Question

2-Methylpropene is obtained when sodium methoxide reacts with

• A. n-butyl bromide
• B. n-propyl bromide
• C. sec-butyl bromide
• D. isopropyl bromide
• E. tert-butyl bromide

Hint:

In Williamson ether synthesis, the alkoxide can be bulky, but the alkyl halide must be primary to avoid alkene formation by elimination.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
When an alkyl halide reacts with a strong base/nucleophile like sodium methoxide, two competing reactions occur: Substitution (\(S_N2\)) and Elimination (\(E2\)).

Step 3: Detailed Explanation:
Sodium methoxide (\(CH_3ONa\)) is a strong base.
- With primary alkyl halides (like n-butyl bromide), substitution (ether formation) is the major product.
- With secondary alkyl halides, both occur, with substitution often still significant.
- With tertiary alkyl halides (like tert-butyl bromide), elimination becomes the exclusive reaction due to steric hindrance preventing the nucleophile from attacking the carbon.
The reaction of \(CH_3ONa\) with tert-butyl bromide results in the formation of an alkene:
\[ (CH_3)_3C-Br + CH_3O^- \to (CH_3)_2C=CH_2 + CH_3OH + Br^- \]
The product is 2-methylpropene (isobutylene).

Step 4: Final Answer:
The reaction uses tert-butyl bromide.