Question #14bad

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Question #14bad

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Answer 1 · 2016-10-12T15:32:01

The molecular formula $C_{3} H_{6} O$ $C_3H_6O$ which follows general molecular formula $C_{n} H_{2 n} O$ $C_nH_(2n)O$ represents either a saturated aldhyde or a saturated ketone.A compound containing ketomethyl group ( $C H_{3} C O -$ $CH_3CO-$ ) linked with H atom or hydrocarbon group responds to iodoform test and a compound containing $- C H O$ $-CHO$ group responds to Tollen's reagent.A reduces this reagent but B does not.So A contains $- C H O$ $-CHO$ group.Again B undergoes iodoform test but A does not.So it contains $C H_{3} C O -$ $CH_3CO-$ group but not $- C H O$ $-CHO$

So $A \to C H_{3} C H_{2} C H O \to propanal$ $" "A->CH_3CH_2CHO->"propanal"$

and $B \to C H_{3} C O C H_{3} \to propanone$ $B->CH_3COCH_3->"propanone"$

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