Can vicinal dihalide produce conjugated alkene in strong base?

Question

In Klein's Organic Chemistry (3rd edition), p. 410, we learn that vicinal dihalide can produce alkyne in strong base. And on Reaxys I see many reactions of this sort. For example:

However, why isn't the conjugated alkene, which should be more stable, the major product under such conditions?

Somebody once gave me an explanation, to the effect of "the hydrogen geminal to the halide is more acidic, hence more easily eliminated." However, Smith's Organic Synthesis (4th edition), p. 121 says the following:

[In the following reaction] $\mathrm{H_a}$ is more acidic than $\mathrm{H_b}$ since it is attached to the less substituted carbon. The base should react with the more acidic hydrogen atom, which would lead to the less substituted alkene, 3-ethylhex-1-ene, but that is not the major product.

This suggests that we cannot base our logic on the acidity of the hydrogen atoms.

Answer 1

Comprehensive Organic Transformations (2nd edition) p. 467 gives examples of converting 2,3-dibromobutane to buta-1,3-diene.

On the other hand, Reaxys has many examples of producing alkynes from vicinal dihalides, even when potassium t-butoxide is used.

According to Carey's Advanced Organic Chemistry (5th edition, volume A), p. 554, the direction of E2 elimination is determined by whether the mechanism is more similar to E1 or to E1cb. If it is similar to E1, then the thermodynamically more stable product dominates; if it is similar to E1cb, then the direction is determined by the more easily removed hydrogen atom.

On page 556, Carey attributes the Hoffman rule of t-butoxide to its stronger basicity, not to its bulky size, and states that stronger bases drive the mechanism toward the E1cb side.

Therefore I think the problem might be understood this way: If a strong base such as $\mathrm{NaNH_2}$ is used, then the elimination direction is determined by which hydrogen atom is the easiest to remove. Since bromine acidifies its geminal hydrogen, the alkyne product dominates. If a weak base is used, the diene product might predominate.

Answer 2

I can't find a pka value for a simple alkyl bromide, but bromoform has a pKa of about 14, while methane is ~50, so clearly having bromine atoms on an alkyl carbon does acidify it. So it looks like the premise of the argument in Smith is somewhat flawed.

Nonetheless, I'm guessing the text goes on to explain that E2 reactions have product-like transition states, and as a result the Saytzev (more-substituted) product is generally formed. So in this case, both the pKa argument and the alternative Saytzev selectivity argument give the same result - H(a) will eliminate preferentially over H(b).