Single Choice

Urea reacts with water to form $$A$$ which will decompose to form $$B$$. $$B$$ when passed through $$Cu^{2+} (aq)$$, deep blue colour solution $$C$$ is formed. What is the formula of $$C$$ from the following?

A$$[Cu(NH_3)_4]^{2+}$$
Correct Answer
B$$Cu(OH)_2$$
C$$CuCO_3\cdot Cu(OH)$$
D$$CuSO_4$$

Solution

Urea reacts with water to give ammonium carbamate

$$NH_2CONH_2 + H_2O \rightarrow NH_2COO^-NH_4^+$$
$$(A)$$

Ammonium carbamate on decomposition gives ammonia.

$$NH_2COO^-NH_4^+ \xrightarrow {\Delta} 2NH_3 +CO_2$$

$$ (B)$$

When $$NH_3$$ passsed through $$Cu^{2+}$$ solution to form a deep blue coloured complex.

The complex formed is $$[Cu(NH_3)_4]^{2+}$$

Option A is correct.

SIMILAR QUESTIONS

Organic Compounds Containing Nitrogen

Urea on slow heating gives

Organic Compounds Containing Nitrogen

An aqueous solution of urea :

Organic Compounds Containing Nitrogen

A concentrated solution of urea gives a white precipitate with:

Organic Compounds Containing Nitrogen

Urea reacts with hydrazine to give.

Organic Compounds Containing Nitrogen

Urea reacts with malonic ester to give:

Organic Compounds Containing Nitrogen

Urea can be distinguished from acetamide using

Organic Compounds Containing Nitrogen

Urea on treatment with sodium hypobromite gives:

Organic Compounds Containing Nitrogen

Urea on hydrolysis gives:

Organic Compounds Containing Nitrogen

What happens when? (i) Urea is heated with acetoacetic ester, (ii) Urea is heated with succinic acid, (iii)Urea is treated with sodium hypobromide in alkaline medium (iv) Urea is treated with acetyl chloride

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