EK #382 Organic Chemistry 1001

[mtravis2190]

Hi guys,

I'd appreciate any help with the following question that you may be able to offer. Also, I've posted a few other others previously and I'd like to direct your attention to those as well. They were several physics questions and I haven't received any feedback on them yet.

But anyway, onto some organic!

The reaction of 1,3-butadiene with HBr is shown below. At 40C, the major product is the 1,4-addition product; however, at-80C, the major product is 1,2-addition product.
Q.382- Why are two products formed?
a. There are two double bonds present.
b. The carbocation intermediate allows delocalization of the second double bond.
c. The fact that the carbocation is planar allows attack from both sides of the plane.
d. There are 2 moles of HBr.

I thought the correct answer is c., but it's b. I don't see how.

HELP!

P.S. There will be more questions from this section of Ek. Stay tuned!

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[brood910]

You dont have to know anything about double bond reactions for MCAT.

 

[mtravis2190]

You dont have to know anything about double bond reactions for MCAT.

No way??? Are you for real??

 

[mtravis2190]

You dont have to know anything about double bond reactions for MCAT.

But we still need to know elimination reactions?

 

[brood910]

But we still need to know elimination reactions?

That's not really "double bond" reaction.
You have to know SN1, SN2, E1, E2, etc.

EK themselves stated in their book that they wont be tested. Benzene + triple bond reactions wont be tested either. If they are, they will give you info in the passages.

 

[DrknoSDN]

Just to answer the why... in 1,3-butadiene (C=C-C=c) +HBr the pi electrons will grab the hydrogen forming a secondary carbocation on Carbon 2.
(H3C-CH+-C=C)

That remaining pi electrons can delocalize the positive charge between C2 and C4 and the Bromine can attack either carbocation.
I believe at lower temperatures the lower kinetic energy of the molecule forces the positive charge to stay more on carbon 2 because a secondary carbocation is much more stable than a primary forcing the 1,2 addition. Steric factors cause 1,4 addition to favor at high temp.
,,,,,
Also this question might not be tested verbatim but electron delocalization is heavily tested with things like Aldol condenstation, and alpha hydrogen acidity etc...

 

[ReggieD6868]

As Drkno said, Diene chemistry may not be directly tested on the MCAT, but I find it hard to believe that the concepts associated with the original question such as conjugation and thermodynamic vs. kinetic control aren't important for the exam.

The question has to do with thermodynamic vs kinetic control and more specifically those differences in activation energy.

The kinetic product in this case, the 1,2 addition, is not the most stable of the two possibilities, but it occurs at the greatest rate (kinetic) because the 1,2 addition is the result of the secondary allylic carbocation that is formed. The secondary allylic carbocation at carbon number 2 is very stable so it will often, but that doesn't mean it will form the most stable product.

The thermodynamic product in this case, the 1,4 addition, forms a primary allylic carbocation that is less readily formed than the aforementioned secondary carbocation, but the product of this reaction is the more substituted alkene, so the more stable (thermodynamic) product is formed, and the energy barrier that was required to overcome the kinetic favor was met when the temperature was raised to 80C.

Although the correct answer choice, B, requires one to draw an inference from the explanation I gave, it's the only answer that has anything to do with the question prompt, kinetic vs. thermodynamic control in the conjugated diene.

Choice A is wrong because two double bonds doesn't necessarily mean the two double bonds are in conjugation, and this question is dealing with the results of addition to a symmetric conjugated diene.

Choice C is a garbled answer that highlights a concept of carbocation formation that adds nothing to do the issue with the conjugated diene. This is the type of answer one would choose if they were guessing but knew general carbocation rules and picked what felt right.

Choice D would mean both double bonds underwent an addition reaction and two Br ions were added to the hydrocarbon, which does not pertain to 1,2 and 1,4 addition.

Again, although choice B may not have been explicitly stated, it's the only answer that pertains to the issue at hand in the question.

I hope that helps.

 

[mtravis2190]

As Drkno said, Diene chemistry may not be directly tested on the MCAT, but I find it hard to believe that the concepts associated with the original question such as conjugation and thermodynamic vs. kinetic control aren't important for the exam.

The question has to do with thermodynamic vs kinetic control and more specifically those differences in activation energy.

The kinetic product in this case, the 1,2 addition, is not the most stable of the two possibilities, but it occurs at the greatest rate (kinetic) because the 1,2 addition is the result of the secondary allylic carbocation that is formed. The secondary allylic carbocation at carbon number 2 is very stable so it will often, but that doesn't mean it will form the most stable product.

The thermodynamic product in this case, the 1,4 addition, forms a primary allylic carbocation that is less readily formed than the aforementioned secondary carbocation, but the product of this reaction is the more substituted alkene, so the more stable (thermodynamic) product is formed, and the energy barrier that was required to overcome the kinetic favor was met when the temperature was raised to 80C.

Although the correct answer choice, B, requires one to draw an inference from the explanation I gave, it's the only answer that has anything to do with the question prompt, kinetic vs. thermodynamic control in the conjugated diene.

Choice A is wrong because two double bonds doesn't necessarily mean the two double bonds are in conjugation, and this question is dealing with the results of addition to a symmetric conjugated diene.

Choice C is a garbled answer that highlights a concept of carbocation formation that adds nothing to do the issue with the conjugated diene. This is the type of answer one would choose if they were guessing but knew general carbocation rules and picked what felt right.

Choice D would mean both double bonds underwent an addition reaction and two Br ions were added to the hydrocarbon, which does not pertain to 1,2 and 1,4 addition.

Again, although choice B may not have been explicitly stated, it's the only answer that pertains to the issue at hand in the question.

I hope that helps.

Thank you both! I was really hoping someone would still give me an explanation just for completion sake!

I think I may need to spend just a tab bit more time to accept this one. I keep getting hung up on carbocation stability to determine the right answer, but it seems we're looking for the more stable alkene?

 

[ReggieD6868]

Thank you both! I was really hoping someone would still give me an explanation just for completion sake!

I think I may need to spend just a tab bit more time to accept this one. I keep getting hung up on carbocation stability to determine the right answer, but it seems we're looking for the more stable alkene?

Although I don't think my prior explanation could have answered your OP any clearer, I will say that both of the two underlined portions in the quote from above from you were factors in choosing the correct answer to EK-Organic Chemistry #382.