Question

Complete the following chemical equation: \(\mathrm{CH_3CHBrCH_3}\ \xrightarrow[\ \Delta\ ]{\mathrm{KOH(alc)}\ (A)\ \xrightarrow[\ \mathrm{ROOR}\ ]{\mathrm{HBr}}\ (B)\).}

Hint:

Alcoholic KOH promotes elimination (E2) to an alkene, whereas HBr with peroxides adds anti-Markovnikov via a radical chain to give the primary bromoalkane.

Answer 1

Step 1 (Dehydrohalogenation, alcoholic KOH, heat):
\[ \mathrm{CH_3{-}CHBr{-}CH_3 \xrightarrow[\ \Delta\ ]{KOH(alc)} CH_3{-}CH{=}CH_2\ +\ KBr\ +\ H_2O} \] Thus, \(\boxed{A=\mathrm{CH_3{-}CH{=}CH_2}\ (\text{propene})}\).
Step 2 (Anti-Markovnikov addition of HBr in presence of peroxides):
\[ \mathrm{CH_3{-}CH{=}CH_2 \xrightarrow[\ \mathrm{ROOR}\ ]{HBr} CH_3{-}CH_2{-}CH_2Br} \] Thus, \(\boxed{B=\mathrm{CH_3{-}CH_2{-}CH_2Br}\ (\text{1-bromopropane})}\).