Question

Complete the following chemical equation: \(\; \mathrm{C_2H_5Br} \xrightarrow[\text{}]{\mathrm{KOH(aq)} (A) \xrightarrow[\Delta]{\mathrm{K_2Cr_2O_7,\;H_2SO_4}} (B)\).}

Hint:

Aqueous KOH leads to substitution ($\mathrm{R{-}X \to R{-}OH}$), while alcoholic KOH favours elimination ($\mathrm{R{-}X \to R{=}Alkene}$). Acidified K$_2$Cr$_2$O$_7$ oxidises primary alcohols to aldehydes under mild conditions and to acids under strong reflux.

Answer 1

Step 1 (Nucleophilic substitution with aqueous KOH):
\[ \mathrm{C_2H_5Br + OH^- \;\longrightarrow\; C_2H_5OH + Br^-} \] Therefore, \(\boxed{A = \mathrm{C_2H_5OH}\;(\text{ethanol})}\).
Step 2 (Oxidation with acidified K$_2$Cr$_2$O$_7$ under heat):
Under reflux/heat, primary alcohols are oxidised to carboxylic acids.
\[ \mathrm{C_2H_5OH \;+\; [O] \;\xrightarrow[\Delta]{\mathrm{K_2Cr_2O_7/H_2SO_4}}\; CH_3COOH \;+\; H_2O} \] Hence, \(\boxed{B = \mathrm{CH_3COOH}\;(\text{ethanoic acid})}\).