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Reducing Aldehydes, Ketones and Acids to Alcohols

Chemistry · Alcohols, Phenols And Ethers · NEET

To make alcohols, you add hydrogen to a C=O group. An aldehyde becomes a primary (1°) alcohol, a ketone becomes a secondary (2°) alcohol, and a carboxylic acid becomes a primary (1°) alcohol. Memory hook: "count the carbons on the C=O carbon" — aldehyde has 1 carbon side, so it stays low (1°); ketone has 2 carbon sides, so it goes to 2°.
Reduction of Carbonyls and Acids to AlcoholsAldehydeR-CHO+ H2 (Ni/Pt/Pd)NaBH4 or LiAlH4R-CH2-OH1° alcoholKetoneR-CO-R'+ H2 or NaBH4 / LiAlH4R-CH(OH)-R'2° alcoholAcidR-COOHLiAlH4 or B2H6(NaBH4 fails)R-CH2-OH1° alcohol
Aldehydes and acids reduce to 1° alcohols; ketones reduce to 2° alcohols. NaBH4 works on aldehydes and ketones only — acids need LiAlH4 or B2H6.

Your doubts, answered

Does an aldehyde give a primary or a secondary alcohol when reduced?

An aldehyde gives a PRIMARY (1°) alcohol. Look at the carbonyl carbon in R-CHO: it is joined to only ONE carbon (the R) and one H. When you add H to the C=O, that carbon ends up bonded to R, two H atoms and the OH, i.e. R-CH2-OH, which is a 1° alcohol. This is a very common NEET clue: aldehyde to 1°, ketone to 2°.

Does NaBH4 reduce carboxylic acids to alcohols?

No. NaBH4 (sodium borohydride) is a MILD reducing agent. It reduces aldehydes and ketones to alcohols, but it CANNOT reduce carboxylic acids (-COOH) or esters. To reduce a -COOH group to a -CH2OH primary alcohol you must use a STRONG reducing agent like LiAlH4 (lithium aluminium hydride) or diborane B2H6. NEET loves this exact trap: if the question shows R-COOH going to R-CH2OH, NaBH4 is the wrong option.

Which reagents can reduce a carboxylic acid straight to a 1° alcohol?

Two work directly: LiAlH4 (strong hydride donor) and B2H6 / BH3 (diborane). Both convert R-COOH to R-CH2OH. Note that B2H6 is fast with acids and slow with esters. A softer, indirect route used in some NEET questions is to first make the ESTER (R-COOH + CH3OH, H+) and then reduce the ester with H2/catalyst — this is why one 2026 match question maps acid to alcohol through esterification then H2.

What is the difference between LiAlH4 and NaBH4?

Both add H to C=O. NaBH4 is mild: it only touches aldehydes and ketones. LiAlH4 is strong: it reduces aldehydes, ketones, AND carboxylic acids and esters down to alcohols. Simple rule for NEET: if the substrate is only an aldehyde or ketone, either reagent works; if a -COOH or -COOR must become an alcohol, only LiAlH4 (or B2H6) works, not NaBH4.

What does H2 with a catalyst (Ni, Pt, Pd) do to a ketone?

Catalytic hydrogenation, H2 with finely divided Ni, Pt or Pd, adds H across the C=O bond and turns a ketone into a secondary (2°) alcohol and an aldehyde into a primary (1°) alcohol. Warning: if the molecule also has a C=C double bond, H2/catalyst will reduce that too. If you want to reduce ONLY the carbonyl and keep a C=C, use NaBH4 or LiAlH4 instead.

Why does a ketone give a 2° alcohol and not a 3° alcohol?

In reduction you only ADD one hydrogen to the carbonyl carbon; you do not add a new carbon. A ketone R-CO-R' has two carbons attached to the carbonyl carbon. After adding H and forming OH you get R-CH(OH)-R', which has the OH carbon bonded to two other carbons = secondary. You only reach a 3° alcohol when a NEW carbon group is added, e.g. by a Grignard reagent, not by reduction.

⚠️ The NEET trap
✗ Picking NaBH4 to convert R-COOH into R-CH2OH because 'NaBH4 makes alcohols'.
✓ NaBH4 cannot reduce a carboxylic acid. Use LiAlH4 or B2H6 (diborane) to turn -COOH into a 1° alcohol.
🧠 NaBH4 = soft hands (aldehydes/ketones only). LiAlH4 / B2H6 = strong hands (also acids and esters).

Real NEET questions

NEET 2023 Phase 2

Identify 'X' in the following reactions: R-COOH -[(i) X][(ii) H2O/HCl]-> R-CH2OH R-CH=CH2 -[(i) X][(ii) H2O, NaOH, H2O2]-> R-CH2-CH2-OH

A · NaBH4
B · H2/Pd
C · B2H6 ✓
D · LiAlH4
Solution: Diborane B2H6 does BOTH jobs. It reduces a carboxylic acid R-COOH to the primary alcohol R-CH2OH, and it also does hydroboration of the alkene, which on oxidation with H2O2/OH- gives the anti-Markovnikov 1° alcohol R-CH2-CH2-OH. NaBH4 cannot reduce -COOH at all. LiAlH4 reduces -COOH but does NOT do the alkene hydroboration step, so it fails the second reaction. Only B2H6 explains both, so X = B2H6.
NEET 2026

Match the transformation with the reagent set. For B: CH3COOH -> CH3CH2OH, which reagent set is correct?

A · (i) O2; (ii) H2O/H+
B · (i) CH3OH, H+; (ii) H2, catalyst ✓
C · (i) conc. H2SO4, heat; (ii) H+/H2O
D · (i) Oleum; (ii) NaOH heat; (iii) H+
Solution: Acetic acid CH3COOH is turned into ethanol CH3CH2OH by an INDIRECT reduction. First the acid is esterified with CH3OH and H+ to give the ester (methyl acetate). Then the ester is reduced with H2 over a catalyst to give the primary alcohol CH3CH2OH. So the reagent set is (i) CH3OH/H+ then (ii) H2/catalyst. The other sets make phenol (O2 route, oleum route) or cause dehydration (conc. H2SO4).
NEET 2023 Phase 2

For 2-(2-oxopropyl)cyclopentan-1-one (a cyclopentanone ring with a -CH2-CO-CH3 side chain), reagents (i) NaBH4 then (ii) H2SO4, heat give what type of first-step change?

A · Only the ring ketone is reduced
B · Only the side-chain ketone is reduced
C · Both carbonyl groups are reduced to alcohols (a diol) ✓
D · No reduction occurs
Solution: NaBH4 reduces BOTH C=O groups because both are ketones, and NaBH4 handles aldehydes and ketones. So step (i) gives a diol: the ring ketone becomes a ring 2° alcohol (cyclopentanol) and the side-chain ketone -CH2-CO-CH3 becomes -CH2-CH(OH)-CH3. Step (ii) then dehydrates both alcohols with H2SO4/heat to give the conjugated alkene product. The key reduction idea: NaBH4 turns every ketone present into a 2° alcohol.

Solved Alcohols, Phenols And Ethers NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 33 Alcohols, Phenols And Ethers NEET PYQs ›
Next concept: Alcohols with HX, PCl5 and PCl3 →Keep learning — 2 minFeeling ready? Solve the Alcohols, Phenols And Ethers NEET PYQs ›Or practice on your phone — get the free MedicNEET app ›

Frequently asked

What alcohol type does each carbonyl give on reduction?

Aldehyde (R-CHO) gives a primary (1°) alcohol R-CH2OH. Ketone (R-CO-R') gives a secondary (2°) alcohol R-CH(OH)-R'. Carboxylic acid (R-COOH) gives a primary (1°) alcohol R-CH2OH. No reduction gives a 3° alcohol; for that you need a Grignard reagent.

Which reducing agents are safe to memorise for NEET?

H2 with Ni/Pt/Pd (catalytic hydrogenation) reduces aldehydes and ketones (and also any C=C present). NaBH4 reduces aldehydes and ketones only. LiAlH4 reduces aldehydes, ketones, carboxylic acids and esters. B2H6 (diborane) reduces carboxylic acids fast and also does hydroboration of alkenes.

Why can't NaBH4 reduce esters and acids but LiAlH4 can?

LiAlH4 is a much stronger hydride donor than NaBH4. The Al-H bond releases hydride more easily than the B-H bond of NaBH4, so LiAlH4 can attack the less reactive -COOH and -COOR carbonyls, while NaBH4 is too mild and stops at aldehydes and ketones.

If a molecule has both C=C and C=O, how do I reduce only the C=O?

Use NaBH4 or LiAlH4. These hydride reagents attack the polar C=O but leave the non-polar C=C double bond untouched. Do NOT use H2 with a metal catalyst, because that would also reduce the C=C.

Related concepts:
Making Alcohols with Grignard ReagentsOxidation of Alcohols to Aldehydes, Ketones and AcidsPreparing Alcohols from Alkenes: Hydration vs HydroborationPrimary, Secondary and Tertiary Alcohols ExplainedAlcohols with HX, PCl5 and PCl3: Making Alkyl Halides
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Alcohols, Phenols And Ethers PYQs