Question:
\(C_{2}H_{2} \xrightarrow[Hg^{2+}/H^{+},\,333K]{H_{2}O} A \xrightarrow{CH_{3}MgBr/H_{3}O^{+}} B \xrightarrow{Cu/573K} C\)
A and C cannot be distinguished by using
\(C_{2}H_{2} \xrightarrow[Hg^{2+}/H^{+},\,333K]{H_{2}O} A \xrightarrow{CH_{3}MgBr/H_{3}O^{+}} B \xrightarrow{Cu/573K} C\)
A and C cannot be distinguished by using
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The only aldehyde giving iodoform test is ethanal. Acetone also gives iodoform test because it contains the \(-COCH_3\) group.
Updated On: Jun 22, 2026
- \(H^{+}/K_{2}Cr_{2}O_{7}\)
- Fehling's reagent
- Tollens' reagent
- Iodoform test \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
We first identify compounds \(A\), \(B\) and \(C\) from the given reaction sequence and then compare their reactions with the given reagents.
Step 1: Identify compound A.
Hydration of acetylene in the presence of \(Hg^{2+}/H^+\) gives acetaldehyde. \[ HC\equiv CH \xrightarrow{H_2O/Hg^{2+}} CH_3CHO \] Hence, \[ A=CH_3CHO \] (Ethanal)
Step 2: Identify compound B.
Ethanal reacts with methyl magnesium bromide followed by hydrolysis. \[ CH_3CHO + CH_3MgBr \longrightarrow CH_3CH(OH)CH_3 \] Thus, \[ B=CH_3CH(OH)CH_3 \] (2-Propanol)
Step 3: Identify compound C.
Passing 2-propanol over heated copper at \(573K\) causes dehydrogenation. \[ CH_3CH(OH)CH_3 \xrightarrow{Cu,573K} CH_3COCH_3 \] Hence, \[ C=CH_3COCH_3 \] (Acetone)
Step 4: Examine the given tests.
Acidified dichromate: \[ CH_3CHO \] is oxidized whereas acetone is not easily oxidized. Hence distinguishes A and C. Fehling's reagent: Ethanal gives positive test. Acetone gives negative test. Hence distinguishes A and C. Tollens' reagent: Ethanal gives silver mirror. Acetone does not. Hence distinguishes A and C. Iodoform test: Both ethanal and acetone contain the required structural unit. \[ CH_3CHO \] and \[ CH_3COCH_3 \] both give positive iodoform test. Hence they cannot be distinguished. \[ \boxed{\text{Answer = (D)}} \]
Step 1: Identify compound A.
Hydration of acetylene in the presence of \(Hg^{2+}/H^+\) gives acetaldehyde. \[ HC\equiv CH \xrightarrow{H_2O/Hg^{2+}} CH_3CHO \] Hence, \[ A=CH_3CHO \] (Ethanal)
Step 2: Identify compound B.
Ethanal reacts with methyl magnesium bromide followed by hydrolysis. \[ CH_3CHO + CH_3MgBr \longrightarrow CH_3CH(OH)CH_3 \] Thus, \[ B=CH_3CH(OH)CH_3 \] (2-Propanol)
Step 3: Identify compound C.
Passing 2-propanol over heated copper at \(573K\) causes dehydrogenation. \[ CH_3CH(OH)CH_3 \xrightarrow{Cu,573K} CH_3COCH_3 \] Hence, \[ C=CH_3COCH_3 \] (Acetone)
Step 4: Examine the given tests.
Acidified dichromate: \[ CH_3CHO \] is oxidized whereas acetone is not easily oxidized. Hence distinguishes A and C. Fehling's reagent: Ethanal gives positive test. Acetone gives negative test. Hence distinguishes A and C. Tollens' reagent: Ethanal gives silver mirror. Acetone does not. Hence distinguishes A and C. Iodoform test: Both ethanal and acetone contain the required structural unit. \[ CH_3CHO \] and \[ CH_3COCH_3 \] both give positive iodoform test. Hence they cannot be distinguished. \[ \boxed{\text{Answer = (D)}} \]
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