Question

Which of the following compounds is obtained when t-butyl bromide is treated with alcoholic ammonia?

• A. CH\(_3\)-C-CH\(_3\)-NHBr
• B. CH\(_3\)-C-CH\(_2\)-NH\(_2\)Br
• C. CH\(_3\)-C=CH\(_2\)
• D. CH\(_3\)-C-CH\(_3\)-NH\(_2\)

Hint:

In the reaction of t-butyl bromide with alcoholic ammonia, elimination occurs to form an alkene, not just substitution.

Answer 1

The Correct Option is C

Step 1: Understanding the reaction.
When t-butyl bromide (C\(_4\)H\(_9\)Br) is treated with alcoholic ammonia (NH\(_3\)), the ammonia replaces the bromine atom (via nucleophilic substitution) to form an amine. However, due to the structure of t-butyl bromide, a double bond is formed through elimination. The resulting compound will be an alkene, specifically isobutene (CH\(_3\)-C=CH\(_2\)).

Step 2: Analyzing the options.
(A) CH\(_3\)-C-CH\(_3\)-NHBr: Incorrect. This is a product of nucleophilic substitution but not elimination.
(B) CH\(_3\)-C-CH\(_2\)-NH\(_2\)Br: Incorrect. This compound contains an amine group, but this is not the result of the reaction described.
(C) CH\(_3\)-C=CH\(_2\): Correct. This is the product of elimination in the presence of alcoholic ammonia, resulting in isobutene.
(D) CH\(_3\)-C-CH\(_3\)-NH\(_2\): Incorrect. This product would be obtained if an amine formed by nucleophilic substitution without elimination, which is not the case here.

Step 3: Conclusion.
The correct product formed is CH\(_3\)-C=CH\(_2\), corresponding to option (C).