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Reaction of tert-butyl bromide with sodium methoxide produces
  • a)
    iso-Butane
  • b)
    iso-Butylene
  • c)
    tert-Butyl methyl ether
  • d)
    Sodium tert-butoxide
Correct answer is option 'B'. Can you explain this answer?
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Reaction of Tert-Butyl Bromide with Sodium Methoxide

Reaction: Tert-butyl bromide reacts with sodium methoxide in methanol to form iso-Butylene.

Explanation:

Tert-butyl bromide is a primary halide that reacts with sodium methoxide in methanol to form iso-Butylene. The reaction mechanism involves the substitution of bromine atom with methoxide ion, which is a nucleophilic substitution reaction.

The reaction takes place in two steps:

Step 1: Formation of tert-butoxide ion

Tert-butyl bromide reacts with sodium methoxide in methanol to form tert-butoxide ion and sodium bromide.

C4H9Br + NaOCH3 → C4H9O–Na+ + NaBr

Step 2: Formation of iso-Butylene

Tert-butoxide ion then eliminates a molecule of methanol to form iso-Butylene.

C4H9O–Na+ → C4H8 + NaOCH3

The final product of the reaction is iso-Butylene, which is a branched alkene. The reaction is an example of an elimination reaction, where a leaving group is eliminated from the substrate to form a double bond.

Conclusion:

The reaction of tert-butyl bromide with sodium methoxide produces iso-Butylene, which is a branched alkene. The reaction mechanism involves the substitution of bromine atom with methoxide ion followed by the elimination of a molecule of methanol.
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