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We have the reaction of the alkyl halide $\ce{CH3-CHBr-CH2Br}$ with 1) $\ce{OH-}$ 2) $\ce{NaNH2}$ The answer is propyne.

I am unable to understand why we have performed an elimination reaction here. Why can't it be a substitution reaction instead? Also, how do we decide which two carbon atoms would lose their hydrogen atoms? (as in elimination, there are two choices for losing the $\ce{H}$ atom - from left and right carbon respectively)

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marked as duplicate by M.A.R., Tyberius, airhuff, Mithoron, Todd Minehardt Mar 14 '18 at 1:02

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Substitution, meaning displacement reactions are Sn1 or Sn2. These reactions are dependent on the leaving group, nucleophilicity vs base of attacking group, and solvent. Currently you have two good leaving groups (Br) and strong bases. Strong bases (have localized electron density, no resonance etc...) favor proton attack due to partial + charge on H, no spatial constrains. So the protons are attacked preferentially. It would be much easier sterically and kinetically compared to either Sn2 (recall l, these also requires specific orientation of attack) or Sn1. These reactions will occur but at much much small amounts, such that on time of reaction they are minimal.

The reason for the -yne product is because there are two adjacent leaving groups. Once the first comes of (which is aided by partial resonance by the other Br), the electron densities around the double bond, the large radii of Br, along with the +ve charge character on Br due to resonance, it easier to remove the next Br, but this requires a strong base. The amine is definitely strong enough, but so is the OH ( less so due to higher electronegativity of Oxygen which doesn't want to share its electrons so to speak ) Because you end up with a -yne it doesn't matter which Br is removed first, regarding the final regiostereochemistry.

If you need more when I get home I can try again.

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