- Joined
- Sep 15, 2007
- Messages
- 254
- Reaction score
- 1
states that:
"when an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon that has the greater number of hydrogen substituents, and the halogen to the carbon have the fewer number of hydrogen substituents"
ok so when u got a free radical like Br* why is it anit-markovnikov?
i understand that the free radical is the most stable state at 3>2>1>CH3 just like the carbocation formed when adding HBr to alkenes.
CH3CH=C(CH3)2 + HBr ---------> CH3CH2--C+(CH3)2 MARKOVNIKOV
CH3CH=C(CH3)2 + *Br ---------> CH3CHBr--C*(CH3)2 ANTI
both of the reactions above added the same way and the position of the carbocation and free radical is on the same carbon, so why is one markie and the other anti-markie????
"when an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon that has the greater number of hydrogen substituents, and the halogen to the carbon have the fewer number of hydrogen substituents"
ok so when u got a free radical like Br* why is it anit-markovnikov?
i understand that the free radical is the most stable state at 3>2>1>CH3 just like the carbocation formed when adding HBr to alkenes.
CH3CH=C(CH3)2 + HBr ---------> CH3CH2--C+(CH3)2 MARKOVNIKOV
CH3CH=C(CH3)2 + *Br ---------> CH3CHBr--C*(CH3)2 ANTI
both of the reactions above added the same way and the position of the carbocation and free radical is on the same carbon, so why is one markie and the other anti-markie????