Question

#3 A and B but i mostly want this question answered. The picture provided is supposed to be the right answer. How come iodination doesn't occur on the benzene ring? And assume it was fluorination, would it be similar to iodination and not "attack" hydrogens on benzene ring?

c) products: 3. The hydrocarbon shown below is subjected to free-radical mono-lodination reaction. The b iodination reaction. The relative rates for mono-iodination of 3, 2, and 1 hydrogens are 3.5, 2.5, and I, respectively. DAC Е И a) Show the structures for all possible products: 9 b) Calculate the percentages for the formation of the products you showed in part "a" above. Halabel type number x و 4 ی لحا لله 2 ส่ง ๗ ง 1 |--NN--- Rel.Rate - 2.5 2. 5A 25 = 2.5a 2.5 soc =5.0D 2.5 = 3.5G + 23.5 2.5 2.5 = 2.56 = 2. 57 3,5 2.A=17,B-%E32F = 2 x 100 = 10.6%, 2C-%D - 5,0x100 21.3% 3.G = 325 x 100 = 14,9) 3.5 23.5

Answer 1

The reaction asked in the question is a radical reaction. Its mechanism involves the formation of radical ions I ^$\cdot$ which combines with the carbon atoms containing hydrogens A, B, C and so on as presented in the question. The hydrogen atoms which get substituted belong to sp³ carbon atoms which are easier to abstract than the hydrogens belonging to sp² carbon atoms in the benzene ring. Therefore, the radical mechanism occurs in the alkyl chain and not on the benzene ring.