Question

A compound with molecular mass 180 is acylated with $C{H_3}COCl$ to get with a compound of molecular mass 390. What is the number of amino groups present per molecule of the former compound?

Answer 1

Hint: First find out the change in molar mass for acylation of 1 amino group only. Then find out the total change in molar mass for the given compound. Now use the unitary method to calculate the total number of amino groups present per molecule.

Complete step by step answer:
-Here we are talking about acylation. This reaction can be represented as:
              $R - N{H_2} + C{H_3} - COCl\xrightarrow[{ - HCl}]{}R - NH - COC{H_3}$
Here we can see that each ($ - N{H_2}$) group is changed to ($ - NH - COC{H_3}$) and also that 1 mole or 1 molecule of $C{H_3}COCl$ will react with only 1 mole or 1 molecules of $R - N{H_2}$.

-We know that the weight of each ($ - N{H_2}$) group is: 14 + 2 = 16 and that of ($ - NH - COC{H_3}$) group is: 14 + 1 + 12 + 16 + 12 +3 = 58.
So, the above given statement can also be said as: each ($ - N{H_2}$) group of weight 16 g is replaced by a ($ - NH - COC{H_3}$) group of weight 58 g.

-Hence for each molecule there is a change of weight by (58 – 16) units = 42 units. So, for every ($ - N{H_2}$) group there is an increase of 42 g after acylation reaction.
The question says that the molecular weight of the compound before acylation is = 180 g and the molecular weight of compound obtained after acylation is 390 g.
So, the difference in the molecular weights after and before acylation is = 390 – 180
                                                                                                                                 = 210 g

-We will now find out the number of amino groups per molecule.
The total increase in mass is 210 g. But for 1 ($ - N{H_2}$) group there is an increase of 42 g only. So, we will use the unitary method to find out the number of amino groups.
              42 g increase → for 1 ($ - N{H_2}$) group
                 1 g increase → for $\dfrac{1}{{42}}$ ($ - N{H_2}$) group
            210 g increase → for $\left[ {\dfrac{1}{{42}} \times 210} \right]$ ($ - N{H_2}$) groups
                                         = $\dfrac{{210}}{{42}}$
                                         = 5 ($ - N{H_2}$) groups
So, we can now tell that there are 5 amino groups per molecule of the former compound.

Note: $C{H_3}COCl$ is a derivative of acetic acid. The acylation reactions using this are done in the presence of some base like pyridine, DMAP, triethylamine, etc. These bases act as catalysts which help in promoting the reaction and also since the bases neutralize the HCl formed as a byproduct along with amide.