- recall the reactions (reagents and conditions) by which alcohols can be produced: (a) electrophilic addition of steam to an alkene, $\text{H}_2\text{O(g)}$ and $\text{H}_3\text{PO}_4$ catalyst (b) reaction of alkenes with cold dilute acidified potassium manganate(VII) to form a diol (c) substitution of a halogenoalkane using $\text{NaOH(aq)}$ and heat (d) reduction of an aldehyde or ketone using $\text{NaBH}_4$ or $\text{LiAlH}_4$ (e) reduction of a carboxylic acid using $\text{LiAlH}_4$ (f) hydrolysis of an ester using dilute acid or dilute alkali and heat
- describe: (a) the reaction with oxygen (combustion) (b) substitution to form halogenoalkanes, e.g. by reaction with $\text{HX(g)}$; or with $\text{KCl}$ and concentrated $\text{H}_2\text{SO}_4$ or concentrated $\text{H}_3\text{PO}_4$; or with $\text{PCl}_3$ and heat; or with $\text{PCl}_5$; or with $\text{SOCl}_2$ (c) the reaction with $\text{Na(s)}$ (d) oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or acidified $\text{KMnO}_4$ to: (i) carbonyl compounds by distillation (ii) carboxylic acids by refluxing (primary alcohols give aldehydes which can be further oxidised to carboxylic acids, secondary alcohols give ketones, tertiary alcohols cannot be oxidised) (e) dehydration to an alkene, by using a heated catalyst, e.g. $\text{Al}_2\text{O}_3$ or a concentrated acid (f) formation of esters by reaction with carboxylic acids and concentrated $\text{H}_2\text{SO}_4$ as catalyst as exemplified by ethanol
- (a) classify alcohols as primary, secondary and tertiary alcohols, to include examples with more than one alcohol group (b) state characteristic distinguishing reactions, e.g. mild oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$, colour change from orange to green
- deduce the presence of a $\text{CH}_3\text{CH(OH)}-$ group in an alcohol, $\text{CH}_3\text{CH(OH)}-\text{R}$, from its reaction with alkaline $\text{I}_2\text{(aq)}$ to form a yellow precipitate of tri-iodomethane and an ion, $\text{RCO}_2^-$
- explain the acidity of alcohols compared with water
Hydroxy compounds
A-Level Chemistry · Topic 16
16.1
Alcohols
Syllabus
Source: Cambridge International syllabus
An alcohol 醇 has the $\text{–OH}$ (hydroxyl) functional group.
Ethanol, the alcohol in hand sanitiser, kills microbes
Making alcohols
| Method | Reagents and conditions |
|---|---|
| addition of steam to an alkene | $\text{H}_2\text{O(g)}$, $\text{H}_3\text{PO}_4$ catalyst (electrophilic addition 亲电加成) |
| an alkene 烯烃 with cold dilute $\text{KMnO}_4$ | gives a diol 二醇 (two $\text{–OH}$ groups) |
| substitution of a halogenoalkane 卤代烷 | $\text{NaOH(aq)}$, heat |
| reduction 还原 of an aldehyde 醛 or ketone 酮 | $\text{NaBH}_4$ or $\text{LiAlH}_4$ |
| reduction of a carboxylic acid 羧酸 | $\text{LiAlH}_4$ |
| hydrolysis 水解 of an ester 酯 | dilute acid or alkali, heat |
Reactions of alcohols
- combustion: alcohols burn in oxygen to give carbon dioxide and water.
- substitution to a halogenoalkane, for example with $\text{HX}$, $\text{PCl}_5$, $\text{PCl}_3$ and heat, or $\text{SOCl}_2$.
- with sodium: alcohols react with sodium metal to give hydrogen and a sodium alkoxide — like water, but more slowly.
- oxidation 氧化 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (or $\text{KMnO}_4$). The product depends on the class of alcohol (see below).
- dehydration 脱水 to an alkene, using a hot $\text{Al}_2\text{O}_3$ catalyst or concentrated acid.
- ester formation: an alcohol reacts with a carboxylic acid (with concentrated $\text{H}_2\text{SO}_4$ catalyst) to make an ester.
An alcohol can burn, dehydrate, esterify or be oxidised
Three classes and how oxidation tells them apart
An alcohol is primary 伯, secondary 仲 or tertiary 叔, depending on how many carbons are joined to the carbon holding the $\text{–OH}$. Some molecules have more than one $\text{–OH}$ group.
The class is set by the carbon holding the –OH: one R group makes it primary, two secondary, three tertiary — which decides how it oxidises
| Class | Oxidation product |
|---|---|
| primary | aldehyde (by distillation 蒸馏), then carboxylic acid (by reflux 回流) |
| secondary | a ketone |
| tertiary | not oxidised |
Oxidation by class: a primary alcohol gives an aldehyde then a carboxylic acid, a secondary gives a ketone, a tertiary is not oxidised
In a quick test, acidified $\text{K}_2\text{Cr}_2\text{O}_7$ turns from orange to green with a primary or secondary alcohol, but stays orange with a tertiary alcohol.
Acidified dichromate turns orange to green with a primary or secondary alcohol, but stays orange with a tertiary alcohol
- distillation removes the aldehyde as it forms, before it can be oxidised further.
- reflux keeps boiling the mixture and returning the vapour, so the alcohol is fully oxidised to the carboxylic acid.
Distillation removes the aldehyde as it forms; reflux keeps boiling and returning the vapour, fully oxidising to the acid
A real distillation set-up: the vapour boils off, cools in the condenser and is collected — the same idea separates ethanol from a fermented mixture
The iodoform test
If you warm an alcohol that contains the $\text{CH}_3\text{CH(OH)}-$ group with alkaline aqueous iodine, you get a pale yellow precipitate of tri-iodomethane 三碘甲烷 ($\text{CHI}_3$) and the ion $\text{RCO}_2^-$. This is a useful test for that group.
Acidity of alcohols
The $\text{–OH}$ group makes alcohols very weakly acidic: they can lose the $\text{H}^+$ to form an $\text{RO}^-$ ion. But their acidity 酸性 is lower than that of water. This is because the alkyl group pushes electron density onto the oxygen, which makes the $\text{RO}^-$ ion less stable, so the alcohol holds onto its $\text{H}^+$ more tightly.
Worked example. Three unlabelled bottles hold butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. How does warming each with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ tell them apart? What matters is how many hydrogens sit on the carbon carrying the $\text{OH}$. Butan-1-ol is primary (two such hydrogens): the orange dichromate turns green, and by distilling you collect an aldehyde (butanal), or by refluxing you get the carboxylic acid (butanoic acid). Butan-2-ol is secondary (one such hydrogen): also green, but the product is a ketone (butanone), which will not oxidise further. 2-methylpropan-2-ol is tertiary (no such hydrogen): there is nothing to remove, so the dichromate stays orange. The colour only separates the tertiary from the other two - to split primary from secondary you must identify the product (an aldehyde gives a silver mirror with Tollens', a ketone does not).
Explore
Alcohol reaction map
Choose what happens to alcohols under different reagents.
Explore
Alcohol oxidation test lab
Classify alcohols by oxidation product and observation.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| alcohol | 醇 | chún |
| electrophilic addition | 亲电加成 | qīn diàn jiā chéng |
| alkene | 烯烃 | xī tīng |
| diol | 二醇 | èr chún |
| halogenoalkane | 卤代烷 | lǔ dài wán |
| reduction | 还原 | huán yuán |
| aldehyde | 醛 | quán |
| ketone | 酮 | tóng |
| carboxylic acid | 羧酸 | suō suān |
| hydrolysis | 水解 | shuǐ jiě |
| ester | 酯 | zhǐ |
| oxidation | 氧化 | yǎng huà |
| dehydration | 脱水 | tuō shuǐ |
| primary | 伯 | bó |
| secondary | 仲 | zhòng |
| tertiary | 叔 | shū |
| distillation | 蒸馏 | zhēng liú |
| reflux | 回流 | huí liú |
| tri-iodomethane | 三碘甲烷 | sān diǎn jiǎ wán |
| acidity | 酸性 | suān xìng |
16.1
Exam tips
- Classify the alcohol as primary, secondary or tertiary first — it decides the oxidation product.
- Oxidation with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (orange → green): primary → aldehyde (distil) → acid (reflux); secondary → ketone; tertiary → no reaction.
- Distinguish the conditions for aldehyde vs acid from a primary alcohol (distillation vs reflux).
- Name esters correctly (the acid part comes second, ending "-oate").