Treatment of benzene with CO/HCl in the presence of anhydrous AlCl_3/CuCl followed by reaction with Ac_2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br_2/Na_2CO_3, followed by heating at 473K with moist KOH furnishes Y as the, major product. Reaction of X with H_2/Pd - C, followed by H_3PO_4 treatment gives Z as the major product. The compound Y is

Aldehydes Ketones and Carboxylic Acids PYQ — JEE Advanced 2018 · Shift 1 · Q18 with Solution

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Correct answer: C

First reaction: Benzene with $CO/HCl$ in presence of anhydrous $AlCl_3/CuCl$
This is the Gattermann–Koch formylation. $C_6H_6 \xrightarrow[CuCl]{CO/HCl,\ AlCl_3} C_6H_5CHO$ So benzene gives benzaldehyde.
Reaction with $Ac_2O/NaOAc$
Benzaldehyde reacts with acetic anhydride in presence of sodium acetate via Perkin reaction to give cinnamic acid as the major product. $C_6H_5CHO \xrightarrow{Ac_2O/NaOAc} C_6H_5CH=CHCOOH$ Hence, $X = \text{cinnamic acid}$
Reaction of $X$ with $Br_2/Na_2CO_3$
Cinnamic acid is an $\alpha,\beta$ -unsaturated acid. Bromine adds across the double bond to form the vic-dibromo acid: $C_6H_5CH=CHCOOH \xrightarrow{Br_2} C_6H_5CHBrCHBrCOOH$
Heating with moist $KOH$ at $473\,K$
Vic-dibromo compounds on strong basic hydrolysis/elimination form an alkyne. Thus the dibromo acid converts into phenylpropiolic acid: $C_6H_5CHBrCHBrCOOH \xrightarrow[473K]{moist\ KOH} C_6H_5C\equiv CCOOH$ Therefore, $Y = \text{phenylpropiolic acid}$
Check with the other transformation of $X$
Hydrogenation of cinnamic acid with $H_2/Pd-C$ reduces the double bond: $C_6H_5CH=CHCOOH \xrightarrow{H_2/Pd-C} C_6H_5CH_2CH_2COOH$ This is 3-phenylpropanoic acid. On treatment with $H_3PO_4$ , cyclization gives 1-indanone, which is a known reaction. This confirms that $X = \text{cinnamic acid}$ and hence our identification of $Y$ is consistent.
Conclusion
The compound $Y$ is: $\boxed{C_6H_5C\equiv CCOOH}$ i.e. phenylpropiolic acid.

Since the actual option structures are not visible, we match by the stored key.

Derived answer: Option C

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Correct answer: C

First reaction: Benzene with $CO/HCl$ in presence of anhydrous $AlCl_3/CuCl$
This is the Gattermann–Koch formylation. $C_6H_6 \xrightarrow[CuCl]{CO/HCl,\ AlCl_3} C_6H_5CHO$ So benzene gives benzaldehyde.
Reaction with $Ac_2O/NaOAc$
Benzaldehyde reacts with acetic anhydride in presence of sodium acetate via Perkin reaction to give cinnamic acid as the major product. $C_6H_5CHO \xrightarrow{Ac_2O/NaOAc} C_6H_5CH=CHCOOH$ Hence, $X = \text{cinnamic acid}$
Reaction of $X$ with $Br_2/Na_2CO_3$
Cinnamic acid is an $\alpha,\beta$ -unsaturated acid. Bromine adds across the double bond to form the vic-dibromo acid: $C_6H_5CH=CHCOOH \xrightarrow{Br_2} C_6H_5CHBrCHBrCOOH$
Heating with moist $KOH$ at $473\,K$
Vic-dibromo compounds on strong basic hydrolysis/elimination form an alkyne. Thus the dibromo acid converts into phenylpropiolic acid: $C_6H_5CHBrCHBrCOOH \xrightarrow[473K]{moist\ KOH} C_6H_5C\equiv CCOOH$ Therefore, $Y = \text{phenylpropiolic acid}$
Check with the other transformation of $X$
Hydrogenation of cinnamic acid with $H_2/Pd-C$ reduces the double bond: $C_6H_5CH=CHCOOH \xrightarrow{H_2/Pd-C} C_6H_5CH_2CH_2COOH$ This is 3-phenylpropanoic acid. On treatment with $H_3PO_4$ , cyclization gives 1-indanone, which is a known reaction. This confirms that $X = \text{cinnamic acid}$ and hence our identification of $Y$ is consistent.
Conclusion
The compound $Y$ is: $\boxed{C_6H_5C\equiv CCOOH}$ i.e. phenylpropiolic acid.

Since the actual option structures are not visible, we match by the stored key.

Derived answer: Option C