Question

Complete the following equations:
(i) $\mathrm{CH_3CH_2CH_2OH + SOCl_2 \;\longrightarrow\; ?}$
(ii) $\mathrm{CH_3CH_2CH{=}CH_2 + HBr \;\longrightarrow\; ?}$
(iii) $\mathrm{(CH_3)_3CBr + KOH \;\longrightarrow\; ?}$
(iv) $\mathrm{CH_3CH_2Br + KCN \xrightarrow{Ethanol\ (aq.)}\; ?}$
(v) $\mathrm{CH_3CH_2Br + AgNO_2 \xrightarrow{alc.}\; ?}$

Hint:

- Alcohol + SOCl$_2$ $\Rightarrow$ alkyl chloride (best method as byproducts are gases).
- Alkene + HX $\Rightarrow$ halogen addition (Markovnikov unless peroxides present).
- Tertiary halides with KOH (alc.) favour elimination $\Rightarrow$ alkene.
- KCN $\Rightarrow$ cyanides (C–C bond formation).
- AgNO$_2$ gives alkyl nitrites ($R{-}ONO$), while KNO$_2$ gives nitroalkanes ($R{-}NO_2$).

Answer 1

(i) Reaction with thionyl chloride (SOCl$_2$):
\[ \mathrm{CH_3CH_2CH_2OH + SOCl_2 \;\longrightarrow\; CH_3CH_2CH_2Cl + SO_2 + HCl} \] (Product: 1-chloropropane).
(ii) Addition of HBr to alkene (Markovnikov's rule):
\[ \mathrm{CH_3CH_2CH{=}CH_2 + HBr \;\longrightarrow\; CH_3CH_2CHBrCH_3} \] (Product: 2-bromobutane).
(iii) Tertiary alkyl halide with KOH (alc., elimination):
\[ \mathrm{(CH_3)_3CBr + KOH \;\longrightarrow\; (CH_3)_2C{=}CH_2 + KBr + H_2O} \] (Product: isobutene, via elimination E2).
(iv) Substitution with KCN (nucleophilic substitution):
\[ \mathrm{CH_3CH_2Br + KCN \xrightarrow{Ethanol} CH_3CH_2CN + KBr} \] (Product: propionitrile).
(v) Reaction with AgNO$_2$ (alc.):
\[ \mathrm{CH_3CH_2Br + AgNO_2 \;\longrightarrow\; CH_3CH_2{-}ONO + AgBr} \] (Product: ethyl nitrite, an isomeric nitro–oxy compound).