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For the free radical bromination of 2-methylbutane show all of the mono-bromination isomers and predict their relative abundance. Clearly show the relative amounts by labeling 1 as the most abundant to the least abundant.

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Answer:

See explanation

Explanation:

We must take into account the order of stability of free radicals. The order of stability of free radicals is;

Tertiary>secondary>primary> methyl.

As a result of this, structure 1 involving a tertiary radical leads to the more abundant product as shown.

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