Question

The correct order of stability for the following alkoxides is:

• A. (C) > (A) > (B)
• B. (C) > (B) > (A)
• C. (B) > (A) > (C)
• D. (B) > (C) > (A)

Answer 1

The correct option is B

(C) > (B) > (A)

Explanation for the correct option:

Option B (C) > (B) > (A):

1. The correct order of stability for the following alkoxides is: (C) > (B) > (A).
2. Higher the delocalization of the negative charge, more will be the stability of the anion.
3. A: The negative charge is stabilized only through the –I effect exhibited by the $–{\mathrm{NO}}_2$ group.
4. B: The negative charge is stabilized by the delocalization of the double bond and the –I effect exhibited by the $–{\mathrm{NO}}_2$ group.
5. C: The negative charge is stabilized by extended conjugation.
6. Hence correct option.

Explanation for the incorrect options:

Option A (C) > (A) > (B):

1. Here C is more stable because in this case, -ve charge on O is in conjugation with a double bond and ${\mathrm{NO}}_2$. So there is an extra double bond in conjugation.
2. In A, there is no conjugation. so, it is the least stable.
3. Hence incorrect option.

Option C (B) > (A) > (C):

1. In the case of B, the -ve charge on oxygen is in conjugation with a double bond and not with ${\mathrm{NO}}_2$_.
2. Hence incorrect option.

Option D (B) > (C) > (A):

1. C is more stable than B because in this case, the -ve charge on O is in conjugation with a double bond and ${\mathrm{NO}}_2$. So there is an extra double bond in conjugation.
2. Hence incorrect option.

The correct order of stability for the following alkoxides is: (C) > (B) > (A).