Question

Heterolysis of propane gives:
(A) methyl and ethyl free radical.
(B) methylium and ethyl free radical.
(C) methyl anion and ethylium cation.
(D) methylium and ethylium cation.

Answer 1

Hint: Consider the difference between homolytic cleavage and heterolytic cleavage of bond. Consider how electrons are distributed in each type of cleavage. In other words, decide if the bond pair of electrons are symmetrically distributed between two atoms or if there is asymmetrical distribution of the electrons.

Complete step by step answer:
In homolytic cleavage of a bond, you equally distribute the shared pair of electrons between two atoms to obtain two free radicals.
In the above example, each X atom gains one electron. An example is the homolytic fission in the hydrogen molecule to produce two hydrogen free radicals.\[{\text{X}} - {\text{X }} \to {\text{ 2X}} \cdot \]
 \[{\text{H}} - {\text{H }} \to {\text{ 2H}} \cdot \]
Homolytic fission of propane gives methyl and ethyl free radicals.
In heterolytic cleavage of a bond, you unequally distribute the shared pair of electrons between two atoms to obtain a cation and an anion.
\[{\text{X}} - {\text{Y }} \to {\text{ }}{{\text{X}}^ + }{\text{ + }}{{\text{Y}}^ - }\]
In the above example, the more electronegative each \[{\text{Y}}\] atom gains two electrons and a negative charge. Less electronegative \[{\text{X}}\] atom gains only a positive charge. An example is the heterolytic fission in the hydrogen chloride molecule to produce hydrogen cation and chloride anion.
 \[{\text{H}} - {\text{Cl }} \to {\text{ }}{{\text{H}}^ + }{\text{ + C}}{{\text{l}}^ - }\]
In propane, the electron density on methylene carbon atoms is greater than the electron density on terminal methyl groups. This is due to electron donating (+I) effect of two methyl groups. In the ethylium cation, the positive charge on methylene carbon atoms is stabilized through electron releasing (+I ) effect of adjacent methyl groups.
Heterolysis of propane gives methyl anion and ethylium cation
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}} \to {\text{CH}}_3^ - {\text{ + C}}{{\text{H}}_{\text{3}}}{\text{CH}}_2^ + \]

Hence, the correct option is (C) methyl anion and ethylium cation.

Note: Heterolytic cleavage of propane could also have formed methylium cation and ethylium anion. But ethylium cation is more stable than methylium cation. Also, methyl anion is more stable than ethyl anion. This is due to a combination of inductive effects and hyperconjugation.