propane chain ($$\mathrm{CH_2-CH(CH_3)-CH_3}$$) with a methyl branch. The $$\mathrm{C_2H_5-O-}$$ group is named as ethoxy.

$$\mathrm{\underset{1}{CH_2(OC_2H_5)}-\underset{2}{CH(CH_3)}-\underset{3}{CH_3}}$$ — ethoxy at C-1, methyl at C-2.

Name: 1-Ethoxy-2-methylpropane.

The larger group here is the 2-carbon chain $$\mathrm{-CH_2CH_2Cl}$$; it is taken as the parent (ethane). The $$\mathrm{CH_3-O-}$$ group is a methoxy substituent and $$\mathrm{Cl}$$ is a chloro substituent.

$$\mathrm{\underset{1}{CH_2Cl}-\underset{2}{CH_2(OCH_3)}}$$ — number to give the lower locant set; chloro at C-1, methoxy at C-2.

Name: 1-Chloro-2-methoxyethane.

The benzene ring is the parent. It carries a methoxy ($$\mathrm{-OCH_3}$$) group and, para to it, a nitro ($$\mathrm{-NO_2}$$) group.

Citing substituents alphabetically, methoxy gets C-1 and nitro gets C-4.

Name: 1-Methoxy-4-nitrobenzene (common name: p-nitroanisole).

This ether joins a propyl group and a methyl group. The larger group, propyl, is taken as the parent (propane); the $$\mathrm{CH_3-O-}$$ group is named as methoxy at C-1.

Name: 1-Methoxypropane.

The parent is a cyclohexane ring. The carbon bearing the ethoxy ($$\mathrm{-OC_2H_5}$$) group is taken as C-1, and the two methyl groups are on the adjacent carbon, C-2.

Numbering this way gives the lowest locant set $$\{1,2,2\}$$.

Name: 1-Ethoxy-2,2-dimethylcyclohexane.

The benzene ring is the parent, carrying one ethoxy ($$\mathrm{-OC_2H_5}$$) group.

Name: Ethoxybenzene (common name: phenetole).

1-Propoxypropane is the symmetrical ether $$\mathrm{CH_3CH_2CH_2-O-CH_2CH_2CH_3}$$.

Reagents: sodium n-propoxide and 1-bromopropane (a primary halide).

$$\mathrm{CH_3CH_2CH_2ONa + CH_3CH_2CH_2Br \rightarrow CH_3CH_2CH_2-O-CH_2CH_2CH_3 + NaBr}$$

Ethoxybenzene is $$\mathrm{C_6H_5-O-C_2H_5}$$.

We must use sodium phenoxide with ethyl bromide — not sodium ethoxide with bromobenzene, because the aryl halide (bromobenzene) does not undergo the $$\mathrm{S_N2}$$ substitution needed here.

$$\mathrm{C_6H_5ONa + C_2H_5Br \rightarrow C_6H_5-O-C_2H_5 + NaBr}$$

2-Methoxy-2-methylpropane is $$\mathrm{(CH_3)_3C-O-CH_3}$$ (tert-butyl methyl ether).

The tertiary group must be supplied as the alkoxide, and the methyl group as the (primary) halide. (Using a tertiary halide would give only elimination.)

$$\mathrm{(CH_3)_3C-ONa + CH_3Br \rightarrow (CH_3)_3C-O-CH_3 + NaBr}$$

1-Methoxyethane is $$\mathrm{CH_3-O-C_2H_5}$$ (ethyl methyl ether). Both groups are primary, so either group may be the alkoxide and the other the halide.

Using sodium ethoxide and bromomethane:

$$\mathrm{C_2H_5ONa + CH_3Br \rightarrow C_2H_5-O-CH_3 + NaBr}$$

(Equally, sodium methoxide and bromoethane: $$\mathrm{CH_3ONa + C_2H_5Br \rightarrow CH_3-O-C_2H_5 + NaBr}$$.)

1-Propoxypropane is the symmetrical ether $$\mathrm{CH_3CH_2CH_2-O-CH_2CH_2CH_3}$$. With HI, the $$\mathrm{C-O}$$ bond is cleaved: one propyl group becomes the iodide and the other the alcohol.

$$\mathrm{CH_3CH_2CH_2-O-CH_2CH_2CH_3 + HI \rightarrow CH_3CH_2CH_2I + CH_3CH_2CH_2OH}$$

(With excess HI on heating, the propan-1-ol formed reacts further to give a second molecule of 1-iodopropane.)

Methoxybenzene (anisole) is $$\mathrm{C_6H_5-O-CH_3}$$, an aryl alkyl ether. The $$\mathrm{O-C(aryl)}$$ bond is not cleaved (the aryl carbon resists nucleophilic substitution), so a phenol is formed; iodide attacks the methyl carbon.

$$\mathrm{C_6H_5-O-CH_3 + HI \rightarrow C_6H_5-OH + CH_3I}$$

Products: phenol and iodomethane (methyl iodide).

Benzyl ethyl ether is $$\mathrm{C_6H_5CH_2-O-C_2H_5}$$. Both groups attached to oxygen are primary, but because one of them is a benzyl group, cleavage with HI follows the $$\mathrm{S_N1}$$ path (just as NCERT shows for tert-butyl methyl ether, where a stable carbocation is possible).

Step 1 — Protonation of the ether: HI protonates the lone pair on the ether oxygen, converting the poor leaving group ($$\mathrm{-OR}$$) into a good one (a neutral alcohol).

$$\mathrm{C_6H_5CH_2-O-C_2H_5 + HI \rightarrow C_6H_5CH_2-\overset{+}{O}(H)-C_2H_5 + I^-}$$

Step 2 — Heterolysis ($$\mathrm{S_N1}$$, slow step): The oxonium ion breaks at the benzyl $$\mathrm{C-O}$$ bond, expelling ethanol and giving the resonance-stabilised benzyl carbocation. Cleavage at the ethyl $$\mathrm{C-O}$$ bond would give an unstable primary ethyl cation, so this path is preferred.

$$\mathrm{C_6H_5CH_2-\overset{+}{O}(H)-C_2H_5 \xrightarrow{\text{slow}} C_6H_5\overset{+}{C}H_2 + C_2H_5OH}$$

Step 3 — Capture by iodide (fast): The iodide ion combines with the benzyl carbocation.

$$\mathrm{C_6H_5\overset{+}{C}H_2 + I^- \xrightarrow{\text{fast}} C_6H_5CH_2I}$$

Overall reaction:

$$\mathrm{C_6H_5CH_2-O-C_2H_5 + HI \rightarrow C_6H_5CH_2I + C_2H_5OH}$$

Products: benzyl iodide and ethanol. (The ethanol is not attacked further by $$\mathrm{I^-}$$ under these conditions; it can react with a second mole of HI only on prolonged heating with excess HI.)

In an aryl alkyl ether (e.g. anisole, $$\mathrm{C_6H_5-OCH_3}$$), the oxygen of the alkoxy group carries lone pairs of electrons.

One of these lone pairs occupies a p-orbital that overlaps with the $$\pi$$ system of the benzene ring. Through this conjugation, oxygen donates electron density into the ring (the $$+R$$ / +M, electron-releasing resonance effect).

This can be shown by resonance structures in which the ring carries a negative charge at the ortho and para positions and oxygen carries a positive charge.

As a result, the electron density of the benzene ring increases. Since electrophiles attack electron-rich centres, the more electron-rich ring reacts faster with electrophiles. Hence the alkoxy group activates the ring towards electrophilic substitution.

When the lone pair of the alkoxy oxygen is delocalised into the ring, the extra electron density does not spread evenly. The resonance structures show that the negative charge appears specifically on the carbon atoms that are ortho and para to the alkoxy group.

That is, the ortho and para positions become the most electron-rich centres of the ring, while the meta positions are not enriched.

An electrophile is attracted to the position of highest electron density. Therefore the incoming electrophile attacks mainly at the ortho and para positions, and the alkoxy group is described as an ortho/para-directing group.

Anisole ($$\mathrm{C_6H_5OCH_3}$$) undergoes Friedel-Crafts alkylation with an alkyl halide in the presence of anhydrous aluminium chloride. The $$\mathrm{-OCH_3}$$ group is activating and ortho/para-directing, so the alkyl group enters mainly at the para position (with some ortho).

$$\mathrm{C_6H_5OCH_3 + CH_3Cl \xrightarrow{anhyd.\,AlCl_3} \underset{\text{(major, para)}}{4\text{-}CH_3\text{-}C_6H_4\text{-}OCH_3} + \underset{\text{(ortho)}}{2\text{-}CH_3\text{-}C_6H_4\text{-}OCH_3} + HCl}$$

Products: 2-methoxytoluene (o-) and 4-methoxytoluene (p-, major).

Anisole is nitrated by a mixture of concentrated nitric acid and concentrated sulphuric acid. The activating, ortho/para-directing $$\mathrm{-OCH_3}$$ group sends the $$\mathrm{-NO_2}$$ group to the ortho and para positions, the para isomer being major.

$$\mathrm{C_6H_5OCH_3 \xrightarrow[conc.\,H_2SO_4]{conc.\,HNO_3} \underset{\text{(ortho)}}{2\text{-}O_2N\text{-}C_6H_4\text{-}OCH_3} + \underset{\text{(para, major)}}{4\text{-}O_2N\text{-}C_6H_4\text{-}OCH_3} + H_2O}$$

Products: 2-nitroanisole and 4-nitroanisole (para major).

Anisole reacts with bromine in ethanoic acid (acetic acid) medium. The reaction does not need a catalyst because $$\mathrm{-OCH_3}$$ strongly activates the ring. In this medium the para product is formed almost exclusively (the bulky para-selectivity arises since the ortho positions are sterically hindered).

$$\mathrm{C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} \underset{\text{(major, para)}}{4\text{-}Br\text{-}C_6H_4\text{-}OCH_3} + HBr}$$

The major product is 4-bromoanisole (p-bromoanisole).

In Friedel-Crafts acetylation, anisole reacts with acetyl chloride ($$\mathrm{CH_3COCl}$$) in the presence of anhydrous aluminium chloride; an acetyl group ($$\mathrm{-COCH_3}$$) is introduced into the ring.

Guided by the ortho/para-directing $$\mathrm{-OCH_3}$$ group, the acetyl group enters mainly at the para position.

$$\mathrm{C_6H_5OCH_3 + CH_3COCl \xrightarrow{anhyd.\,AlCl_3} \underset{\text{(major, para)}}{4\text{-}CH_3CO\text{-}C_6H_4\text{-}OCH_3} + \underset{\text{(ortho)}}{2\text{-}CH_3CO\text{-}C_6H_4\text{-}OCH_3} + HCl}$$

Products: 2-methoxyacetophenone (o-) and 4-methoxyacetophenone (p-, major).

In 1-methylcyclohexan-1-ol the $$\mathrm{-OH}$$ is on the ring carbon that also carries the $$\mathrm{-CH_3}$$ group. The suitable alkene is 1-methylcyclohex-1-ene (a ring double bond, with the methyl on one of the doubly-bonded carbons).

On acid-catalysed hydration, water adds by Markovnikov's rule: protonation gives the more stable tertiary carbocation on the methyl-bearing carbon, so $$\mathrm{-OH}$$ goes to that carbon.

$$\mathrm{1\text{-methylcyclohex-1-ene} \xrightarrow[H^+]{H_2O} 1\text{-methylcyclohexan-1-ol}}$$

The target is 3-ethylhexan-3-ol, $$\mathrm{CH_3CH_2-C(OH)(C_2H_5)-CH_2CH_2CH_3}$$ — a tertiary alcohol with $$\mathrm{-OH}$$ on C-3.

The suitable alkene is 3-ethylhex-2-ene, $$\mathrm{CH_3-CH=C(C_2H_5)-CH_2CH_2CH_3}$$. On acid-catalysed hydration, Markovnikov addition places $$\mathrm{H}$$ on C-2 and $$\mathrm{-OH}$$ on the more substituted C-3 (via the stable tertiary carbocation).

$$\mathrm{CH_3-CH=C(C_2H_5)-CH_2CH_2CH_3 \xrightarrow[H^+]{H_2O} CH_3CH_2-C(OH)(C_2H_5)-CH_2CH_2CH_3}$$

The target is pentan-2-ol, $$\mathrm{CH_3-CH(OH)-CH_2CH_2CH_3}$$ — a secondary alcohol with $$\mathrm{-OH}$$ on C-2.

The suitable alkene is pent-1-ene, $$\mathrm{CH_2=CH-CH_2CH_2CH_3}$$. On acid-catalysed hydration, Markovnikov addition puts $$\mathrm{H}$$ on the terminal C-1 and $$\mathrm{-OH}$$ on the more substituted C-2 (via the secondary carbocation).

$$\mathrm{CH_2=CH-CH_2CH_2CH_3 \xrightarrow[H^+]{H_2O} CH_3-CH(OH)-CH_2CH_2CH_3}$$

(Pent-1-ene is preferred over pent-2-ene, which would give a mixture of pentan-2-ol and pentan-3-ol.)

In 1-ethylcyclohexan-1-ol the $$\mathrm{-OH}$$ is on the ring carbon that also carries the $$\mathrm{-C_2H_5}$$ group. The suitable alkene is 1-ethylcyclohex-1-ene (a ring double bond, with the ethyl group on one of the doubly-bonded carbons).

On acid-catalysed hydration, Markovnikov addition places $$\mathrm{-OH}$$ on the more substituted carbon (the one bearing the ethyl group), via the stable tertiary carbocation.

$$\mathrm{1\text{-ethylcyclohex-1-ene} \xrightarrow[H^+]{H_2O} 1\text{-ethylcyclohexan-1-ol}}$$