Question

Identify the compound which gives an optically active haloalkane on reaction with \(H_2/Ni\) on heating.

• A. A
• B. B
• C. D
• D. C

Hint:

For optical activity, check whether the final product has a carbon attached to four different groups. Hydrogenation changes \(C=C\) to \(C-C\), but does not remove halogen.

Answer 1

The Correct Option is B

Step 1: Understand the reaction.
The reagent \(H_2/Ni\) reduces carbon-carbon double bonds by hydrogenation.
\[ C=C \xrightarrow{H_2/Ni} C-C \]

Step 2: Check optical activity condition.
A haloalkane is optically active if it contains a chiral carbon atom.
A chiral carbon is attached to four different groups.

Step 3: Analyse compound B.
In compound B, after hydrogenation of the double bond, the carbon bearing chlorine becomes attached to four different groups.
These groups are:
\[ Cl,\quad H,\quad CH_3,\quad \text{alkyl chain} \]

Step 4: Formation of chiral haloalkane.
Since the carbon attached to chlorine has four different substituents, the product is optically active.

Step 5: Check other options.
In the other compounds, after hydrogenation, the carbon attached to chlorine does not become a proper chiral centre due to symmetry or identical groups.

Step 6: Identify correct compound.
Therefore, only compound B gives an optically active haloalkane after hydrogenation.

Step 7: Final conclusion.
\[ \boxed{B} \]