Q.Arrange the following compounds in decreasing order of acidity.
$H_2O$, ROH, $HC \equiv CH$

The key idea is that acidity depends on the stability of the conjugate base formed on losing a proton — for these three, the electronegativity of the atom bearing the negative charge dominates over hybridisation effects.

Reasoning:

1. Comparing conjugate bases: $OH^-$ (from $H_2O$), $RO^-$ (from ROH), and $HC\equiv C^-$ (from $HC\equiv CH$).
2. Oxygen is far more electronegative than carbon, so an oxygen-centred anion ($OH^-$ or $RO^-$) is much more stable than a carbon-centred anion ($HC\equiv C^-$) — this outweighs the fact that the alkyne's sp-hybridised carbanion has higher s-character. …

Why Acidity Trends Matter

Acidity in organic chemistry is not about memorising pKa values — it’s about asking one question: How happy is the molecule to give up a proton? The happier it is, the stronger the acid. And that happiness depends entirely on what’s left behind: the conjugate base.

A stable conjugate base means a stronger acid. So to compare $H_2O$, $ROH$ (a generic alcohol), and $HC \equiv CH$ (terminal alkyne), we need to compare the stability of $OH^-$, $RO^-$, and $HC \equiv C^-$.

Three factors dominate here:

1. Electronegativity of the atom bearing the negative charge — oxygen is more electronegative than carbon, so $O^-$ is more stable than $C^-$.
2. Inductive effects — alkyl groups ($R$) are electron-donating, which destabilises a negative charge.
3. Hybridisation — $sp$ carbon (as in alkyne) holds negative charge more tightly than $sp^3$ carbon because $s$-character is higher.

Let’s walk through each compound.

1. Water ($H_2O$)

   When water loses a proton, it forms the hydroxide ion $OH^-$. The negative charge sits on a highly electronegative oxygen atom. There are no electron-donating alkyl groups to push extra electron density onto the oxygen — in fact, the hydrogen atoms are slightly electron-withdrawing compared to alkyl groups. So $OH^-$ is a relatively stable anion.

   This makes water the strongest acid among the three.

2. Alcohol ($ROH$)

   An alcohol loses a proton to give the alkoxide ion $RO^-$. Here, the negative charge is still on oxygen, but the alkyl group $R$ is electron-donating (through the inductive effect). That pushes electron density toward the already negative oxygen, making the anion less stable than $OH^-$.

   So $ROH$ is a weaker acid than water.

   Watch out

   A common mistake is to think that because $R$ is “bigger”, it stabilises the charge. In fact, alkyl groups are electron-donating relative to hydrogen, so they destabilise a nearby negative charge. Water is actually more acidic than a typical alcohol.

3. Terminal alkyne ($HC \equiv CH$)

   This is the trickiest. The acidic proton is on an $sp$-hybridised carbon. The conjugate base is the acetylide ion $HC \equiv C^-$, where the negative charge is on carbon. …

Method: Relative Acidity via Conjugate Base Stability

This method uses the principle: stronger acid → more stable conjugate base.

We compare the stability of the anion formed after losing $H^+$.

Step 1: Write the conjugate bases

Step 2: Compare conjugate base stability

• $OH^-$ vs $RO^-$:

  Both are negatively charged oxygen anions.

  The alkyl group ($R$) in $RO^-$ is electron-donating (inductive effect), which destabilises the negative charge.

  → $OH^-$ is more stable than $RO^-$.

• $RO^-$ vs $HC \equiv C^-$:

  The acetylide carbanion places the negative charge on an $sp$-hybridised carbon, which is far less electronegative than the oxygen of an alkoxide. Oxygen holds the charge much better than carbon, so $RO^-$ is more stable than $HC \equiv C^-$.

Key fact: The measured $\text{p}K_a$ values fix the order:

$H_2O$ (pKa ≈ 15.7) < $ROH$ (pKa ≈ 16–18) < $HC \equiv CH$ (pKa ≈ 25). …

Mistake 1: Confusing Acidity with Boiling Point or Reactivity

What students do:

They see $H_2O$ and think "water has hydrogen bonding, so it's more acidic." Or they see $HC \equiv CH$ and think "alkynes are reactive, so they must be more acidic."

Why it's wrong:

Acidity depends on the stability of the conjugate base after losing $H^+$, not on boiling point or general reactivity.

How to avoid:

Always ask: "After losing $H^+$, which anion is most stable?"

• $H_2O \rightarrow OH^-$ (moderately stable)
• $ROH \rightarrow RO^-$ (less stable than $OH^-$ due to +I effect of alkyl group)
• $HC \equiv CH \rightarrow HC \equiv C^-$ (very stable — negative charge on an $sp$-hybridised carbon)

Correct order:

Mistake 2: Forgetting the Role of Hybridisation

What students do:

They compare $O-H$ vs $C-H$ bond polarity and think "oxygen is more electronegative, so $H_2O$ is most acidic."

Why it's wrong:

Bond polarity is not the only factor. The hybridisation of the atom bearing the negative charge in the conjugate base matters more.

• $sp$-carbon (alkyne) holds negative charge closer to nucleus → more stable
• $sp^3$-oxygen (water/alcohol) holds charge less tightly

How to avoid:

Remember the hybridisation trend for anion stability:

So $HC \equiv C^-$ (sp) is more stable than $OH^-$ or $RO^-$ (both sp³ oxygen).

Mistake 3: Ignoring the Inductive Effect in Alcohols vs Water

What students do:

They rank $ROH$ and $H_2O$ as equal, or put $ROH$ first because "alcohols are acidic."

Why it's wrong:

The alkyl group ($R$) in $ROH$ is electron-donating (+I effect). This destabilises the alkoxide ion ($RO^-$) by pushing electron density toward the negative charge.

• $OH^-$ has no such destabilisation → more stable → $H_2O$ is more acidic than $ROH$

How to avoid:

Compare conjugate bases:

• $OH^-$ (no alkyl group)
• $RO^-$ (alkyl group pushes electrons → less stable)

So:

Mistake 4: Memorising Without Understanding the Trend

What students do:

They memorise "alkynes are more acidic than water" but can't explain why, so they mix up the order in exams.

How to avoid:

Learn the two key reasons for alkyne acidity: …