A carbonyl compound with molecular weight 86 does not reduce Fehling's solution but forms crystalline bisulphite derivatives and gives an iodoform test. The possible compounds can be:

2-pentanone and 3-pentanone

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2-pentanone and 3-methyl-2-butanone

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2-pentanone and pentanal

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3-pentanone and 3-methyl-2-butanone

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Solution

The correct option is B 2-pentanone and 3-methyl-2-butanone
Given: Molecular formula of compound is 86.

Since compound does not reduce Fehling’s solution i.e. it must be a ketone.

Iodoform test given by compound containing $C H_{3} - O | | C -$ group.

2-pentanone and 3-methyl-2-butanone are isomers, they have the same molecular mass.

Molecular mass of 2 – pantanone : 5× 12 + 1 × 10 + 1 × 16 = 86

Molecular mass of 3 – methyl – 2 – butanone : 5 × 12 + 1 × 10 + 1× 16 = 86

Hence correct option is (b).

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