Question

Identify the set of reagent/reaction conditions 'X' and 'Y' in the following set reaction:

• A. X = dilute aqueous NaOH, ${20}^{\circ }C,$ Y = $B{r}_2/CHC{l}_3,0^{\circ }C$
• B. X = concentrated alcoholic NaOH, ${80}^{\circ }C;Y=B{r}_2/CHC{l}_3,0^{\circ }C$
• C. X = concentrated alcoholic NaOH, ${80}^{\circ }C;$ Y = HBr/acetic acid, ${20}^{\circ }C$
• D. X = dilute aqueous NaOH, ${20}^{\circ }C;$ Y = HBr/acetic acid, ${20}^{\circ }C$

Answer 1

The correct option is C X = concentrated alcoholic NaOH, ${80}^{\circ }C;$ Y = HBr/acetic acid, ${20}^{\circ }C$


First dehydrohalogenation reaction proceed by reagent alc. NaOH forming propene and then by adding HBr, the addition of $HBr$ proceeds by Markovnikov's rule.
Aqueous NaOH gives substitution products and alkene with $B{r}_2$ gives disubstituted haloalkanes.
Hence, option (b) is correct.