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Topic: Monohalogenation via radicals  (Read 9841 times)

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Offline coolguy99

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Monohalogenation via radicals
« on: November 16, 2009, 08:34:51 PM »
quick question, doing some OWL homework and I keep getting this question wrong.

Quote
Is it possible to prepare (R)-2-bromo-3-methylbutane in high yield by halogenation of an alkane?

How many monohalo isomers will be formed upon radical halogenation of the parent alkane?

Consider stereoisomers as well.

So for the first one, the answer is no, because bromine will almost always go to the tertiary carbon; that much I get.

Now for the isomers..

The only chiral center that is formed by bromination is the 2nd carbon on the chain, which could be R or S, so that's two right there. The bromine could go on the first carbon, not chiral, so 1 more there. It could go on the tertiary carbon, still wouldn't be chiral, 1 more there, and it could go on either of the identical methyl groups, still wouldn't be chiral, so only 1 more there. For a total of 5, which is wrong according to OWL, so

whattttt am I missing? :(

Offline coolguy99

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Re: Monohalogenation via radicals
« Reply #1 on: November 16, 2009, 10:57:38 PM »
Either I'm blind or the edit button is missing...

but, just punching in random numbers, I found that the answer is 6.

I still can only find 5 isomers. Any help?

here is my work so far...

« Last Edit: November 16, 2009, 11:09:20 PM by coolguy99 »

Offline orgopete

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Re: Monohalogenation via radicals
« Reply #2 on: November 17, 2009, 01:36:16 AM »
Your 5th structure has a chirality center, so there is one more isomer to it.
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