Which of the following statement is correct for the stability of ions of ethyl alcohol and phenol:

Question

The most suitable reagent for the conversion of \( \text{R-CH}_2\text{-OH} \) to \( \text{R-CHO} \) is:

Answer 1

To determine the stability of ions formed from ethyl alcohol and phenol, we need to understand the nature of electronic distribution in these ions.

Phenoxide Ion Stability:
- Phenol, when it loses a proton, forms a phenoxide ion. This ion benefits from the delocalization of \( \pi \)-electrons. The resonance structures in phenoxide ion allow the negative charge to be distributed over the aromatic ring, enhancing stability. This phenomenon is known as resonance stabilization.
Ethoxide Ion Stability:
- Ethyl alcohol forms an ethoxide ion upon deprotonation. Ethoxide ion does not have such resonance stabilization, so its stability relies more on the inductive effect, which is less significant compared to resonance.

Let us analyze the given options to determine the correct statement:

Option 1: Delocalisation of \( \pi \)-electrons in phenoxide ion:
- This is correct. The stability of the phenoxide ion results from the delocalization of \( \pi \)-electrons which allows the negative charge to be spread out over the aromatic ring. This is a classical example of resonance.
Option 2: Delocalisation of electrons in ethoxide ion:
- This is incorrect. There is no \( \pi \)-electron delocalization in the ethoxide ion. Any stabilization would be minor and due to inductive effects, which are relatively weak.
Option 3: Inductive effect of ethyl and phenyl group:
- Inductive effects play a role but are significantly less effective than resonance effects. Therefore, this is not the primary reason for the phenoxide or ethoxide ion stability.
Option 4: Localisation of \( \pi \)-electrons in phenoxide ion:
- This is incorrect. It is the delocalization, not localization, of \( \pi \)-electrons that provides stability to the phenoxide ion.

Thus, the correct answer is: Delocalisation of \( \pi \)-electrons in phenoxide ion.

Answer 2

To determine the stability of ions formed from ethyl alcohol and phenol, we need to understand the nature of electronic distribution in these ions.

Phenoxide Ion Stability:
- Phenol, when it loses a proton, forms a phenoxide ion. This ion benefits from the delocalization of \( \pi \)-electrons. The resonance structures in phenoxide ion allow the negative charge to be distributed over the aromatic ring, enhancing stability. This phenomenon is known as resonance stabilization.
Ethoxide Ion Stability:
- Ethyl alcohol forms an ethoxide ion upon deprotonation. Ethoxide ion does not have such resonance stabilization, so its stability relies more on the inductive effect, which is less significant compared to resonance.

Let us analyze the given options to determine the correct statement:

Option 1: Delocalisation of \( \pi \)-electrons in phenoxide ion:
- This is correct. The stability of the phenoxide ion results from the delocalization of \( \pi \)-electrons which allows the negative charge to be spread out over the aromatic ring. This is a classical example of resonance.
Option 2: Delocalisation of electrons in ethoxide ion:
- This is incorrect. There is no \( \pi \)-electron delocalization in the ethoxide ion. Any stabilization would be minor and due to inductive effects, which are relatively weak.
Option 3: Inductive effect of ethyl and phenyl group:
- Inductive effects play a role but are significantly less effective than resonance effects. Therefore, this is not the primary reason for the phenoxide or ethoxide ion stability.
Option 4: Localisation of \( \pi \)-electrons in phenoxide ion:
- This is incorrect. It is the delocalization, not localization, of \( \pi \)-electrons that provides stability to the phenoxide ion.

Thus, the correct answer is: Delocalisation of \( \pi \)-electrons in phenoxide ion.

The Correct Option is A