Question

Identify the product formed when tertiary butyl bromide reacts with alcoholic $\text{NH}_3$ solution?

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

Hint:

Whenever a tertiary ($3^\circ$) alkyl halide is treated with basic reagents (like alcoholic $\text{NH}_3$ or alcoholic KOH), steric hindrance makes nucleophilic substitution extremely difficult.
Instead, elimination wins out, and an alkene is always the principal product!

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 main product.
Step 2: Recall the reagent's role.
Alcoholic ammonia acts as a base, especially on heating. A tertiary halide is too crowded for a back-side substitution, so elimination wins.
Step 3: Identify the elimination.
The base removes a hydrogen from a carbon next to the C-Br carbon while bromide leaves. A double bond forms between those carbons.
Step 4: Write the reaction.
\[ (CH_3)_3CBr \xrightarrow{\text{alc. } NH_3} CH_2=C(CH_3)_2 + HBr \]
Step 5: Name the alkene.
The product $CH_2=C(CH_3)_2$ has a three carbon chain, a double bond at carbon 1, and a methyl branch at carbon 2. This is 2-methylpropene.
Step 6: Choose the answer.
The major product is 2-methylpropene, which is option 1. \[ \boxed{\text{2-Methylpropene}} \]