The compound \( \text{Cl-(CH}_2\text{)}_4-\text{Cl} \) reacts with excess ammonia (\( \text{NH}_3 \)) to form an intermediate \( \text{A} \), which is \( \text{NH}_3^+-(\text{CH}_2)_4-\text{NH}_3^+ \cdot 2\text{Cl}^- \). This intermediate compound is a diammonium salt. Upon treatment with \( \text{NaOH} \), it undergoes deprotonation to yield compound \( \text{B} \), which is \( \text{H}_2\text{N}-(\text{CH}_2)_4-\text{NH}_2 \), also known as 1,4-diaminobutane or putrescine. Therefore, the correct answer is \( \text{H}_2\text{N}-(\text{CH}_2)_4-\text{NH}_2 \).
The Correct answer is: \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\)
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).