羟基化合物

An alcohol 醇 has the $\text{–OH}$ (hydroxyl) functional group.

Making alcohols

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.

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.

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.

• 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.

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.

Worked example. Which of these give a yellow precipitate with alkaline iodine: ethanol, propan-1-ol, propan-2-ol, butan-2-ol, 2-methylpropan-2-ol? Look for a $\text{CH}_3$ on the carbon that carries the OH, and an H on that same carbon.

• ethanol, $\text{CH}_3\text{CH}_2\text{OH}$: yes ($\text{CH}_3\text{CH(OH)}-$ with H in place of R). It gives $\text{CHI}_3$ and methanoate, $\text{HCO}_2^-$.
• propan-2-ol, $\text{CH}_3\text{CH(OH)CH}_3$: yes, giving $\text{CHI}_3$ and ethanoate, $\text{CH}_3\text{CO}_2^-$.
• butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$: yes, giving $\text{CHI}_3$ and propanoate, $\text{CH}_3\text{CH}_2\text{CO}_2^-$. The $\text{CH}_3$ becomes $\text{CHI}_3$; the rest of the molecule leaves as the carboxylate ion, one carbon shorter.
• propan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$: no. The OH carbon carries $\text{CH}_2\text{CH}_3$, not $\text{CH}_3$.
• 2-methylpropan-2-ol, $(\text{CH}_3)_3\text{COH}$: no. The OH carbon has $\text{CH}_3$ groups but no H, so it cannot be oxidised first.

Dehydration can give more than one alkene

In dehydration the OH leaves together with an H from a carbon next to the OH carbon, and a C=C forms between those two carbons. If the two neighbouring carbons are different, you get different alkenes.

Worked example. Butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$, is heated with $\text{Al}_2\text{O}_3$. The OH is on C2.

• An H lost from C1 gives $\text{CH}_2\text{=CHCH}_2\text{CH}_3$, but-1-ene.
• An H lost from C3 gives $\text{CH}_3\text{CH=CHCH}_3$, but-2-ene. But-2-ene has two different groups on each carbon of the C=C, so it exists as cis and trans isomers.

So the answer is three alkenes: but-1-ene, cis-but-2-ene and trans-but-2-ene. Butan-1-ol, with only one neighbouring carbon, gives only but-1-ene.

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).

Common mistakes.

• Classifying the alcohol by the number of carbons in the whole molecule. Count only the carbons joined to the carbon carrying the OH.
• Writing $[\text{O}]$ for a tertiary alcohol and giving a product. A tertiary alcohol has no H on the OH carbon, so it is not oxidised.
• Saying distillation "makes" the aldehyde. Both set-ups make the aldehyde first; distillation simply removes it before it is oxidised further.
• Missing the cis and trans isomers of but-2-ene when you list dehydration products.
• Saying any alcohol with a $\text{CH}_3$ gives the iodoform test. The $\text{CH}_3$ must be on the same carbon as the OH, and that carbon needs an H.

• 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").