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Propionic acid with $Br_2$/P yields a dibromo product. Its structure would be
A
$CH_3-CBr_2-COOH$
B
$CH_2Br-CHBr-COOH$
C
$Br_2CH-CH_2COOH$
D
$CH_2Br-CH_2-COBr$
Detailed Solution
Carboxylic acids having α-hydrogen atoms are halogenated at the α-position by chlorine or bromine in the presence of a small amount of red phosphorus (Hell–Volhard–Zelinsky reaction).
Propionic acid, $CH_3-CH_2-COOH$, has two α-hydrogen atoms on the $-CH_2-$ group next to –COOH.
First: $CH_3CH_2COOH \xrightarrow{Br_2/P} CH_3CHBrCOOH$
With more bromine the second α-hydrogen is also replaced: $CH_3CHBrCOOH \xrightarrow{Br_2/P} CH_3CBr_2COOH$
The β-hydrogens are not substituted in this reaction.
Hence the dibromo product is $CH_3-CBr_2-COOH$ (2,2-dibromopropanoic acid).
Propionic acid, $CH_3-CH_2-COOH$, has two α-hydrogen atoms on the $-CH_2-$ group next to –COOH.
First: $CH_3CH_2COOH \xrightarrow{Br_2/P} CH_3CHBrCOOH$
With more bromine the second α-hydrogen is also replaced: $CH_3CHBrCOOH \xrightarrow{Br_2/P} CH_3CBr_2COOH$
The β-hydrogens are not substituted in this reaction.
Hence the dibromo product is $CH_3-CBr_2-COOH$ (2,2-dibromopropanoic acid).