Question

The stability of carbanions in the following is in the order of:

• A. $a>c>b>d$
• B. $a>b>c>d$
• C. $b>c>d>a$
• D. $d>b>c>a$

Answer 1

The correct option is B $a>b>c>d$

Among a,b,c and d q will be most stable and d will be least stable because in negative charge is kept on ${Sp}^2$ hybridised carbon but in d negative charge is kept on ${Sp}^3$ hybridized carbon atom. Again b is more stable than c because in b electron $\Pi$ cloud is delocalized between 6 atom. So, correct order of stability : a>b>c>d.