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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: caseyh on October 19, 2009, 01:20:37 AM

Title: Bromination of dimethylcyclohexane?
Post by: caseyh on October 19, 2009, 01:20:37 AM

When cis-1,2-dimethylcyclohexane undergoes free radical bromination and the product is subjected to fractional distillation, 2 fractions are collected neither of which is optically active.

When trans-1,2-dimethylcyclohexane undergoes free radical bromination and the product is subjected to fractional distillation, 2 fractions are collected and both are optically active.

How many products are there for each? And why are there only 2 fractions? We've been trying to figure this out all weekend without any luck. I know that because the cis compound is meso it is achiral and therefore its products are achiral...but I'm still really confused. Any help would be greatly appreciated!

Thanks,

Casey

Title: Re: Bromination of dimethylcyclohexane?
Post by: KritikalMass on October 19, 2009, 07:13:51 PM

If I remember correctly the bromination can occur either in the methyl group or on the ring where the methyl group attaches to the ring. That would account for their being two different fractions for both the cis and the trans conformations.

In regards to the number of products, it seems like there would be two for cis and two for trans since you got two fractions in each case.

I'll do a little more research on it- it seems like I came across a website that answered this exact question awhile back but have been unable to locate it at this time.

Title: Re: Bromination of dimethylcyclohexane?
Post by: macman104 on October 20, 2009, 06:51:19 PM

Which position(s) in the molecule forms the most stable radical?