(c) Why racemisation occurs in \( \mathrm{S_N1} \) reaction ?
Show Hint
Br2 with heat brominates the benzylic carbon through a radical path. SN1 gives a planar carbocation that is attacked from both faces equally, so a racemic mixture forms.
Step 1: Read the three parts. Part (a) gives a reaction of 1-ethyl-3-nitrobenzene with bromine and heat. Part (b) asks for the meaning of chirality. Part (c) asks why an \(\mathrm{S_N1}\) reaction gives a racemic mixture.
Step 2: Part (a) - choose the site of attack. Bromine with heat and no Lewis acid acts through free radicals. The ring is also deactivated by the \(\mathrm{-NO_2}\) group, so ring substitution is not preferred. The reaction therefore happens at the side chain. The hydrogen on the carbon next to the ring (the benzylic carbon) is removed. The benzylic radical formed is stabilised by resonance with the ring.
Step 3: Part (a) - write the product. One benzylic hydrogen of the \(\mathrm{-CH_2CH_3}\) group is replaced by Br. \[ m\text{-}\mathrm{O_2N{-}C_6H_4{-}CH_2CH_3} + \mathrm{Br_2} \xrightarrow{\text{heat}} m\text{-}\mathrm{O_2N{-}C_6H_4{-}CH(Br)CH_3} + \mathrm{HBr} \] The major product is 1-(1-bromoethyl)-3-nitrobenzene. The \(\mathrm{-NO_2}\) group stays at the meta position on the ring.
Step 4: Part (b) - define chirality. A molecule is chiral when it cannot be superimposed on its mirror image. Such a molecule and its mirror image are called enantiomers. The usual cause is a carbon bonded to four different groups (an asymmetric or stereocentre carbon). Chiral molecules rotate plane polarised light.
Step 5: Part (c) - the \(\mathrm{S_N1}\) route. The \(\mathrm{S_N1}\) reaction goes in two steps. First, the halide leaves slowly and a carbocation forms. This carbocation is \(\mathrm{sp^2}\) hybridised and planar.
Step 6: Part (c) - why a racemic mixture forms. The nucleophile can attack the planar carbocation from the front face or the back face. Both faces are equally open, so both attacks are equally likely. Attack from one face gives one enantiomer, and attack from the other face gives its mirror image. The two form in equal amounts (50:50). This racemic mixture has no net optical rotation, so racemisation is seen.
Final Answer: (a) \(\mathrm{m\text{-}O_2N{-}C_6H_4{-}CH(Br)CH_3}\), that is 1-(1-bromoethyl)-3-nitrobenzene. (b) Chirality is the property of a molecule that cannot be superimposed on its mirror image. (c) The planar carbocation is attacked equally from both faces, giving equal amounts of the two enantiomers.
