Question

Question 28 and 29 with explanation

Answer 1

28) homolysis is equal lysis of bond like in a bond 2 e-s are present so both the intermediates will get 1 e-
In the question only option 1) and 2) represent homolysis.

That one will be more stable which have more hyperconjugating structures(note:- hydrogen free radical is common so only think about carbon free radical). In option 2) there are 9 hyperconjugating structures so it is more stable. Hence option 2) is the correct answer.

29) In this question you just need to count the number of hyperconjugating structures.
In Option 1) there are 6 conjugating structures.
In Option 2) there are 5 conjugating structures.
In Option 3) there are 5 conjugating structures.
In Option 4) there are 4 conjugating structures.

So on the basis of no. of hyperconjugating structures option 1) is correct.

Hope it helps.