Polarization of electrons in acrolein may be written as

  • A
    $\mathop{C}\limits^{\delta^-}H_2 = CH - \mathop{C}\limits^{\delta^+}H = O$
  • B
    $\mathop{C}\limits^{\delta^-}H_2 = CH - CH = \mathop{O}\limits^{\delta^+}$
  • C
    $\mathop{C}\limits^{\delta^-}H_2 = \mathop{C}\limits^{\delta^+}H - CH = O$
  • D
    $\mathop{C}\limits^{\delta^+}H_2 = CH - CH = \mathop{O}\limits^{\delta^-}$

Explore More

Similar Questions

The total number of contributing structures showing hyperconjugation (involving $C-H$ bonds) for the following carbocation is

Explain the delocalization of electrons in $CH_3-CH_2^+$ and $CH_3-CH=CH_2$.

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 $O^-$)

In which $C-C$ bond of $CH_3-CH_2-CH_2-Br$,the inductive effect is expected to be the least?

Which of the following represents the correct order of stability?

Difficult
View Solution

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