Question

An equimolar mixture of toluene and chlorobenzene is treated with a mixture of conc. ${H_2}S{O_4}$ and conc. $HN{O_3}$. Indicate the correct statement from the following:
A) o-nitrotoluene is formed in excess
B) Equimolar amounts of p-nitrotoluene and p-nitrochlorobenzene are formed
C) p-nitrotoluene benzene is formed in excess
D) m-nitrochlorobenzene is formed in excess

Answer 1

Hint: Try to recall that toluene has more tendency to undergo electrophilic substitution compared to chlorobenzene and the electrophile is nitronium ion. Now, by using this you can easily answer the given question.

Complete step by step solution:

It is known to you that when concentrated sulphuric acid is mixed with concentrated nitric acid, then it gives nitronium ion ($N{O_2}^ +$) which act as an electrophile.
Also, Toluene has more electron density compared to chlorobenzene. This is because of the +I effect of - $C{H_3}$ group in toluene, benzene ring gets activated for electrophilic substitution whereas in chlorobenzene, due to –I effect of chlorine benzene ring gets deactivated.
So, when an equimolar mixture of toluene and chlorobenzene is treated with a mixture of concentrated sulphuric acid and nitric acid, then the first toluene will undergo electrophilic substitution and o-nitrotoluene is formed in excess.
In toluene, the electron density is more at ortho position compared to para position due to the electron releasing effect of methyl group. So, o-nitrotoluene is formed in excess compared to p-nitrotoluene.

Therefore, from above we can easily conclude that option A is the correct option to the given question.

Note: It should be remembered to you that chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 673 K and a pressure of 300 atmospheres.
Also, you should remember that the presence of an electron-withdrawing group at the ortho- and para-positions increases the reactivity of haloarenes.