DAT Achiever Test 1 #82

[moblongata]

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I have always understood that peroxides lead to anti-Markovnikov products while tertiary radicals are more stable than secondary radicals and are added to by Markovnikov.

SO in the question, I am confused why HBr, hv adds "Br" anti-Markovnikov when there are no peroxides present in the question.

Am I missing something? Is it a mistake in the key?

Thanks in advance!🙂

 

[g23armani]

you need hv (uv light) OR peroxides for anti mark

 

[moblongata]

Thanks for your quick response. That makes sense for a lot of the questions, I have been doing but there was another question from Achiever that messed me up.

It says that:
(CH3)2CHCH2CH3 + Cl2 (hv/35 degrees)--> (CH3)2CClCH2CH3 + HCl

The Cl added markovnikov in the presence of hv for this question?

Thanks for the help!

 

[g23armani]

HBr is the only hydrogen halide that consistently works for anti markovnikov addition to an alkene.

 

[moblongata]

Thanks for explaining.I'm still a little confused. 🙁

So Cl2 even in the presence of hv will be markovnikov? Doesn't that contradict the idea that the presence of hv results in anti-markovnikov?

 

[jlt0530]

Thanks for your quick response. That makes sense for a lot of the questions, I have been doing but there was another question from Achiever that messed me up.

It says that:
(CH3)2CHCH2CH3 + Cl2 (hv/35 degrees)--> (CH3)2CClCH2CH3 + HCl

The Cl added markovnikov in the presence of hv for this question?

Thanks for the help!

Whenever I see hv or peroxide, i think of 4 scenarios which are

1. alkane + Br2/hv
2. alkane + Cl2/hv
3. alkene + HBr/hv
4. alkene + HCl/hv

(here, it can be either hv or peroxide)

1. major product is Br on most substitued carbon because free radical bromination occurs slowly, so that it seeks for the most stable carbon. So, Br is usually attached to tertiary or secondary carbon. (even with the hv, add markovnikov)

2. major product is Cl on most accessible carbon (least substitued) because free radical chlorination occurs so fast that it just wants to settle on the most easily approacheable carbon. So, Cl is usually attached to primary carbon. (so, anti-markovnikov)

3. here, hv or peroxide add Br on less substituted carbon of the alkene, following anti-markovnikov. This is only true for HBr/peroxide or hv on alkene.

4. peroxide or hv doesn't add clorine anti-markovnikov. Even with the hv, Cl adds to the most stable carbon.

This is what i think (at least what i think it is). let me know if i made any mistakes. I hope this helped you someway. btw, that little puppy is so cute! 🙂

 

[moblongata]

Whenever I see hv or peroxide, i think of 4 scenarios which are

1. alkane + Br2/hv
2. alkane + Cl2/hv
3. alkene + HBr/hv
4. alkene + HCl/hv

(here, it can be either hv or peroxide)

1. major product is Br on most substitued carbon because free radical bromination occurs slowly, so that it seeks for the most stable carbon. So, Br is usually attached to tertiary or secondary carbon. (even with the hv, add markovnikov)

2. major product is Cl on most accessible carbon (least substitued) because free radical chlorination occurs so fast that it just wants to settle on the most easily approacheable carbon. So, Cl is usually attached to primary carbon. (so, anti-markovnikov)

3. here, hv or peroxide add Br on less substituted carbon of the alkene, following anti-markovnikov. This is only true for HBr/peroxide or hv on alkene.

4. peroxide or hv doesn't add clorine anti-markovnikov. Even with the hv, Cl adds to the most stable carbon.

This is what i think (at least what i think it is). let me know if i made any mistakes. I hope this helped you someway. btw, that little puppy is so cute! 🙂

Thanks for your help! Sorry I didn't write back earlier. I didn't check this thread until now and ended up taking the DAT like the next day or two. I got lucky on my ochem questions that day 🙂 I am a fan of cute puppies too!

 

[iDreamofDent]

Hi Jlt05, rethink your answer look at this http://www.cartage.org.lb/en/themes/sciences/chemistry/Organicchemistry/CommonReaction/FreeRadical/FreeRadical.htm

when there is X-X radicalized with peroxide or light it the first radical X* will abstract the hydrogen that leads to the most stable radical then the second X* will add to this part and result in the most substituted product

 

[iDreamofDent]

Hi Jlt05, rethink your answer look at this http://www.cartage.org.lb/en/themes/sciences/chemistry/Organicchemistry/CommonReaction/FreeRadical/FreeRadical.htm

when there is X-X radicalized with peroxide or light it the first radical X* will abstract the hydrogen that leads to the most stable radical then the second X* will add to this part and result in the most substituted product.

If it was HX radicalized then it would add in the least substituted position