Question
Borax on treatment with NaOH and $H_2O_2$ 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
The correct answer is
$Na_2B_2(O_2)_2(OH)_4 \cdot nH_2O$ and $PhCONH_2$
To solve this question, we first need to understand the chemical reactions and products involved in the transformation of borax through various reagents.
- Borax, chemically known as sodium tetraborate decahydrate, is represented by the formula Na_2B_4O_7 \cdot 10H_2O.
- When borax is treated with NaOH and H_2O_2, the mixture forms the perborate compound X, represented as Na_2B_2(O_2)_2(OH)_4 \cdot nH_2O.
- This process involves the conversion of the boron in borax into the perborate form, which is achieved due to the oxidative action of H_2O_2.
- The compound X is identified as Na_2B_2(O_2)_2(OH)_4 \cdot nH_2O.
Next, let's analyze the reaction when compound X reacts with PhCN at 60^\circ C in a methanol-water mixture:
- The compound X, with its peroxy linkage, facilitates a nucleophilic attack on the cyanide group in PhCN.
- This results in the partial hydrolysis of PhCN to form benzamide, represented as PhCONH_2, which is identified as compound Y.
Thus, the major product of this reaction is PhCONH_2, known as benzamide.
In conclusion, the correct pairing of compounds X and Y is X = Na_2B_2(O_2)_2(OH)_4 \cdot nH_2O and Y = PhCONH_2.
The correct answer is:
- Na_2B_2(O_2)_2(OH)_4 \cdot nH_2O and PhCONH_2
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Important Questions from p Block Elements
- The phosphazene compound that acts as a superbase is
- In cyclophosphazenes, $(NPX_2)_3$ (X = F, Cl, Br and Me), the strength of P-N $\pi$-bond varies with X in the order
- The ease of formation of the adduct, $NH_3 \cdot BX_3$ (where, $X = F, Cl, Br$) follows the order
- Ammonolysis of $S_2Cl_2$ in an inert solvent gives
- The reaction of $P_2O_5$ with $HNO_3$ and $HClO_4$, respectively, gives
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