Q.Complete the reaction and identify compounds A and B:
CH3–CH(=O) (i.e. CH3CHO, drawn with the aldehyde H explicit) + NH2NH2 → A --KOH/453K, GlyCol--> B
Himachal HpboseHPBOSE Plus Two Board 2025Subjective· 2mImportance★★★★★
Imagine you have a ketone or an aldehyde — a molecule with a carbonyl group (C=O) — and you want to remove that oxygen entirely, replacing it with two hydrogens. You want to go from C=O to CH₂. That's a reduction, but not the kind that gives you an alcohol. You want to strip the oxygen out completely.
The problem is that simply adding hydrogen gas (H₂) with a metal catalyst usually stops at the alcohol stage. You need a different strategy.
The Wolff-Kishner reduction solves this by first converting the carbonyl into a hydrazone, then using a strong base to break it apart, releasing nitrogen gas (N₂) and leaving behind the CH₂ group. The nitrogen gas bubbles away, driving the reaction forward.
The Intuition: Why It Works
The key insight is that nitrogen is an excellent leaving group when it can form N₂. The reaction proceeds in two main stages:
Formation of a hydrazone. Hydrazine (NH₂–NH₂) attacks the carbonyl carbon. After losing water, you get a C=N–NH₂ group — the hydrazone. This is like making an imine, but with an extra NH₂ attached.
Base-induced decomposition. A strong base (like KOH or NaOEt) in a high-boiling solvent (like ethylene glycol) deprotonates the hydrazone. The resulting anion rearranges, kicking out N₂ gas. The carbon that was double-bonded to oxygen now picks up two protons from the solvent, becoming CH₂.
The driving force is the formation of the extremely stable N≡N triple bond. That's why the reaction works so cleanly under basic conditions.
The carbonyl compound is treated with hydrazine (NH₂NH₂) to form a hydrazone. This intermediate is then heated with a strong base (typically KOH) in a high-boiling solvent like diethylene glycol or ethylene glycol. The product is the corresponding alkane (CH₂ group in place of C=O), along with nitrogen gas and water.
Step-by-Step Mechanism (Simplified)
Step 1: Hydrazone formation (acid-catalysed, but the overall conditions are basic)
The carbonyl oxygen is protonated (if acid is present), making the carbon more electrophilic. Hydrazine attacks, then water is eliminated to give the hydrazone.
RX2C=O+NHX2NHX2RX2C=N−NHX2+HX2O
Step 2: Deprotonation and elimination
The base abstracts a proton from the NH₂ group of the hydrazone, generating a negatively charged nitrogen. This anion undergoes a series of rearrangements, ultimately expelling N₂ and forming a carbanion intermediate.
RX2C=N−NHX2+OHX−RX2C=N−NHX−+HX2O
RX2C=N−NHX−RX2CX−+NX2+HX+
Step 3: Protonation
The carbanion picks up a proton from the solvent (or from water present), giving the final alkane.
RX2CX−+HX2ORX2CHX2+OHX−
Important Exam Points
Watch out
The Wolff-Kishner reduction requires strongly basic conditions and high temperatures (around 200°C). It is not suitable for base-sensitive compounds. If your molecule has other base-labile groups (esters, nitro groups, etc.), this reaction will destroy them.
Acetaldehyde first forms an intermediate with hydrazine, which is then decomposed under strongly basic, high-boiling conditions to remove the carbonyl oxygen entirely — identifying A and B means tracking that carbonyl group's fate through both steps. …
Acetaldehyde first forms a hydrazone (A) with hydrazine, which is then reduced all the way to the hydrocarbon ethane (B) under Wolff-Kishner conditions (KOH, ethylene glycol, heat).
Step 1 — Formation of A:
Acetaldehyde reacts with hydrazine (NH2NH2) via nucleophilic addition-elimination at the carbonyl carbon to form the hydrazone:
CH3CHO+H2N−NH2→CH3CH=N−NH2(A)+H2O
Step 2 — Formation of B (Wolff-Kishner Reduction):
Heating the hydrazone with KOH in a high-boiling solvent like ethylene glycol at ~453 K decomposes it, releasing nitrogen gas and reducing the carbonyl-derived carbon fully to a CH2/CH3 group (i.e., the C=O is converted all the way to CH2): …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set ANNUAL2 marks
Q.Complete the reaction and identify compounds A and B:
CH3–CH(=O) (i.e. CH3CHO, drawn with the aldehyde H explicit) + NH2NH2 → A --KOH/453K, GlyCol--> B
›Reveal solutionSolution
Acetaldehyde first forms a hydrazone (A) with hydrazine, which is then reduced all the way to the hydrocarbon ethane (B) under Wolff-Kishner conditions (KOH, ethylene glycol, heat).
Step 1 — Formation of A:
Acetaldehyde reacts with hydrazine (NH2NH2) via nucleophilic addition-elimination at the carbonyl carbon to form the hydrazone:
CH3CHO+H2N−NH2→CH3CH=N−NH2(A)+H2O
Step 2 — Formation of B (Wolff-Kishner Reduction):
Heating the hydrazone with KOH in a high-boiling solvent like ethylene glycol at ~453 K decomposes it, releasing nitrogen gas and reducing the carbonyl-derived carbon fully to a CH2/CH3 group (i.e., the C=O is converted all the way to CH2): …