Question

If the relative reactivity of $1^{\circ },2^{\circ }$ and $3^{\circ }$ hydrogens towards free redical chlorination is $1:4.5:5.5,$ what would be the percentage of 1-chloro-2-methycyclohexane in the free radical chlorination of methyl cyclohexane?

• A. 25%
• B. 40%
• C. 60%
• D. 75%

Answer 1

The correct option is A 25%

If the relative reactivity of $1^{\circ },2^{\circ }$ and $3^{\circ }$ hydrogens towards free radical chlorination is $1:4.5:5.5,$, the percentage of 1-chloro-2-methycyclohexane in the free radical chlorination of methyl cyclohexane would be $\frac{45}{53.5}\times \frac{2}{5}\times 100=$ 33%
There are 3 primary hydrogen atoms, 10 secondary atoms and 1 tertiary atom. The number of hydrogen atoms of each type is multiplied with their reactivity. The resulting numbers are then added to obtain 53.5. This is shown in the table. This explains the fraction 45/53.5 in the calculation.
Total 5 different products can be obtained depending on the position of Cl atom. Out of these one product is 1-chloro-2-methycyclohexane which can be obtained when Cl is attached in two different ways. This explains the fraction 2/5 in the calculations.