Question

Hell-Volahrd-Zelinsky (HVZ) reaction involves the reaction of aliphatic carboxylic acid with bromine in the presence of a trace of $PBr _3$ to form $\alpha$-bromocarboxylic acid
$CH _3 CH _2 COOH \xrightarrow{ Br _2 / PBr _3} CH _3 CH _2 COBr \rightarrow CH _3 CHBrCOOH \xleftarrow{ H _2 O } CH _3 CHBrCOBr ^2$
$\alpha$-halogenated acids are infact good stating material for preparing other $\alpha$-halogenated acids as ;

When two equilvalents of the reagent are used the product formed in the above reaction will be,

• A.
• B.
• C.
• D. Both (b) and (c)

Answer 1

Answer: (C)

Maleic acid is stronger for the first ionisation but weaker for the second. The reason for this is believed to be due to the fact that hydrogen bonding can occur in the maleate anion whereas this is not possible for the corresponding fumarate anion. Thus, the former anion is stabilised with respect to its acid and so there is a driving force for ionisation which is absent in the case of Fumaric acid. Furthermore, since the maleate anion is stabilised by $H$-bonding, the furmarate anion ionises to the dianion more readily than the former

Presence of $- CH _3$ group in acetate ion shows+I. Effect and therby intensifying charge on $0^{-}$of acetate ion than formate ion or acetate ion is destablised. Thus formate ion is more stable than acetate ion and $HCOOH$ loses proton more easily The HVZ reaction brings about halogenation only at the $\alpha$-carbon of the fatty acid in case the acid has two $\alpha$-hydrogens, $\alpha$-dihalo product will be formed when 2 equivalents of halogen are used