[timchai]

i want to ask for Q1 , this is the ans , the otheer , i don't hv , i jusk want to ask for ans
for Q1 , i think Br and H should be interchanged , because of stability of carbocation , and i want to ask the use of benzoyl peoxide
for Q2 , i, think this is addition Rx , am i correct
for Q3 , i want to askwhich 1 is correct, or both wrong , then wt is the answer ?can u draw?
because i think H+ can both attach to C= C or C=O , i don't klnow how to choose

[P]

Not sure if I have read your post correctly, but here is what I make of it:

For Q1 you have Styrene in the presence of HBr and BPO initiator right?  Depending on the temperature, the BPO will dissosciate into 2 radicals which will initiate a Free Radical Polymerisation by breaking the double bond in the styrene and starting a chain reaction (polymerisation)  -  You would then be left with polystyrene.   Not sure about the presence of the HBr in there  -  it might cause some bromination to occur in the polymerisation -  resulting in brominated polystyrene.

[cundi]

Q1. Your are mistaked. In this case you have a radical reaction, not cationic one.
Q2. Attached you can find my solution proposal.
Q3. I can't see which is the starting material of the reaction. Please draw it better.

[typhoon2028]

Q1 is correct.

Reasoning:  (2-bromoethyl)benzene can be synthesized by Anti-Markonikov addition of gaseous HBr to Styrene in presence of peroxide catalyst.  Without peroxide catalyst, standard Markonikov addition occurs resulting in (1-bromoethyl)benzene.

[P]

Thank you Typhoon. I'm happy now. I still think that the question should include the temperature at which this takes place, because if it was over 40C then the BPO would dissociate and start the FRP I mentioned above would it not?