Q.The following is not an appropriate reaction for the preparation of t-butyl ethyl ether. $\mathrm{C_2H_5ONa \; + \; CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-Cl \longrightarrow CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OC_2H_5}$

The Williamson Ether Synthesis is an $S_N2$ reaction where an alkoxide attacks a primary alkyl halide to form an ether.

In the given reaction, sodium ethoxide ($C_2H_5ONa$) is a strong base and a good nucleophile, while t-butyl chloride ($(CH_3)_3C-Cl$) is a tertiary alkyl halide. Tertiary alkyl halides are sterically hindered, making them very poor substrates for $S_N2$ reactions. Instead, the strong base will abstract a $\beta$-hydrogen, leading to an E2 elimination reaction.

(i) The major product of this reaction will be the alkene formed via E2 elimination.

$C_2H_5ONa + (CH_3)_3C-Cl \xrightarrow{\text{E2 elimination}} CH_2=C(CH_3)_2 + C_2H_5OH + NaCl$ …

The Williamson ether synthesis is a classic method for making ethers: an alkoxide ion attacks an alkyl halide in an SN2 reaction. The key constraint is that the alkoxide must be the stronger base (the one derived from the alcohol you want as the alkoxy part), and the alkyl halide should be primary or methyl to avoid elimination. When you mix a strong base with a tertiary halide, elimination almost always wins over substitution.

Let’s see why the given reaction fails and how to fix it.

1. Identify the reactants and the intended product The reaction is:

Sodium ethoxide ($C_2H_5ONa$) is a strong base and a good nucleophile. t-Butyl chloride ($(CH_3)_3C-Cl$) is a tertiary alkyl halide. The target is t-butyl ethyl ether.

2. Why this reaction fails — the elimination problem Tertiary halides cannot undergo SN2 reactions because the backside attack is blocked by the three bulky methyl groups. With a strong base like ethoxide, the only viable pathway is E2 elimination. The base abstracts a β-hydrogen from t-butyl chloride, and the chloride leaves, forming isobutylene (2-methylpropene) as the major product:

So the major product is isobutylene, not the ether.

3. Answer to part (i): major product

   The major product of the given reaction is isobutylene ($(CH_3)_2C=CH_2$). A small amount of substitution might occur via SN1 (if the carbenium ion forms), but under typical conditions with a strong base, elimination dominates.

4. Part (ii): correct preparation of t-butyl ethyl ether …

Williamson Ether Synthesis — Method & Analysis

Method: Williamson Ether Synthesis

This is an SN2 reaction between an alkoxide ion and an alkyl halide (or tosylate). The alkoxide acts as a nucleophile, attacking the electrophilic carbon of the alkyl halide.

Key rule: The alkoxide must come from the less hindered alcohol, and the alkyl halide must be primary (or methyl) for a clean SN2 reaction. If the alkyl halide is tertiary, elimination dominates.

(i) Major product of the given reaction

Given:

$C_2H_5ONa + (CH_3)_3C-Cl \longrightarrow$

• Alkoxide: ethoxide ($C_2H_5O^-$) — a strong base and nucleophile
• Alkyl halide: tert-butyl chloride — a tertiary halide

Since the halide is tertiary, SN2 is impossible (steric hindrance blocks backside attack). Instead, the strong base eliminates H⁺ from the β-carbon, forming an alkene.

Major product:

Minor product: Some substitution may occur via SN1, but elimination dominates.

(ii) Suitable preparation of t-butyl ethyl ether

To get the ether cleanly, swap the roles:

• Use ethyl halide (primary — good for SN2)
• Use sodium tert-butoxide (alkoxide from the more hindered alcohol)

Reaction: …

Why This Reaction Fails (The Core Concept)

Williamson ether synthesis is an SN2 reaction between an alkoxide ion and an alkyl halide.

For it to work well, the alkyl halide must be primary (or methyl) — because SN2 needs a back-side attack with no steric hindrance.

Here, the alkyl halide is t-butyl chloride — a tertiary halide.

Tertiary halides cannot undergo SN2 (too crowded). Instead, they undergo E2 elimination (with a strong base like alkoxide) to give an alkene.

(i) Major Product of the Given Reaction

The reaction:

• $C_2H_5O^-$ is a strong base and a bulky nucleophile.
• $(CH_3)_3C-Cl$ is a tertiary halide.
• Result: E2 elimination dominates.

Major product: Isobutylene (2-methylpropene)

Plus: $C_2H_5OH$ and $NaCl$ as byproducts.

Key point: No ether forms here.

(ii) Suitable Reaction for t-Butyl Ethyl Ether

To make an unsymmetrical ether like t-butyl ethyl ether, choose the correct combination:

• Alkoxide from the smaller alkyl group (less steric hindrance)
• Alkyl halide that is primary (good for SN2)

Correct approach:

Use sodium ethoxide ($C_2H_5ONa$) with primary alkyl halide — but here the t-butyl group must come from the alkoxide.

So the correct reaction is:

• $(CH_3)_3C-ONa$ = sodium t-butoxide (alkoxide from the bulky group)
• $C_2H_5Br$ = ethyl bromide (primary halide — perfect for SN2)

Common Mistakes & How to Avoid Each