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}{|}}{\u

Question

The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

$\mathrm{C_2H_5ONa \; + \; CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-Cl \longrightarrow CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OC_2H_5}$

(i) What would be the major product of this reaction?

(ii) Write a suitable reaction for the preparation of t-butylethyl ether.

Accepted Answer

> [!TLDR]
> The given reaction fails because t-butyl chloride undergoes elimination (E2) instead of substitution (SN2) with ethoxide, giving isobutylene as the major product. The correct preparation uses t-butoxide with ethyl bromide via Williamson ether synthesis.

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.

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1. **Identify the reactants and the intended product**  
   The reaction is:  
   $$C_2H_5ONa + (CH_3)_3C-Cl \longrightarrow (CH_3)_3C-OC_2H_5$$  
   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:  
   $$C_2H_5O^- + (CH_3)_3C-Cl \longrightarrow (CH_3)_2C=CH_2 + C_2H_5OH + Cl^-$$  
   So the major product is **isobutylene**, not the ether.

> [!WARNING]  
   > A common mistake is to assume that any alkoxide + alkyl halide pair gives an ether. Williamson synthesis works only when the alkyl halide is primary (or methyl/secondary with care). Tertiary halides always eliminate with strong bases.

3. **Answer to part (i): major product**  
   The major product of the given reaction is **isobutylene** ($(CH_3)_2

Alkyl halide	Works?	Why
Methyl ( $CH_3X$ )	Yes	Least hindered, fastest SN2
Primary (1°)	Yes	Clean SN2
Secondary (2°)	Sometimes	Works if not too bulky; elimination competes
Tertiary (3°)	No	Elimination dominates
Aryl (e.g., bromobenzene)	No	SN2 impossible on $sp^2$ carbon

Alkyl Halide Type	Steric Hindrance	$\text{S}_\text{N}2$ Reactivity
Methyl ( $\text{CH}_3\text{X}$ )	Minimal	Very fast
Primary ( $\text{RCH}_2\text{X}$ )	Low	Fast
Secondary ( $\text{R}_2\text{CHX}$ )	Moderate	Slow
Tertiary ( $\text{R}_3\text{CX}$ )	High	Does not occur

Halide	Leaving Group Ability	Reason
$\text{I}^-$	Excellent	Large, polarizable, weak base
$\text{Br}^-$	Good	Moderate size, weak base
$\text{Cl}^-$	Fair	Smaller, stronger base
$\text{F}^-$	Poor	Small, strong base, holds tightly

Mistake	Why It's Wrong	How to Avoid
Using a tertiary alkyl halide (like t-butyl chloride)	SN2 impossible; elimination dominates	Always check: alkyl halide must be primary (or methyl) for SN2
Thinking any alkoxide + any alkyl halide works	Ignores steric hindrance and elimination side reactions	Identify which group is bulky — put it in the alkoxide, not the halide

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