JEE Main 2026 — Hydrocarbons Question with Solution

Step 1: The first reaction is a Friedel-Crafts alkylation. Benzene reacts with $C{H}_{3}Cl$ in the presence of anhydrous $AlC{l}_3$ to form toluene.

Step 2: The second reaction is electrophilic aromatic substitution. Toluene reacts with $C{l}_2$ in the presence of $FeC{l}_3$ to form a mixture of o-chlorotoluene and p-chlorotoluene. Due to steric hindrance at the ortho position, the major product is p-chlorotoluene.

Step 3: The third reaction is the oxidation of the alkyl side chain. p-chlorotoluene is oxidized by $K_{2}C{r}_2{O}_7/H_{2}S{O}_4$ to form p-chlorobenzoic acid, which is the major product $(X)$.

The molecular formula of p-chlorobenzoic acid is $C_7{H}_{5}Cl{O}_2$.
Molar mass of $(X)=(7\times 12)+(5\times 1)+35.5+(2\times 16)=84+5+35.5+32=156.5{\text{ g mol}}^{-1}$.

Step 4: Reaction of p-chlorobenzoic acid with $NaHC{O}_3$ liberates $C{O}_2$ gas:
$C_6{H}_{4}ClCOOH+NaHC{O}_3\to C_6{H}_{4}ClCOONa+H_{2}O+C{O}_2\uparrow$

From the stoichiometry, $1$ mole of p-chlorobenzoic acid produces $1$ mole of $C{O}_2$ gas.
Volume of $C{O}_2$ liberated at STP = $11.2{\text{ dm}}^3$.
Moles of $C{O}_2$ liberated = $\frac{11.2}{22.4}=0.5\text{ mol}$.

Therefore, moles of p-chlorobenzoic acid reacted = $0.5\text{ mol}$.
Mass of $(X)$ reacted, $P=0.5\text{ mol}\times 156.5{\text{ g mol}}^{-1}=78.25\text{ g}$.

Answer: $78.25$