dentistry2011

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i keep getting confused with reactions where halogens are added.

i thought when a halogen (ie. Br2) + UV light is added to something - the halogen attaches to the least substituted atom.....

but sometimes it doesn't, it adds to the most substituted.....

Is it different depending onthe whether the molecule is an alkene/a ring/ etc?

can someone clarify this?? ill try to go find some exact examples in the meantime....
 

allstardentist

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i think if the free radical halogenation involves alkene or alkyne, then it follows anti-Markovnikovs rule. But with alkane, I think it goes to the most substituted. Thats my best guess based on kaplan book. So im not 100% sure.
 

allstardentist

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Ok, I just read in the princeton review that in free radical bromination alkane, the Br goes to the most substituted carbon as the major product and less substituted ones as minor. But for chlorine, it can go anywhere, meaning that the substituion of a Carbon doesnt matter.
 
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dat_student

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dentistry2011 said:
i keep getting confused with reactions where halogens are added.

i thought when a halogen (ie. Br2) + UV light is added to something - the halogen attaches to the least substituted atom.....

but sometimes it doesn't, it adds to the most substituted.....

Is it different depending onthe whether the molecule is an alkene/a ring/ etc?

can someone clarify this?? ill try to go find some exact examples in the meantime....
instead of trying to memorize what goes where and when most subst. and when least >> 1st try to find the most stable intermediate. That's the best way in my opinion and it has no exceptions. good luck.
 

dat_student

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allstardentist said:
Ok, I just read in the princeton review that in free radical bromination alkane, the Br goes to the most substituted carbon as the major product and less substituted ones as minor. But for chlorine, it can go anywhere, meaning that the substituion of a Carbon doesnt matter.
r u sure about this? Cl & Br usually follow the same rules. What page?
 

Resonance

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dat_student said:
r u sure about this? Cl & Br usually follow the same rules. What page?
page 287 (kaplan blue book) 3rd paragraph. Wade 5th edition page 220 (not that you necessarily have this text)
 

DonExodus

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Chlorine still is subject to anti-markovnikov addition- the intermediate stability is not affected by the type of halogen, its the location of the C+.
(For the most part)

-Don
 

dat_student

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allstardentist said:
Ok, I just read in the princeton review that in free radical bromination alkane, the Br goes to the most substituted carbon as the major product and less substituted ones as minor. But for chlorine, it can go anywhere, meaning that the substituion of a Carbon doesnt matter.
allstardentist said:
pg 502&503. look at the practice problems. I didnt really read(just skimmed through) the text, but thats what i inferred from the example problems.
ok look at the bottom of pg. 501 (MCAT Biological Sciences review)
it says "Bromination of alkanes is much more selective than chlorination" but it doesn't say that it can go anywhere. In fact, for chlorination it says the following:

.............................reaction rate...........
........................R3CH......R2CH2.....RCH3
Chlorination........5.3.........3.9.........1


P.S. selectivity = (reactivity) / (# of sites available to react)


Look at the example on p. 501

selectivity of tertiaty....37% / 1......37
_________________ = ________ = ____ = 5.3
selectivity of primary.....63% / 9......7

I also double checked with McMurry. :)

McMurry 4th edition, pg. 348 says:

"Thus, a tertiary hydrogen is 35/7.2 = 5 times as reactive as a primary hydrogen toward chlorination"
 

allstardentist

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That is true. But Chlorination is not as selective as bromination as to specify the products as major/minor product as in bromination. That's why I was saying chlorine can basically be attached any of the carbons in a alkane thats attached to at least one hydrogen. But I know what ure saying. There is a little bit of preference for more substituted carbon.
 

dat_student

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allstardentist said:
...There is a little bit of preference for more substituted carbon.
yes, tertiary hydrogen is ~5 times as reactive (see above) :) you can kinda deduce major and minor product based on that. Just look at how many primary and tertiary sites you have..etc etc All I am saying is that there is a preference and it's not random. You still get major and minor products.
 

dat_student

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allstardentist said:
But look at the practice problems, they didnt label major/minor for chlorination. You think the book is wrong?
According to that book: "Bromination of alkanes is much more selective than chlorination." The book says chlorination is still selective. "5 times" is still a lot in my opinion. Let's say you have one primary site and one tertiaty site and you do chlorination what do you think the major product would be?
 

allstardentist

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I think if each answer choice has one product i would put down tertiary. But if there was another answer choice that has both tertiary and primary i would pick that instead. dentistry2011, most likely chlorination wont be on the test, so just know how bromination works for the DAT. I think that should suffice.
 

dat_student

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allstardentist said:
...But for chlorine, it can go anywhere, meaning that the substituion of a Carbon doesnt matter.
I just didn't agree with that statement.

allstardentist said:
I think if each answer choice has one product i would put down tertiary. ...
If you have 9 primary sites and 1 tertiary site (as princeton review and mcmurry say) the major product is the primary...(i think ratio is 63%:37%)

If you have 1 primary site and 1 tertiary site the major product is tertiary (5 to 1 ratio).

P.S. I agree with you. "Chlorination" is a bit tricky. The OP probably won't see it on the DAT.
 

allstardentist

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dat_student said:
I just didn't agree with that statement.



If you have 9 primary sites and 1 tertiary site (as princeton review and mcmurry say) the major product is the primary...(i think ratio is 63%:37%)

If you have 1 primary site and 1 tertiary site the major product is tertiary (5 to 1 ratio).

P.S. I agree with you. "Chlorination" is a bit tricky. The OP probably won't see it on the DAT.
ah I think that's what they were saying...yeah.
 

DonExodus

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Youre also forgetting about the fact that youre mentioning two different mechanisms... just because its more selective for substitution / addition doesnt mean its the same for free radical halogenation...

The DAT is testing to see if you know it goes anti-markov, etc- not the super fine details.
 

Golddanboy

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Hey, i was actually just looking this up. Randomly DAT Achiever (question 82, test 1 of sci) says that free radical bromination is anti-mark. This is a clear mistake and it probably confuses a lot of people who use DAT Achiever. The overall rule is that Br2 and light goes by the Markovnikov rule. As i remember it, free radical stability is Allyl > Tertiary > Secondary > Primary > Methyl > Vinyl.

The ONLY time i know where bromination clearly is anti-Markovnikov is when you use the peroxide effect. I've made a nice little habit of checking all the questions i got wrong, since every practice test i took so far had at least 1 major error.
 
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