Question

Given below are some nitrogen containing compounds:

Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ....... mg of HCl.
(Given Molar mass in g mol\(^{-1}\): C: 12, H: 1, O: 16, Cl: 35.5.)

Hint:

When determining the amount of HCl consumed in such reactions, remember that the basicity of the compound is the key factor. A primary amine like methyleneamine will consume more HCl due to its higher basicity.

Answer 1

In order to solve this, we need to identify the most basic compound from the given list. The basicity is typically determined by the ability of the compound to donate an electron pair to form a salt with HCl. In this case, primary amines are generally the most basic. 
Aniline (\( {C}_6{H}_5{NH}_2 \)) is a primary amine and will react with HCl to form an ammonium salt. 
Methyleneamine (\( {CH}_2{NH}_2 \)) is another primary amine and more basic than aniline. 
Acetanilide (\( {N-COCH}_3 {NH}_2 \)) is an amide derivative of aniline and will be less basic due to the electron-withdrawing nature of the acetyl group. 
Nitroaniline (\( {NO}_2 {C}_6{H}_4 {NH}_2 \)) is less basic due to the strong electron-withdrawing group (-NO2) attached to the aromatic ring. Thus, methyleneamine (\( {CH}_2{NH}_2 \)) is the most basic compound. Now we calculate how much HCl will be consumed by 1.0 g of methyleneamine. 
Molar mass of methyleneamine: \[ {Molar mass of methyleneamine} = 12 + 2 \times 1 + 14 = 29 \, {g/mol} \] The number of moles of methyleneamine in 1.0 g: \[ \frac{1.0 \, {g}}{29 \, {g/mol}} = 0.03448 \, {mol} \] Since methyleneamine is a primary amine, it will react with 1 mole of HCl per mole of amine group: \[ {HCl required} = 0.03448 \, {mol} \times 36.5 \, {g/mol} = 1.26 \, {g of HCl} \] Thus, 1.0 g of the most basic compound will consume 1.26 g of HCl. Since the question asks for the answer in mg, we convert: \[ 1.26 \, {g} = 1260 \, {mg} \] 
Correct Answer: The most basic compound, methyleneamine, will consume 1260 mg of HCl.