Question #cd21d

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Answer 1 · 2015-05-14T21:40:00

The main product is propyne, and the minor product is propadiene.

This is an $E2$ $"E2"$ elimination that occurs in stages.

The product of the first elimination is 2-chloropropene.

${CH}_{3} {CCl}_{2} {CH}_{3} -HCl - - \to {CH}_{3} {CCl=CH}_{2}$ $"CH"_3"CCl"_2"CH"_3 stackrel(color(blue)("-HCl"))→ "CH"_3"CCl=CH"_2$

For the second step, we can eliminate the β-hydrogen at either $C-1$ $"C-1"$ or $C-3$ $"C-3"$ .

If we eliminate the $H$ $"H"$ at $C-1$ $"C-1"$ , the product is propyne.

${CH}_{3} {CCl=CH}_{2} -HCl - - \to {CH}_{3} C\equivCH$ $"CH"_3"CCl=CH"_2 stackrel(color(blue)("-HCl"))→ "CH"_3"C≡CH"$

If we eliminate the $H$ $"H"$ at $C-3$ $"C-3"$ , the product is propadiene.

${CH}_{3} {CCl=CH}_{2} -HCl - - \to {CH}_{2} {=C=CH}_{2}$ $"CH"_3"CCl=CH"_2 stackrel(color(blue)("-HCl"))→ "CH"_2"=C=CH"_2$

Alkynes are more stable than 1,2-dienes, so the major product is propyne.

I would expect the two products to form in about a 2:1 ratio, because a commercial product called MAPP gas contains the two gases in that proportion.