What is the product of this reaction

[rippinitez]

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If the compound (see attachment) was treated with KOH in ethanol and heat. I'm having a hard time figuring out how stereochemistry comes into play with reactions like this.

 

[TexasOMFS]

Well first of all ask yourself what type of reactant you have and look at your reaction conditions. You have a secondary alkyl halide with OH (strong base/nucleophile) in EtOH (polar protic solvent). I believe the reaction is going to go E1. Note that the carbocation intermediate would have the possibility to immediately rearrange to a tertiary carbocation by a 1,2-hydride shift. Of course with a carbocation intermediate, you usually have a mix of Sn1/E1. Here, we treat the reaction with heat, which favors the trisubstituted alkene product. Not much stereochemistry here, just note that for the substitution product we'll have the formation of a stereogenic center where the OH nucleophile attacks and therefore a equal amount of R & S products.

 

[UndergradGuy7]

Well first of all ask yourself what type of reactant you have and look at your reaction conditions. You have a secondary alkyl halide with OH (strong base/nucleophile) in EtOH (polar protic solvent). I believe the reaction is going to go E1. Note that the carbocation intermediate would have the possibility to immediately rearrange to a tertiary carbocation by a 1,2-hydride shift. Of course with a carbocation intermediate, you usually have a mix of Sn1/E1. Here, we treat the reaction with heat, which favors the trisubstituted alkene product. Not much stereochemistry here, just note that for the substitution product we'll have the formation of a stereogenic center where the OH nucleophile attacks and therefore a equal amount of R & S products.

KOH is OH-. OH- is a strong base isn't it? Wouldn't it favor E2 then? E2 uses a strong base. E1 uses a weak base.

Not sure if they product will be cis or trans.

Just my opinion. I barely started reviewing organic chemistry again.

 

[sfoksn]

Yeah.. It should lose its stereochemistry because it will mostly go E2.. I think.

I Just hate these kind of questions because it seems like it might be able to go E1 due to the fact that the carbocation formed will be resonance stablized by the benzene.

So it probably won't shift as one of the posters said.

But then again it might shift!

You see why I don't like questions like this?

 

[TexasOMFS]

E2 conditions are strong base, tertiary>secondary>primary alkyl halide, and polar aprotic solvent. Usually the alkyl halide is the main consideration and I'm thinking it would favor E1 because it has the possibility to rearrange to a tertiary carbocation, which would be further stabilized by the polar protic solvent. This question is definitely one of those gray area ones: it would be a mix of all four (Sn1, Sn2, E1, E2).

I don't think many textbooks address stereochemistry of E2 product but if I recall correctly, the positions of the substituents with respect to each other remains the same. The reason being that E2 can only happen when the halide and the hydrogen are anticoplanar (antiperiplanar) to each other. The hydrogen gets attacked at the same time as the halide leaves and the double bond forms before there is any potential for rotation around the single bond.

 

[sfoksn]

yes, the carbocation will rearrange.

Thank you texasomfs, I am so careless! haha.

edited

 

[TexasOMFS]

Well notice there's a benzene coming off both central carbons. If it rearranges (ie: the lower left hydrogen shifting to I's former position), you'll get a tertiary benzylic carbocation. A tertiary benzylic carbocation is more stable than a secondary benzylic carbocation right?

 

[RSD2014]

KOH is OH-. OH- is a strong base isn't it? Wouldn't it favor E2 then? E2 uses a strong base. E1 uses a weak base.

Not sure if they product will be cis or trans.

Just my opinion. I barely started reviewing organic chemistry again.

KOH strong base in aprotic solvent.
I see your logic. I did think it is SN2 as well; however, do you think KOH would have a problem deprotonating the H anticoplanar to LG Iodide? Since the C2H5 group is on the same carbon bearing the Anticoplanar Beta Hydrogen. Wouldn't this group hinder KOH?

Wouldn't E1 and E2 give the same product? tri-substituted alkene.

Well notice there's a benzene coming off both central carbons. If it rearranges (ie: the lower left hydrogen shifting to I's former position), you'll get a tertiary benzylic carbocation. A tertiary benzylic carbocation is more stable than a secondary benzylic carbocation right?

How do you account for E1, with Strong base KOH in aprotic solvent?
E1 does run with Weak base. The part about forming a better carbocation makes sense to me.
Perhaps it is E1 and E2, but since OH is hindered by C2H5, E1 would predominate?

 

[rippinitez]

After some extensive research, it looks like the reaction will go E2 because of a STRONG base and our halide is secondary. I'm almost certain that the strength of the base and nucleophile is one of the first things to look at when determining whether a reaction will go E1, E2, SN1 and SN2. Obviously because it is a tertiary carbon it can't go SN2.

I think the solvent plays a bigger role in SN1 and E1 (can only take place in protic) reactions as opposed to SN2 and E2 which can take place in protic solvents.

 

[rippinitez]

My bad, it's a secondary carbon. But still, E2 will go faster than SN2 because SN2 prefer CH3>1*>2*

 

[UndergradGuy7]

My bad, it's a secondary carbon. But still, E2 will go faster than SN2 because SN2 prefer CH3>1*>2*

Does the answer key doesn't explain it or you dont have the answer?

I am confused too, what RSD said makes sense. So you can argue Sn1 or Sn2.

 

[rippinitez]

The answer just says that it goes E2. So I had to refer to chad's lectures to really understand why.