Arrange the following in increasing order of reactivity towards nitration
A. p-xylene
B. bromobenzene
C. mesitylene
D. nitrobenzene
E. benzene
Choose the correct answer from the options given below

We need to arrange the given compounds in increasing order of reactivity towards electrophilic aromatic nitration. The reactivity of a benzene ring towards nitration depends on the nature of the substituents attached to it — electron-donating groups (EDGs) activate the ring (increase reactivity), while electron-withdrawing groups (EWGs) deactivate the ring (decrease reactivity).

Let us analyze each compound:

D. Nitrobenzene: The nitro group ($$-NO_2$$) is a strong electron-withdrawing group that deactivates the benzene ring significantly. This makes nitrobenzene the least reactive towards nitration.

B. Bromobenzene: The bromo group ($$-Br$$) is a deactivating group (due to its electronegativity and $$-I$$ effect, which outweighs its weak $$+M$$ effect). So bromobenzene is less reactive than benzene itself, but more reactive than nitrobenzene.

E. Benzene: Benzene has no substituent and serves as the reference. It is more reactive than bromobenzene and nitrobenzene but less reactive than compounds with electron-donating groups.

A. p-Xylene: p-Xylene has two methyl groups ($$-CH_3$$) on the ring at the para positions. Methyl groups are electron-donating groups (through hyperconjugation and $$+I$$ effect), which activate the ring. Two methyl groups provide more activation than a single one, so p-xylene is more reactive than benzene.

C. Mesitylene: Mesitylene (1,3,5-trimethylbenzene) has three methyl groups on the ring. The cumulative electron-donating effect of three methyl groups makes mesitylene the most activated and therefore the most reactive towards nitration.

So the increasing order of reactivity towards nitration is:

$$D \lt B \lt E \lt A \lt C$$

(Nitrobenzene < Bromobenzene < Benzene < p-Xylene < Mesitylene)

Hence, the correct answer is Option B.