Question 68

Consider the following compound (X):

The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C-H bond are:

Solution

Position I ($$sp$$ hybridized carbon): Homolytic cleavage at this position forms an acetylenic radical ($$-\text{C}\equiv\dot{\text{C}}$$). Because $$sp$$ hybridized carbons are highly electronegative, they hold the unshared electron very tightly, making this radical highly unstable (least stable).

Position II (Propargylic position): The radical formed here ($$-\text{C}\equiv\text{C}-\dot{\text{C}}\text{H}-$$) is stabilized by resonance with the adjacent triple bond, similar to an allylic radical. Resonance stabilization is significantly stronger than hyperconjugation, making this radical the most stable.

Position III ($$3^\circ$$ carbon): This forms a tertiary free radical. It is stabilized via hyperconjugation by $$8\ \alpha\text{-hydrogens}$$, but it lacks resonance stabilization, making it less stable than the propargylic radical at position II.

Position IV ($$1^\circ$$ carbon): This forms a primary free radical stabilized by only $$1\ \alpha\text{-hydrogen}$$, making it less stable than III but more stable than I.

The correct option is D (II, I).

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Consider the following compound (X): The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C-H bond are: