Q.Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal with primary, secondary and tertiary alcohols.

The key idea is that the reactivity of an alcohol with sodium depends on the acidity of the O–H bond — the more acidic the hydrogen, the faster the reaction.

1. Acidity is determined by the stability of the alkoxide ion formed after losing $H^+$.
2. Alkoxide stability decreases as the alkyl group becomes more bulky and electron-donating (inductive effect). …

Why does sodium react with alcohols at all?

Sodium metal is a strong reducing agent. When it meets an alcohol, it donates an electron to the proton of the hydroxyl group. This breaks the O–H bond, releasing hydrogen gas and leaving behind the alkoxide ion ($RO^-$) paired with $Na^+$.

The key point: the reaction rate depends on how easily the O–H bond breaks. That ease is directly linked to the acidity of the alcohol — the more acidic the O–H hydrogen, the faster it reacts with sodium.

What controls the acidity of an alcohol?

In the gas phase, the acidity order is actually: tertiary > secondary > primary. But in the liquid phase (which is what we deal with in a lab), the order flips. Why? Because solvation effects dominate.

The alkoxide ion ($RO^-$) that forms after deprotonation is stabilised by the solvent (usually the alcohol itself, or any polar medium). A smaller, less bulky alkoxide ion can be better solvated — the solvent molecules can pack around it more tightly, stabilising the negative charge more effectively.

• Primary alkoxide ($CH_3CH_2O^-$): small, compact, easily solvated → very stable → reaction is fast.
• Secondary alkoxide ($(CH_3)_2CHO^-$): bulkier, solvation is less efficient → less stable → reaction is slower.
• Tertiary alkoxide ($(CH_3)_3CO^-$): very bulky, solvation is poor → least stable → reaction is slowest.

So the reactivity order with sodium metal in the liquid phase is:

Primary > Secondary > Tertiary

Step-by-step reasoning

1. Identify the reaction type.

   This is an acid–base reaction where sodium acts as a base (it accepts a proton). The rate depends on the acidity of the O–H bond in the liquid phase.

2. Recall the liquid-phase acidity trend for alcohols.

   Due to solvation effects, the acidity decreases as the alkyl group becomes more substituted: …

Method: Steric Hindrance & Carbanion Stability Approach

This method uses two key concepts to predict reactivity:

• Steric hindrance around the O–H bond
• Stability of the alkoxide ion formed after H⁺ removal

Steps

1. Identify the site of reaction

   Sodium metal reacts with the O–H group of alcohols:

   $2 \text{ROH} + 2 \text{Na} \rightarrow 2 \text{RO}^- \text{Na}^+ + \text{H}_2 \uparrow$

2. Analyze steric hindrance

   • Primary alcohol: –OH is attached to a carbon with only one alkyl group → least crowded, easiest for Na to approach.
   • Secondary alcohol: –OH attached to a carbon with two alkyl groups → more crowded.
   • Tertiary alcohol: –OH attached to a carbon with three alkyl groups → most crowded, hardest for Na to approach.

3. Analyze alkoxide ion stability

   • Alkoxide ion ($\text{RO}^-$) is stabilised by electron-withdrawing groups.
   • Alkyl groups are electron-donating (hyperconjugation + inductive effect).
   • More alkyl groups → more destabilisation of the negative charge on oxygen. …

Here are the common mistakes students make when answering this question, along with how to avoid each.

Mistake 1: Getting the Reactivity Order Backwards

The Error:

Students often assume that because tertiary carbocations are more stable, tertiary alcohols must be more reactive with sodium. They write:

Tertiary > Secondary > Primary (✗)

Why it’s wrong:

Reaction with sodium is an acid-base reaction, not a carbocation-forming reaction. The alcohol acts as a weak acid ($\text{ROH}$), and sodium acts as a base. The key factor is the stability of the alkoxide ion ($\text{RO}^-$) formed.

• Primary alkoxides are the most stable (least steric hindrance, negative charge is more exposed and better solvated).
• Tertiary alkoxides are the least stable (bulky alkyl groups hinder solvation and destabilize the negative charge).

Correct Order:

How to Avoid:

Always ask: “Which alkoxide ion is most stable?” Not “Which carbocation is most stable?” Draw the alkoxide ions and compare steric hindrance around the negative oxygen.

Mistake 2: Confusing This with Dehydration or SN1 Reactivity

The Error:

Students mix up the reactivity order for:

• Reaction with Na (acid-base, alkoxide stability)
• Acid-catalyzed dehydration (carbocation stability)
• SN1 reactions (carbocation stability)

They apply the tertiary > secondary > primary order from dehydration to this reaction.

How to Avoid:

Make a mental checklist:

Tip: If the reaction involves a metal (Na, K, Mg), think alkoxide stability → primary first.

Mistake 3: Ignoring the Role of Steric Hindrance

The Error:

Students only think about electronic effects (inductive effect) and forget that steric hindrance matters.

Why it matters:

• In a tertiary alcohol, the bulky alkyl groups physically block the Na atom from approaching the O–H bond.
• This slows down the reaction even if the O–H bond were equally polar.

How to Avoid:

Visualize the reaction: Na atom needs to get close to the O–H hydrogen. In a tertiary alcohol, three alkyl groups crowd the oxygen. In a primary alcohol, only one small alkyl group is present. Less crowding = faster reaction.

Mistake 4: Forgetting That This Is an Acid-Base Reaction

The Error:

Students treat it as a redox or substitution reaction and look for leaving groups or oxidation states. …