Question

i) Explain the Hell-Volhard-Zelinsky reaction with example. (2)
ii) Explain the Hofmann bromamide degradation reaction with example. (2)

Hint:

HVZ: carboxylic acid with an alpha-H plus \(X_2\) and red phosphorus gives an alpha-halo acid. Hofmann bromamide: primary amide plus \(Br_2\) and NaOH gives a primary amine with one carbon less.

Answer 1

Step 1 (HVZ, what it does): It is a way to put a halogen specifically at the carbon next to the \(-COOH\) group. The acid must have at least one \(\alpha\)-H; red phosphorus acts as catalyst (it forms \(PBr_3/PCl_3\), converting the acid to acid halide whose enol is halogenated).
Step 2 (HVZ example): With propanoic acid, \(CH_3CH_2COOH + Br_2 \xrightarrow{\text{red P}} CH_3CHBrCOOH + HBr\), giving 2-bromopropanoic acid. (For acetic acid the product is \(ClCH_2COOH\).) The \(\alpha\)-halo acids are useful for making \(\alpha\)-hydroxy and \(\alpha\)-amino acids.

Step 3 (Hofmann bromamide, what it does): A primary amide \(R-CONH_2\) is heated with \(Br_2\) in strong alkali. The reaction shortens the chain by one carbon and produces a primary amine \(R-NH_2\); it is a standard step-down (descent of series) method.
Step 4 (Hofmann example): \(C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + 2KBr + K_2CO_3 + 2H_2O\), i.e. benzamide gives aniline. Equivalently acetamide gives methylamine.
\(\boxed{\text{HVZ adds }X\text{ at }\alpha\text{-C; Hofmann removes one C to give an amine}}\)