An organic compound X (C5H10O) forms a phenylhydrazone but gives a negative iodoform test and a negative Tollens’ test. On Clemmensen reduction, X gives n-pentane. What is X?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Pentan-3-one
Each test removes some of the C5H10O candidates:
Forms a phenylhydrazone: X has a carbonyl group; it is an aldehyde or a ketone.
Negative Tollens’ test: X is not an aldehyde, so it is a ketone. Pentanal is ruled out.
Negative iodoform test: X has no CH3CO group. Pentan-2-one (CH3COCH2CH2CH3) and 3-methylbutan-2-one (CH3COCH(CH3)2) are ruled out.
Clemmensen reduction gives n-pentane: C=O becomes CH2, so X has an unbranched chain of five carbon atoms.
The only straight-chain C5 ketone without a CH3CO group is pentan-3-one, CH3CH2COCH2CH3:
\( \displaystyle \ce{(C2H5)2C=O + H2NNHC6H5 -> (C2H5)2C=NNHC6H5 + H2O} \)
\( \displaystyle \ce{CH3CH2COCH2CH3 + 4[H]} \xrightarrow[\ce{conc. HCl}]{\ce{Zn-Hg}} \ce{CH3CH2CH2CH2CH3 + H2O} \)
NCERT (2026-27 edition):
8.4 Chemical Reactions – “(ii) Reduction to hydrocarbons: The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc-amalgam and concentrated hydrochloric acid [Clemmensen reduction] or with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol (Wolff-Kishner reduction).”
“Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.”