Apr 12, 2013 · Addition of HBr to alkenes in the presence of ROOR (peroxides) and light / heat (of course) results in "anti-Markovnikov" addition of HBr. But why? Read on!

Anti-Markovnikov addition of HBr to alkenes. We saw in section 10.4 that under normal conditions, HBr adds to an unsymmetrical alkene to form an alkyl halide where the H goes onto the less substituted carbon, and the Br goes onto the more substituted carbon – thus, it obeys Markovnikov’s Rule.

This organic chemistry video tutorial explains what happens when Alkene reacts with HBr in the presence of organic peroxides such as ROOR. This video provid...

Nov 25, 2013 · Take toluene and treat with either Br 2 in the presence of light, as per this procedure, or N-Bromosuccinimide (NBS) in the presence of a radical initiator + heat (or light) and one of the benzylic C–H bonds is replaced with C–Br.

Jan 21, 2024 · With the organic peroxides present you get a free radical chain reaction. The chain is initiated by free radicals produced by an oxygen-oxygen bond in the organic peroxide breaking. These free radicals extract a hydrogen atom from a hydrogen bromide molecule to produce bromine radicals.

Dec 1, 2005 · Br2 with ROOR (or HV or heat) + an alkene = bromine is added across the double bond in an anti fashion. Note that this is not anti-markovnikov. Because we are adding two bromines across the double bond, this reaction is not regioselective so we don't even think about mark or anti-mark.

Jan 23, 2023 · Free radical addition to a carbon-carbon double bond. If you have read the introductory page (see above), you will know that hydrogen bromide adds to the carbon-carbon double bond in alkenes via a free radical mechanism in the presence of organic peroxides or oxygen from the air.

Apr 14, 2006 · The peroxide effect only occurs with Br2. Anti mark addition with Br2 and peroxide. Mark addition with any other halogens and peroxide. This all relates to the stability of the transition state.

When hydrobromic acid (HBr) and any peroxide (R-O-O-R, ie H2O2) react with an alkene, the reaction favors the anti-Markovnikov product: If the carbons have the same ranking (ie both are Secondary), a racemic mixture of 2 products will be formed, with the functional group being added to each carbon: The reaction mechanism is depicted below:

These free radicals extract a hydrogen atom from a hydrogen bromide molecule to produce bromine radicals. A bromine radical joins to the ethene using one of the electrons in the pi bond. That creates a new radical with the single electron on the other carbon atom.