Question

1. The free radical chlorination reaction of 2,3-dimethylbutane leads to the production of two major monochlorinated products 1-choloro-2,3-dimethylbutane and 2-chloro-2,3-dimethylbutane. When run through a gas chromatograph,it was determined that the product distrubution was 51.7% 1-chloro-2,3-dimethylbutane and 48.3% 2-chloro-2,3-dimethylbutane. If we assign the primary H's a reactivity of 1.0, what is the relative reactivity of the tertiary H's.

2. A free radical bromination of 2,3, dimethybutane( same as above) will give almost exclusively the 2-bromo-2,3-dimethylbutane. Use the Hammond Postulate to explain why free radical brominations are more selective than free radical chlorinations.

Answer 1

If the reactivity of primary hydrogen's is 1.0, and relative reactivity of secondary hydrogen's is x.

Then, total reactivity of primary hydrogen's = 12(1.0) = 12

and that of tertiary hydrogens = 2(x) = 2x

% of 1-chloro product = [12/(12+2x)].100 = 51.7

or, 23.2 = 12 + 2x

Hence, x = 5.61

Therefore relative reactivity of the tertiary H's is 5.61.

Bromine is less reactive than chlorine. The reactivity of primary H's with bromine radical is 1.0 whereas that for tertiary H's is 1640. Due to this very big ratio of 1:1640, the bromination at tertiary carbon is almost exclusively formed.

According to Hammond's postulate, intermediate in bromination resembles reactants whereas that in chlorination resembles products.

Also, the activation energy difference for bromination of tertiary hydrogen to primary hydrogen is high. So, bromination of tertiary hydrogen is the major product.