Question

Identify X, the final product formed when 2 moles of ethanal undergoes the following series of reactions with reagents (i) to (iv).

\[ \mathrm{CH_3CHO \xrightarrow[(ii)\ HCl]{(i)\ dil.\ NaOH} \xrightarrow{(iii)\ NaBH_4} \xrightarrow{(iv)\ Cr_2O_7^{2-}/H^+}{} [X]} \]

• A. But-3-enoic acid
• B. But-2-enoic acid
• C. Propanoic acid
• D. Ethanoic acid

Hint:

Aldol condensation followed by dehydration gives \(\alpha,\beta\)-unsaturated aldehyde. NaBH\(_4\) reduces only the aldehyde group, while strong oxidizing agents convert alcohols into carboxylic acids.

Answer 1

The Correct Option is B

Step 1: Aldol condensation of ethanal.
Two molecules of ethanal undergo aldol condensation in presence of dilute NaOH.
\[ 2CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO \] This product is aldol (3-hydroxybutanal).

Step 2: Dehydration using HCl.
The aldol undergoes dehydration in acidic medium to form an \(\alpha,\beta\)-unsaturated aldehyde.
\[ CH_3CH(OH)CH_2CHO \xrightarrow{HCl} CH_3CH=CHCHO \] This compound is crotonaldehyde (but-2-enal).

Step 3: Reduction using NaBH\(_4\).
NaBH\(_4\) reduces aldehyde group to alcohol without affecting the double bond.
\[ CH_3CH=CHCHO \xrightarrow{NaBH_4} CH_3CH=CHCH_2OH \] The product formed is but-2-en-1-ol.

Step 4: Oxidation using \(Cr_2O_7^{2-}/H^+\).
Primary alcohol gets oxidized to carboxylic acid using strong oxidizing agent.
\[ CH_3CH=CHCH_2OH \xrightarrow{Cr_2O_7^{2-}/H^+} CH_3CH=CHCOOH \] The product formed is but-2-enoic acid.

Step 5: Checking structure of final product.
The final compound contains:
- Four carbon atoms
- A double bond between C2 and C3
- A carboxylic acid group
Thus, it is but-2-enoic acid.

Step 6: Matching with options.
Among the given options, but-2-enoic acid corresponds to option (B).

Step 7: Final Answer.
\[ \boxed{\text{But-2-enoic acid}} \] Hence, the correct answer is option (B).