Question

What is the order of reactivity of hydrogen atoms attached to the carbon atom in an alkane for free radical substitution?
Options-
A.${3^ \circ } > {1^ \circ } > {2^ \circ }$
B.${2^ \circ } > {1^ \circ } > {3^ \circ }$
C.${3^ \circ } > {2^ \circ } > {1^ \circ }$
D.${1^ \circ } > {2^ \circ } > {3^ \circ }$

Answer 1

The reactivity orders associated with hydrogens atoms attached to primary, secondary or tertiary carbon is related to the ease with which the carbon-hydrogen bond undergoes a homolytic bond cleavage and forms a free radical at that particular position.

Complete answer:
A free radical reaction requires the cleavage of a carbon-hydrogen bond which should be homolytic in nature. A homolytic cleavage is a method of breaking a covalent bond in which the bonded pair of electrons are redistributed to the individual atoms in an equal ratio. Thus both carbon and hydrogen get a single electron on segregation.
The reactivity of such reactions is dependent upon the stability of the free radical intermediate formed during the reaction. A reaction with a more stable intermediate would take place quickly and more efficiently.
A carbon radical can be stabilized by the phenomenon of hyperconjugation or resonance. Since alkanes do not contain any double bonds, only hyperconjugation can be used to stabilize the radical. Thus, the carbon atom which is more substituted i.e. has more number of $\alpha$ hydrogens attached to it, is more stable due to the possibility of a higher number of hyperconjugative structures associated with it.
Tertiary carbon atoms have the highest number of $\alpha$ hydrogens followed by secondary and primary carbons respectively.
Therefore, the order of reactivity is given as:
${3^ \circ } > {2^ \circ } > {1^ \circ }$

Hence option (c) is correct.

Note:
A carbon free radical is an electron deficient species in which the carbon atom has only seven electrons in its outermost shell i.e. has an incomplete octet due to which the radical is very unstable and has a tendency to combine with another free radical so to form a stable bond and complete its octet.