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

Step 1: Find alkene X from ozonolysis.
Ozonolysis splits the double bond into two carbonyls. Butanone, $CH_3COCH_2CH_3$, and methanal, $HCHO$, rejoin to give 2-methyl-1-butene, $CH_3CH_2C(CH_3){=}CH_2$, with formula $C_5H_{10}$.
Step 2: Add HBr with peroxide to get Y.
With peroxide the addition is anti-Markovnikov, so bromine goes to the terminal carbon, giving $Y = CH_3CH_2CH(CH_3)CH_2Br$ (a $C_5H_{11}Br$).
Step 3: Do the Wurtz reaction to get Z.
Wurtz couples two of these $C_5H_{11}$ pieces at the former C-Br positions, giving \[ Z = CH_3CH_2CH(CH_3)CH_2{-}CH_2CH(CH_3)CH_2CH_3 \] a $C_{10}H_{22}$ alkane.
Step 4: Mark the primary carbons.
Primary carbons (joined to one carbon) are the two chain-end $CH_3$ groups and the two branch $CH_3$ groups, giving four $1^\circ$ carbons.
Step 5: Mark the secondary carbons.
Secondary carbons (joined to two carbons) are the two $CH_2$ groups next to the ends and the two central $CH_2$ groups, giving four $2^\circ$ carbons.
Step 6: Mark the tertiary carbons and conclude.
The two $CH$ centres each bearing a methyl branch are joined to three carbons, giving two $3^\circ$ carbons. So the counts are $4, 4, 2$, which is option 1.
\[ \boxed{4, 4, 2} \]