Question

A compound 'A' having the molecular formula $C_5H_{12}O$ on oxidation gives a compound 'B' with molecular formula $C_5H_{10}O$. Compound 'B' gave a 2, 4-DNP derivative but did not give the haloform test. The structure of the compound A is

• A. $CH_3COCH_2CH_2CH_3$
• B. $CH_3CHOHCH_2CH_2CH_3$
• C. $CH_3CH_2COCH_2CH_3$
• D. $CH_3CH_2CHOHCH_2CH_3$

Answer 1

Answer: (D)

$CH_3CH_2CHOHCH_2CH_3$

Answer 2

Since compound $A(C_5H_{12}O)$ on oxidation gives compound $B(C_5H_{10}O)$ with two H-atoms less, therefore, A must be an alcohol. Further since compound B gives a 2, 4-DNP but does not give haloform test, therefore, B cannot be a methyl ketone. In other words, B is $CH_3CH_2COCH_2CH_3$ and hence A must be $CH_3CH_2CHOHCH_2CH_3$. Thus, option (d) is correct.