Question

Both reactions (i) and (ii) give the same compound \(X\) as the major product. Identify \(X\).
\[ \text{(i) } 3\text{-Methylbut-1-ene} + HCl \rightarrow X \] \[ \text{(ii) Neopentyl alcohol} + HCl \ (\text{anh. } ZnCl_2) \rightarrow X \]

• A. \( (CH_3)_2CH-CHCl-CH_3 \)
• B. \( (CH_3)_2CCl-CH_2-CH_3 \)
• C. \( CH_3-CH_2-CH(CH_3)-CH_2Cl \)
• D. \( (CH_3)_2CH-CH_2-CH_2Cl \)

Hint:

In reactions involving carbocations, always check for rearrangement because the major product usually forms through the most stable carbocation.

Answer 1

The Correct Option is B

Step 1: Analyse reaction (i).
3-Methylbut-1-ene has the structure:
\[ CH_2=CH-CH(CH_3)-CH_3 \]
When it reacts with \(HCl\), electrophilic addition takes place according to Markovnikov's rule.

Step 2: Formation of carbocation in reaction (i).
Protonation at the terminal carbon gives a secondary carbocation:
\[ CH_3-\overset{+}{CH}-CH(CH_3)-CH_3 \]
This carbocation can rearrange to form a more stable tertiary carbocation.

Step 3: Rearrangement in reaction (i).
A hydride shift occurs from the adjacent carbon to form a tertiary carbocation:
\[ CH_3-CH_2-C^+(CH_3)-CH_3 \]
Then \(Cl^-\) attacks this tertiary carbocation.
\[ X=(CH_3)_2CCl-CH_2-CH_3 \]

Step 4: Analyse reaction (ii).
Neopentyl alcohol has the structure:
\[ (CH_3)_3C-CH_2OH \]
With \(HCl\) in presence of anhydrous \(ZnCl_2\), the hydroxyl group is replaced by chlorine through carbocation rearrangement.

Step 5: Carbocation rearrangement in reaction (ii).
The initially formed neopentyl carbocation is highly unstable.
It rearranges by methyl shift to form a more stable tertiary carbocation.
\[ (CH_3)_2C^+-CH_2-CH_3 \]

Step 6: Chloride attack.
The chloride ion attacks the tertiary carbocation to form:
\[ (CH_3)_2CCl-CH_2-CH_3 \]

Step 7: Compare both reactions.
Both reactions form the same major product due to carbocation rearrangement.
Thus,
\[ X=(CH_3)_2CCl-CH_2-CH_3 \]
Final Answer:
\[ \boxed{(CH_3)_2CCl-CH_2-CH_3} \]