Question

Which of the following substituents at para-position is most effective in stabilizing the phenoxide ion?
A. $\text{-C}{{\text{H}}_{3}}$
B. $\text{-OC}{{\text{H}}_{3}}$
C. $\text{-COC}{{\text{H}}_{3}}$
D. $\text{-C}{{\text{H}}_{3}}\text{OH}$

Answer 1

For this problem, we have to identify which group is an electron withdrawing group and which is an electron donating group. Then the group will be the strong electron-withdrawing group and will stabilise the phenoxide ion more.

Complete step by step answer:
- In the given question, we have to explain which group will be responsible to increase the stability of the phenoxide ion.
- As we know that there are two types of groups based on the donation or acceptance of the electrons that is an electron-withdrawing group and electron-donating group.
- Now, firstly we will see what is an electron-withdrawing group and how it will affect the stability of the phenoxide ion.
- The electron-withdrawing group are the groups which are responsible for attracting the pair of an electron from the adjacent group.

- Now, in the phenoxide ion when we will attach an electron-withdrawing group such as $\text{-COC}{{\text{H}}_{3}}$ then it will stable the negative charge by delocalised it.
- As the electron will undergo delocalisation in the benzene ring it will make the ion more stable.
- Now, among $\text{-COC}{{\text{H}}_{3}}$ and $\text{-OC}{{\text{H}}_{3}}$, the acetyl group is a strong electron-withdrawing group than the methoxy group.
- So, the acetyl group will stabilise the phenoxide ion more.
So, the correct answer is “Option C”.

Note: In the given problem, $\text{-C}{{\text{H}}_{3}}$ and $\text{-C}{{\text{H}}_{3}}\text{OH}$ are the electron-donating groups due to which they will donate the pair of electrons to the phenoxide ion instead of withdrawing. Thus, the negative charge will increase and the stability of the phenoxide ion will decrease.