Question

Predict the ratios of products that result from chlorination of isopentane.$(2-methylbutane)$

Short answer

The products formed will be in the ratio$38.3 \% : 38.3 \% :23.4 \%$

Step-by-step solution

Primary hydrogen  (10)

Primary hydrogen is hydrogen that resides on a carbon which is only attached to one carbon atom.

Secondary hydrogen  (20)

Secondary hydrogen is hydrogen that resides on a carbon which is only attached two carbon atoms.

Tertiary hydrogen(30)

Tertiary hydrogen is hydrogen that resides on a carbon which is only attached to three carbon atoms.

Primary and secondary hydrogens in isopentane (2-methylbutane)

Primary and secondary hydrogen in isopentane

$(2-\mathrm{methylbutane})$

There are nine primary $\left(1^0\right)$ hydrogen, two secondary $\left(2^0\right)$ hydrogen, and one tertiary $\left(3^0\right)$ hydrogen in isopentane .$(2-\mathrm{methylbutane})$

Reaction of chlorination of isopentane(2-methylbutane)

chlorination of isopentane

Calculating the ratios of products formed

$(9 \mathrm{primary} \mathrm{hydrogens})\times (\mathrm{reactivity} 1.0)= 9.0 \mathrm{relativeamountofreaction}$

$(2 \mathrm{secondary} \mathrm{hydrogens})\times (\mathrm{reactivity} 4.5)= 9.0 \mathrm{relativeamountofreaction}$

$(1 \mathrm{tertiary} \mathrm{hydrogen})\times (\mathrm{reactivity} 5.5)= 5.5 \mathrm{relativeamountofreaction}$

Now,

$\mathrm{fractions} \mathrm{ofprimary} = \frac{9.0}{9.0+9.0+5.5}=\frac{9.0}{23.5}=38.3 \%$

$\mathrm{fractions} \mathrm{ofsecondary} = \frac{9.0}{9.0+9.0+5.5}=\frac{9.0}{23.5}=38.3 \%$

$\mathrm{fractions} \mathrm{oftertiary} = \frac{5.5}{9.0+9.0+5.5}=\frac{5.5}{23.5}=23.4 \%$

Hence, the products formed will be in the ratio$38.3 \% : 38.3 \% :23.4 \%$