Organic - Isolated Diene reacts with Br2

[Gogogirl]

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Hi guys.. I have a question

what is the major product of reacting 2-methyl-1,4-hexadiene (isolated diene) with Br2. Doe sthe Br2 add to the two secondary carbons or the two primary and tertiary carbons?...My answer would be the adding to the two secondary ones...what do you think?

 

[PiBond]

Hi guys.. I have a question

what is the major product of reacting 2-methyl-1,4-hexadiene (isolated diene) with Br2. Doe sthe Br2 add to the two secondary carbons or the two primary and tertiary carbons?...My answer would be the adding to the two secondary ones...what do you think?

My guess would be to add to the primary and tertiary carbons.

The reason why is due to the more stable carbocation intermediate at the tertiary carbon. More stable intermediates lead to lower activation energy and thus a faster reaction. Also, the terminal pi bond is more reactive because it's not substituted--thus it's a better nucleophile.

Just my two cents.

 

[boaz]

The product depends on whether it is under (A) under kinetic control, or (B) under thermodynamic control.

(A) Since the tertiary carbocation is most likely to form, the nucleophilic attack of Br- will occur at the tertiary carbon. Since there is not enough heat to reverse the reaction, the Br will remain there. So the major product will be 1,2-dibromo-2-methyl-hex-3-ene.

(B) Since there is enough heat to reverse the reaction, the tertiary carbocation will re-form and there will be resonance with the other pi-bond. This provides opportunity to the Br- to attack C4, leaving the pi bond between C2 and C3. Since this is the more substituted alkene, it is most stable. So the major product will be 1,4-dibromo-2-methyl-hex-2-ene.

This, of course, is all assuming that there is only ONE mole of Br2 per mole of alkene.

Hope that helps!

 

[BennieBlanco]

The product depends on whether it is under (A) under kinetic control, or (B) under thermodynamic control.

(A) Since the tertiary carbocation is most likely to form, the nucleophilic attack of Br- will occur at the tertiary carbon. Since there is not enough heat to reverse the reaction, the Br will remain there. So the major product will be 1,2-dibromo-2-methyl-hex-3-ene.

(B) Since there is enough heat to reverse the reaction, the tertiary carbocation will re-form and there will be resonance with the other pi-bond. This provides opportunity to the Br- to attack C4, leaving the pi bond between C2 and C3. Since this is the more substituted alkene, it is most stable. So the major product will be 1,4-dibromo-2-methyl-hex-2-ene.

This, of course, is all assuming that there is only ONE mole of Br2 per mole of alkene.

Hope that helps!

hmmm.

(A) Since the tertiary carbocation is most likely to form, the nucleophilic attack of Br- will occur at the tertiary carbon.

Why would a cation form? The nucleophile initially would be the pi bond. I guess if you are saying that a pi bond attacked a Br2 and one of the Br's was floating around to attack a alkene, but really the pi bond does the initial attack forming a bromonium ion.

(B) Since there is enough heat to reverse the reaction, the tertiary carbocation will re-form and there will be resonance with the other pi-bond.

Please redraw the molecule. It will not have resonance with the other pi bond, it is too far away.

leaving the pi bond between C2 and C3.

There will not be a pi bond between C2 and C3.

Since this is the more substituted alkene, it is most stable. So the major product will be 1,4-dibromo-2-methyl-hex-2-ene.

I think you are thinking of starting with 2-methyl-1,3-hexadiene and not 2-methyl-1,4-hexadiene (what he asked).

 

[BennieBlanco]

I'll go with Pi Bond on this one. The intermediate is actually a bromonium ion (3-membered ring with bromine) not a carbocation but is somewhat carbocation-like and therefore adding to carbons 1 and 2 (the primary and tertiary carbons) forms a more stable intermediate and so has a lower activation energy.

agreed with pibond and chemnerd.

There is no advantage of going to the 4th and 5th carbons. Not sterics, not stability, neither kinetic/thermodynamics.

Yet, in the real world both form. Likely the 1st and 2nd carbons will be favored, and some with both 1 and 2 with 4 and 5.

 

[boaz]

hmmm.

Why would a cation form? The nucleophile initially would be the pi bond. I guess if you are saying that a pi bond attacked a Br2 and one of the Br's was floating around to attack a alkene, but really the pi bond does the initial attack forming a bromonium ion.

I figured it would be obvious that I was talking about step 2, where Br- attacks the carbocation.

Please redraw the molecule. It will not have resonance with the other pi bond, it is too far away.

There will not be a pi bond between C2 and C3.

I think you are thinking of starting with 2-methyl-1,3-hexadiene and not 2-methyl-1,4-hexadiene (what he asked).

Yes, I misread the name.

My guess would be to add to the primary and tertiary carbons.

The reason why is due to the more stable carbocation intermediate at the tertiary carbon. More stable intermediates lead to lower activation energy and thus a faster reaction. Also, the terminal pi bond is more reactive because it's not substituted--thus it's a better nucleophile.

Just my two cents.

Correct.

I'll go with Pi Bond on this one. The intermediate is actually a bromonium ion (3-membered ring with bromine) not a carbocation but is somewhat carbocation-like and therefore adding to carbons 1 and 2 (the primary and tertiary carbons) forms a more stable intermediate and so has a lower activation energy.

Correct.