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0.44 g of a monohydric alcohol when adde...

0.44 g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at S.T.P., `112 cm^(3)` of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is

A

`H_3C - underset(OH)underset(|)CH-CH_2 - CH_3`

B

`(CH_3)_3C - CH_2OH`

C

`H_3C - underset(OH)underset(|)CH-CH_2-CH_2-CH_3`

D

`(CH_3)_2CH - CH_2OH`

Text Solution

Verified by Experts

The correct Answer is:
C

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