Question

What are A and C in the following sequence of reactions? \(C_2H_2 \xrightarrow{A} B \xrightarrow{\text{polymerisation}} C\)

• A. \(Pd/H_2\); LDP
• B. \(Pd/H_2\); HDP
• C. \(H_2, Pd/C\)-quinoline; LDP
• D. \(H_2, Pd/C\)-quinoline; HDP

Hint:

Lindlar catalyst stops hydrogenation at alkene stage; full hydrogenation would give alkane.

Answer 1

The Correct Option is D

Step 1: Read the reaction sequence.
We start from ethyne $C_2H_2$, treat it with reagent A to get B, then polymerise B to get C. We must name A and C.
Step 2: Decide what B should be.
Since B then polymerises into a polyethylene type plastic, B must be ethene $C_2H_4$. So step A has to convert the triple bond of ethyne into the double bond of ethene, that is a partial hydrogenation.
Step 3: Choose the reagent A for partial hydrogenation.
To stop at the alkene and not go all the way to ethane, we use a poisoned catalyst. The choice here is $H_2$ with Pd/C poisoned by quinoline (Lindlar type), so $A = H_2, Pd/C$-quinoline.
Step 4: Reject the plain Pd/H2 options.
Ordinary $Pd/H_2$ without a poison would fully reduce ethyne to ethane, which cannot polymerise, so those options are wrong for A.
Step 5: Identify the polymer C.
Ethene polymerises to polyethylene; under high pressure the dense form is High Density Polyethylene, written HDP. So $C = $ HDP.
Step 6: Combine A and C.
Therefore A is $H_2, Pd/C$-quinoline and C is HDP.
\[ \boxed{H_2, Pd/C\text{-quinoline; HDP}} \]