Adding H and Br, Anti-Markovnikov

Finally, the chloride ion now attacks the tertiary carbocation to give our product  

PROBLEMS Draw the mechanism for each of the following reactions  

In the previous section, we saw how to add H and X, placing X at the more substituted carbon (Markovnikov addition). There is another reaction that will allow us to add H and X a h-Markovnikov, but it only works well with HBr (not any other H—X). 

If we use HBr in the presence of peroxides (ROOR), Br ends up on the less substituted carbon  

Why does the presence of peroxides cause the addition to be anti-Markovnikov In order to understand the answer to this question, we will need to explore the mechanism in detail. This reaction follows a mechanism that involves radical intermediates (such as Br ), rather than ionic intermediates (such as Br ). Peroxides are used to generate bromine radicals, in the following way  

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Answer HBr indicates that we will be adding H and Br across the double bond. The presence of peroxides indicates that the regiochemistry will be anti-Markovnikov. To determine whether stereochemistry is relevant in this particular case, we need to look at whether we are creating two new stereocenters. When we place the Br on the less substituted carbon (and the H on the more substituted carbon), we will only be creating one new stereocenter. With only one stereocenter, there are not four possible stereoisomers but just two possible products (a pair of enantiomers). And we will get this pair of enantiomers regardless of whether the reaction was syn or anti ... 

As we saw in the protonation of an alkene, the electrophile (in this case, Br -) adds to the less substituted end of the double bond, and the unpaired electron appears on the more substituted carbon to give the more stable free radical. This intermediate reacts with HBr to give the anti-Markovnikov product, in which H has added to the more substituted end of the double bond the end that started with fewer hydrogens. 

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