which of the following reaction is NOT correctly represented?

The first three reactions are routine electrophilic substitutions on the aromatic ring and are therefore depicted correctly, whereas the fourth reaction is a case of free-radical halogenation of an alkyl side chain; the product shown in the option is the ring-substituted bromide, which is wrong. Let us analyse them one by one.

Reagent set : $$conc.\; HNO_3 + conc.\; H_2SO_4$$ on benzene.
These are the mixed acids used for nitration. The electrophile $$NO_2^+$$ substitutes one $$H$$ of the ring to give nitrobenzene. The option shows nitrobenzene, hence A is correctly represented.

Reagent set : $$conc.\; H_2SO_4$$ (fuming) at 323-333 K on benzene.
The electrophile is $$SO_3H^+$$ and the product is benzene-sulphonic acid. The structure drawn in option B matches this, so B is correct.

Reagent set : $$CH_3Cl/AlCl_3$$ (Friedel-Crafts alkylation) on benzene.
The electrophile is the methyl carbocation $$CH_3^+$$ generated in situ; substitution yields toluene. The option shows toluene, therefore C is correct.

Reagent set : $$Br_2$$, $$hv$$ or $$Br_2$$, $$peroxide$$ or $$Br_2$$ in $$CCl_4$$ with light on toluene.
• Under these radical conditions, benzylic C-H bonds are far more reactive than aromatic C-H bonds.
• The radical chain mechanism gives benzyl bromide $$C_6H_5CH_2Br$$ (and on further bromination $$C_6H_5CHBr_2$$, $$C_6H_5CBr_3$$).
• Bromination of the ring (o-/p-bromotoluene) needs $$Br_2/FeBr_3$$ and absence of light/free-radical initiators.
The option, however, depicts ring-brominated toluene under radical conditions, which is impossible. Hence option D is NOT correctly represented.

Therefore the incorrect depiction is
Option D which is: the radical bromination of toluene shown as o-/p-bromotoluene instead of benzyl bromide.