What do you guys think about this question?
Thanks.
Thanks.
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Ochem question...
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Answer choice D for question 1.
Answer choice C for question 2.
Btw, are these from achiever?
Answer choice C for question 2.
Btw, are these from achiever?
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No, first one is free radical halogenation, it is Markonikov with a radical intermediate stabilized in the same order as carbocations (3>2>1). Thus, B would be the more likely intermediate, with the radical electron on the middle carbon.
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Yeah I was mistaken. The first answer choice looks to be "B". I can confirm what Ivtran said based on the alkanes chapter of KBB.
The bromination reaction occurs in a way to create the most stable alkyl radical. Unlike, chlorine radicals, bromine radicals are very selective. In this example, the most stable radical is a secondary radical.
The bromination reaction occurs in a way to create the most stable alkyl radical. Unlike, chlorine radicals, bromine radicals are very selective. In this example, the most stable radical is a secondary radical.
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Having peroxides or light makes it add anti-markovnikov. Light or peroxides have the same effect. This is for HBr only. HCl or HI do not react with peroxides because the reaction is unfavorable. They always add by markovnikov's rule by electrophilic addition.
Anyway my kaplan book and my organic chemistry textbook say the Br adds first. So the answer should be E.
The mechanism using peroxides is...
Peroxide is ROOR.
1. Peroxide ROOR breaks apart to make 2 OR radicals
2. RO reacts with HBr by radicals to make ROH and Br radical.
3. Alkene + Br radical adds the Br to the less substituted carbon, making the more stable radical.
4. Termination by adding HBr to the alkene radical (with the Br). A hydrogen adds to the alkene radical.
Also for the second question, I think the answer might be I and III only. So D.
The Br2 adds by anti-addition. You make one Br on C2 coming towards you and the other going away from you on C3, but there is no reason why the opposite cannot happen where a Br on C2 is going away from you and a Br on C3 is coming towards you. So both enantiomers are made and the product mixture is racemic? So choice two is wrong?
Post the correct answer if possible.
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Free radical with hv (light) is Anti-Markovnikov.
Br added first, forming a more stable free radical in the middle.
Radical should be in the middle, but it's missing a Br on Carbon#1.
UndergradGuy7 you're right about the products being racemic (optically inactive). my bad.
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Radical reactions involve electrophilic radicals combining to form a bond. Br*, being more electronegative and electrophilic than H*, is much more likely to jump in to form the initial bond that disrupts the alkene. The intermediate will have the Br already attached and will form the most stable radical. That's what sets up the reaction as anti-markovnikov with the next-step addition of H* to the intermediate. In this question, E will be the intermediate. This is different from the normal electrophilic addition of HBr to an alkene, in which the alkene bond attacks an electrophile to form the intermediate, and since HBr is more apt to form H+/Br- than H-/Br+, the alkene forms the intermediate by addition of H+.
If you knew that Br* adds first and will be in the intermediate, you narrow your choices down to C and E. A secondary radical is more stable than a primary radical.
Oh yes yes, I stand corrected, sorry! I knew UV and peroxide makes it anti-mark, so this question was still nagging me.
Free radical halogenation start with Br2, Cl2 or such, not HX. And with UV light or ROOR the Br would add first. Choice E is the correct answer. My apologies.
Free radical halogenation start with Br2, Cl2 or such, not HX. And with UV light or ROOR the Br would add first. Choice E is the correct answer. My apologies.
my professor said for the second question..
Electrophilic addition of bromine to alkenes is an anti addition. In this case, a racemic (optically inactive) mixture of products will be formed. Carbons 2 and 3 are chiral centers (It is not clear what is meant by "both.") The best answer would be "D."
Electrophilic addition of bromine to alkenes is an anti addition. In this case, a racemic (optically inactive) mixture of products will be formed. Carbons 2 and 3 are chiral centers (It is not clear what is meant by "both.") The best answer would be "D."
Crap, I tried to add another attachment and when I clicked on save changes it removed all the attachments..
So I guess I got confused.. the answers were E for the first one, and I and III for the second one.
How do we know if it will be racemic or not?
Is there a good way of knowing how?
I see how this one can be racemic, but in general, what can I do to realize these racemic mixtures? I seem to fall into their traps every time.
Thank you so much, everyone
So I guess I got confused.. the answers were E for the first one, and I and III for the second one.
How do we know if it will be racemic or not?
Is there a good way of knowing how?
I see how this one can be racemic, but in general, what can I do to realize these racemic mixtures? I seem to fall into their traps every time.
Thank you so much, everyone
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sfoskn, can you repost the questions?
Perhaps just keep in mind how things happen. In the second question, you were being tested on both whether you knew the products of the reaction, and whether or not you knew that the anti addition could have occurred on either the top or bottom side of the alkene, leading to optically inactive product. Racemic mixtures happen when there's no preference on the direction of attack like in this case, or in Sn1 reactions.
