Question

why is it difficult to prepare grignard reagents of allyl halides? ​

Answer 1

In order to form the Grignard reagent, a carbanion (negatively charged species) has to form from the alkyl group (methyl, ethyl or propyl in the given cases).

Now methyl is the most stable carbanion followed by ethyl. This is because of inductive effect. Because of this effect alkyl groups donate electron density to an attached carbon. This increased electron density makes the formation of a negatively charged species more difficult. Hence methyl with no attached carbons is most stable and propyl with most attached carbons is least stable.

Now more stable the carbanion, the easier the reaction and hence higher the reactivity.

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Answer 2

Explanation:

This compound cannot be used to form Grignard reagent as it contains an alcoholic group which immediately react with the Grignard reagent.

The alcoholic group is slightly acidic in nature and the hydrogen atom present on the alcohol group react with Grignard's reagent.