miznaz

Two different radicals can be produced when an initiator radical adds to styrene- a primary radical and a secondary radical. Would the secondary radical be more stable due to something called hyperconjugation? If so, then why do they http://www.meta-synthesis.com/webbook/14_radical/radical.html show it as a primary one? (scroll to the near bottom of the link above)

Only allowed to use molecules that are 3 or less carbons to make the attached molecule. So I've started making this molecule and im stuck. I started with propane. Propane + Br2 + light --> bromopropane---> Mg, ether---> grignard + propanal + Hcl --> Ch3CH2CH(Ch3)OH---> PBr3 --> CH3CH2CH(CH3)Br.... added more Br, then NaOH +H20 to get BrCH(OH)CH(Ch3)Br--> PCC--> BrC=OCH(Ch3)Br + Ch3Ch2Ch2MgBr + Hcl---> see 2nd molecule attached. and now im stuck. i have a feeling theres an easier/simpler way to do this and i cant think of it. and if someone could help to get the last CH3CH2 on there and replace the H (red) to a CH3, that would be awesome!

protonating the O of the formaldehyde (by HCl) causes the carbon to be more vulnerable to attack by one of the pi electrons from the 1,3,5-trimethylbenzene.

To make 2,4,6-trimethylbenzyl chloride, I have to use 1,3,5-trimethylbenzene and react it with formaldehyde and HCl. Ive gotten started but now im stuck. So the H-Cl protonates the formaldehyde making that carbon delta positive. this carbon can now be attacked by one of the carbons of the 1,3,5-trimethylbenzene at any of the available spots. creating an OH where the Cl needs to be. How do i get rid of the OH now in order for the Cl to attack? I was thinking of protonating the OH with another molecule of HCl and then allowing an SN2 reaction to take place so the Cl can attack and replace the OH. But i have a feeling that I've done something wrong?

what would happen if imidazole acted as an acid to react with Ch3MgBr (a base). im not too sure how this acid base rxn would go further?