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Question: Consider the following carbanions. I. \({{C}}{{{H}}_3}{{ - C}}{{{H}}_2}^ - \) II. \({{C}}{{{H}}_...

Consider the following carbanions.
I. CH3−CH2−{{C}}{{{H}}_3}{{ - C}}{{{H}}_2}^ -
II. CH2=CH−{{C}}{{{H}}_2}{{ = C}}{{{H}}^ - }
III. CH≡C−{{CH}} \equiv {{{C}}^ - }
Correct order of stability of these carbanions in decreasing order is:
A. I>II>III
B. III>II>I
C. I>III>II
D. III>I>II

Explanation

Solution

As we know that anions are formed by gaining electrons. They have a negative charge on them. Similarly, when a carbon atom has an unshared pair of electrons and a negative charge, then it is known as carbanion.

Complete step by step answer:
When a heterolytic cleavage occurs in a compound, it forms a positive charge and a negative charge. The positive charged species are termed as carbocations and the negative charged species are termed as carbanions.
When a proton is abstracted from carbon with the help of a base, carbanions are formed. Thus it has a very high nucleophilic character.
Stability of the carbanions depends upon several factors. In the given question, the compounds differ in the number of bonds to which the carbanion is attached, i.e. carbanion I has a single bond, II has a double bond and III has a triple bond. In carbanion I, the percentage of s character is only 25%25\% . While in carbanion II, the percentage of s character is 33%33\% . In carbanion III, the percentage of s character is 50%50\% . When s character is increased, stability is also increased. The carbanion having more s character will have greater electronegativity on carbanionic carbon.
Thus the order is CH≡C−>CH2=CH−>CH3−CH2−{{CH}} \equiv {{{C}}^ - } > {{C}}{{{H}}_2}{{ = C}}{{{H}}^ - } > {{C}}{{{H}}_3}{{ - C}}{{{H}}_2}^ - .

So, the correct option is B.

Note: Stability of carbanions is dependent on the following factors:
1.Resonance effect
2.Inductive effect
3.Electron-donating groups
4.Amount of s-character at the carbanionic carbon
5.Field effects
6.Stabilization by sulfur or phosphorus