100k Followers100k500k Followers500k+1 (510) 706-9331+1 (510) 706-9331
Schedule Your Free Exam Readiness Analysis Session!
Testprepkart Logo
Sign InEnroll NowEnroll
Select an exam to view its content.
  • Blog
  • Download
  • Course
  • Result
  • Video Library
  • Pages
  • Notifications

Loading...

Preparing content

Testprepkart Logo

Enabling students prepare and crack toughest examinations worldwide for over a decade with problem solving aptitude!

Contact Us

Useful Links

  • Connect With Counselor
  • University Admissions
  • Prime Videos
  • Enrollment Form
  • Online Fee Payment
  • Testprepkart Operations
  • Faculty Registration

Our Company

  • Contact Us
  • Work With Us
  • Blogs
  • Facultie
  • Partner

Contact Details

  • Phone: +91 0120 4525484
  • Whatsapp: +1 (510) 706-9331
  • Admission: +91 8800123492
  • E-mail: info@testprepkart.com
  • Head Office: F 377, Sector 63, Noida, Uttar Pradesh, India

Copyright Ā© 2024 CounselKart Educational Services Pvt. Ltd.. All Rights Reserved

Terms of service|Privacy policy|Refund Policy|Login & Register

Aldehydes and Ketones

NEET > Chemistry > Organic Compounds Containing Oxygen

Unit Progress

0%

Overview content

Chapter Snapshot - Aldehydes and Ketones

This chapter covers preparation of aldehydes (oxidation of primary alcohols with PCC/K2Cr2O7, Gattermann-Koch synthesis, Stephen reduction, Rosenmund reduction, Reimer-Tiemann), preparation of ketones (oxidation of secondary alcohols, Friedel-Crafts acylation, acetoacetic ester synthesis), physical properties (boiling points, solubility, dipole moment), chemical reactions (nucleophilic addition with HCN/Grignard/ammonia derivatives, Cannizzaro reaction, aldol condensation, Perkin condensation, benzoin condensation, Baeyer-Villiger oxidation, Beckmann rearrangement, pinacol-pinacolone rearrangement), and analytical tests (Tollen silver mirror, Fehling red precipitate, Schiff test, 2,4-DNP, iodoform test). NEET heavily tests nucleophilic addition mechanisms and distinguishing tests.

āœ“ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
3-5
Aldehyde/ketone chemistry is a core NEET topic with multiple named reactions tested each year.
Time Required (Practical)
ā±
8-10 hrs
Dense chapter with 17+ named reactions and mechanisms.
Difficulty Level
⚔
Hard
Numerous named reactions and subtle mechanistic distinctions require thorough study.
Most Asked Style: Name-the-reaction MCQs, product identification from nucleophilic addition, and aldehyde vs ketone distinguishing tests.Biggest Trap: Confusing which compounds undergo Cannizzaro reaction (no alpha-hydrogen) vs aldol condensation (needs alpha-hydrogen). Both use base but give completely different products.Fast Win: Memorise the ammonia derivative table (hydroxylamine gives oxime, hydrazine gives hydrazone, phenylhydrazine gives phenylhydrazone, semicarbazide gives semicarbazone, 2,4-DNP gives orange precipitate). This table appears as a direct question.Revision-Friendly: Create a single flowchart branching from C=O into nucleophilic addition (HCN, Grignard, NH2-G), condensation (aldol, Perkin, benzoin), and oxidation/reduction paths.

Subtopics - Aldehydes and Ketones (NEET)

Carbonyl chemistry: preparation, nucleophilic addition, named reactions, and analytical tests

Revision tip: Group reactions by type: oxidation (Tollen, Fehling, KMnO4), reduction (LiAlH4, Clemmensen, Wolff-Kishner, MPV), nucleophilic addition (HCN, Grignard, NH2-G, alcohol), C-C bond formation (aldol, Perkin, benzoin, Reformatsky), and rearrangements (Baeyer-Villiger, Beckmann, pinacol-pinacolone).
NCERT LinesMCQsQuick Test

1) Preparation of Aldehydes and Ketones

Aldehydes: oxidation of 1 degree alcohols (PCC for selective aldehyde), Gattermann-Koch (CO + HCl + AlCl3 on ArH), Stephen reduction (RCN + SnCl2/HCl), Rosenmund reduction (RCOCl + H2/Pd-BaSO4), Reimer-Tiemann (phenol + CHCl3/KOH), dry distillation of Ca salts. Ketones: oxidation of 2 degree alcohols, Friedel-Crafts acylation (RCOCl + ArH/AlCl3), organocadmium with acid chloride, acetoacetic ester synthesis.

PCC oxidationRosenmund reductionGattermann-KochFriedel-Crafts acylationAcetoacetic ester
›
Aldehyde preparation methodsPCC in CH2Cl2 selectively oxidises primary alcohol to aldehyde (stops at aldehyde, unlike K2Cr2O7 which goes to acid). Rosenmund: RCOCl + H2 over Pd/BaSO4 poisoned with quinoline gives RCHO. Gattermann-Koch: ArH + CO + HCl with AlCl3 gives ArCHO. Stephen: RCN + SnCl2/HCl gives aldimine then RCHO.
›
Ketone preparation methodsFriedel-Crafts acylation: ArH + RCOCl/AlCl3 gives ArCOR (no rearrangement unlike alkylation). Organocadmium (R2Cd) with RCOCl gives ketone without tertiary alcohol overreaction. Acetoacetic ester synthesis: alkylation then hydrolysis and decarboxylation gives methyl ketone.

2) Nucleophilic Addition Reactions

General mechanism: nucleophile attacks electrophilic carbonyl carbon, then protonation. HCN addition: base-catalysed, gives cyanohydrin (racemic if chiral centre formed). Grignard: gives alcohol (1/2/3 degree depending on carbonyl). Ammonia derivatives: hydroxylamine gives oxime, hydrazine gives hydrazone, phenylhydrazine gives phenylhydrazone, semicarbazide gives semicarbazone, 2,4-DNP gives orange precipitate. Alcohol addition: hemiacetal then acetal (acid-catalysed). Tischenko reaction: Al(OEt)3 converts aldehyde to ester.

HCN cyanohydrinGrignard additionOximeSemicarbazone2,4-DNPAcetal
›
HCN and Grignard additionHCN adds in base-catalysed manner: CN- attacks C=O, then protonation gives cyanohydrin. Aldehydes and aliphatic ketones react well; ArCOAr does not react. Grignard: RMgX adds to C=O, acid hydrolysis gives alcohol. HCHO gives 1 degree, RCHO gives 2 degree, RCOR gives 3 degree.
›
Ammonia derivatives and acetal formationGeneral: C=O + H2N-G gives C=N-G + H2O. Key derivatives: oxime (NH2OH), hydrazone (NH2NH2), phenylhydrazone (NH2NHPh), semicarbazone (NH2NHCONH2), 2,4-DNP (orange/yellow ppt, used for identification). Alcohol addition: RCHO + 2 ROH with H+ gives acetal. Ketones need 1,2-diol for cyclic ketal.

3) Condensation and Named Reactions

Aldol condensation: requires alpha-hydrogen, base-catalysed, gives beta-hydroxy aldehyde then alpha-beta-unsaturated aldehyde. Crossed aldol with different carbonyl compounds. Cannizzaro reaction: aldehydes without alpha-hydrogen undergo disproportionation (one molecule oxidised to acid, other reduced to alcohol). Crossed Cannizzaro (HCHO preferentially reduced). Perkin condensation: ArCHO + acid anhydride gives cinnamic acid derivative. Benzoin condensation: 2 ArCHO + KCN gives alpha-hydroxy ketone. Reformatsky: alpha-bromoester + Zn + C=O gives beta-hydroxy ester.

Aldol condensationCannizzaroPerkinBenzoinReformatsky
›
Aldol and crossed aldol condensationTwo molecules of aldehyde or ketone with alpha-hydrogen in NaOH give beta-hydroxy carbonyl (aldol) which dehydrates to alpha-beta-unsaturated carbonyl. Crossed aldol: one component without alpha-H (like PhCHO or HCHO) accepts, other provides alpha-H.
›
Cannizzaro, Perkin, benzoin, and Reformatsky reactionsCannizzaro: no alpha-H aldehydes (PhCHO, HCHO, CCl3CHO) disproportionate with NaOH. Crossed Cannizzaro: HCHO + PhCHO gives PhCH2OH + HCOONa. Perkin: PhCHO + (CH3CO)2O/CH3COONa gives cinnamic acid. Benzoin: 2 PhCHO + NaCN gives PhCHOHCOPh. Reformatsky: C=O + BrCH2COOEt + Zn gives beta-hydroxy ester.

4) Oxidation, Reduction, and Rearrangements

Oxidation: aldehydes by KMnO4 to acids, Tollen test (silver mirror), Fehling test (red Cu2O). Haloform reaction for methyl ketones. Reduction: to alcohols (LiAlH4, NaBH4, H2/Pt), to hydrocarbons (Clemmensen Zn-Hg/HCl, Wolff-Kishner NH2NH2/KOH). MPV reduction with Al(OiPr)3. Baeyer-Villiger: ketone + peracid gives ester (oxygen insertion). Beckmann: ketoxime + acid gives N-substituted amide (anti migration). Pinacol-pinacolone: 1,2-diol + H+ gives ketone via methyl migration.

Tollen testFehling testClemmensenWolff-KishnerBaeyer-VilligerBeckmann
›
Oxidation tests and haloform reactionTollen: RCHO + Ag(NH3)2+ gives silver mirror (specific for RCHO). Fehling: RCHO + Cu2+ gives red Cu2O precipitate. Neither works for ketones (except alpha-hydroxy ketones). Haloform: CH3COR + NaOI gives CHI3 (yellow) + RCOONa. Iodoform test identifies methyl ketones.
›
Reduction and rearrangement reactionsClemmensen: C=O + Zn-Hg/conc. HCl gives CH2 (acidic conditions). Wolff-Kishner: C=O + NH2NH2/KOH at high temperature gives CH2 (basic conditions). MPV: Al(OiPr)3 reduces C=O selectively (does not touch C=C or NO2). Baeyer-Villiger: RCOR + RCO3H gives ester RCOOR (migratory aptitude: aryl > 3 > 2 > 1 > methyl). Beckmann: oxime + H+ gives amide (anti group migrates). Pinacol-pinacolone: vicinal diol + H+ gives ketone.

5) Analysis and Distinguishing Tests

2,4-DNP test positive for all aldehydes and ketones (orange/yellow precipitate). Tollen test positive for aldehydes only (silver mirror). Fehling test positive for aliphatic aldehydes (red Cu2O). Schiff test positive for aldehydes (magenta colour). Iodoform test positive for methyl ketones and ethanol. Comparison table for systematic identification.

2,4-DNPTollen testFehling testSchiff testIodoform
›
Systematic identification schemeStep 1: 2,4-DNP test to confirm carbonyl. Step 2: Tollen or Fehling test to distinguish aldehyde from ketone. Step 3: Iodoform test to identify methyl ketones. Fehling solution is a mixture of copper sulphate (A) and alkaline tartrate (B); aldehyde reduces Cu2+ to Cu2O (red precipitate).

Aldehydes and Ketones Download Notes & Weightage Plan

For each topic in the Aldehydes and Ketones chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Preparation of Aldehydes and Ketones

PCC selective oxidation, Rosenmund reduction, Gattermann-Koch, Stephen, Friedel-Crafts acylation, acetoacetic ester synthesis.

PCCRosenmundGattermann-KochFriedel-Crafts

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Aldehyde: PCC/CH2Cl2 (stops at RCHO). Rosenmund: RCOCl + H2/Pd-BaSO4 (poisoned). Gattermann-Koch: ArH + CO + HCl/AlCl3. Stephen: RCN + SnCl2/HCl. Ca salt distillation: (RCOO)2Ca + (HCOO)2Ca gives RCHO. Ketone: 2 degree ROH + K2Cr2O7. Friedel-Crafts: ArH + RCOCl/AlCl3. Organocadmium + RCOCl. Acetoacetic ester: alkylate, hydrolyse, decarboxylate gives CH3COCH2R.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Tabulate each method with reagent, product, and mechanism type. Highlight PCC vs K2Cr2O7 selectivity difference.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Named reaction identification for aldehyde/ketone preparation.
Time Required2 hrsMultiple preparation methods to memorise.
DifficultyMediumMostly factual recall of reagents and conditions.
  • Scoring Focus: Rosenmund reduction and Gattermann-Koch are favourite named reaction questions. PCC selectivity is a common comparison question.
  • High-risk Area: Students forget that Rosenmund reduction uses poisoned catalyst (Pd-BaSO4/quinoline). Without poison, further reduction occurs.
  • Best Practice Style: Write equations with all reagents and conditions from memory.
Priority rule: High priority for named preparation methods.

Nucleophilic Addition Reactions

HCN cyanohydrin, Grignard addition, ammonia derivatives table, acetal formation, Tischenko reaction.

HCNGrignardAmmonia derivativesAcetal

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)HCN + C=O (base cat.) gives cyanohydrin (racemic at new stereocentre). Works for RCHO and aliphatic RCOR; fails for ArCOAr. Grignard: HCHO gives 1 degree, RCHO gives 2 degree, RCOR gives 3 degree alcohol. NH2-G derivatives: oxime (NH2OH), hydrazone (NH2NH2), phenylhydrazone, semicarbazone, 2,4-DNP (orange ppt). Acetal: RCHO + 2ROH/H+ (protects -CHO). Ketones need 1,2-diol for ketal.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make the ammonia derivative table with reagent, product name, and product structure. Practice Grignard product prediction.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Product identification from nucleophilic addition is a NEET staple.
Time Required2 hrsMultiple addition types with mechanisms.
DifficultyMediumSystematic once the general mechanism is understood.
  • Scoring Focus: Ammonia derivative names are directly tested. Grignard product class (1/2/3 degree) is a standard question.
  • High-risk Area: Students confuse semicarbazone with hydrazone. Remember: semicarbazide = H2NNHCONH2 (has extra CONH2).
  • Best Practice Style: Memorise the table by writing it out three times without looking.
Priority rule: Highest priority. Nucleophilic addition is the most tested section.

Condensation and Named Reactions

Aldol, Cannizzaro, crossed variants, Perkin, benzoin, Reformatsky.

AldolCannizzaroPerkinBenzoinReformatsky

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Aldol: 2 RCHO (with alpha-H) + NaOH gives beta-hydroxy aldehyde, then dehydrates to alpha-beta unsaturated aldehyde. Cannizzaro: no alpha-H aldehydes (PhCHO, HCHO, (CH3)3CCHO, CCl3CHO) + conc. NaOH gives alcohol + acid salt. Crossed Cannizzaro: HCHO always reduces. Perkin: ArCHO + (CH3CO)2O/CH3COONa gives ArCH=CHCOOH. Benzoin: 2 ArCHO + NaCN gives ArCH(OH)COAr. Reformatsky: Zn + BrCH2COOEt + C=O gives beta-hydroxy ester.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make a decision tree: does the aldehyde have alpha-H? Yes = aldol. No = Cannizzaro. Then learn the crossed variants.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Aldol/Cannizzaro identification and product prediction.
Time Required2 hrsMultiple named reactions with mechanisms.
DifficultyHardRequires understanding of alpha-hydrogen concept and multiple mechanisms.
  • Scoring Focus: Aldol vs Cannizzaro distinction based on alpha-hydrogen is the single most tested concept in this section.
  • High-risk Area: Students apply aldol conditions to formaldehyde (no alpha-H) or Cannizzaro conditions to acetaldehyde (has alpha-H).
  • Best Practice Style: Practice categorising aldehydes: which undergo aldol, which undergo Cannizzaro?
Priority rule: Highest priority. Aldol and Cannizzaro are tested almost every year.

Oxidation, Reduction, and Rearrangements

Tollen, Fehling, haloform. Clemmensen, Wolff-Kishner, MPV. Baeyer-Villiger, Beckmann, pinacol-pinacolone.

TollenFehlingClemmensenWolff-KishnerBaeyer-VilligerBeckmann

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Tollen: RCHO + Ag(NH3)2+ gives Ag mirror. Fehling: RCHO + Cu2+ gives Cu2O (red). Neither for ketones. Haloform: CH3COR + I2/NaOH gives CHI3. Clemmensen: C=O + Zn-Hg/HCl gives CH2 (acid). Wolff-Kishner: C=O + NH2NH2/KOH/high T gives CH2 (base). MPV: Al(OiPr)3 selective. Baeyer-Villiger: RCOR + mCPBA gives ester (migration: Ar > 3 > 2 > 1 > Me). Beckmann: oxime/H+ gives amide (anti migrates). Pinacol-pinacolone: 1,2-diol/H+ gives ketone.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Pair complementary reactions: Clemmensen (acid) vs Wolff-Kishner (base) for the same transformation. Write Baeyer-Villiger migratory aptitude order.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Reduction method selection or rearrangement product identification.
Time Required2 hrsMultiple named reactions and rearrangement mechanisms.
DifficultyHardRearrangements require understanding of migratory aptitude and stereospecificity.
  • Scoring Focus: Clemmensen vs Wolff-Kishner conditions are directly asked. Baeyer-Villiger migratory aptitude is a conceptual favourite.
  • High-risk Area: Using Clemmensen (strongly acidic) on acid-sensitive substrates, or Wolff-Kishner (strongly basic) on base-sensitive substrates.
  • Best Practice Style: For each substrate, decide: use Clemmensen (acid-stable) or Wolff-Kishner (base-stable)?
Priority rule: High priority for Clemmensen/Wolff-Kishner and Baeyer-Villiger.

Analysis and Distinguishing Tests

2,4-DNP for all carbonyls, Tollen for aldehydes, Fehling for aliphatic aldehydes, Schiff for aldehydes, iodoform for methyl ketones.

2,4-DNPTollenFehlingSchiffIodoform

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)2,4-DNP: positive for all RCHO and RCOR (orange/yellow ppt). Tollen: RCHO gives Ag mirror (also alpha-hydroxy ketones and HCOOH). Fehling: aliphatic RCHO gives Cu2O red ppt (benzaldehyde does not reduce Fehling). Schiff: RCHO gives magenta (highly sensitive). Iodoform: CH3COR + I2/NaOH gives CHI3 (yellow). Systematic: 2,4-DNP first, then Tollen/Fehling, then iodoform.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write a flowchart: unknown compound tested with 2,4-DNP (carbonyl?), then Tollen (aldehyde?), then iodoform (methyl ketone?).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Test identification or exception-based question.
Time Required1 hrFactual recall of test reagents and results.
DifficultyEasyDirect recall once the test table is memorised.
  • Scoring Focus: Fehling test fails for aromatic aldehydes (benzaldehyde). This exception is frequently tested.
  • High-risk Area: Students think all aldehydes give positive Fehling test. Benzaldehyde does not reduce Fehling solution.
  • Best Practice Style: Memorise the exceptions: benzaldehyde fails Fehling but passes Tollen.
Priority rule: Medium priority. Distinguishing tests are commonly asked but straightforward.

Aldehydes and Ketones Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Aldehydes and Ketones chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
Aldol vs Cannizzaro: Alpha-Hydrogen Rule
aldolCannizzaroalpha-hydrogencondensation

Mistake Snapshot (What Students Do Wrong)

  • Applying aldol to no-alpha-H aldehydes: Aldol condensation requires at least one component with alpha-hydrogen. Formaldehyde, benzaldehyde, and trimethylacetaldehyde have no alpha-H and undergo Cannizzaro reaction instead.
  • Applying Cannizzaro to acetaldehyde: Acetaldehyde has alpha-hydrogen and undergoes aldol condensation with NaOH, not Cannizzaro reaction.
2–3 Line Example (Typical Error)

Benzaldehyde (no alpha-H) + conc. NaOH gives benzyl alcohol + sodium benzoate (Cannizzaro). Acetaldehyde (has alpha-H) + dil. NaOH gives 3-hydroxybutanal (aldol).

How NEET Frames The Trap

NEET gives NaOH with an aldehyde and asks the product. Students must first check for alpha-hydrogen to decide between aldol and Cannizzaro.

NEET-Style Trap Question Format

Q. Formaldehyde on treatment with concentrated NaOH gives:
A. Aldol product   B. Methanol + sodium formate   C. Cannizzaro product and aldol product both   D. No reaction  
Trick: Methanol + sodium formate via Cannizzaro reaction because formaldehyde has no alpha-hydrogen. One HCHO is oxidised to HCOONa and the other is reduced to CH3OH.

Quick rule: Alpha-H present = aldol condensation. No alpha-H = Cannizzaro reaction. Never mix the two.
Fehling Test: Benzaldehyde Exception
Fehling testbenzaldehydeTollen testdistinguishing

Mistake Snapshot (What Students Do Wrong)

  • Thinking all aldehydes reduce Fehling: Benzaldehyde does NOT reduce Fehling solution. Only aliphatic aldehydes and formic acid give a positive Fehling test (red Cu2O precipitate). Benzaldehyde does reduce Tollen reagent (silver mirror).
  • Confusing Tollen and Fehling selectivity: Tollen test is positive for all aldehydes (including benzaldehyde). Fehling is positive only for aliphatic aldehydes. This distinction is a classic NEET question.
2–3 Line Example (Typical Error)

Benzaldehyde + Tollen reagent gives silver mirror (positive). Benzaldehyde + Fehling solution gives no reaction. Acetaldehyde gives positive result with both.

How NEET Frames The Trap

A question asks which aldehyde does NOT reduce Fehling solution. Students who assume all aldehydes pass Fehling choose incorrectly.

NEET-Style Trap Question Format

Q. Which of the following does NOT give a positive Fehling test?
A. Formaldehyde   B. Acetaldehyde   C. Benzaldehyde   D. Propanal  
Trick: Benzaldehyde does not reduce Fehling solution because aromatic aldehydes cannot be oxidised by the mild alkaline Cu2+ reagent. However, it does give a positive Tollen test.

Quick rule: Tollen is universal for aldehydes. Fehling fails for aromatic aldehydes (benzaldehyde).
Clemmensen vs Wolff-Kishner Conditions
ClemmensenWolff-Kishnerreductionconditions

Mistake Snapshot (What Students Do Wrong)

  • Using the wrong reduction for acid/base sensitive substrates: Clemmensen uses strongly acidic conditions (Zn-Hg/conc. HCl). Wolff-Kishner uses strongly basic conditions (NH2NH2/KOH/high temperature). Acid-sensitive groups survive Wolff-Kishner; base-sensitive groups survive Clemmensen.
  • Thinking both give different products: Both convert C=O to CH2 (same product). The choice depends on substrate compatibility with acid or base, not on the desired product.
2–3 Line Example (Typical Error)

For a substrate with an acid-labile protecting group, use Wolff-Kishner (basic). For a substrate with a base-labile ester group, use Clemmensen (acidic).

How NEET Frames The Trap

NEET gives a substrate with an acid- or base-sensitive group and asks which reduction method to use. Students who do not consider side reactions choose incorrectly.

NEET-Style Trap Question Format

Q. To reduce a carbonyl group to CH2 in a compound containing an ester group, the preferred method is:
A. Wolff-Kishner reduction   B. Clemmensen reduction   C. Catalytic hydrogenation   D. LiAlH4 reduction  
Trick: Clemmensen reduction because esters are base-sensitive and would hydrolyse under Wolff-Kishner conditions. Clemmensen uses acidic conditions that leave esters intact.

Quick rule: Clemmensen = acid conditions (Zn-Hg/HCl). Wolff-Kishner = base conditions (NH2NH2/KOH). Choose based on what the substrate can tolerate.
Baeyer-Villiger Migratory Aptitude
Baeyer-Villigermigratory aptituderearrangement

Mistake Snapshot (What Students Do Wrong)

  • Wrong group migrates: In Baeyer-Villiger oxidation, the group with higher migratory aptitude migrates to oxygen. Order: aryl > tertiary > secondary > primary > methyl. Students often assume the smaller group migrates.
  • Forgetting ring expansion in cyclic ketones: Cyclic ketones give lactones (ring expanded by one atom) via Baeyer-Villiger. Cyclopentanone gives a six-membered lactone.
2–3 Line Example (Typical Error)

PhCOCH3 + mCPBA gives PhOCOCH3 (phenyl migrates, not methyl) because aryl has higher migratory aptitude than methyl.

How NEET Frames The Trap

A question gives an unsymmetrical ketone and asks for the Baeyer-Villiger product. Students who pick the wrong migrating group get the wrong ester.

NEET-Style Trap Question Format

Q. Baeyer-Villiger oxidation of methyl phenyl ketone (acetophenone) with mCPBA gives:
A. Methyl benzoate   B. Phenyl acetate   C. Benzoic acid + methanol   D. No reaction  
Trick: Phenyl acetate (CH3COOPh) because the phenyl group has higher migratory aptitude than methyl and migrates to oxygen, placing it next to the ester oxygen.

Quick rule: The better migrator (aryl > 3 degree > 2 degree > 1 degree > Me) inserts between C=O and the migrating group.
Crossed Cannizzaro with Formaldehyde
Cannizzaroformaldehydecrossed reaction

Mistake Snapshot (What Students Do Wrong)

  • Wrong molecule gets oxidised/reduced: In crossed Cannizzaro between HCHO and another no-alpha-H aldehyde, HCHO is always the one that gets oxidised to formate. The other aldehyde gets reduced to its alcohol.
  • Thinking both undergo self-Cannizzaro: In a mixture, crossed Cannizzaro is faster than self-Cannizzaro. HCHO is a better hydride donor, so it preferentially gets oxidised.
2–3 Line Example (Typical Error)

PhCHO + HCHO + NaOH gives PhCH2OH (benzyl alcohol) + HCOONa (sodium formate). HCHO is oxidised; PhCHO is reduced.

How NEET Frames The Trap

NEET gives a mixture of HCHO and another aldehyde with NaOH. Students must identify which is oxidised and which is reduced.

NEET-Style Trap Question Format

Q. When a mixture of benzaldehyde and formaldehyde is treated with concentrated NaOH, the products are:
A. Benzyl alcohol + sodium formate   B. Sodium benzoate + methanol   C. Benzyl alcohol + sodium benzoate   D. Methanol + sodium formate  
Trick: Benzyl alcohol + sodium formate because in crossed Cannizzaro, formaldehyde acts as the hydride donor (oxidised to formate) and benzaldehyde is reduced to benzyl alcohol.

Quick rule: In crossed Cannizzaro, formaldehyde always gets oxidised to formate. The other aldehyde gets reduced to its alcohol.
Previous
Alcohol, Phenol and Ether
Next
Carboxylic Acid and Their Derivatives

Loading tests...

NEET > Chemistry > Organic Compounds Containing Oxygen Chapters

Review your status and progress for each chapter in this unit. Use the slider to set progress or click "Mark as Done" to complete.

ChapterStatusProgress

Alcohol, Phenol and Ether

Weightage: 02.2K
0%

Aldehydes and Ketones

Weightage: 02.2K
0%

Carboxylic Acid and Their Derivatives

Weightage: 02.2K
0%

Comments

Leave a comment

0/2000Comments are moderated

You can comment without logging in. We'll ask for your name and email before submitting.

Comments (0)

No comments yet. Be the first to comment!