Question

Consider the following transformations of a compound P.

Choose the correct options.

• A.

  

• B. $\mathrm{X}$ is $\mathrm{Pd}-\mathrm{C}$/quinoline/${\mathrm{H}}_2$
• C.

  

• D.

  

Answer 1

The correct option is C

Explanation for the correct options:

B $\mathrm{X}$ is $\mathrm{Pd}-\mathrm{C}$/quinoline/${\mathrm{H}}_2$

1. Compound P given in question reacts in presence of ${\mathrm{H}}_2/\mathrm{Pt}$ to form $1$-propane cyclohexane.
2. We know that alkynes react with ${\mathrm{H}}_2/\mathrm{Pt}$ to form alkanes hence compound P is an alkyne.
3. Now for$\mathrm{X}$: When alkynes react with $\mathrm{Pd}-\mathrm{C}$/quinoline/${\mathrm{H}}_2$ (lindlar's catalyst) then alkenes are formed, as given in the question.

C

1. As we know that the product formed in the reaction, of compound P with ${\mathrm{H}}_2/\mathrm{Pt}$ is $1$-propane cyclohexane.
2. We also know that compounds $\mathrm{X}$ react with alkynes to form alkene.
3. From the above two reactions, we can estimate that product P is $1$-ethyne cyclohex-$2$-ene.

Explanation for the incorrect option:

A

1. As we know that the product formed in the reaction, of compound P with ${\mathrm{H}}_2/\mathrm{Pt}$ is $1$-propane cyclohexane.
2. We also know that compound $\mathrm{X}$ react with alkynes to form alkene.
3. From the above two reactions, we can estimate that product P is $1$-ethyne cyclohex-$2$-ene.
4. Hence option A is incorrect.

D

1. As we know that product P is $1$-ethyne cyclohex-$2$-ene.
2. So when $1$-ethyne cyclohex-$2$-ene react with ${\mathrm{NaNH}}_2$ in presence of acid undergo elimination reaction to form optically active compounds.
3. The compound given in option D is not possible to accept because there is the presence of triple and double bonds in continuity, which is not possible.
4. Hence option D is incorrect.

Hence options B and C are correct, $\mathrm{X}$ is $\mathrm{Pd}-\mathrm{C}$/quinoline/${\mathrm{H}}_2$ and product P is $1$-ethyne cyclohex-$2$-ene.