Question

Identify the alkyne in the following sequence of reactions: Alkyne \(\xrightarrow[\text{Lindlar's catalyst}]{H_2}\) A \(\xrightarrow[\text{only}]{Ozonolysis}\) \(\longrightarrow\) \(\text{Wacker Process} \longrightarrow CH_2=CH_2\)

• A. \(H_3C- C \equiv C-CH_3\)
• B. \(H_3C-CH_2-C \equiv CH\)
• C. \(H_2C=CH-C \equiv CH\)
• D. \(HC \equiv C-CH_2-C \equiv CH\)

Hint:

If ozonolysis gives only one product, the alkene (and hence original alkyne) must be symmetric.

Answer 1

The Correct Option is A

Step 1: Understand Lindlar reduction.
Lindlar catalyst reduces an alkyne to a cis-alkene.
So starting alkyne becomes an alkene \(A\).
Step 2: Use ozonolysis clue.
Ozonolysis breaks the double bond to give carbonyl compounds.
If ozonolysis gives only one type of product, the alkene must be symmetric.
Step 3: Check which alkyne gives symmetric alkene.
(A) \(CH_3-C\equiv C-CH_3\) is symmetric, gives cis-2-butene.
cis-2-butene on ozonolysis gives only one kind of product: acetaldehyde (2 moles).
Other options are unsymmetrical, giving two different products.
Step 4: Hence correct alkyne is option (A).
Final Answer:
\[ \boxed{H_3C-C\equiv C-CH_3} \]