Free Radical Bromination

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sap1622

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Attached is the question on the free practice OAT (its similar to the DAT). According to Kaplan I thought free radicals, esp bromine, since it reacts slowly would attack the carbon that would produce the most stable free radical. I thought it was A, the tertiary one, but the answer says C, the methane. HELP PLZ!! :scared:
 
Odd, I thought radicals were more stable when allylic, tertiary, or secondary. I guess maybe sterics win here? Methane is the smallest molecule and the bromine can just go right up and be added.
 
Your source is wrong. Ther answer is A. Its impossible for the answer to be C. radical bromination has a ratio of 1600:80:1 for tertiary:secondary😛rimary carbons. Theres waaaay too much unstable energy in the transition state when the radical methane is being made. tertiary radical is more stable thus more likely to form fastest.

on a side note, if your doing radical chlorination then your ratios of products are bit more even.


remember ppl at kaplan are dumbas*es
 
Your source is wrong. Ther answer is A. Its impossible for the answer to be C. radical bromination has a ratio of 1600:80:1 for tertiary:secondary😛rimary carbons. Theres waaaay too much unstable energy in the transition state when the radical methane is being made. tertiary radical is more stable thus more likely to form fastest.

on a side note, if your doing radical chlorination then your ratios of products are bit more even.


remember ppl at kaplan are dumbas*es

HAHA. Funny thing is that this was from a sample OAT. pretty ridiculous. 🙄
 
Your source is wrong. Ther answer is A. Its impossible for the answer to be C. radical bromination has a ratio of 1600:80:1 for tertiary:secondary😛rimary carbons. Theres waaaay too much unstable energy in the transition state when the radical methane is being made. tertiary radical is more stable thus more likely to form fastest.

on a side note, if your doing radical chlorination then your ratios of products are bit more even.

+1

Thats what i learned
 
This is a tricky question. One of the option, cyclopropane reacts even more faster. However by electrophilic substitution method.

Regarding the correct option; A is correct.
 
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