Question

Write short notes on the following:
(i) Hunsdiecker reaction
(ii) Frankland's reaction
(iii) Dehydrohalogenation

Hint:

Remember: Hunsdiecker = silver carboxylate \(\rightarrow\) halide \(-\,\mathrm{CO_2}\) (chain shortens by 1 C); Frankland = Zn couples two R–X to R–R; Dehydrohalogenation = \( \beta \)-elimination with alcoholic base giving the more substituted alkene (Zaitsev).

Answer 1

(i) Hunsdiecker (Hunsdiecker–Borodin) reaction:
Decarboxylative halogenation of the silver salt of a carboxylic acid with halogen (\(\mathrm{Br_2}\) or \(\mathrm{Cl_2}\)) in \(\mathrm{CCl_4}\) under light/heat to give an alkyl halide with \(\,1\)\,carbon less. Radical mechanism.
\[ \mathrm{RCOOAg} + \mathrm{X_2} \xrightarrow[\text{hv or }\Delta]{\mathrm{CCl_4}} \mathrm{R{-}X} + \mathrm{CO_2} + \mathrm{AgX} \] Example: \(\mathrm{C_2H_5COOAg + Br_2 \rightarrow C_2H_5Br + CO_2 + AgBr}\).
(ii) Frankland's reaction (zinc coupling):
Two alkyl halides couple in presence of zinc to form a higher alkane (symmetrical), with zinc halide as by-product.
\[ 2\,\mathrm{R{-}X} + \mathrm{Zn} \longrightarrow \mathrm{R{-}R} + \mathrm{ZnX_2} \] Example: \(\mathrm{2\,CH_3I + Zn \rightarrow C_2H_6 + ZnI_2}\).
(iii) Dehydrohalogenation:
Base-induced elimination of \(\mathrm{HX}\) from an alkyl halide to form an alkene; carried out with alcoholic KOH/NaOEt and heat; follows Zaitsev's rule (more substituted alkene major, via E2).
\[ \mathrm{RCH_2CHXCH_3} \xrightarrow[\Delta]{\mathrm{alc.\ KOH}} \mathrm{RCH{=}CHCH_3} + \mathrm{KX} + \mathrm{H_2O} \] Example: \(\mathrm{CH_3CH_2CH_2Br \xrightarrow[\,\Delta\,]{alc.\ KOH} CH_3CH{=}CH_2 + KBr + H_2O}\).