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????

All I know is that for a free radical, it wants to form the most stable free radical (branched)... I didn't know you could classify it as anti-/markovnikov....

hmmm, I should look into this....

totally forgot about hydride shifts....

when there is a secondary or tertiary carbon next to the double bond, a hydride switch will occur so that the bromine can be on the most stable carbon. this is a common exception so always look to see if there are is a more stable carbon next to a double bond.

ooooh i think i got it guys.... this is totally retarded

its becuz the first reaction the H (from the HBr) is the one that attaches to the carbon with the most hydrogens.... but in the second reaction the *Br attaches to it , and then the H (from the radical) attaches to the tertiary carbon

they gave this guys his own rule for stating the obvious? lol

hey look the sky is blue, now wheres my noble prize?

hey look the sky is blue, now wheres my noble prize?

lol what? it's not that simple. its useful in practical matters where you want to create a compound with antimarkovnikov addition....
BH3 THF = anti mark of alcohol
H+/H20 = markovnikov addition of alcohol

Antimarkovnikov will occur when a peroxide and HBR (any ROOR and HBr)
This is because the ROOR causes the HBr to become a radical (doubt the mechanism is necessary for the exam).... the radical BR- ATTACKS the double bond...... so since its doing the attacking, it will attack the LEAST substituted area.


Markovnikov on the other hand is with just the HBr.... in this case the H attaches first (attacked by double bond).... it will attack to the least substituted..... this is because it wants to create THE MOST STABLE CARBOCATION.... then the Br will attack that carbocation and attach the most substituted.


This is NOT an example of a hydride shift. That occurs during SN1 reactions. This is just difference in the order in which things are attaching so they attach differently.

ya i know that, but the mechanism helps me understand it so that it'll stick in my head. just memorizing that ROOR makes free radicals and its antimarkovniv doesnt put me at ease.

and i guess in the late 1800's figuring that out in practical ways is a b*tch lol, this poor guy was in the lab mixing stuff till he came up with something he can claim as his own

Br prefers the more subsituted - which does not deal with Markovnikov/AntiMarkovnikov.

For Markovnicov addition- remember that the rich get richer - H goes to the side with more Hs.

For antimarkovnikov, its the opposite.

yea you are definitely right. without knowing the mechanism.. what attacks what.. its difficult to know where everything will end up unless you just straight up memorize reactions.

just thought i'd explain the mechanism of attacking incase it helped... but i meant i dont think you need to know how
ROOR + HBR = radical...

good luck :)

Markovnikov and Zaytsev rule in ORGO lol