Question

Identify electron donating groups in resonance among the following
(A) $-CON{H}_2$
(B) $-N{O}_2$
(C) $-OCOC{H}_3$
(D) $-COC{H}_3$
(E) $-CHO$
(F) $-NHCOC{H}_3$

Answer 1

Hint: Resonance effect explains polarity induced in a molecule by the reaction between a lone pair of electrons and a pi-bond. Electron donating group is an atom or functional group that donates some of its electron density into a conjugated $\pi$ system via resonance.

Complete step by step solution:
Resonance effect occurs when there is a pi-bond in a molecule. In this situation, the lone pair of electrons on a group next to the pi-bond reacts with it and changes the polarity in the molecule. This effect is called the resonance effect. The changing polarity due to movement of electrons is called delocalization.
There are two types of resonance effects:
1. Positive resonance effect: If a group next to a pi-bond donates electrons to the other molecule through the process of delocalization. Then such a resonance effect is called positive resonance effect. And such a group is called the electron donating group (EDG).
Examples of electron donating groups are: $-OH$, $-OR$, $-NHCOR$
2. Negative resonance effect: If a group next to a pi-bond withdraws electrons from the other molecule through the process of localization. Then such a resonance effect is called a negative resonance effect. And such a group is called electron withdrawing group (EWG)
Examples of electron withdrawing groups are: $-CON{H}_2$, $-N{O}_2$, $-COR$, $-CHO$

Therefore, from the above explanation, we can observe that, the correct answer is, option (C) $-OCOC{H}_3$ and option (F) $-NHCOC{H}_3$

Note:
Usually, if a group next to a molecule which has pi-bond, is more electron negative, then it is an electron withdrawing group. And if the group next to a molecule which has pi-bond is less electron negative, then it is an electron donating group. But there are some exceptions in this rule. But, if you have a general idea about reactivity of groups, then this rule is enough to find the correct answer for this question.