Question

Write the products of the following reactions:

Hint:

Remember: plain HBr follows Markovnikov’s rule, whereas HBr with peroxides (ROOR) adds anti-Markovnikov via a radical pathway. Aryl + alkyl halides with Na in dry ether give Wurtz–Fittig coupling to an alkylbenzene. Aniline forms a stable diazonium salt only in cold acidic solution (273–278 K).

Answer 1

(i) Addition of HBr to styrene (Markovnikov):
\[ \mathrm{Ph{-}CH{=}CH_2 \xrightarrow[\text{}]{\;HBr\;} \; Ph{-}CH(Br){-}CH_3} \] Product: 1-bromo-1-phenylethane (Markovnikov addition via benzylic carbocation).
(ii) Addition of HBr to allylbenzene (Markovnikov):
\[ \mathrm{Ph{-}CH_2{-}CH{=}CH_2 \xrightarrow[\text{}]{\;HBr\;} \; Ph{-}CH_2{-}CH(Br){-}CH_3} \] Product: 3-phenyl-2-bromopropane.
(iii) Addition of HBr to 1-phenylpropene in the presence of peroxides (anti-Markovnikov, radical):
\[ \mathrm{Ph{-}CH{=}CH{-}CH_3 \xrightarrow[\text{ROOR}]{\;HBr\;} \; Ph{-}CH_2{-}CH_2{-}CH_2Br} \] Product: 3-phenyl-1-bromopropane (Br adds to the less substituted end to form a benzylic radical intermediate).
(iv) Wurtz–Fittig coupling (dry ether):
\[ \mathrm{C_6H_5Br + 2\,Na + Br{-}(CH_2)_4CH_3 \xrightarrow[\Delta]{\text{dry ether}} C_6H_5{-}(CH_2)_4CH_3 + 2\,NaBr} \] Product: \(\mathrm{n}\)-butylbenzene.
(v) Diazotisation of aniline at 273–278 K:
\[ \mathrm{Ph{-}NH_2 \xrightarrow[\;273{-}278\,K\;]{NaNO_2/HX} Ph{-}N_2^+X^- + 2\,H_2O} \] Product: benzenediazonium halide (X = Cl/Br, as per HX used).