How Do You Convert Tertiary Haloalkanes Using Free Radical Substitution?

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Homework Statement


can someone teach me how do you convert this reactant to get the product?

IMG_20141005_180344[1].jpg


I consider this as free radical substituition reaction and in the presence of UV , then the H atom in the 2 CH3 will also be substituted... i want ony the H atom on the carbon bonded to C to be substituted only.

Homework Equations

The Attempt at a Solution

 
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Why would you assume that the rate of addition of bromine to the terminal carbons will be so high as to make free radical bromination not useful for this conversion? Think about the mechanism and selectivity of free radical bromination. What type of carbons are favored to form radicals in these mechanisms? What type of carbon is the carbon of interest and what type are the terminal carbons?

Sometimes it is better to do a one step lower yielding synthesis than it is to go through several steps which may individually appear better but all that handling and loss etc makes the synthesis less efficient and more time/energy consuming etc.
 
Yanick said:
Why would you assume that the rate of addition of bromine to the terminal carbons will be so high as to make free radical bromination not useful for this conversion? Think about the mechanism and selectivity of free radical bromination. What type of carbons are favored to form radicals in these mechanisms? What type of carbon is the carbon of interest and what type are the terminal carbons?

Sometimes it is better to do a one step lower yielding synthesis than it is to go through several steps which may individually appear better but all that handling and loss etc makes the synthesis less efficient and more time/energy consuming etc.
i was told that The H on the side chain is a "Benzylic hydrogen atom" it is very reactive and even chlorine will react mostly by attacking at that position. Bromine, as shown, is much more selective.
does it mean benzylic hydrogen atom more reactive than aliphatic hydrogen atom... why is it so?
 
A majority of introductory O-Chem is explained by charge/electron density as influenced by the Inductive Effect and Resonance Effects. Although I'm not exactly sure why radicals become more stable when they are "spread out," I have just come to accept this fact.

http://www.chemgapedia.de/vsengine/.../radi_brom_benzyl/radi_brom_benzyl.vscml.html

For the record, I googled this in about 30 seconds (I certainly hope you took initiative yourself and haven't been waiting for someone to google for you).