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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.
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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Reaction of tert-butyl bromide with sodium methoxide producesa)iso-Butaneb)iso-Butylenec)tert-Butyl methyl etherd)Sodium tert-butoxideCorrect answer is option 'B'. Can you explain this answer?
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