# Alcohols

> Chemistry · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9701-u14-alcohols/

This subtopic covers nomenclature, classification, preparation, key reactions, and chemical identification of alcohols, a core functional group in CIE A-Level organic chemistry. We break down all common exam reactions and highlight frequent pitfalls.

**Prerequisites:** [Organic functional groups and IUPAC nomenclature](https://www.owlsprep.com/study/cie-9701-u12-organic-nomenclature/); [Halogenoalkanes](https://www.owlsprep.com/study/cie-9701-u14-halogenoalkanes/)

## Learning objectives

- Name alcohols according to IUPAC rules and classify them by degree of substitution
- Describe common methods for laboratory and industrial preparation of alcohols
- Predict products of combustion, oxidation, substitution and elimination reactions of alcohols
- Use chemical tests to distinguish between different classes of alcohols

## Nomenclature and Classification

**Hydroxyl Group** — The defining functional group of alcohols, bonded to a saturated sp³-hybridised carbon atom

*Notation:* -OH

*Example:* Ethanol ($CH_3CH_2OH$) has one hydroxyl group

Alcohols are named by replacing the -e suffix of the parent alkane with -ol. The position of the hydroxyl group is indicated by a number before the suffix. For alcohols with multiple hydroxyl groups, use suffixes like -diol or -triol.

**Alcohol Classification** — Alcohols are classified by the number of alkyl groups bonded to the carbon that bears the hydroxyl group

*Example:* 1° = 1 alkyl group, 2° = 2 alkyl groups, 3° = 3 alkyl groups

**Worked example:** Classify 2-methylpropan-2-ol as primary, secondary or tertiary

1. Step 1: Identify the carbon bonded to the hydroxyl group (the second carbon in the parent chain)
2. Step 2: Count how many other carbon atoms are bonded to this carbon:
3. $$(CH_3)_3COH$$
4. The hydroxyl-bearing carbon is bonded to 3 separate methyl groups (3 other carbon atoms)
5. Step 3: Conclusion: 3 alkyl groups = tertiary (3°) alcohol

## Preparation of Alcohols

There are four common preparation routes for alcohols regularly tested in CIE exams:

- Nucleophilic substitution (hydrolysis) of halogenoalkanes with aqueous sodium hydroxide
- Electrophilic addition (hydration) of alkenes with steam and acid catalyst (industrial)
- Reduction of aldehydes/ketones with $NaBH_4$ to 1°/2° alcohols
- Anaerobic fermentation of glucose to produce ethanol

**Worked example:** State reagents, conditions and the product for preparation of ethanol from ethene

1. Step 1: This is an industrial hydration reaction of the alkene double bond
2. Step 2: Reagents are ethene and steam. Reaction conditions:
3. Temperature = 300°C, Pressure = 60-70 atm, Catalyst = concentrated $H_3PO_4$ on silica
4. $$CH_2=CH_2(g) + H_2O(g) \rightleftharpoons CH_3CH_2OH(g)$$
5. Step 3: Unreacted ethene is recycled to increase overall yield of ethanol

> **tip**
>
> Fermentation requires anaerobic conditions, yeast enzymes, and ~30°C temperature, this is a common short exam question

## Key Reactions of Alcohols

Alcohols undergo four core reaction classes tested in exams: combustion, oxidation, substitution to form halogenoalkanes, and elimination (dehydration) to form alkenes. Oxidation is the most frequently examined, with products dependent on alcohol classification.

**Oxidation of Alcohols** — Uses acidified potassium dichromate(VI) ($K_2Cr_2O_7$) as oxidising agent. Products depend on alcohol class and reaction conditions

*Example:* 1° → aldehyde (distillation) → carboxylic acid (reflux); 2° → ketone (reflux); 3° no oxidation

**Worked example:** What product forms when butan-1-ol is heated under reflux with excess acidified $K_2Cr_2O_7$?

1. Step 1: Butan-1-ol is a primary alcohol, with the hydroxyl group on the terminal carbon
2. Step 2: Heating under reflux with excess oxidising agent causes full oxidation to the carboxylic acid
3. $$CH_3CH_2CH_2CH_2OH \xrightarrow[reflux]{H^+ / K_2Cr_2O_7} CH_3CH_2CH_2COOH$$
4. If the product was distilled off as it formed, the intermediate aldehyde (butanal) would be collected instead
5. Observation: Orange dichromate(VI) ions reduce to green chromium(III) ions

Other key reactions: Substitution with $PCl_5$, $SOCl_2$ or $HCl/ZnCl_2$ replaces -OH with -Cl to form a halogenoalkane. Dehydration (elimination) with concentrated acid catalyst eliminates water to form an alkene, following Zaitsev's rule (more substituted alkene = major product).

## Chemical Identification of Alcohols

Alcohols can be identified via simple chemical tests, as well as spectroscopic methods covered in other subtopics. The standard test for a hydroxyl group is reaction with sodium metal.

**Worked example:** Describe a test using sodium to distinguish between ethanol and ethane

1. Step 1: Add a small piece of clean sodium metal to each test sample
2. Step 2: Ethanol contains a reactive hydroxyl group that reacts with sodium
3. $$2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa + H_2$$
4. Result for ethanol: Effervescence of hydrogen gas is observed, and sodium dissolves
5. Result for ethane: No reaction occurs, with no visible change, confirming it is not an alcohol

> **info**
>
> Acidified potassium dichromate distinguishes alcohol classes: 1° and 2° turn orange to green, 3° shows no color change

## Common pitfalls

- **Wrong:** Claiming tertiary alcohols can be oxidized by acidified dichromate
  - Why it fails: Tertiary alcohols have no C-H bond on the hydroxyl-bearing carbon, so oxidation cannot occur without breaking the carbon skeleton
  - Correct: State that tertiary alcohols do not react with acidified potassium dichromate(VI) under standard conditions
- **Wrong:** Drawing an aldehyde as product when 1° alcohol is refluxed with excess oxidant
  - Why it fails: Reflux with excess oxidising agent gives full oxidation to carboxylic acid; aldehyde is only collected if distilled off immediately
  - Correct: Product = carboxylic acid for 1° alcohol under reflux, aldehyde for distillation of 1° alcohol
- **Wrong:** Confusing conditions for hydrolysis of halogenoalkanes vs elimination
  - Why it fails: Different conditions give completely different products, a common 1-2 mark exam question
  - Correct: Aqueous NaOH = substitution (alcohol product); ethanolic NaOH = elimination (alkene product)
- **Wrong:** Claiming all alcohols turn acidified dichromate from orange to green
  - Why it fails: Only oxidisable alcohols cause the color change; tertiary alcohols do not react
  - Correct: Only 1° and 2° alcohols give an orange to green color change with acidified dichromate

## Cheatsheet

| Property | Primary (1°) | Secondary (2°) | Tertiary (3°) |
| --- | --- | --- | --- |
| OH-C bonded to | 1 other C | 2 other C | 3 other C |
| K₂Cr₂O₇ (distill) | Aldehyde | Ketone | No reaction |
| K₂Cr₂O₇ (reflux) | Carboxylic acid | Ketone | No reaction |
| Reaction with Na | H₂ + sodium alkoxide | H₂ + sodium alkoxide | H₂ + sodium alkoxide |
| Dehydration product | Follows Zaitsev rule | Follows Zaitsev rule | Follows Zaitsev rule |

## What's next

Alcohols are a foundational functional group that connects to almost all other organic chemistry topics in CIE A-Level. The oxidation reactions of alcohols covered here are the first step in learning about carbonyl compounds and carboxylic acids, while substitution reactions of alcohols are the basis for ester formation. Alcohols also feature prominently in multi-step organic synthesis questions, which make up a large share of extended response marks in Papers 2 and 4.

- [Carbonyl compounds and carboxylic acids](https://www.owlsprep.com/study/cie-9701-u15-overview/)
- [Aldehydes and Ketones](https://www.owlsprep.com/study/cie-9701-u15-aldehydes-and-ketones/)
- [Carboxylic acids](https://www.owlsprep.com/study/cie-9701-u15-carboxylic-acids/)

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