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Discussion

Borax on treatment with NaOH and H2O2 forms X. The compound X on reaction with PhCN at 60 °C in methanol-water mixture gives Y as the major product.

X and Y, respectively, are

  • A.\(\mathrm{NaB(O)(OH)_{2} \cdot}\)n\(\mathrm{H_{2}O}\) and \(\mathrm{PhCONH_{2}}\)
  • B.\(\mathrm{NaB(O)(OH)_{2} \cdot}\)n\(\mathrm{H_{2}O}\) and \(\mathrm{PhCOOH}\)
  • C.\(\mathrm{Na_{2}B_{2}(O_{2})_{2}(OH)_{4} \cdot}\)n\(\mathrm{H_{2}O}\) and \(\mathrm{PhCONH_{2}}\)
  • D.\(\mathrm{Na_{2}B_{2}(O_{2})_{2}(OH)_{4} \cdot}\)n\(\mathrm{H_{2}O}\) and \(\mathrm{PhCOOH}\)

Answer: C

X is Na2B2(O2)2(OH)4·nH2O and Y is PhCONH2.

Borax treated with NaOH and hydrogen peroxide gives sodium perborate, whose anion is a six-membered ring of two boron atoms bridged by two peroxo (O-O) groups, each boron also carrying two OH groups: [B2(O2)2(OH)4]2-. The formula NaB(O)(OH)2 has no peroxide and is not perborate.

In water, perborate releases hydroperoxide, which adds to nitriles and converts them to primary amides (as in the Radziszewski reaction), so benzonitrile gives benzamide. Hydrolysis all the way to benzoic acid would need much harsher conditions, which rules out the PhCOOH options.

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