Destroyer ochem

[osimsDDS]

alright this might be stupid but its really annoying the hell out of me and i need to know:

on the road map #3 page i dont understand why its going from a methyl-cyclopentane and adding Br2 with UV markonikov, isnt it supposed to add anti-mark to the least substituted part of the alkane which would be the methyle, its basically adding it to the tertiary methyl on the ring...doesnt make sense to me can someone explain??? is it an exception, thanks a lot

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[TeamGuo]

Br2 uv is selective in radical reactions, it really doesn't relate to Mark's rule since we are dealing with alkane, not alkene. Br2 with uv selects most stable position = which is tertiary here. But with Cl2 and uv, you pick the most available spot given on a compound = so in this case with methyl-cyclopentane, you would chlorinate any one of the secondary positions.

Hope that helps. You just have to remember, Br2 radical reaction = selective, Cl2 = is not.

 

[DRHOYA]

It there was a cyclopentane ring, with a CH2 double bonded to the ring instead of the CH3.........then it would add to the CH2 to give a five-membered ring, with CH2Br bonded to it.

 

[osimsDDS]

Br2 uv is selective in radical reactions, it really doesn't relate to Mark's rule since we are dealing with alkane, not alkene. Br2 with uv selects most stable position = which is tertiary here. But with Cl2 and uv, you pick the most available spot given on a compound = so in this case with methyl-cyclopentane, you would chlorinate any one of the secondary positions.

Hope that helps. You just have to remember, Br2 radical reaction = selective, Cl2 = is not.

So you would not chlorinate the primary methyl position at the top of the ring?You would just select any secondary position?? thanks

 

[TheGreenWave]

So you would not chlorinate the primary methyl position at the top of the ring?You would just select any secondary position?? thanks

Like someone already said, Br2 / hv is a radical reaction. The most stable radical is tertiary. The carbon within the ring where the methyl is coming off is tertiary and a Br will add there. Which is markovnikov I guess.

To get a Br on a carbon coming off a cyclopentane there would have to be a double bond b/w a carbon on the ring and the carbon coming off the ring (sorry if thats confusing). Then you would use HBr w/ ROOR which would give anti-Markovnikov addition.

Hope that helped some. Does anyone know if the road maps in destroyer basically cover all the reactions we should know for the DAT????

 

[TeamGuo]

So you would not chlorinate the primary methyl position at the top of the ring?You would just select any secondary position?? thanks

Yes you would chlorinate the secondary position since you have 4 secondary positions available. When you are given Cl2 uv for radical reaction, you chlorinate the most available spot, meaning which positions has most vacancy. There's only 1 primary position so secondary wins in this case.

Hope it makes sense.