(a) Based on what you know about the relative stabilities of alkyl cations and benzylic cations, predict the product of addition of HBr to 1-phenylpropene.

(b) Propose a mechanism for this reaction.

(c) Based on what you know about the relative stabilities of alkyl radicals and benzylic radicals, predict the product of addition of HBr to 1-phenylpropene in the presence of a free-radical initiator.

(d) Propose a mechanism for this reaction.

An atom, molecule, or ion that comprises at least one unpaired valence electron is known as a free radical. Unpaired electrons present in radicals make them highly chemically reactive.

The Benzylic cations are stabilized by the resonance (+R effect) which makes them more stable than alkyl cations. The alkyl cations on the other hand are stabilized by the +I (inductive) effect only.

The product of the addition of HBr to 1-phenylpropene is 1-bromo-1-phenylpropane.

The hydrohalogenation of an alkene with HBr is shown below.

Here, the benzylic cation is more stable than the secondary cation. Therefore, the major product formed is 1-bromo-1-phenylpropane.

The benzylic radicals are more stable than alkyl radicals because of resonance (+R effect).

The benzylic radical is formed when bromine gets added first to the second carbon.

Later, the free H-radical goes to carbon radical to form 2-bromo-1-phenylpropane.

Hence, the product of the addition of HBr to 1-phenylpropene in presence of a free radical initiator is 2-bromo-1-phenylpropane.

In the initiation step, the free radicals are formed in presence of light.

The free radicals are reactive and unstable. As soon as the benzylic radical is formed, the hydrogen radical goes to it to stabilize the compound and decrease the energy.

Firstly, bromine is added and then H-radical comes in. The reaction goes as;