Which one of the following ions is the most resonance stabilized?
(a) ethoxide
(b) phenoxide
(c) tertiary-butoxide
(d) iso-propoxide
(e) n-butoxide

Answer

To determine which ion is the most resonance stabilized among the given options, we need to analyze the resonance structures of each ion. Let's go through each option step by step.

### Step 1: Analyze Ethoxide Ion (C2H5O-)
- Structure: Ethoxide ion is represented as CH3CH2O-.
- Analysis: In this structure, there is a negative charge on the oxygen atom. However, there are no double bonds or conjugated systems present that would allow for resonance stabilization. Therefore, ethoxide does not exhibit resonance.

**Hint:** Look for the presence of double bonds or conjugated systems to determine if resonance is possible.

### Step 2: Analyze Phenoxide Ion (C6H5O-)
- Structure: Phenoxide ion can be represented as C6H5O-.
- Analysis: The negative charge on the oxygen can delocalize into the aromatic ring. This allows for multiple resonance structures:
1. The negative charge can be placed on different carbon atoms of the benzene ring, creating several resonance forms.
2. This delocalization significantly stabilizes the phenoxide ion.

**Hint:** Consider how the negative charge can move into a conjugated system, such as an aromatic ring, to enhance resonance.

### Step 3: Analyze Tertiary Butoxide Ion (C4H9O-)
- Structure: Tertiary butoxide is represented as (CH3)3CO-.
- Analysis: The negative charge is on the oxygen, and there are no double bonds or adjacent conjugated systems. Thus, there is no resonance stabilization in this ion.

**Hint:** Check for the presence of double bonds or lone pairs that can participate in resonance.

### Step 4: Analyze Isopropoxide Ion (C3H7O-)
- Structure: Isopropoxide can be represented as (CH3)2CHO-.
- Analysis: Similar to tertiary butoxide, there are no double bonds or conjugated systems present. The negative charge on the oxygen does not have resonance stabilization.

**Hint:** Look for structural features that allow for resonance; lack of such features means no resonance.

### Step 5: Analyze N-Butoxide Ion (C4H9O-)
- Structure: N-butoxide is represented as CH3CH2CH2CH2O-.
- Analysis: The negative charge is localized on the oxygen atom, with no adjacent double bonds or conjugated systems. Therefore, there is no resonance stabilization in n-butoxide.

**Hint:** Identify if there are any conjugated systems or double bonds that could allow for resonance.

### Conclusion
After analyzing all the ions, we find that the only ion that exhibits resonance stabilization is the phenoxide ion. The ability of the negative charge to delocalize into the aromatic ring provides significant resonance stabilization compared to the other ions.

**Final Answer:** (b) Phenoxide ion is the most resonance stabilized ion among the options given.