[carotis]

Greetings to you
Which stereochemical product would you expect to happen?
Hence I guess that the rear attack of the nucklephile isn't hampered by the cis-methyl group and hence is cis to chloride, a SN2 Substitution would occur. What you think?

[Dan]

LDA is very bulky, which makes it a poor nucleophile. Are there any reactions other than substitution that you think might be possible?

[carotis]

Well, LDA is used for selctive deprotonation of Hydrid-kations. But I guess it could also eliminate the Cl-
i dunno, there are no infos about solvations or conditions. It is asked merely about the stereochmical correct product. More precisly, a E2 reaction, because Cl- ion is primer positional, hence not stable in an interim state. Isn't this reaction similar with potassium-tert Butoxide?

Then, if I assumed correctly, question is, which of the two products, Sayzeff or Hoffman will result.

[Dan]

E2 is a good idea.

There are 3 possible isomeric products of E2 elimination of chloride this molecule (one pair of enantiomers and one achiral compound), which do you think will be the major product(s)?

[carotis]

I know two products (as stated in the edited post upon), Sayzeff and Hoffmann. I think (EDIT: correction) Sayzeff product would be the main product, by forming an allen.

I think this way: the chloride goes off, so we have a negative charge right? So this would mean a hydrid-kation goes off... but hence the molecule prefere tertiary positions, because it's more stable we would have  an Allen between the right Methyl group and the top C atom above or not?

[carotis]

so it would me give this: http://www.chemspider.com/Search.aspx?rid=2c7cdbb2-0afc-42cc-9134-ad471e4d1ea1
correct?

[orgopete]

I agree this is a Hoffman v Zaitsev elimination question. I also understand the kind of thinking being applied to this question. As I think about these reactions, I prefer to further delineate concerted reactions. If time could be infinitely divisible, then no two events could occur at exactly the same time, one will always occur befor the other. If this were so, the E2-elimination reactions may be SN1-like (Zaitsev) and SN2-like (Hoffman).

If you accept this concept, then we can analyze whether the reaction is going to be more SN1-like or SN2-like. The classic Zaitsev products are often found with NaOEt/EtOH. I'd argue these are SN1-like conditions in which the C-X bond lengthens and begins to break. At the same time, electrons from a tertiary center begin attraction, increase acidity, and favor Zaitsev product formation.

Hoffman reactions then favor a semi-initial deprotonation of the most acidic hydrogen and give Hoffman products. I say semi-initial as no anion is formed by deprotonation, simply that hydrogen acidity is a factor leading to product.

Are these conditions SN1-like or SN2-like? Will it give Hoffman or Zaitsev products?

[carotis]

Sorry, but not am I the one with the questions here? I thought we came to the aggreement that it is a E2 elimination. I dont frankly get your point here, so..which is the correct main product now?

[carotis]

here

[orgopete]

Hoffman