The acid shown is treated with $Br X_{2}$ /red P and then heated with aqueous KOH (substitution conditions) followed by acidification. Which structure is the final product?

AStructure A
BStructure B
CStructure C
DStructure D

Answer: A

The drawn compound is 3-methylbutanoic acid. HVZ brominates the alpha CH2 giving 2-bromo-3-methylbutanoic acid; hot aqueous KOH replaces Br by OH (alcoholic KOH would favour elimination). Acidification gives 2-hydroxy-3-methylbutanoic acid. The tertiary beta carbon is not brominated.

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The acid shown is treated with BrX2​/red P and then heated with aqueous KOH (substitution conditions) followed by acidification. Which structure is the final product?

Answer: A

More questions on Reduction, oxidation and HVZ products of drawn carbonyls and acids

The acid shown is treated with $Br X_{2}$ /red P and then heated with aqueous KOH (substitution conditions) followed by acidification. Which structure is the final product?