Question

Consider the following species:
A. ${OH}^ - $
B. $C{H_3}\,O^ - $
B. ${C{H_3}}^ - $
C. ${N{H_2}}^ - $
Arrange these species in their decreasing order of nucleophilicity.
1.C > D > A > B
2.B > A > C > D
3.A > B > C > D
4.C > A > B > D

Answer 1

Hint: We know that as electronegativity increases, nucleophilicity decreases. So, across a row in the periodic table nucleophilicity that is the lone pair donation for \[{C^ - } > {N^{ - \;}} > {O^ - }\; > {F^ - }\] since increasing electronegativity decreases the lone pair availability. This is also the same order as for the basicity.

Complete step by step solution:
- Nucleophile means "nucleus loving" that describes the tendency of an electron rich species to be attracted to the positive nuclear charge of an electron poor species, that is the electrophile.
- Nucleophilicity is the donation of an electron pair. The less electronegative the atom, the less tightly held those electrons will be. The key factor for determining how “tightly held” an electron pair is bound is the familiar concept of electronegativity.
- There, of all the nucleophiles, ${C{H_3}}^ - $, will have the highest nucleophilicity because carbon atoms are less electronegative than Oxygen (O) and Nitrogen (N). So, ${N{H_2}}^ - $ be on number two.
- Also, condition for nucleophilicity, the conjugate base is always a stronger nucleophile. So, ${OH}^ - $ will be on number three. Hence, $C{H_3}\,O^ - $ will have the least nucleophilicity.
So, the correct order of these species in their decreasing order of nucleophilicity is C > D > A > B.

The correct answer is option 1.

Note: For nucleophilicity, when we are discussing reactions at carbon, we have to take into account that orbitals at carbon that participate in reactions are generally less accessible than protons are. An effect called “steric hindrance” comes into play. So, the bulkier a given nucleophile is, the slower the rate of its reactions and therefore the lower its nucleophilicity.