The compound \( \ce{HOOC-CH2-CH2-COOC2H5} \) has a carboxyl group and an ester group. It is treated with diborane (B2H6) and then with H3O+. Which of the following is the main product?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \ce{HOCH2-CH2-CH2-COOC2H5} \)
Diborane reduces a carboxylic acid to a primary alcohol, but it does not easily reduce an ester group. So only the –COOH end of the molecule changes to –CH2OH, and the –COOC2H5 end stays:
\( \displaystyle \ce{HOOC-CH2-CH2-COOC2H5} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{B2H6}} \ce{HOCH2-CH2-CH2-COOC2H5} \)
Why the other options are incorrect
Option (1) \( \ce{HOOC-CH2-CH2-CH2OH} \) — Incorrect. This would need the ester group to be reduced and the acid group left alone, the reverse of what diborane does.
Option (2) \( \ce{HOCH2-CH2-CH2-CH2OH} \) — Incorrect. This diol needs both groups reduced; diborane leaves the ester group unchanged.
Option (3) \( \ce{OHC-CH2-CH2-COOC2H5} \) — Incorrect. The reduction of the –COOH group does not stop at the aldehyde; it goes on to the primary alcohol.
NCERT (2026-27 edition):
8.9.3 Reactions Involving –COOH Group – “Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or better with diborane. Diborane does not easily reduce functional groups such as ester, nitro, halo, etc.”
\( \displaystyle \ce{R-COOH} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{LiAlH4}\text{/ether or }\ce{B2H6}} \ce{R-CH2OH} \)