Question

Identify the product formed when tertiary butyl bromide reacts with alcoholic NH\(_3\) solution?

• A. 2-Methylpropene
• B. But-2-ene
• C. But-1-ene
• D. 2-Methylpropan-1-ol

Hint:

With alcoholic NH\(_3\), tertiary alkyl halides mainly undergo elimination. Check the carbon skeleton carefully.

Answer 1

The Correct Option is A

Step 1: Understand the question.
Tertiary butyl bromide, $(CH_3)_3CBr$, reacts with alcoholic ammonia. We must find the product.
Step 2: Recall how the reagent behaves.
Alcoholic ammonia acts as a base, especially with heating. A tertiary halide is too crowded for simple substitution, so it prefers elimination.
Step 3: Identify the elimination.
The base pulls a hydrogen off a carbon next to the C-Br carbon, and bromide leaves. A carbon double bond forms. This is dehydrohalogenation.
Step 4: Build the product.
\[ (CH_3)_3CBr \xrightarrow{\text{alc. } NH_3} (CH_3)_2C=CH_2 + HBr \]
Step 5: Name the alkene.
The product $(CH_3)_2C=CH_2$ has a three carbon chain with a double bond at carbon 1 and a methyl branch at carbon 2. This is 2-methylpropene.
Step 6: Choose the answer.
Because tert-butyl bromide has only four equivalent carbons, the elimination can only give 2-methylpropene, not but-2-ene. So the correct option is 1. \[ \boxed{\text{2-Methylpropene}} \]