Question

Consider the following reaction sequences and choose the correct option. \[ Ph-C\equiv C-CH_3 \] On reduction with \[ H_2/Pd-C \; (Lindlar's\; catalyst) \] gives K. On reduction with \[ Na/Liq.NH_3 \] gives L. Further reaction with \[ HBr/benzoyl\; peroxide \] gives M and N respectively.

• A. M and N are stereoisomers
• B. K and L are geometrical isomers
• C. K and L are enantiomers
• D. M and N are geometrical isomers

Hint:

Remember reduction of alkynes: Lindlar catalyst \[ \rightarrow cis\; alkene \] Sodium + Liquid ammonia \[ \rightarrow trans\; alkene \] Very important organic chemistry reaction rule.

Answer 1

The Correct Option is B

Concept: Alkynes on partial reduction can produce alkenes with different stereochemistry depending on reagent used. Important reactions:

• Lindlar catalyst gives cis alkene

• Sodium in liquid ammonia gives trans alkene
Thus reagent choice determines geometry of final product.

Step 1: Reaction with Lindlar catalyst.
The compound is \[ Ph-C\equiv C-CH_3 \] Hydrogenation with Lindlar catalyst gives syn addition. Both hydrogen atoms add from same side. Hence product K is cis alkene. \[ K = cis\; Ph-CH=CH-CH_3 \]

Step 2: Reaction with sodium in liquid ammonia.
Reduction with \[ Na/Liq.NH_3 \] causes anti addition. Hydrogen atoms add from opposite sides. Hence product L is trans alkene. \[ L = trans\; Ph-CH=CH-CH_3 \]

Step 3: Compare K and L.
K and L have same molecular formula. Connectivity of atoms remains same. Difference exists only in arrangement around double bond. One is cis and other is trans. This means K and L are geometrical isomers.

Step 4: Check remaining options.
They are not mirror images. Therefore not enantiomers. Hence option (3) is wrong. \[ \boxed{K\; and\; L\; are\; geometrical\; isomers} \] \[ \boxed{\text{Correct Answer = Option (2)}} \]