Question

An alkene (X) with formula \(C_5H_{10}\) on ozonolysis gives butanone and methanal. X with HBr in the presence of organic peroxide gives Y as major product. When Y is subjected to Wurtz reaction gives Z. The number of \(1^{\circ}, 2^{\circ}\) and \(3^{\circ}\) carbons in Z respectively are:

• A. 4, 4, 2
• B. 4, 2, 4
• C. 3, 3, 4
• D. 5, 3, 2

Hint:

Use ozonolysis cleavage to reconstruct the double bond by removing oxygen atoms from the carbonyl products.

Answer 1

The Correct Option is A

Concept: Ozonolysis of an alkene yields carbonyl compounds. Anti-Markovnikov addition occurs with HBr in the presence of peroxide.

Step 1: Determine the structure of X.
Butanone (\(CH_3COCH_2CH_3\)) + Methanal (\(HCHO\)) come from 2-methyl-1-butene (\(CH_3CH_2C(CH_3)=CH_2\)).

Step 2: Determine Y and Z.
X (\(CH_3CH_2C(CH_3)=CH_2\)) + \(HBr/peroxide \rightarrow\) Y (Anti-Markovnikov: \(CH_3CH_2C(CH_3)(H)-CH_2Br\)). Wurtz reaction on Y (\(C_5H_{11}Br\)) involves joining two \(C_5H_{11}\) chains: \(CH_3CH_2CH(CH_3)CH_2-CH_2CH(CH_3)CH_2CH_3\).

Step 3: Analyze Z for carbon types.
Counting the structure \(C_{10}H_{22}\): There are four \(1^\circ\), four \(2^\circ\), and two \(3^\circ\) carbons. \[ \boxed{4, 4, 2} \]