Chemistry - Hydrocarbon Question with Solution | TestHub

The given starting material is 1-phenylpropyne, $\text{Ph}-\text{C}\equiv \text{C}-\text{Me}$.

Reduction of the alkyne with ${\text{H}}_2$ in the presence of Lindlar's catalyst ($\text{Pd/C}$ poisoned with ${\text{BaSO}}_4$ or quinoline) proceeds via syn-addition. This yields the cis-alkene. Therefore, $\text{K}$ is cis-1-phenylpropene.

Reduction of the alkyne with $\text{Na}$ in liquid ${\text{NH}}_3$ (Birch reduction conditions) proceeds via anti-addition. This yields the trans-alkene. Therefore, $\text{L}$ is trans-1-phenylpropene.

Since $\text{K}$ is the cis-isomer and $\text{L}$ is the trans-isomer of the same compound, they are geometrical isomers.

Further reactions:
Addition of $\text{HBr}$ to $\text{K}$ is an electrophilic addition that proceeds via the more stable benzylic carbocation ($\text{Ph}-{\text{C}}^{+}\text{H}-{\text{CH}}_2-\text{Me}$). This gives 1-bromo-1-phenylpropane as product $\text{M}$.

Addition of $\text{HBr}$ to $\text{L}$ in the presence of benzoyl peroxide is a free radical addition (anti-Markovnikov). The bromine radical adds to form the more stable benzylic radical ($\text{Ph}-{\text{C}}^{\bullet }\text{H}-\text{CH}(\text{Br})-\text{Me}$). This gives 2-bromo-1-phenylpropane as product $\text{N}$.

$\text{M}$ and $\text{N}$ are positional isomers (constitutional isomers), not stereoisomers.

Thus, the only correct statement is that $\text{K}$ and $\text{L}$ are geometrical isomers.

Answer: $\text{K}$ and $\text{L}$ are geometrical isomers