Which of the following orders correctly depic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ that disperse the negative charge through inductive $(-I)$ or res

Vedclass · Accepted Answer

$(CHO)_2\overset{\Theta}{C}H > CHO\overset{\Theta}{C}H_2 > \overset{\Theta}{C}H_3$

Vedclass · Answer

(B) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ that disperse the negative charge through inductive $(-I)$ or resonance $(-M)$ effects. Conversely,electron-donating groups $(EDG)$ decrease stability by intensifying the negative charge.$1$. In option $B$,the stability order is $(CHO)_2\overset{\Theta}{C}H > CHO\overset{\Theta}{C}H_2 > \overset{\Theta}{C}H_3$. This is correct because the $-CHO$ group exerts a strong $-M$ effect,and having two such groups provides more stabilization than one.$2$. In option $C$,the correct order of stability is $\overset{\Theta}{C}H_2NO_2 > \overset{\Theta}{C}H_2CN > \overset{\Theta}{C}H_2Cl > \overset{\Theta}{C}H_2CH_3$. This is based on the strength of the electron-withdrawing groups: $-NO_2$ (strong $-M$) > $-CN$ $(-I, -M)$ > $-Cl$ $(-I)$ > $-CH_3$ ($+I$,destabilizing).

Which of the following orders correctly depicts the decreasing order of stability of carbanion?

Similar Questions

Which among the following is an incorrect statement?

Explain hyperconjugation or no-bond resonance with an example.

The correct stability order of the following species is:

In which of the following molecules do all the effects,namely inductive,mesomeric,and hyperconjugation,operate?

The higher stabilities of $tert$-butyl cation over isopropyl cation and $trans-2-butene$ over propene,respectively,are due to orbital interactions involving

Vedclass Test Series

Exam Paper Generator

Online Exam Module