The decreasing order of stability of the following anions is:
$(P)$ $p-OCH_3-C_6H_4-CH_2^-$
$(Q)$ $p-CHO-C_6H_4-CH_2^-$
$(R)$ $p-Cl-C_6H_4-CH_2^-$
$(S)$ $p-CH_3-C_6H_4-CH_2^-$

  • A
    $Q > R > S > P$
  • B
    $R > Q > P > S$
  • C
    $S > P > R > Q$
  • D
    $P > Q > R > S$

Explore More

Similar Questions

State True or False for the following statements:
$(i)$ The stability is explained by resonance effect and hyperconjugation.
$(ii)$ The resonance structures are drawn in resonance and hyperconjugation.
$(iii)$ Hyperconjugation is a bondless resonance.
$(iv)$ In resonance structure,there is movement of electron pair of only $\pi$ bond.

What is the difference between the mesomeric effect and the resonance effect?

Arrange the following according to the instructions:
$(i)$ Arrange in descending order of acidic strength:
$CH_3COOH, (CH_3)_3CCOOH, (CH_3)_2CHCOOH, CH_3CH_2COOH$
(ii) Arrange in descending order of stability:
$\stackrel{+}{CH}_3, (CH_3)_3\stackrel{+}{C}, CH_3CH_2^+, (CH_3)_2\stackrel{+}{CH}$
(iii) Arrange in increasing order of acidic strength:
$CCl_3COOH, CH_3COOH, CHCl_2COOH, CH_2ClCOOH$
(iv) Arrange in increasing order of stability:
$(a)$ $CH_3CH=CH-CHO$
$(b)$ $CH_3\stackrel{+}{CH}-CH=C-O:^-$
$(c)$ $\stackrel{+}{CH}_2-CH=CH-C=O$
$(v)$ Arrange in decreasing order of stability:
$(I)$ $CH_2=CH-CHO$
$(II)$ $\stackrel{+}{CH}_2-CH=C-H$ (with $O^-$)
$(III)$ $:CH_2-CH=C-H$ (with $O^-$)

Identify ortho and para directing groups from the following: $I. -CHO$,$II. -NHCOCH_3$,$III. -OCH_3$,$IV. -SO_3H$.

The electron displacement in a covalent bond of a molecule is produced by which type of effects?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo