Given below are two statements : Statement (I) : On nitration of m-xylene with $$HNO_{3}$$, $$H_{2}SO_{4}$$ followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. Statement (II) : $$-CH_{3}$$ group is o/p-directing while $$-NO_{2}$$ group is m-directing group. In the light of the above statements, choose the correct answer from the options given below :

We need to evaluate two statements about the nitration of m-xylene and the directing effects of substituents.

On nitration of m-xylene with $$HNO_3/H_2SO_4$$ followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. m-Xylene has two $$-CH_3$$ groups at positions 1 and 3 of the benzene ring, and both groups are ortho/para directing and activating. During nitration, the incoming $$NO_2^+$$ electrophile preferentially attacks the ring positions that are ortho or para to both methyl groups simultaneously. Position 2 is ortho to both methyl groups but is sterically hindered; positions 4 and 5 are each para to one methyl and ortho to the other. The major nitration product therefore has $$-NO_2$$ at position 4 (equivalently position 5 by symmetry), giving 1,3-dimethyl-4-nitrobenzene. Oxidation of both $$-CH_3$$ groups with a strong oxidizing agent such as $$KMnO_4$$ converts them to $$-COOH$$ groups, producing 4-nitrobenzene-1,3-dicarboxylic acid, which shows that Statement I is true.

The methyl group is an electron-donating group via hyperconjugation and the inductive effect, activating the ring and directing incoming electrophiles to ortho and para positions, whereas the nitro group is a strong electron-withdrawing group by resonance and inductive effects, deactivating the ring and directing electrophiles to the meta position. Thus Statement II is also true.

Since both statements are true, the correct answer is Option (4): Both Statement I and Statement II are true.