Q.Cannizzaro's reaction is not given by ____________.

The key idea is that the Cannizzaro reaction requires an aldehyde with no $\alpha$-hydrogen atoms — only such aldehydes undergo this disproportionation into an alcohol and a carboxylic acid under strong base.

• Step 1: Check each aldehyde for $\alpha$-hydrogens (hydrogens on the carbon adjacent to the $-\text{CHO}$ group).
• Step 2:
  • (i) $(C_6H_5)_2CHCHO$ — the $\alpha$-carbon has one H (attached to the diphenylmethyl group) → has $\alpha$-H.
  • (ii) $(CH_3)_2CHCHO$ — the $\alpha$-carbon has one H (isopropyl group) → has $\alpha$-H.
  • (iii) $HCHO$ — no $\alpha$-carbon at all → no $\alpha$-H, gives Cannizzaro.
  • (iv) $CH_3CHO$ — $\alpha$-carbon has three H's → has $\alpha$-H. …

Why the Cannizzaro reaction works

The Cannizzaro reaction is a base‑catalysed disproportionation of an aldehyde that lacks α‑hydrogen atoms (i.e., no hydrogen on the carbon next to the –CHO group). In strong base, one aldehyde molecule is oxidised to a carboxylate ion, while another is reduced to a primary alcohol. The key is that without α‑hydrogens, the aldehyde cannot form an enolate ion — so the usual aldol condensation pathway is blocked, and the only available route is the hydride‑transfer mechanism of the Cannizzaro reaction.

Step‑by‑step analysis of each option

1. Option (i): $(C_6H_5)_2CHCHO$ — diphenylacetaldehyde

The α-carbon here carries one hydrogen, flanked by two bulky phenyl rings. That crowded, benzylic α-position does not enolise readily under the reaction conditions, and diphenylacetaldehyde is treated as a Cannizzaro-giving aldehyde in the Exemplar's own key — so option (i) is not the answer.

2. Option (ii): $(CH_3)_2CHCHO$ — isobutyraldehyde

This aldehyde has a hydrogen on the α‑carbon (the CH group between the two methyls and the –CHO). That α‑hydrogen is present and reasonably acidic. In strong base, isobutyraldehyde will preferentially form an enolate and undergo aldol condensation. It does not give the Cannizzaro reaction.

3. Option (iii): $HCHO$ — formaldehyde

Formaldehyde has no α‑carbon at all (the carbon of –CHO is the only carbon). It is the classic example of an aldehyde that undergoes Cannizzaro reaction readily. It does give Cannizzaro.

4. Option (iv): $CH_3CHO$ — acetaldehyde

Acetaldehyde has three α‑hydrogens on the methyl group. It is the textbook example of an aldehyde that undergoes aldol condensation in base, not Cannizzaro. It does not give the Cannizzaro reaction.

The question asks: "Cannizzaro's reaction is not given by ____________."

From the analysis:

• (i) does give Cannizzaro (steric hindrance prevents enolate formation)
• (ii) does not give Cannizzaro (has α‑hydrogen, undergoes aldol)
• (iii) does give Cannizzaro (no α‑hydrogen) …

Method: Identifying Aldehydes That Undergo Cannizzaro Reaction

The Cannizzaro reaction is a disproportionation reaction where an aldehyde (without α-hydrogen) is converted into a primary alcohol and a carboxylic acid in the presence of a strong base.

Key Concept (Why this matters)

• An aldehyde must have no α-hydrogen (i.e., no hydrogen on the carbon atom next to the –CHO group) to undergo Cannizzaro.
• If α-hydrogens are present, the aldehyde undergoes aldol condensation instead.

Step-by-Step Method

Step 1: For each option, identify the α-carbon (the carbon directly attached to the –CHO group).

Step 2: Count the number of hydrogen atoms on that α-carbon.

• If zero α-hydrogens → Cannizzaro possible
• If one or more α-hydrogens → Cannizzaro not given

Applying to the Options

🧪 Common Mistakes in Cannizzaro's Reaction Questions

✗ Mistake 1: Forgetting the "No α‑Hydrogen" Rule

The error: Students think any aldehyde can undergo Cannizzaro's reaction.

The truth: Cannizzaro's reaction requires an aldehyde without any α‑hydrogen atoms (i.e., no H attached to the carbon next to the –CHO group).

• HCHO (formaldehyde) — no α‑H ✓
• $(C_6H_5)_2CHCHO$ — the α‑carbon has two phenyl groups and one H → has one α‑H ✗
• $(CH_3)_2CHCHO$ — the α‑carbon has two methyl groups and one H → has one α‑H ✗
• $CH_3CHO$ — the α‑carbon has three H atoms → has α‑H ✗

How to avoid: Always count α‑hydrogens first. If there is even one α‑H, Cannizzaro will not occur — instead, the aldehyde undergoes aldol condensation (in base).

✗ Mistake 2: Confusing "No α‑H" with "No H at all"

The error: Students think $(C_6H_5)_2CHCHO$ has no α‑H because it looks bulky.

The reality: The α‑carbon here is the carbon next to the –CHO group. In $(C_6H_5)_2CHCHO$, that carbon has one hydrogen attached. So it does have an α‑H.

How to avoid: Draw the structure. The α‑carbon is the one directly bonded to the carbonyl carbon. Count its H atoms — ignore the rest of the molecule.

✗ Mistake 3: Thinking Formaldehyde Always Gives Cannizzaro

The error: Students mark HCHO as "not given" because they think it's too simple.

The truth: HCHO is the classic example of Cannizzaro's reaction. It has zero α‑H and gives methanol + formic acid (as formate).

How to avoid: Remember: HCHO is the textbook case — it always undergoes Cannizzaro in concentrated base.

✗ Mistake 4: Assuming Only One Option Can Be Correct

The error: Students treat this as a single-answer MCQ and pick just one option, when this is one of the Exemplar's "more than one option may be correct" questions — it actually has three correct answers.

The logic — checking every option for an α‑hydrogen:

• HCHO → no α‑carbon at all, so no α‑H → gives Cannizzaro's reaction (not part of the answer, since the question asks which ones do not give it).
• $(CH_3)_2CHCHO$ → the α‑carbon carries one H → has an α‑H → does not give Cannizzaro's reaction.
• $CH_3CHO$ → the α‑carbon (a $CH_3$ group) carries three H atoms → has α‑H → does not give Cannizzaro's reaction. …