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Question

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product through the following sequence of reactions in which Q is an intermediate organic compound.

The structure of the compound P is:


A
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B
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C

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D
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Solution

The correct option is D

Clearly all the options have the same formula: C6H10

A corresponding saturated alkane would have had the formula C6H14, which means the double bond equivalent for P is = 2 (since two equivalents are lesser than the corresponding saturated compound). This could also be verified by using the formula for DBE or degree of unsaturation.
Also, it is given that the compound is acyclic. So the given compound has no rings or cycles, and this further implies that it is an alkadiene or an alkyne. Look at the first half of the scheme given for conversion of P into Q.

From here, let us work backwards from Q, which is the secondary alcohol - 3,3-dimethylbutan-2-ol.

In the above transformation from P to Q, steps 2 and 3 imply a reduction and work-up,although the work is not entirely necessary. What on reduction + work-up will give Q?


Which alkyne having the formula C6H10 on oxymercuration - de-mercuration gives the above methyl ketone?



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