Question

What are \(A\) and \(B\) in the following set of reactions respectively? \[ Y \xrightarrow[\mathrm{AlCl_3}]{B} \text{Benzene} \xrightarrow[\mathrm{AlCl_3}]{A} X \] \[ X \text{ gives positive iodoform test.} \]

• A. CO, HCl ; \(\mathrm{CH_3CH_2COCl}\)
• B. CO, HCl ; \(\mathrm{CH_3COCl}\)
• C. \(\mathrm{CH_3COCl}\) ; \(\mathrm{CH_3CH_2COCl}\)
• D. \(\mathrm{CH_3COCl}\) ; CO, HCl \bigskip

Hint:

• Gattermann--Koch reaction: \(\mathrm{CO+HCl}\) introduces the \(-\mathrm{CHO}\) group into benzene.
• Friedel--Crafts acylation uses acyl chlorides in the presence of \(\mathrm{AlCl_3}\).
• Compounds containing the \(\mathrm{-COCH_3}\) or oxidizable \(\mathrm{-CH(OH)CH_3}\) group give the iodoform test.

Answer 1

The Correct Option is A

Step 1: Identify \(B\). The conversion of benzene to benzaldehyde is carried out by the Gattermann--Koch reaction using \[ \boxed{\mathrm{CO + HCl}} \] in the presence of anhydrous \(\mathrm{AlCl_3}\) and \(\mathrm{CuCl}\). Thus, \[ B=\mathrm{CO,\ HCl} \]

Step 2: Identify \(A\). The product \(X\) gives the iodoform test. Among Friedel--Crafts acylation products, propiophenone \[ \mathrm{C_6H_5COCH_2CH_3} \] contains the \(\mathrm{-COCH_2CH_3}\) group, which gives the iodoform test. It is obtained using \[ \boxed{\mathrm{CH_3CH_2COCl}} \] Hence, \[ A=\mathrm{CH_3CH_2COCl} \]

Step 3: Final conclusion. Therefore, \[ \boxed{A=\mathrm{CO,\ HCl},\qquad B=\mathrm{CH_3CH_2COCl}} \] Hence, the correct option is \(\boxed{(A)}\).