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| DAT Discussions Discuss study tips and resources to excel on the Dental Admissions Test. |
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#1 |
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Junior Member
Join Date: Jun 2006
Posts: 57
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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.... |
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#2 |
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All-Star
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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.
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#3 |
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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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#4 | |
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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#5 | |
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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#6 | |
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#7 |
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Senior Member
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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 |
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#8 | |
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#9 | ||
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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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" |
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#10 |
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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.
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#11 | |
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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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.
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#12 |
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All-Star
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yea, as opposed to ~1680 times for bromination.
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#13 |
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All-Star
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But look at the practice problems, they didnt label major/minor for chlorination. You think the book is wrong?
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#14 | |
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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#15 |
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All-Star
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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.
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#16 | ||
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Junior Member
Join Date: Apr 2005
Location: Weyburn Terrace :-)
Posts: 2,079
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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. |
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#17 | |
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#18 |
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Senior Member
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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. |
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#19 |
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Member
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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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