An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives a mixture of but-1-ene and but-2-ene, both of which on reduction give (D). (C) on oxidation with pyridinium chlorochromate (PCC) gives (E), which is an aldehyde.
What is the name of the compound (E) in the reaction sequence given in the passage?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Butanal
First find the alcohol (C) from the reaction sequence in the passage:
Compound (C) has 4 carbon atoms: the ester C8H16O2 splits into an acid (B) and an alcohol (C), and (C) is oxidised to (B), so each has half of the 8 carbon atoms: 4 carbons.
Compound (C) is butan-1-ol: it is oxidised to an acid with the same chain, so it is a primary alcohol, and its dehydration gives unbranched alkenes (a mixture of but-1-ene and but-2-ene).
Pyridinium chlorochromate (PCC) oxidises a primary alcohol only as far as the aldehyde, with no change in the number of carbon atoms:
\( \displaystyle \ce{CH3CH2CH2CH2OH} \xrightarrow{\text{PCC}} \ce{CH3CH2CH2CHO} \)
So (E) is butanal.
Why the other options are incorrect
Option (1) Propanal — Incorrect. It has 3 carbon atoms; PCC oxidation does not remove a carbon from butan-1-ol.
Option (3) Pentanal — Incorrect. It has 5 carbon atoms; oxidation does not add a carbon.
Option (4) Ethanal — Incorrect. It has only 2 carbon atoms; it would come from ethanol, not from a four-carbon alcohol.
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “… A better reagent for oxidation of primary alcohols to aldehydes in good yield is pyridinium chlorochromate (PCC), a complex of chromium trioxide with pyridine and HCl.”