Would 2-chloropropane or 1-chloro-2,2-dimethylpropane undergo substitution faster with #"HC≡C"^-"#, #"Na"^+#?

Question

Ryan Andrews · Accepted Answer

2-Chloropropane will undergo substitution faster than 1-chloro-2,2-dimethylpropane.

The acetylide anion, HC≡C⁻, is the conjugate base of the extremely weak acid, acetylene.

So it is both a powerful nucleophile and an extremely strong base — about a billion times as strong as hydroxide ion.

It undergoes #"S"_"N"2"# reactions with methyl and primary alkyl halides.

But acetylide ion is such a strong base that it undergoes mainly #"E2"# eliminations with 2° alkyl halides.

And it undergoes only #"E2"# elimination with 3° alkyl halides.

An #"S"_"N"2"# reaction goes by backside attack at the α carbon atom.

1-Chloro-2,2-dimethylpropane is completely hindered to backside attack.

The only product is the elimination product.

2-Chloropropane is less hindered.

The acetylide ion can attack more readily, so its substitution reaction is faster than on the 3° halide.

Isabella Acosta · Answer

undefined 2-chloropropane would undergo substitution faster with "HC≡C"^-", "Na"^+ because it is a primary alkyl halide, meaning the carbon attached to the halogen (chlorine) is only bonded to one other carbon. In contrast, 1-chloro-2,2-dimethylpropane is a tertiary alkyl halide, where the carbon attached to the halogen is bonded to three other carbons. Primary alkyl halides generally undergo substitution reactions faster than tertiary alkyl halides due to the increased steric hindrance in tertiary alkyl halides, which hinders the approach of the nucleophile and slows down the reaction rate. Therefore, 2-chloropropane would react faster with "HC≡C"^-", "Na"^+.