JEE Main 2026 — Hydrocarbons Question with Solution

Statement I:
When $3$-phenylpropene reacts with $HBr$, protonation of the alkene occurs to form a secondary carbocation.

$C_6{H}_5-C{H}_2-CH=C{H}_2+H^{+}\to C_6{H}_5-C{H}_2-C{H}^{+}-C{H}_3$

This secondary carbocation undergoes a $1,2$-hydride shift to form a more stable, resonance-stabilized benzylic carbocation.

$C_6{H}_5-C{H}_2-C{H}^{+}-C{H}_3\to C_6{H}_5-C{H}^{+}-C{H}_2-C{H}_3$

The bromide ion then attacks this benzylic carbocation to form the major product, $1$-bromo-$1$-phenylpropane.

$C_6{H}_5-C{H}^{+}-C{H}_2-C{H}_3+B{r}^{-}\to C_6{H}_5-CH(Br)-C{H}_2-C{H}_3$

The carbon atom bonded to the bromine is attached to four different groups: a phenyl group ($-C_6{H}_5$), an ethyl group ($-C{H}_{2}C{H}_3$), a hydrogen atom ($-H$), and a bromine atom ($-Br$). Thus, it is a chiral carbon. The product is a secondary alkyl bromide. Therefore, Statement I is true.

Statement II:
The Sandmeyer reaction involves treating a diazonium salt with $Cu(I)$ salts ($CuCl$, $CuBr$, $CuCN$) to prepare aryl chlorides, aryl bromides, and aryl cyanides, respectively.

The Gattermann reaction involves treating a diazonium salt with copper powder ($Cu$) and halogen acids ($HCl$, $HBr$) to prepare aryl chlorides and aryl bromides. Aryl cyanides are not prepared using the Gattermann reaction. Therefore, Statement II is false.

Conclusion: Statement I is true but Statement II is false.

Answer: Statement I is true but Statement II is false