Question

Which of the following has the highest $p{K_b}$ value?
A) ${(C{H_3})_3}CN{H_2}$
B) $N{H_3}$
C) ${(C{H_3})_2}NH$
D) $C{H_3}N{H_2}$

Answer 1

Hint:Use the concept of basic strength, after calculating the order of basic strength make it reverse for $p{K_b}$ order. This is because $p{K_b}$ value is inversely proportional to the basic strength, so if we have two compounds in which the basic strength of one compound is more than that of another. It will have lesser $p{K_b}$ value.

Complete solution:
According to the options given above, all have one atom in common which is nitrogen. Nitrogen is having its lone pair of electrons for donation, in these types of questions where any element is having lone pairs to donate. Firstly, check what their basic strength is. Donation of electrons of nitrogen makes it a very good base in nature. Now if we have different options where nitrogen is common, in those the approach will be like this.
As we know if an electron donating group is attached with nitrogen, it will make nitrogen more basic and makes higher electron density on it while if an electron withdrawing group will attach with nitrogen or any other element which makes electron density lower then it will become less basic.
Among the options given above, we have ammonia, ${1^ \circ }$ amine primary amine $C{H_3}N{H_2}$ and ${(C{H_3})_3}CN{H_2}$ . As its visible that both primary amines contains EDG electron donating group but option A. have a tertiary group while option D. contains only primary EDG so therefore option A. is more basic than D. \[\left( {Order:{\text{ }}A{\text{ }} > {\text{ }}D} \right)\]
Now let’s see other options, B. is simple ammonia having lone pairs of electrons on nitrogen. Option C. is secondary amine having two methyl groups which electron are donating in nature so order becomes at last as- \[\left( {Order:{\text{ C}} > {\text{ A}} > \,D\, > \,B} \right)\]
Therefore, from above compounds option C. is having highest basic strength and lowest $p{K_b}$ value. Similarly we can say option B. ammonia is having least basic strength but have highest $p{K_b}$ value.
Hence, the Option B. is correct.

Note:ake a point to remember that whenever you are taking order of amines and its derivatives having methyl groups order is such that - ${3^ \circ }\, > \,{2^ \circ }\, > \,{1^ \circ } > \,N{H_3}$ but in case of amines compounds having groups other than methyl group then the order is such as ${2^ \circ } > {3^ \circ }\,\, > \,{1^ \circ } > \,N{H_3}$ . In the above example if you got the concept of EDG and EWG then try to solve it directly by using the above order rule.