7.1 Classification, Nomenclature and Preparation
Put an –OH group on a saturated carbon and you have an alcohol; put it on a benzene ring and you have a phenol; put an oxygen between two carbon groups and you have an ether. This section classifies and names them, looks at the shape of the C–O–H and C–O–C groups, shows how alcohols and phenols are made, and explains their boiling points and solubility through hydrogen bonding.
The hand sanitiser by the classroom door is about 70% ethanol or propan-2-ol. The antifreeze in a car radiator is ethane-1,2-diol, and the moisturiser in a tube of toothpaste or a bar of soap is glycerol, propane-1,2,3-triol. Menthol cools your throat, cholesterol builds your cell membranes, and the sugars in your food are covered with –OH groups. Phenol was the first surgical antiseptic, its relatives are in disinfectants such as the chloroxylenol of household antiseptic liquids, and the ether ethoxyethane put patients to sleep in the first operations under general anaesthesia. All of these compounds carry an oxygen atom held by single bonds to carbon.
Chemically they belong to three families. In alcohols and phenols an $\mathrm{{-}OH}$ (hydroxyl) group is bonded to carbon; in ethers an oxygen atom joins two carbon groups. The family depends on which carbon carries the oxygen, and that decides how the compound reacts, as Sections 7.2 and 7.3 will show. This section lays the groundwork: how the three families are classified and named, what their functional groups look like, how alcohols and phenols are prepared, and why hydrogen bonding gives alcohols such high boiling points.
Classification: Which Carbon Carries the Oxygen?
An alcohol contains one or more hydroxyl groups, $\mathrm{{-}OH}$, each bonded to an $sp^{3}$ (saturated) carbon atom: $\mathrm{R{-}OH}$, e.g. $\mathrm{CH_{3}CH_{2}OH}$. A phenol contains an $\mathrm{{-}OH}$ group bonded directly to a carbon atom of an aromatic ring: $\mathrm{Ar{-}OH}$, e.g. $\mathrm{C_{6}H_{5}OH}$. An ether contains an oxygen atom bonded to two alkyl or aryl groups: $\mathrm{R{-}O{-}R'}$, $\mathrm{R{-}O{-}Ar}$ or $\mathrm{Ar{-}O{-}Ar'}$, e.g. $\mathrm{CH_{3}OCH_{3}}$.
Formally, an alcohol or a phenol is a hydrocarbon with one hydrogen replaced by $\mathrm{{-}OH}$, and an ether is water with both hydrogens replaced by carbon groups (an alcohol is water with one replaced).
(a) By the number of –OH groups: monohydric (one), dihydric (two, e.g. ethane-1,2-diol) and trihydric (three, e.g. propane-1,2,3-triol) alcohols and phenols.
(b) Alcohols with –OH on an $sp^{3}$ carbon, $\mathrm{C}(sp^{3}){-}\mathrm{OH}$:
- Primary (1°), secondary (2°) and tertiary (3°) alcohols: the carbon carrying $\mathrm{OH}$ is bonded to one, two or three other carbon atoms ($\mathrm{RCH_{2}OH}$, $\mathrm{R_{2}CHOH}$, $\mathrm{R_{3}COH}$). Methanol, $\mathrm{CH_{3}OH}$, is counted with the primary alcohols.
- Allylic alcohols: $\mathrm{OH}$ on an $sp^{3}$ carbon next to a $\mathrm{C{=}C}$ double bond, e.g. $\mathrm{CH_{2}{=}CH{-}CH_{2}OH}$.
- Benzylic alcohols: $\mathrm{OH}$ on an $sp^{3}$ carbon attached directly to an aromatic ring, e.g. $\mathrm{C_{6}H_{5}CH_{2}OH}$.
Allylic and benzylic alcohols are also 1°, 2° or 3°, by the same carbon count.
(c) –OH on an $sp^{2}$ carbon, $\mathrm{C}(sp^{2}){-}\mathrm{OH}$: vinylic alcohols (enols), $\mathrm{CH_{2}{=}CH{-}OH}$, which change at once into the isomeric aldehyde or ketone, and phenols, which are stable.
(d) Ethers are symmetrical (simple) when the two groups on oxygen are the same, e.g. $\mathrm{C_{2}H_{5}OC_{2}H_{5}}$, and unsymmetrical (mixed) when they differ, e.g. $\mathrm{CH_{3}OC_{2}H_{5}}$, $\mathrm{C_{6}H_{5}OCH_{3}}$.
propan-1-ol
propan-2-ol
2-methylpropan-2-ol
prop-2-en-1-ol
phenylmethanol
phenol
ethoxyethane
methoxybenzene
Nomenclature and the Shape of the Functional Groups
Simple alcohols have common names made from the alkyl group plus the word "alcohol" (methyl alcohol, isopropyl alcohol), and simple ethers from the two alkyl or aryl groups in alphabetical order plus "ether" (ethyl methyl ether). IUPAC names are systematic: the $\mathrm{OH}$ group is shown by the suffix -ol, and an ether is named as an alkoxy substituent on a hydrocarbon.
- Alcohols. Take the longest carbon chain that contains the carbon bearing $\mathrm{OH}$ (and any $\mathrm{C{=}C}$). Replace the final -e of the alkane by -ol, and number the chain so that the carbon bearing $\mathrm{OH}$ gets the lowest locant, even if a double bond or a substituent then gets a higher one: $\mathrm{CH_{3}CH(OH)CH_{3}}$ is propan-2-ol, $\mathrm{CH_{2}{=}CHCH_{2}OH}$ is prop-2-en-1-ol.
- Polyhydric alcohols. Keep the -e of the alkane and add -diol, -triol with a locant for every $\mathrm{OH}$: $\mathrm{HOCH_{2}CH_{2}OH}$ is ethane-1,2-diol.
- Cyclic alcohols take the prefix cyclo-; the carbon bearing $\mathrm{OH}$ is C1: cyclohexanol, 2-methylcyclohexan-1-ol.
- Phenols. The parent is phenol itself, with the carbon bearing $\mathrm{OH}$ as C1: 2-methylphenol ($o$-cresol). The three dihydroxybenzenes are benzene-1,2-diol, benzene-1,3-diol and benzene-1,4-diol.
- Ethers are named as alkoxyalkanes: the larger group is the parent hydrocarbon and the smaller group with the oxygen, $\mathrm{RO{-}}$, is the alkoxy prefix (methoxy, ethoxy, phenoxy). $\mathrm{CH_{3}OC_{2}H_{5}}$ is methoxyethane; $\mathrm{C_{6}H_{5}OCH_{3}}$ is methoxybenzene.
| Alcohol | Common name | IUPAC name |
|---|---|---|
| $\mathrm{CH_{3}CH_{2}CH_{2}OH}$ | $n$-propyl alcohol | propan-1-ol |
| $\mathrm{(CH_{3})_{2}CHOH}$ | isopropyl alcohol | propan-2-ol |
| $\mathrm{(CH_{3})_{3}COH}$ | tert-butyl alcohol | 2-methylpropan-2-ol |
| $\mathrm{CH_{2}{=}CHCH_{2}OH}$ | allyl alcohol | prop-2-en-1-ol |
| $\mathrm{C_{6}H_{5}CH_{2}OH}$ | benzyl alcohol | phenylmethanol |
| $\mathrm{HOCH_{2}CH(OH)CH_{2}OH}$ | glycerol | propane-1,2,3-triol |
| Phenol or ether | Common name | IUPAC name |
|---|---|---|
| $o$-$\mathrm{HOC_{6}H_{4}OH}$ | catechol | benzene-1,2-diol |
| $m$-$\mathrm{HOC_{6}H_{4}OH}$ | resorcinol | benzene-1,3-diol |
| $p$-$\mathrm{HOC_{6}H_{4}OH}$ | quinol (hydroquinone) | benzene-1,4-diol |
| $\mathrm{C_{2}H_{5}OC_{2}H_{5}}$ | diethyl ether | ethoxyethane |
| $\mathrm{C_{6}H_{5}OCH_{3}}$ | anisole | methoxybenzene |
| $\mathrm{C_{6}H_{5}OC_{2}H_{5}}$ | phenetole | ethoxybenzene |
The shape of the functional groups. In all three families the oxygen atom is $sp^{3}$ hybridised: two of its four hybrid orbitals form $\sigma$ bonds and two hold lone pairs. The bond angle is close to tetrahedral, but each family bends it a little differently.
- Alcohols. In methanol the $\mathrm{C{-}O{-}H}$ angle is $108.9^{\circ}$, slightly less than the tetrahedral $109.5^{\circ}$, because the two lone pairs on oxygen repel the bonding pairs more strongly than bonding pairs repel each other. The $\mathrm{C{-}O}$ bond length is $142\ \mathrm{pm}$.
- Phenols. The $\mathrm{C{-}O}$ bond in phenol is shorter, $136\ \mathrm{pm}$, for two reasons: (i) a lone pair on oxygen is partly delocalised into the ring (resonance), giving the bond some double-bond character; (ii) the ring carbon is $sp^{2}$, which holds its bonding electrons closer than an $sp^{3}$ carbon.
- Ethers. In methoxymethane the $\mathrm{C{-}O{-}C}$ angle is $111.7^{\circ}$, slightly more than tetrahedral, because the two bulky methyl groups repel each other. The $\mathrm{C{-}O}$ bond length ($141\ \mathrm{pm}$) is almost the same as in alcohols.
Preparing Alcohols
Alcohols are made from three kinds of starting material: alkenes (adding water across the double bond), carbonyl compounds (adding hydrogen across $\mathrm{C{=}O}$), and Grignard reagents (adding a carbon group across $\mathrm{C{=}O}$, which also builds a longer carbon chain). The first two decide where the $\mathrm{OH}$ ends up; the third decides what the carbon skeleton around it looks like.
(a) Acid-catalysed hydration. An alkene adds water in the presence of an acid catalyst (dilute $\mathrm{H_{2}SO_{4}}$). With an unsymmetrical alkene the addition follows Markovnikov's rule: $\mathrm{H}$ goes to the doubly bonded carbon that already has more hydrogens, $\mathrm{OH}$ to the other one.
$$\mathrm{CH_{3}CH{=}CH_{2}+H_{2}O\xrightarrow{H^{+}}CH_{3}CH(OH)CH_{3}}$$The mechanism has three steps. Step 1 (slow): the $\pi$ electrons take a proton from $\mathrm{H_{3}O^{+}}$, giving the more stable (here secondary) carbocation. Step 2: a water molecule, a nucleophile, attacks the carbocation. Step 3: water removes a proton, giving the alcohol and regenerating $\mathrm{H_{3}O^{+}}$.
$$\mathrm{CH_{3}CH{=}CH_{2}+H_{3}O^{+}\rightarrow CH_{3}\overset{+}{C}HCH_{3}+H_{2}O}$$ $$\mathrm{CH_{3}\overset{+}{C}HCH_{3}+H_{2}O\rightarrow CH_{3}CH(\overset{+}{O}H_{2})CH_{3}}$$ $$\mathrm{CH_{3}CH(\overset{+}{O}H_{2})CH_{3}+H_{2}O\rightarrow CH_{3}CH(OH)CH_{3}+H_{3}O^{+}}$$(b) Hydroboration–oxidation. Diborane, $\mathrm{(BH_{3})_{2}}$, adds across the double bond with boron going to the carbon with more hydrogens (the less crowded end). Three additions give a trialkylborane, which hydrogen peroxide in aqueous sodium hydroxide oxidises to the alcohol:
$$\mathrm{CH_{3}CH{=}CH_{2}+(H{-}BH_{2})_{2}\rightarrow CH_{3}CH_{2}CH_{2}{-}BH_{2}}$$ $$\mathrm{\xrightarrow{2\,CH_{3}CH=CH_{2}}(CH_{3}CH_{2}CH_{2})_{3}B}$$ $$\mathrm{(CH_{3}CH_{2}CH_{2})_{3}B+3H_{2}O_{2}\xrightarrow{OH^{-}}3CH_{3}CH_{2}CH_{2}OH+B(OH)_{3}}$$The net result is the addition of water opposite to Markovnikov's rule ("anti-Markovnikov"): propene gives propan-1-ol, not propan-2-ol. No carbocation is formed, and the yields are excellent.
(a) Reduction of aldehydes and ketones. Hydrogen with a finely divided metal catalyst ($\mathrm{Pt}$, $\mathrm{Pd}$ or $\mathrm{Ni}$), sodium borohydride ($\mathrm{NaBH_{4}}$) or lithium aluminium hydride ($\mathrm{LiAlH_{4}}$) adds $\mathrm{H_{2}}$ across $\mathrm{C{=}O}$. Aldehydes give primary alcohols, ketones give secondary alcohols:
$$\mathrm{RCHO+H_{2}\xrightarrow{Pd}RCH_{2}OH},\qquad \mathrm{RCOR'\xrightarrow{NaBH_{4}}RCH(OH)R'}$$(b) Reduction of carboxylic acids and esters. $\mathrm{LiAlH_{4}}$, a very strong reducing agent, reduces acids to primary alcohols, $\mathrm{RCOOH\xrightarrow{(i)\,LiAlH_{4},\ (ii)\,H_{2}O}RCH_{2}OH}$. It is expensive, so industry first converts the acid into an ester and then reduces the ester with hydrogen over a catalyst (catalytic hydrogenation): $\mathrm{RCOOR'+2H_{2}\xrightarrow{catalyst}RCH_{2}OH+R'OH}$.
(c) From Grignard reagents. A Grignard reagent $\mathrm{RMgX}$ (Chapter 6) adds across the $\mathrm{C{=}O}$ bond: R becomes bonded to the carbonyl carbon and $\mathrm{MgX}$ to the oxygen. Hydrolysis of this adduct gives the alcohol, e.g.
$$\mathrm{HCHO+RMgX\rightarrow RCH_{2}OMgX\xrightarrow{H_{2}O}RCH_{2}OH+Mg(OH)X}$$Methanal gives a primary alcohol, any other aldehyde a secondary alcohol, and a ketone a tertiary alcohol. The new alcohol has the carbons of the Grignard reagent plus those of the carbonyl compound, so this is a way of building carbon chains.
- Find the carbinol carbon (the one carrying $\mathrm{OH}$) in the target alcohol. It was the carbonyl carbon.
- Choose one of the groups on the carbinol carbon (not $\mathrm{H}$, not $\mathrm{OH}$) to come from $\mathrm{RMgX}$. Cut the bond between that group and the carbinol carbon.
- The rest of the molecule, with $\mathrm{C{-}OH}$ turned back into $\mathrm{C{=}O}$, is the carbonyl compound. Count: carbinol carbon with two H's ⇒ methanal; with one H ⇒ an aldehyde; with no H ⇒ a ketone.
- Every different group on the carbinol carbon gives another possible route. Pick the one with the cheapest or most available starting materials.
Preparing Phenols; Physical Properties
Phenol was first obtained from coal tar. Today almost all of it is made from benzene, by one of four routes. In each, the hard part is getting an oxygen atom onto a ring carbon, because a benzene ring does not easily give up a hydrogen to a nucleophile.
(a) From haloarenes (Dow process). Chlorobenzene is fused with sodium hydroxide at $623\ \mathrm{K}$ and $300\ \mathrm{atm}$; the sodium phenoxide formed is acidified:
$$\mathrm{C_{6}H_{5}Cl+2NaOH\xrightarrow[300\ atm]{623\ K}C_{6}H_{5}ONa+NaCl+H_{2}O}$$ $$\mathrm{C_{6}H_{5}ONa+HCl\rightarrow C_{6}H_{5}OH+NaCl}$$(b) From benzenesulphonic acid. Benzene is sulphonated with oleum; the benzenesulphonic acid is fused with molten sodium hydroxide, and the phenoxide is acidified:
$$\mathrm{C_{6}H_{6}\xrightarrow{oleum}C_{6}H_{5}SO_{3}H}$$ $$\mathrm{C_{6}H_{5}SO_{3}H+3NaOH\rightarrow C_{6}H_{5}ONa+Na_{2}SO_{3}+2H_{2}O}$$(c) From diazonium salts. Aniline is diazotised with nitrous acid ($\mathrm{NaNO_{2}}$ + $\mathrm{HCl}$) at $273$–$278\ \mathrm{K}$; warming the benzenediazonium chloride with water replaces $\mathrm{{-}N_{2}^{+}}$ by $\mathrm{{-}OH}$ (more in Chapter 9):
$$\mathrm{C_{6}H_{5}NH_{2}\xrightarrow[273\text{–}278\ K]{NaNO_{2}+HCl}C_{6}H_{5}N_{2}^{+}Cl^{-}}$$ $$\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+H_{2}O\xrightarrow{warm}C_{6}H_{5}OH+N_{2}+HCl}$$(d) From cumene (the main industrial route). Cumene (isopropylbenzene, made from benzene and propene) is oxidised by air to cumene hydroperoxide, which dilute acid splits into phenol and propanone:
$$\mathrm{C_{6}H_{5}CH(CH_{3})_{2}+O_{2}\rightarrow C_{6}H_{5}C(CH_{3})_{2}OOH}$$ $$\mathrm{C_{6}H_{5}C(CH_{3})_{2}OOH\xrightarrow{H^{+}/H_{2}O}C_{6}H_{5}OH+CH_{3}COCH_{3}}$$Propanone (acetone), a valuable solvent, is obtained as a co-product, which is why this route dominates world production.
Physical properties. The lower alcohols are colourless liquids; the higher ones are waxy solids. Pure phenols are colourless solids or liquids that slowly turn pink or brown in air (they are oxidised). Ethers of low molar mass are volatile, sweet-smelling, very flammable liquids. What sets alcohols and phenols apart is the $\mathrm{O{-}H}$ bond: the hydrogen on a strongly electronegative oxygen can form hydrogen bonds to the oxygen of a neighbouring molecule.
- Boiling points rise with molar mass (stronger dispersion forces) and fall with branching (smaller surface area): butan-1-ol $\gt$ 2-methylpropan-1-ol $\gt$ butan-2-ol $\gt$ 2-methylpropan-2-ol.
- Alcohols and phenols boil far higher than hydrocarbons, ethers and haloalkanes of similar molar mass, because their molecules are held together by intermolecular hydrogen bonds, $\mathrm{O{-}H\cdots O}$, which must be broken on boiling. Ethanol (46) boils at about $351\ \mathrm{K}$, methoxymethane (46) at about $249\ \mathrm{K}$ and propane (44) at about $231\ \mathrm{K}$.
- Ethers boil close to alkanes of similar mass: the $\mathrm{C{-}O}$ bonds make them weakly polar, but they have no $\mathrm{O{-}H}$ and cannot hydrogen-bond to one another.
- Solubility in water: an alcohol forms hydrogen bonds with water, so the lower alcohols mix with water in all proportions. As the hydrocarbon part grows, solubility falls (pentan-1-ol and beyond are only slightly soluble). An ether can accept hydrogen bonds from water through its oxygen, so ethoxyethane and butan-1-ol dissolve to a similar extent, about $7$–$9\ \mathrm{g}$ per $100\ \mathrm{g}$ of water at room temperature.
1. Counting the wrong carbon. 1°/2°/3° refers to the carbon carrying $\mathrm{OH}$ only. $\mathrm{(CH_{3})_{3}CCH_{2}OH}$ is a primary alcohol although it contains a quaternary carbon.
2. Locants. The $\mathrm{OH}$ group gets the lowest possible number, ahead of $\mathrm{C{=}C}$ and alkyl groups: but-3-en-2-ol, not but-1-en-3-ol. In diols keep the e: ethane-1,2-diol, not "ethan-1,2-diol".
3. Mixing up the two hydrations. $\mathrm{H^{+}/H_{2}O}$ is Markovnikov (through a carbocation); $\mathrm{B_{2}H_{6}}$ then $\mathrm{H_{2}O_{2}/OH^{-}}$ is anti-Markovnikov. Only hydroboration turns a terminal alkene into a primary alcohol.
4. Grignard products. Methanal $\rightarrow$ 1°, other aldehydes $\rightarrow$ 2°, ketones $\rightarrow$ 3°. The carbon count of the product is the sum of the two reagents' carbons.
5. "Ethers have high boiling points because they are polar." They are only weakly polar and form no hydrogen bonds with one another; ethoxyethane boils at about the same temperature as pentane.
In JEE Main, NEET and JEE Advanced 2026 (mechanism of addition of water; carbocation rearrangements); not required by the CBSE syllabus.
Because acid-catalysed hydration goes through a carbocation, the cation can rearrange to a more stable one before water attacks. 3,3-Dimethylbut-1-ene, $\mathrm{(CH_{3})_{3}CCH{=}CH_{2}}$, is protonated to a secondary cation; a methyl group shifts with its electron pair (a 1,2-methyl shift) to give a tertiary cation, and the product is 2,3-dimethylbutan-2-ol, $\mathrm{(CH_{3})_{2}C(OH)CH(CH_{3})_{2}}$, not 3,3-dimethylbutan-2-ol. Hydroboration–oxidation forms no carbocation, so the skeleton never changes: the same alkene gives 3,3-dimethylbutan-1-ol. Hydroboration also adds H and B to the same face of the double bond (syn addition), so 1-methylcyclohexene gives trans-2-methylcyclohexan-1-ol, with the $\mathrm{CH_{3}}$ and $\mathrm{OH}$ on opposite faces of the ring.
Exercises
- 1 Classify each as a 1°, 2° or 3° alcohol (say also if it is allylic or benzylic), a vinylic alcohol or a phenol: (a) $\mathrm{(CH_{3})_{2}CHCH_{2}OH}$; (b) $\mathrm{(CH_{3})_{2}C(OH)CH_{2}CH{=}CH_{2}}$; (c) $\mathrm{C_{6}H_{5}C(CH_{3})_{2}OH}$; (d) cyclohex-2-en-1-ol; (e) 3-methylphenol; (f) $\mathrm{CH_{3}CH{=}CHOH}$.
- 2 Give IUPAC names: (a) $\mathrm{CH_{3}CH(CH_{3})CH(OH)CH_{3}}$; (b) $\mathrm{CH_{3}CH_{2}C(CH_{3})(OH)CH_{2}CH_{3}}$; (c) $\mathrm{HOCH_{2}CH_{2}CH_{2}OH}$; (d) $\mathrm{CH_{3}CH{=}CHCH_{2}OH}$.
- 3 Give IUPAC names of the ethers: (a) $\mathrm{C_{6}H_{5}OCH_{2}CH_{2}CH_{3}}$; (b) $\mathrm{CH_{3}CH_{2}OCH(CH_{3})CH_{2}CH_{3}}$; (c) $\mathrm{CH_{3}OCH_{2}CH_{2}OCH_{3}}$.
- 4 Name the phenols: (a) phenol with $\mathrm{NO_{2}}$ groups at both ortho positions and the para position; (b) a benzene ring with two $\mathrm{OH}$ groups meta to each other; (c) phenol with a methoxy group ortho to $\mathrm{OH}$.
- 5 Write condensed structures for: (a) 2,3-dimethylbutan-2-ol; (b) cyclopentanol; (c) 3-phenylprop-2-en-1-ol; (d) 1-ethoxy-2-methylpropane.
- 6 Each name is wrong. Draw the structure it describes and give the correct name: (a) 2-ethylpropan-1-ol; (b) 3-butanol; (c) 1,1-dimethylethanol; (d) ethoxymethane.
- 7 Give IUPAC names for: isobutyl alcohol $\mathrm{(CH_{3})_{2}CHCH_{2}OH}$, sec-butyl alcohol, ethyl isopropyl ether, $o$-cresol and phenetole.
- 8 Phenylmethanol and 4-methylphenol are both $\mathrm{C_{7}H_{8}O}$. Explain why one is an alcohol and the other a phenol.
- 1 Draw and name all the structural isomers of $\mathrm{C_{3}H_{8}O}$.
- 2 Draw and name all the alcohols of formula $\mathrm{C_{5}H_{12}O}$, and say how many are 1°, 2° and 3°.
- 3 Draw and name all the ethers of formula $\mathrm{C_{5}H_{12}O}$.
- 4 How many isomers of $\mathrm{C_{7}H_{8}O}$ contain a benzene ring? Name them and say which are phenols.
- 5 The $\mathrm{C{-}O{-}H}$ angle in methanol is $108.9^{\circ}$, but the $\mathrm{C{-}O{-}C}$ angle in methoxymethane is $111.7^{\circ}$. Explain both values.
- 6 Explain why the $\mathrm{C{-}O}$ bond in phenol ($136\ \mathrm{pm}$) is shorter than in methanol ($142\ \mathrm{pm}$).
- 7 Ethenol, $\mathrm{CH_{2}{=}CHOH}$, has never been isolated. What does it turn into, and what is the relationship between the two?
- 1 Give the main product of acid-catalysed hydration of (a) propene; (b) 2-methylbut-2-ene; (c) cyclohexene.
- 2 Give the product of hydroboration–oxidation of (a) but-1-ene; (b) 2-methylbut-2-ene; (c) 2-methylpropene.
- 3 Give the alcohol formed: (a) butanal with $\mathrm{H_{2}/Pd}$; (b) cyclohexanone with $\mathrm{NaBH_{4}}$; (c) benzoic acid with $\mathrm{LiAlH_{4}}$, then water; (d) ethyl ethanoate with $\mathrm{H_{2}}$ over a catalyst.
- 4 Give the alcohol formed after hydrolysis: (a) $\mathrm{CH_{3}MgBr+HCHO}$; (b) $\mathrm{C_{6}H_{5}MgBr+CH_{3}CHO}$; (c) $\mathrm{CH_{3}MgI+CH_{3}COCH_{3}}$.
- 5 Suggest two Grignard routes to 2-phenylpropan-2-ol, $\mathrm{C_{6}H_{5}C(CH_{3})_{2}OH}$.
- 6 Which carbonyl compound would you treat with butylmagnesium bromide to make pentan-1-ol? Why can a ketone never give a primary alcohol with a Grignard reagent?
- 7 Why is hydroboration–oxidation, rather than acid-catalysed hydration, used to turn hex-1-ene into hexan-1-ol?
- 8 Write the three steps of the mechanism for the acid-catalysed hydration of ethene, and say which step is slow.
- 9 $4.2\ \mathrm{g}$ of propene is converted into propan-1-ol by hydroboration–oxidation with an $80\%$ yield. What mass of propan-1-ol is obtained?
- 1 Write equations for the preparation of phenol from chlorobenzene, with conditions.
- 2 Write equations for the preparation of phenol from aniline, with conditions.
- 3 Name the intermediate in the cumene process and the co-product, and explain why this route is the cheapest.
- 4 Arrange in increasing order of boiling point, with a reason: ethanol, propane, methoxymethane.
- 5 Arrange in increasing order of boiling point: pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol. Explain the position of butan-2-ol.
- 6 Ethanol mixes with water in all proportions. Why? Why are ethoxyethane and butan-1-ol only partly soluble, and to a similar extent?
- 7 Ethoxyethane, pentane and butan-1-ol have nearly the same molar mass. Explain why two of them boil near $308\ \mathrm{K}$ and the third at $391\ \mathrm{K}$.
- 8 Health. A $1.00\ \mathrm{L}$ bottle of hand sanitiser is $70\%$ (v/v) propan-2-ol. Taking the density of propan-2-ol as $0.786\ \mathrm{g\,mL^{-1}}$, find the mass and the amount (mol) of propan-2-ol in the bottle.
- 9 Phenol ($94\ \mathrm{g\,mol^{-1}}$) boils about $70\ \mathrm{K}$ higher than toluene ($92\ \mathrm{g\,mol^{-1}}$). Explain.
E1. Board · 2 marks Write the structure and IUPAC name of the main product when propene reacts with (a) water in the presence of dilute sulphuric acid; (b) diborane, followed by hydrogen peroxide in aqueous sodium hydroxide.
E2. JEE Main Which combination, followed by hydrolysis, gives 2-methylbutan-2-ol? (A) $\mathrm{CH_{3}CH_{2}MgBr}$ + propanal (B) $\mathrm{CH_{3}MgBr}$ + butanal (C) $\mathrm{CH_{3}CH_{2}MgBr}$ + propanone (D) $\mathrm{(CH_{3})_{2}CHMgBr}$ + methanal
E3. NEET The correct order of boiling points is (A) propane $\gt$ methoxymethane $\gt$ ethanol $\gt$ propan-1-ol (B) propan-1-ol $\gt$ ethanol $\gt$ methoxymethane $\gt$ propane (C) ethanol $\gt$ propan-1-ol $\gt$ propane $\gt$ methoxymethane (D) methoxymethane $\gt$ propan-1-ol $\gt$ ethanol $\gt$ propane
E4. JEE Main Numerical value. Each of the following is treated with $\mathrm{CH_{3}MgBr}$ and the product hydrolysed: methanal, ethanal, propanone, butan-2-one, benzaldehyde, 1-phenylethan-1-one, cyclohexanone, propanal. How many give a tertiary alcohol?
E5. JEE Advanced One or more correct. In the manufacture of phenol from cumene: (A) the co-product is propanone; (B) the intermediate is cleaved by dilute aqueous acid; (C) the oxygen atom of phenol comes from atmospheric oxygen; (D) cumene is made by nitrating benzene.
E6. Board · Case study Read the passage and answer. Ethanol and methoxymethane have the same molecular formula, $\mathrm{C_{2}H_{6}O}$. Ethanol is a liquid that boils at about $351\ \mathrm{K}$; methoxymethane is a gas at room temperature and boils at about $249\ \mathrm{K}$. Ethanol is made industrially by hydrating ethene and by fermentation. (a) Write the structures of the two compounds and explain the difference in their boiling points. (2 marks) (b) What type of isomers are they? (1 mark) (c) Write the equation for the acid-catalysed hydration of ethene. (1 mark)
Research question. Why do alcohol hand rubs contain 60–80% alcohol, a little water and a trace of glycerol, and not pure alcohol?
Why it matters. Hand hygiene is one of the cheapest ways of preventing infections in hospitals. During outbreaks many hospital pharmacies, schools and community groups have had to make their own hand rub, using published recipes based on ethanol or propan-2-ol. Getting the concentration right matters: too little alcohol does not kill germs quickly, and almost-pure alcohol evaporates before it can act and dries the skin.
What to do.
- A widely used recipe for $10\ \mathrm{L}$ of hand rub mixes $8333\ \mathrm{mL}$ of $96\%$ (v/v) ethanol, $417\ \mathrm{mL}$ of $3\%$ hydrogen peroxide and $145\ \mathrm{mL}$ of $98\%$ glycerol, made up to $10\ \mathrm{L}$ with water. Calculate the final percentage (v/v) of ethanol, glycerol and hydrogen peroxide.
- A second recipe uses $7515\ \mathrm{mL}$ of $99.8\%$ propan-2-ol in place of the ethanol. Find its final alcohol percentage.
- Use the structures and hydrogen bonding of glycerol (three $\mathrm{OH}$ groups, high boiling point) to explain why it is added. Look up the role of the trace of hydrogen peroxide.
- Ethanol (bp about $351\ \mathrm{K}$) and propan-2-ol (bp about $355\ \mathrm{K}$) evaporate at similar rates. Find out why a little water makes the alcohol work better against bacteria than pure alcohol does.
Tools: Calculator; reliable public-health sources; the classification and boiling-point ideas in this section.
Hand in: Your three percentages for each recipe, and a half-page explanation of the role of each ingredient written for a school notice board.
Going further. Some products sold during shortages were found to contain methanol. Find out why methanol is dangerous even through the skin, and how its toxicity is linked to its oxidation in the body (Section 7.2).
🧪 Self-check quiz
Q1. Which compound is a benzylic alcohol?
Q2. Hydroboration–oxidation of propene gives
Q3. Methylmagnesium bromide reacts with propanone and the product is hydrolysed. The alcohol formed is
Q4. The $\mathrm{C{-}O{-}C}$ angle in methoxymethane ($111.7^{\circ}$) is larger than the tetrahedral angle because
Q5. Which of these compounds of similar molar mass has the highest boiling point?