$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{NaOH} \xrightarrow{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}$ Product 'A' Consider the above reactions, identify product B and product C .

(B) $\mathrm{B}=2$-Propanol $\mathrm{C}=1$-Propanol

(A) $\mathrm{B}=\mathrm{C}=2$-Propanol

(C) $\mathrm{B}=1$-Propanol $\mathrm{C}=2$-Propanol

(D) $\mathrm{B}=\mathrm{C}=1$-Propanol

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Haloalkanes and Haloarenes

CH_{3} - CH_{2} - CH_{2} - Br + NaOH C_{2} H_{5} OH

Product 'A'

Consider the above reactions, identify product B and product C .

Answer: (B)

C H_{3} C H_{2} C H_{2} B r

with alcoholic

N a O H

undergoes E2 elimination to give propene (Product A).
Product A (propene) is then treated two ways:

Acid-catalysed hydration ( $H_{2} O, H^{+}$ ) follows Markovnikov's rule, adding $- O H$ to the more substituted carbon, giving 2-propanol (Product B).
Hydroboration–oxidation ( $B_{2} H_{6}$ , then $H_{2} O_{2} / O H^{-}$ ) is anti-Markovnikov, adding $- O H$ to the terminal carbon, giving 1-propanol (Product C).

So B = 2-Propanol, C = 1-Propanol.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option