JEE Main 2026 — 23 January, Evening Shift — Question 66

The reaction of benzene with $\mathrm{CH_3COCH_2CH_2COCl}$ in presence of $\mathrm{anhyd.\ AlCl_3}$ is Friedel–Crafts acylation.

The $\mathrm{-COCl}$ group forms an acylium ion and electrophilic substitution occurs on benzene. Therefore,

$\mathrm{C_6H_6 + CH_3COCH_2CH_2COCl \xrightarrow{AlCl_3} C_6H_5COCH_2CH_2COCH_3}$

Now the compound formed is a 1,5-diketone.

On heating with aqueous $\mathrm{NaOH}$, intramolecular aldol condensation occurs. Since the two carbonyl groups are separated by three carbon atoms, a six-membered ring is formed.

After dehydration, an $\alpha,\beta$-unsaturated ketone is obtained.

Thus compound $(Q)$ has molecular formula $\mathrm{C_{11}H_{10}O}$.

Molar mass of $(Q)$ $= 11 \times 12 + 10 \times 1 + 16$ $= 132 + 10 + 16$ $= 158$

Percentage of oxygen $= \dfrac{16}{158} \times 100$ $= 10.12\%$

Nearest integer $= 10\%$