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

Step 1: Recall the two reduction routes.
An internal alkyne can be reduced to either a cis or a trans alkene depending on the reagent.
Step 2: Reduce with Lindlar catalyst.
Starting from \(Ph-C\equiv C-CH_3\), \(H_2/Pd-C\) (Lindlar) adds both H atoms from the same face (syn addition), giving the cis alkene. So \(K = cis\,Ph-CH=CH-CH_3\).
Step 3: Reduce with sodium in liquid ammonia.
\(Na/liq.NH_3\) gives anti addition, placing the H atoms on opposite faces, so we obtain the trans alkene. So \(L = trans\,Ph-CH=CH-CH_3\).
Step 4: Compare K and L.
K and L have the same molecular formula and the same connectivity; they differ only in the arrangement of groups about the \(C=C\) double bond, one cis and one trans.
Step 5: Name the relationship.
Differing only in geometry about a double bond means K and L are geometrical (cis-trans) isomers, not enantiomers.
Step 6: Conclude.
The correct statement is that K and L are geometrical isomers.
\[ \boxed{\text{K and L are geometrical isomers}} \]