Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
Both change C=O into CH2 (they make a hydrocarbon), but the conditions are opposite. Clemmensen uses zinc-amalgam (Zn-Hg) with concentrated HCl, so it is done in ACIDIC medium. Wolff-Kishner uses hydrazine (NH2-NH2) first, then heating with strong base KOH/NaOH in a high-boiling solvent like ethylene glycol, so it is BASIC medium. Rule for NEET: if the molecule has an acid-sensitive group (like an alcohol or a group that reacts with acid), use Wolff-Kishner (base). If it has a base-sensitive group, use Clemmensen (acid). Result is the same CH2, only the medium differs.
NaBH4 reduces C=O only to an alcohol (C-OH), NOT to CH2. It gives a hydride ion (H-) that adds to the carbonyl carbon. A ketone gives a secondary (2 degree) alcohol and an aldehyde gives a primary (1 degree) alcohol. To go all the way to CH2 (a hydrocarbon) you need Clemmensen or Wolff-Kishner instead. This is a very common NEET trap: NaBH4 stops at the alcohol stage.
NaBH4 is a mild (weak) reducing agent. It reduces only the easy targets: aldehydes and ketones (to alcohols). It CANNOT reduce esters, carboxylic acids, amides or nitriles. LiAlH4 is a much stronger reducing agent, so it reduces esters, acids, amides and nitriles as well. NEET tip: if a question shows a molecule with BOTH a ketone and an ester and uses NaBH4, only the ketone becomes an alcohol; the ester stays untouched.
An aldehyde (R-CHO) has one carbon and one H on the carbonyl carbon. Adding H gives R-CH2-OH, a PRIMARY (1 degree) alcohol. A ketone (R-CO-R') has two carbons on the carbonyl carbon. Adding H gives R-CH(OH)-R', a SECONDARY (2 degree) alcohol. Memory line: aldehyde to primary, ketone to secondary. This holds for NaBH4, LiAlH4 and catalytic H2.
To alcohol: NaBH4 (mild), LiAlH4 (strong, then H3O+), or H2 with Ni/Pt/Pd (catalytic hydrogenation). To hydrocarbon (CH2): Clemmensen = Zn-Hg + concentrated HCl (acidic); Wolff-Kishner = hydrazine NH2-NH2, then heat with KOH/NaOH in ethylene glycol (basic). Extra NEET point: DIBAL-H reduces an ester only up to the aldehyde stage; NaBH4 and H2/Pd-BaSO4 do NOT reduce esters at all.
With H2 at low pressure (like 1 atm) over Pd/C, the carbon-carbon double bond (C=C) is reduced first, while the C=O is left alone. So cyclohex-2-en-1-one gives cyclohexanone, not an alcohol. This appeared in NEET 2016. Remember: mild catalytic hydrogenation prefers the C=C over the C=O.
A compound bearing two -COCH3 (acetyl) groups is treated with Zn-Hg / conc. HCl. Identify the product.
4-methyl-2-(2-methoxy-2-oxoethyl)cyclohexan-1-one (a cyclohexanone ring carrying a -CH2COOCH3 methyl-ester side chain) is treated with NaBH4 in C2H5OH. What forms?
Identify the final product [D]: CH3CHO -(i)LiAlH4 (ii)H3O+-> [A] -H2SO4, heat-> [B] -HBr-> [C] -C6H5Br, Na, dry ether-> [D]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
LiAlH4 is much stronger. It reduces aldehydes, ketones, esters, carboxylic acids, amides and nitriles. NaBH4 is milder and reduces only aldehydes and ketones to alcohols. Both give alcohols from carbonyls, not CH2.
Yes. Both convert C=O to CH2 (a hydrocarbon). The only difference is the medium: Clemmensen is acidic (Zn-Hg/HCl) and Wolff-Kishner is basic (NH2-NH2 then hot KOH). Choose based on which groups in the molecule are acid-safe or base-safe.
First hydrazine (NH2-NH2) reacts with the carbonyl to form a hydrazone. Then this hydrazone is heated with a strong base like KOH or NaOH in a high-boiling solvent such as ethylene glycol. The C=O ends up as CH2.
No. H2 with Ni/Pt/Pd reduces C=O only to an alcohol (C-OH). To reach CH2 you need Clemmensen or Wolff-Kishner. Also, mild H2/Pd reduces a C=C double bond before it touches C=O.
Because of the medium. If the molecule has a group that gets destroyed by acid, use Wolff-Kishner (base). If it has a group that gets destroyed by base, use Clemmensen (acid). The reduction result (CH2) is the same either way.
