NEET Chemistry Question (2026) — Solution

Step-by-step solution

The given starting material is 1-phenylpropyne, Ph - C C - Me . Reduction of the alkyne with H _2 in the presence of Lindlar's catalyst ( Pd/C poisoned with BaSO _4 or quinoline) proceeds via syn-addition. This yields the cis-alkene. Therefore, K is cis-1-phenylpropene. Reduction of the alkyne with Na in liquid NH _3 (Birch reduction conditions) proceeds via anti-addition. This yields the trans-alkene. Therefore, L is trans-1-phenylpropene. Since K is the cis-isomer and L is the trans-isomer of the same compound, they are geometrical isomers. Further reactions: Addition of HBr to K is an electrophilic addition that proceeds via the more stable benzylic carbocation ( Ph - C ^+ H - CH _2- Me ). This gives 1-bromo-1-phenylpropane as product M . Addition of HBr to 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 ( Ph - C ^ H - CH ( Br )- Me ). This gives 2-bromo-1-phenylpropane as product N . M and N are positional isomers (constitutional isomers), not stereoisomers. Thus, the only correct statement is that K and L are geometrical isomers. Answer: K and L are geometrical isomers