There's this problem that I saw in my Organic Chemistry book:

The reaction between $\ce{(CH3)3CH}$ and $\ce{Cl2}$ forms two products: $\ce{(CH3)2CHCH2Cl}$ (63%) and $\ce{(CH3)3CCl}$ (37%). Why is the major product formed by the cleavage of the stronger primary $\ce{C-H}$ bond?

I understand that chlorine is more reactive than bromine, so it is unselective and its reaction would yield a mixture of products. However, from what I have understood, radicals are more stable when they are more substituted, so why is the major product the less substituted compound and not the more substituted one?


1 Answer 1


The experimental relative rates of chlorination to primary, secondary, and tertiary positions are $1$, $3.8$, and $5$, respectively. In the given compound $\ce{(CH3)3CH}$, there are $9$ identical primary hydrogens and $1$ tertiary hydrogen.

Roughly speaking, the rate of chlorination at primary carbon is proportional to the product of relative rate of chlorination at primary position and the number of identical primary hydrogens and similarly to the rate of chlorination at tertiary carbon.

So theoretical yield of $\ce{(CH3)2CHCH2Cl}$ is nearly $64.3\%$ $\left(=\dfrac{1\times9}{1\times9 + 5\times 1} \times 100 \right)$ and theoretical yield of $\ce{(CH3)3CCl}$ is nearly $35.7\%$ $\left(=\dfrac{5\times1}{1\times9 + 5\times 1}\times 100 \right)$. These percentages are very close to the experimental results. So the yield of monochlorinated compound also depends on the number of hydrogen atoms besides the bond energy and other factors.

  • $\begingroup$ I have corrected the formula of monochlorinated product at tertiary carbon. $\endgroup$ May 12 at 19:16

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