Answer
Verified
28.5k+ views
Hint: Strength of the conjugate acid decides the stability of a carbanion. The conjugate base which is weak results in higher stability of carbanion. Other factors are aromaticity, resonance and inductive effect.
Complete step-by-step solution:
Aromatic carbanions are considered to be most stable. Although Benzyl carbanion (\[{C_6}{H_5} - C{H_2}^ - \]) is resonance stabilized, yet in three out of five resonating structures, aromatic stabilisation of the ring is lost.
Another factor that can determine the stability is:
\[ - I{\text{ }}and{\text{ }} - M\]groups are used to define the stability of carbanions. Those carbanions which are stabilised by conjugated highly electronegative atoms are less stable than the aromatic ones. Like option (B).
Stability order decreases as we approach tertiary anion from primary one. This is because of the inductive effect of methyl groups which increases the intensity of negative charge on central carbon in tertiary carbanion. This makes it less stable as in option (D).
Therefore, the carbanion bonded to triple bond is most stable here because it has 50% s-character being sp hybridised.
Hence, the correct option is (A).
Note: Electronegativity and hybridisation are the two important factors along with these to find the most stable carbanion. The more polarizable the atom, the more stable the anion. Carbanions prefer a lesser degree of alkyl group substitution.
Complete step-by-step solution:
Aromatic carbanions are considered to be most stable. Although Benzyl carbanion (\[{C_6}{H_5} - C{H_2}^ - \]) is resonance stabilized, yet in three out of five resonating structures, aromatic stabilisation of the ring is lost.
Another factor that can determine the stability is:
\[ - I{\text{ }}and{\text{ }} - M\]groups are used to define the stability of carbanions. Those carbanions which are stabilised by conjugated highly electronegative atoms are less stable than the aromatic ones. Like option (B).
Stability order decreases as we approach tertiary anion from primary one. This is because of the inductive effect of methyl groups which increases the intensity of negative charge on central carbon in tertiary carbanion. This makes it less stable as in option (D).
Therefore, the carbanion bonded to triple bond is most stable here because it has 50% s-character being sp hybridised.
Hence, the correct option is (A).
Note: Electronegativity and hybridisation are the two important factors along with these to find the most stable carbanion. The more polarizable the atom, the more stable the anion. Carbanions prefer a lesser degree of alkyl group substitution.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
An air capacitor of capacity C10 mu F is connected class 12 physics JEE_Main
The dimensions of stopping potential V in photoelectric class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Differentiate between mass and inertia class 11 physics JEE_Main
Mulliken scale of electronegativity uses the concept class 11 chemistry JEE_Main