organic chemistry whiz i need your help big time, final is tomorrow

[Raptor]

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i got some questions can someone help me out.

 

[jlee9531]

well if you post your questions im sure there will be various people who will answer them with their thoughts and ideas...

 

[ixitixl]

Post away 🙂

 

[Raptor]

How many products would you expect if (3R, 4R)-3-bromo-4-methylhexane were reacted with KOCH2CH3 in ethanol? Clearly write All products in order of increasing stability.

Thank you guys soo much.

 

[Rendar5]

2 elimination products (both E2), depending on which H the ethoxide steals. one is more stable than the other because one has a double bond at a tertiary carbon and the other has a double bond at a secondary carbon.

There are also 2 possible substitution products: one being an Sn1 path, the other being an Sn2 path. One of those two is a more stable pathway.

Then all you have to do is decide if this is more likely to be an elimination reaction or a substitution reaction.

 

[ncalcate]

I think you would get four possible E2 products when you factor in stereoisomerism:

In order of stability (most stable first):

E-4-methyl-3-hexene
Z-4-methyl-3-hexene
E-4-methyl-2-hexene
Z-4-methly-2-hexene

Clearly, the first two products would be the major ones, as predicted by Zaitsevs rule, since these are trisubstituted alkenes, while the last two are disubstituted. And always, trans (E) is more stable than cis (Z).

 

[HollyJ]

Careful, ncalcate -- you're forgetting that this is an E2 reaction, which needs anticoplanar conformation to run. 😉

 

[Raptor]

so which one is it

 

[Jugador75]

We're all such dorks.

But HollyJ is correct.

 

[ncalcate]

I agree we're all dorks. But I still think you can get 4 products.

You have two possible beta protons on C2. You can arrange them in such a way so that you can get both cis and trans 2-hexene products. One of the pathways may require the bromine and beta hydrogen to be syn to each other, but E2 can proceed via either syn or anti elimination.

Regarding the 3-hexenes, I would need a model for this, but I think you can get both cis and trans if you consider both syn and anti elimination. And since your prof seemed to want ALL possible products, the alkenes generated from the syn elimination pathway are fair game (although they will be seen in very small quantities)

Best way to do this is to build a model and see for yourself.

Don't you love a good old fashioned organic debate?

 

[Rendar5]

how can it proceed by syn? The molecular orbitals don't overlap that well in that orientation. Well, go by whatever your chem book says since I don't have mine around.

 

[Raptor]

thanks ncalcate for your great help u are a genius

 

[ncalcate]

As long as the MOs are parallel, then you can get the overlap to form the pi bond (at least that's what my organic book says). In both syn and anti, the sigma bonds of your leaving group and your beta hydrogen are parallel, which leads to the overlap of your p orbitals needed for your pi bond.

Syn elimination occurs far less frequently because it requires an eclipsed conformation (as opposed to staggered for anti), but it can occur nonetheless.

 

[Rendar5]

I wasn't talking about pi bonds. I was talking about an overlap of hybridized orbitals. Give me a minute. I might have my notes from my advanced orgo class. But again, if your textbook says syn works, go by it. I took graduate level orgo, but it's rusty in my head which is why I'm not sure on this.

 

[Rendar5]

well, as luck would have it, I didn't copy down notes on E2, just E1 and E1cb. But in one of my diagrams I copied, there is a syn E2 picture. May have been an accident, but until I find my Adv. Orgo Text, you're probably right that syn is allowed.

 

[MaybeMD]

Originally posted by HollyJ
Careful, ncalcate -- you're forgetting that this is an E2 reaction, which needs anticoplanar conformation to run. 😉

I thought that anticoplanar arrangment was prefered, but not required.

 

[NubianPrincess]

What is an anticoplanar conformation? I have never seen that word before 😕

 

[ATPase]

God, this thread is a nightmare! 😕

 

[Xega]

Originally posted by ATPase
God, this thread is a nightmare! 😕

What's scarier is I still remember and understand all of what was said.

 

[jlee9531]

Originally posted by NubianPrincess
What is an anticoplanar conformation? I have never seen that word before 😕

too bad we cant draw a pic on here...

but its basically where your beta Hydrogen needs to be ANTI and on the same plane to the leaving group so that an E2 reaction can occur.

 

[jlee9531]

Originally posted by MaybeMD
I thought that anticoplanar arrangment was prefered, but not required.

i think you are right maybemd....it is preferred because with a high enough T all the products can still be formed...just in different amounts obviously.

 

[elias514]

Given the fact that you have a secondary alkyl halide (prone to methyl shift), a strong base/great nucleophile, and a polar protic solvent, the most probable reactions involve an E2 mechanism.

The possible E2 products are (E)-3-methyl-3-hexene, (trans)-4-methyl-2-hexene, and (cis)-4-methyl-2-hexene. These products are arranged in order of decreasing stability. All of these products reflect the antiperiplanar geometry restriction for E2.

The fact that this reaction involves a polar protic solvent and a substrate, which is capable of undergoing a methyl shift and forming a tertiary carbocation, creates the possibility of Sn1/E1 reactions. However, the strong basicity of potassium ethoxide (in my opinion) would minimize the contribution from these mechanisms.

Hope this helps.

 

[Raptor]

thanks you guys are geniuses

 

[HollyJ]

I believe that E2 can go syncoplanar, but only when the leaving group is locked into place, like in a bicyclic ring system. Otherwise, the other configuration is so much lower in energy that it eliminates (heh -- I'm a dork) the other product.

So three elimination products. Cis and trans for the Hoffman, and E for the russian guy with variant spellings.

Good luck on your test!