---
title: "Primary alkyl halide with alcoholic and aqueous alkali"
subject: "Engineering Chemistry"
chapter: 10
key: "semester_2/engineering_chemistry:primary-halide-alc-aq-alkali"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/primary-halide-alc-aq-alkali
site: BCT Engineering
---

# Primary alkyl halide with alcoholic and aqueous alkali

Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions

- Asked in: 2067 Chaitra, 5 marks
- Video: [Reaction of Haloalkane with Aqueous KOH & Alcoholic KOH | Mechanism](https://www.youtube.com/watch?v=tlrRl0QB0CQ) — Your Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/primary-halide-alc-aq-alkali
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10

## Question

Explain the reaction mechanism involved when primary alkyl halide react with alcoholic alkali and aqueous alkali.

## Short answer

With aqueous alkali, primary alkyl halides undergo SN2 substitution: OH- attacks from backside, forming an alcohol (e.g., CH3CH2OH) via a bimolecular mechanism. With alcoholic alkali, ethoxide ion acts as a strong base, leading to E2 elimination: removal of β-H and departure of Br- yields an alkene (e.g., CH2=CH2).

## Answer

A primary alkyl halide gives different products with the two reagents.

#### With aqueous alkali: substitution (SN2)
Hydroxide acts as a nucleophile and replaces Br.
```text
HO⁻ + CH₃CH₂—Br → [HO···CH₂(CH₃)···Br]⁻‡ → CH₃CH₂OH + Br⁻
```
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr. Rate = k[RX][OH⁻] (bimolecular).

#### With alcoholic alkali: elimination (E2)
In alcohol, ethoxide forms (OH⁻ + C₂H₅OH ⇌ C₂H₅O⁻ + H₂O). It is a strong base and removes the β-hydrogen as the C–Br bond breaks.
```text
EtO⁻  H—CH₂—CH₂—Br → [EtO···H···CH₂···CH₂···Br]⁻‡ → CH₂=CH₂ + EtOH + Br⁻
```
CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O. Rate = k[RX][base] (bimolecular).

Thus aqueous alkali favours substitution, whereas alcoholic alkali and heat favour elimination.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does a primary halide give with aqueous alkali?
   A. Alkene, by E2
   B. Alcohol, by SN2 substitution
   C. Alkane
   D. Ether by SN1
   Answer: B. OH- acts as a nucleophile and replaces Br. (Key point: Aqueous alkali product)

2. What does a primary halide give with alcoholic alkali and heat?
   A. Alkene, by E2 elimination
   B. Alcohol, by SN2
   C. Carbocation, by SN1
   D. Alkyne
   Answer: A. Ethoxide formed in alcohol is a strong base and removes beta-H. (Key point: Alcoholic alkali product)

3. Why is a stronger base available in alcoholic KOH?
   A. Alcohol makes OH- a better nucleophile
   B. Alcohol converts the halide to a carbocation
   C. Water is completely removed
   D. OH- and ethanol form ethoxide (C2H5O-), a strong base
   Answer: D. OH- + C2H5OH <-> C2H5O- + H2O. (Key point: Why alcoholic medium gives a stronger base)

4. Which rate law applies to both reactions of a primary halide?
   A. Rate = k[RX], unimolecular
   B. Rate = k[base] only
   C. Rate = k[RX][base or nucleophile], bimolecular
   D. Rate = k[RX]^2
   Answer: C. Both SN2 and E2 are second order. (Key point: Rate laws)

## Related in Engineering Chemistry

- [Mechanism for primary alkyl halide in aqueous medium; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-primary-solvent) — Engineering Chemistry, chapter 10
- [3° alkyl halide with alcoholic KOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-alc-koh-solvent) — Engineering Chemistry, chapter 10
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- [Tertiary alkyl halide with aq. NaOH / alc. KOH](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-halide-naoh-koh) — Engineering Chemistry, chapter 10
- [SN reactions; hydrolysis of 3° alkyl halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn-tertiary-hydrolysis) — Engineering Chemistry, chapter 10, asked 2×
- [Hydrolysis of 3° alkyl halide by aq. NaOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-hydrolysis-solvent) — Engineering Chemistry, chapter 10

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