Q.Which one of the following undergoes reaction with 50% sodium hydroxide solution to give the corresponding alcohol and acid? a) Phenylmethanal b) ethanal c) ethanol d) methanol
Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions
These two reactions are often grouped together in textbooks because they both involve carbonyl compounds (C=O), but they do entirely different things. Let's take them one at a time.
Clemmensen Reduction
Intuition first. Imagine you have a ketone or aldehyde — a molecule with a C=O group. You want to remove that oxygen entirely and replace the C=O with two hydrogen atoms, turning it into a simple hydrocarbon chain. That's a reduction (adding hydrogen, removing oxygen). The Clemmensen reduction is a brute-force way to do this using a strongly acidic, reducing environment.
The precise reaction:
A ketone or aldehyde is heated with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The C=O group is reduced to a CH₂ group.
R−C(=O)−RX′+4[H]Zn(Hg),HCl,heatR−CHX2−RX′+HX2O
Aldehyde or KetoneZn(Hg), conc. HCl, ΔHydrocarbon
Key points for exams:
Works only for ketones and aldehydes that are stable in strong acid.
Does not work for compounds that get destroyed by conc. HCl (e.g., acid-sensitive groups like esters, nitriles).
The mechanism is complex and not usually tested in detail — just know it's a reductive removal of C=O.
The product is always a saturated hydrocarbon (alkane).
Watch out
Clemmensen reduction cannot reduce carboxylic acids, esters, or amides. Only aldehydes and ketones.
Intuition first. This is a disproportionation reaction — one molecule of aldehyde gets oxidised (to a carboxylic acid) while another gets reduced (to an alcohol). It happens only with aldehydes that have no alpha-hydrogen atoms (i.e., the carbon next to the C=O has no H). Why? Because if there were alpha-hydrogens, the aldehyde would undergo aldol condensation instead.
The precise reaction:
An aldehyde without α-hydrogen is treated with concentrated aqueous or alcoholic base (NaOH/KOH). Two molecules of aldehyde react: one becomes a carboxylate salt, the other becomes a primary alcohol.
2R−CHO+OHX−R−COOX−+R−CHX2OH
After acidification, the carboxylate salt gives the carboxylic acid.
2HCHOconc. NaOHHCOONa+CH3OH
(Formaldehyde → sodium formate + methanol)
Key points for exams:
Only works for aldehydes with no α-hydrogen: formaldehyde, benzaldehyde, trimethylacetaldehyde, etc.
The base must be concentrated (dilute base won't work).
Formaldehyde is the most common example — it gives formic acid (as formate) and methanol.
Crossed Cannizzaro: When formaldehyde is mixed with another aldehyde (like benzaldehyde), formaldehyde is always the one that gets oxidised (to formate), and the other aldehyde gets reduced (to alcohol). This is because formaldehyde is the strongest reducing agent among aldehydes.
Tip
In a crossed Cannizzaro, formaldehyde always becomes the carboxylate. The other aldehyde becomes the alcohol. This is a common exam question.
Example: …
Why this formula?
Cannizzaro Reaction: Why the Key Formulas Hold
The Cannizzaro reaction is a disproportionation reaction of aldehydes (without α-hydrogens) in the presence of a strong base. Let's build the understanding from the ground up.
1. What Happens in the Reaction?
An aldehyde (like formaldehyde or benzaldehyde) reacts with concentrated base to give:
One molecule is oxidized to a carboxylic acid (or its salt)
Another molecule is reduced to a primary alcohol
General equation (for two identical aldehydes):
2RCHO+OH−→RCOO−+RCH2OH
2. Why Does Disproportionation Occur?
The Key Insight: No α-Hydrogen
Aldehydes with α-hydrogens undergo aldol condensation instead.
Without α-hydrogens, the only available reaction path is hydride transfer.
The Mechanism (Step-by-Step Reasoning)
Nucleophilic attack: OH− attacks the carbonyl carbon of one aldehyde molecule.
Forms a tetrahedral intermediate (a gem-diolate).
Hydride shift: The intermediate acts as a hydride donor (H−) to a second aldehyde molecule.
This is the rate-determining step.
The hydride comes from the C–H bond of the intermediate (not from the OH).
Products:
The donor aldehyde becomes a carboxylate ion (oxidized).
The acceptor aldehyde becomes an alkoxide ion (reduced).
Protonation: In workup, the carboxylate gives the acid, and the alkoxide gives the alcohol.
3. The Key Formula(e) and Their Derivation
Formula 1: Stoichiometry
2RCHO+OH−→RCOO−+RCH2OH
Why this holds:
One aldehyde loses a hydride (H−) → gains an oxygen → oxidation state increases by 2.
The other aldehyde gains a hydride → oxidation state decreases by 2.
The base (OH−) is consumed stoichiometrically (one per two aldehydes).
Formula 2: Oxidation State Change
For an aldehyde carbon (carbonyl carbon):
In RCHO: oxidation state = +1
In RCOO−: oxidation state = +3 (gain of +2)
In RCH2OH: oxidation state = -1 (loss of -2)
Net change: +2 (oxidation) + (−2) (reduction) = 0 — consistent with disproportionation.
Formula 3: Rate Law (for the hydride transfer step)
Rate=k[aldehyde]2[OH−]
Why:
First aldehyde reacts with OH− to form the hydride donor (first order in each).
Second aldehyde accepts the hydride (first order in aldehyde).
Overall: second order in aldehyde, first order in base.
4. Why Only Certain Aldehydes Work?
Condition: Aldehyde must have no α-hydrogen atoms.
Step 1. Reacting with 50% (concentrated) NaOH to give BOTH an alcohol AND an acid from the SAME starting compound is the defining signature of the Cannizzaro reaction (Section 12.5.F.2).
Step 2. The Cannizzaro reaction specifically requires an aldehyde with NO alpha-hydrogen -- phenylmethanal (benzaldehyde), C6H5-CHO, has its 'alpha-position' on the aromatic ring itself, which cannot be deprotonated by base the way an aliphatic alpha-carbon could, so it qualifies.
Step 3. Benzaldehyde + 50% NaOH gives benzyl alcohol (C6H5CH2OH) + sodium benzoate (C6H5COONa) -- exactly one alcohol and one acid salt, matching the question's description. …
Picking ethanal simply because it 'is an aldehyde too' -- ethanal DOES have alpha-hydrogens, so dilute/concentrated base instead drives it toward aldol condensation …