2
$\begingroup$

I am getting confused between the type of mesomeric effect (+M or -M) shown by compounds containing $\ce{-COX}$ where $\ce{X}$ is an electronegative element.

Examples: $\ce{-CONHCH3}$, $\ce{-COOCH3}$, $\ce{-COOH}$.

According to me, all these compounds must show both a +M as well as a -M effect. Since the group (e.g. $\ce{-COX}$) can accept as well as donate $\pi$ electrons (as in $\ce{CH_2=CH-COOH}$).

However, the book says these compounds show only -M effect. Please explain me the reason behind it. enter image description here

In the above example, isn't the group showing +M effect, since it's donating the electrons?

$\endgroup$
1
  • 2
    $\begingroup$ carbonyl group does not show +M effect because in it carbon is highly electron-deficient. In it oxygen cannot donate electron to pi-system from electron pair and donating electron from double bond will leave oxygen with electron sextet, which will not happen under any circumstances, period. $\endgroup$
    – permeakra
    Commented Mar 20, 2014 at 16:22

1 Answer 1

3
$\begingroup$

Following up on Permeakra's comment, the relative importance of resonance structures is typically determined by:

1) Maximize atoms that fulfill the octet rule. 2) Electronegative atoms have negative formal charges and avoid positive charges. 3) Minimize charge separation.

Carboxylic acid derivatives (esters, amides, etc.) have resonance structures that break #2, but since all atoms fulfill the octet rule, they are important contributors to the overall electronic picture of the compounds. Hopefully the image below is legible.

Resonance structures of carboxylic acid derivatives

In general, resonance (mesomeric) structures that contain an electronegative atom that both breaks the octet rule AND bears a formal positive charge are not reasonable resonance structures and do not contribute to the overall electronic picture of the compound. The resonance structure of the carboxylic acid drawn in the original question falls into this category as the 'top' oxygen both breaks the octet rule and contains a positive charge.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.