Question

Which one of the following reactions does not give the correct combination of reactants and the major products formed in the reaction?

• A. \(\mathrm{CH_3COCH_3 + NH_2NHCONH_2 \, (weakly \ basic) \rightarrow (CH_3)_2C=NNHCONH_2}\)
• B. \(\mathrm{CH_3CHO + 2C_2H_5OH \, (dry \ HCl \ gas) \rightleftharpoons CH_3CH(OC_2H_5)_2}\)
• C. \(\mathrm{CH_3CH=CHCH_2CH_2CN \xrightarrow[(ii)\ H_2O]{(i)\ DIBAL-H} CH_3(CH_2)_4CHO}\)
• D. \(\mathrm{C_6H_5COCH_3 + I_2 + Na_2CO_3 \, (Heat) \rightarrow CHI_3 + C_6H_5COO^-}\)

Hint:

DIBAL-H selectively reduces nitriles to aldehydes after hydrolysis, but it does not reduce an isolated \(C=C\) double bond under normal conditions.

Answer 1

The Correct Option is C

Step 1: Understanding the question.
The question asks for the reaction which does not give the correct major product.
So, we have to check each reaction carefully and identify the incorrect product combination.

Step 2: Checking option (A).
In option (A), acetone reacts with semicarbazide.
\[ \mathrm{CH_3COCH_3 + NH_2NHCONH_2 \rightarrow (CH_3)_2C=NNHCONH_2} \] This is the formation of semicarbazone.
Ketones react with semicarbazide to form semicarbazones.
Therefore, option (A) is correct.

Step 3: Checking option (B).
In option (B), acetaldehyde reacts with ethanol in presence of dry HCl gas.
\[ \mathrm{CH_3CHO + 2C_2H_5OH \rightleftharpoons CH_3CH(OC_2H_5)_2} \] This is acetal formation.
Aldehydes react with alcohols in acidic medium to give acetals.
Therefore, option (B) is also correct.

Step 4: Checking option (C).
In option (C), the given compound is an unsaturated nitrile:
\[ \mathrm{CH_3CH=CHCH_2CH_2CN} \] DIBAL-H reduces nitriles to aldehydes after hydrolysis.
But DIBAL-H does not normally hydrogenate or reduce the isolated carbon-carbon double bond.
So, the double bond should remain unchanged.

Step 5: Correct product for option (C).
The correct product should be:
\[ \mathrm{CH_3CH=CHCH_2CH_2CHO} \] But the given product is:
\[ \mathrm{CH_3(CH_2)_4CHO} \] This product is saturated, which means the double bond has also been reduced.
Since DIBAL-H does not reduce the \(C=C\) bond under these conditions, the given product is incorrect.

Step 6: Checking option (D).
In option (D), acetophenone reacts with iodine and base.
\[ \mathrm{C_6H_5COCH_3 + I_2 + Na_2CO_3 \rightarrow CHI_3 + C_6H_5COO^-} \] Acetophenone contains the group \(\mathrm{COCH_3}\), so it gives iodoform reaction.
The products are iodoform \(\mathrm{CHI_3}\) and benzoate ion \(\mathrm{C_6H_5COO^-}\).
Therefore, option (D) is correct.

Step 7: Final conclusion.
Only option (C) gives an incorrect major product because DIBAL-H reduces the nitrile group to aldehyde but does not reduce the carbon-carbon double bond.
Thus, the reaction which does not give the correct product is:
\[ \boxed{\text{Option (C)}} \]