Hydroxy compounds · ⁨ヒドロキシ化合物⁩

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.

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).

• 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").