What is the order of reactivity of `1^(@), 2^(@)and 3^(@)` alcohols with sodium metal ?

Explanation

To determine the order of reactivity of primary (1°), secondary (2°), and tertiary (3°) alcohols with sodium metal, we need to analyze the stability of the alkoxide ions (R-O⁻) formed during the reaction. Here’s a step-by-step breakdown of the reasoning:

### Step 1: Understand the Reaction
When alcohols react with sodium metal, they form alkoxide ions (R-O⁻) and hydrogen gas (H₂). The general reaction can be represented as:
\[ \text{R-OH} + \text{Na} \rightarrow \text{R-O}^- \text{Na}^+ + \text{H}_2 \]

### Step 2: Analyze Alkoxide Ion Stability
The stability of the alkoxide ion (R-O⁻) is crucial in determining the reactivity of the alcohol. The more stable the alkoxide ion, the more reactive the corresponding alcohol will be.
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