6E.37-6E.38 - Alcohol Classification And Reactions

One Functional Group, Several Routes

A bottle labelled "propanol" is not enough information for a chemist. Propan-1-ol and propan-2-ol have the same molecular formula, but they do not give the same oxidation products, because the OH group sits in a different carbon environment. Alcohol chemistry therefore begins with a structural question: what is the carbon attached to the OH group bonded to?

[DIAGRAM: asset_name: Alcohol Reaction Map; asset_slug: edexcel_a_level_chemistry_l038_alcohol_reaction_map; recommended_method: deterministic_chart; description: Four alcohol reaction families with the correct reagents and product families, separating oxidation from substitution and dehydration.]

The pattern is not "alcohols do lots of reactions"; it is "the OH group can be used as a handle for predictable functional-group conversions". The same starting alcohol can be:

The reaction name alone is not enough. Questions usually expect the reagent, condition and product to match each other: for example, $\ce{K2Cr2O7}/\ce{H2SO4}$ is an oxidation condition, while concentrated $\ce{H3PO4}$ is an elimination condition.

Classifying The OH Carbon

Alcohols are classified by looking only at the carbon atom directly bonded to the OH group. Do not classify from the longest carbon chain, the total number of carbons, or the position number in the name unless that position tells you the environment of the OH carbon.

Methanol, $\ce{CH3OH}$, has no carbon group attached to the OH carbon, but it behaves with the primary-alcohol pattern in this topic because the OH carbon still has C-H bonds that can be oxidised. For larger alcohols, read the class directly from the displayed or condensed structural formula.

Common-error contrast

$\ce{CH3CH2CH2OH}$ and $\ce{CH3CH(OH)CH3}$ both contain three carbon atoms, but only one is primary. In $\ce{CH3CH2CH2OH}$, the OH carbon is at the end and is bonded to one other carbon; in $\ce{CH3CH(OH)CH3}$, the OH carbon is in the middle and is bonded to two other carbons. The class follows the OH carbon, not the molecular formula.

Combustion And Halogenation

In complete combustion, an alcohol reacts with oxygen in air to form carbon dioxide and water. For a saturated monohydric alcohol with formula $\mathrm{C}_{n}\mathrm{H}_{2n+2}\mathrm O$, the balanced pattern is:

For ethanol:

$\ce{C2H5OH + 3O2 -> 2CO2 + 3H2O}$

The chemical reasoning is atom accounting. All carbon atoms in the alcohol become $\ce{CO2}$, all hydrogen atoms become $\ce{H2O}$, and the coefficient of $\ce{O2}$ is then chosen to balance oxygen. If a question simply says "combustion in oxygen in air", assume complete combustion unless the stem gives evidence of limited oxygen, carbon monoxide or soot.

For a gas-volume calculation, first use the balanced equation to convert alcohol amount to oxygen amount, then use $V=nV_{\mathrm m}$ with the supplied molar gas volume. For example, 0.100 mol ethanol needs $3\times0.100=0.300\,\mathrm{mol}$ oxygen. At a supplied room-temperature molar gas volume of $24.0\,\mathrm{dm^{3}\,mol^{-1}}$, this is $0.300\times24.0=7.20\,\mathrm{dm^{3}}$. The final unit is a volume, and the equation coefficient supplies the ratio.

Halogenation converts an alcohol into a halogenoalkane by replacing OH with a halogen atom. Three reagent routes replace the OH group:

For the iodoalkane route, red phosphorus and iodine generate phosphorus(III) iodide in the reaction mixture, so you may see $\ce{PI3}$ used as the reacting halogenating agent. The starting reagents are red phosphorus and iodine; the phosphorus(III) iodide is formed in situ.

Worked example

Write an equation for the reaction of propan-1-ol with phosphorus(V) chloride, and name the organic product.

Start from the functional group conversion: $\ce{ROH}$ becomes $\ce{RCl}$. Propan-1-ol is $\ce{CH3CH2CH2OH}$, so the organic product is $\ce{CH3CH2CH2Cl}$.

The product is 1-chloropropane. The equation also shows why this reaction can give steamy acidic fumes: hydrogen chloride is a by-product. Hydrogen chloride forms acidic fumes in moist air, so this reaction needs suitable ventilation and eye protection.

Guided practice

Complete the product for each conversion.

The mechanism is not required here. The assessed decision is the functional-group replacement and the correct reagent-product pairing.

Oxidation Depends On Class

Oxidation of alcohols in this lesson uses potassium dichromate(VI) in dilute sulfuric acid. In equations, the oxidising agent may be represented as [O], which keeps the focus on the organic product rather than the full ionic redox equation. The usual observation is that acidified dichromate(VI) changes from orange to green when it is reduced while the alcohol is oxidised.