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NEET Organic Compounds Containing Oxygen MCQs with answers

313 questions on Organic Compounds Containing Oxygen for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2021

1. Test that distinguishes aldehydes from ketones:

  1. Lucas test
  2. Tollens test
  3. Iodoform test
  4. Baeyer's test
Show answer and explanation

Answer: (B) Tollens test

Tollens (ammoniacal AgNO₃) → silver mirror with aldehydes only.

Quick tip: Tollens=silver mirror=aldehyde. Iodoform=CHI₃=CH₃CO- group.

NCERT Class 12, Chapter 12, page 344

easy

2. Which of the following compounds will not give a yellow precipitate when treated with sodium hypoiodite (I₂/NaOH)?

  1. Acetophenone
  2. Pentan-2-one
  3. Pentan-3-one
  4. Ethanal
Show answer and explanation

Answer: (C) Pentan-3-one

The iodoform test is given by compounds containing a CH₃-C=O or CH₃-CH(OH)- group. Pentan-3-one has the structure CH₃CH₂COCH₂CH₃, which lacks a methyl carbonyl group, and therefore fails to give the yellow precipitate of CHI₃.

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3. An alkene 'A' (C₆H₁₂) upon reductive ozonolysis with O₃ followed by Zn/H₂O gives an equimolar mixture of methanal and an aldehyde 'B'. Aldehyde 'B' does not undergo self-aldol condensation due to lack of α-hydrogens. What is the IUPAC name of alkene 'A'?

  1. 4-Methylpent-1-ene
  2. 2,3-Dimethylbut-1-ene
  3. 3,3-Dimethylbut-1-ene
  4. 2,2-Dimethylbut-3-ene
Show answer and explanation

Answer: (C) 3,3-Dimethylbut-1-ene

Since reductive ozonolysis produces methanal (HCHO) and an aldehyde B, the alkene must have a terminal =CH₂ group (hex-1-ene skeleton). Because B lacks α-hydrogens and contains 5 carbons (total 6 - 1 = 5), B must be 2,2-dimethylpropanal, (CH₃)₃C-CHO. Therefore, alkene 'A' is 3,3-dimethylbut-1-ene, (CH₃)₃C-CH=CH₂.

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4. When 2,6-dimethylphenol is subjected to the Reimer-Tiemann reaction conditions (CHCl₃ + aqueous NaOH, followed by acidification), what is the major organic product formed?

  1. 3,5-Dimethylbenzoic acid
  2. 2-Hydroxy-3,5-dimethylbenzaldehyde
  3. 4-Hydroxy-3,5-dimethylbenzaldehyde
  4. 2,6-Dimethylbenzyl alcohol
Show answer and explanation

Answer: (C) 4-Hydroxy-3,5-dimethylbenzaldehyde

In 2,6-dimethylphenol, both ortho positions relative to the phenolic -OH group are occupied by methyl groups. Consequently, the electrophilic attack of dichlorocarbene (:CCl₂) is directed exclusively to the vacant para position, yielding 4-hydroxy-3,5-dimethylbenzaldehyde upon alkaline hydrolysis and acidification.

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5. Given below are two statements: Assertion (A): Fehling's test cannot be used to distinguish benzaldehyde from acetophenone. Reason (R): Both benzaldehyde and acetophenone fail to reduce Fehling's solution.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
Show answer and explanation

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Benzaldehyde is an aromatic aldehyde and acetophenone is an aromatic ketone. Aromatic aldehydes are too weak as reducing agents to reduce Cu²⁺ complexes in Fehling's solution, and ketones do not react either. Since both give a negative Fehling's test, the test cannot distinguish between them.

easy

6. Which of the following carboxylic acids has the highest value of pKa?

  1. 4-Nitrobenzoic acid
  2. 4-Methoxybenzoic acid
  3. 4-Cyanobenzoic acid
  4. Benzoic acid
Show answer and explanation

Answer: (B) 4-Methoxybenzoic acid

A higher pKa corresponds to a weaker acid. The methoxy group (-OCH₃) at the para position acts as a strong electron-donating group through resonance (+R > -I), which destabilizes the carboxylate anion and decreases the acidity compared to unsubstituted benzoic acid and acids with electron-withdrawing groups (-NO₂, -CN).

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7. When cyclohexylmethanol is heated with concentrated H₂SO₄, it undergoes dehydration accompanied by ring expansion. The IUPAC name of the major product is:

  1. Cycloheptene
  2. Methylenecyclohexane
  3. 1-Methylcyclohex-1-ene
  4. Cyclohexene
Show answer and explanation

Answer: (A) Cycloheptene

Loss of water from protonated cyclohexylmethanol generates a primary carbocation attached to the cyclohexyl ring. A 1,2-bond shift (ring expansion) occurs to relieve ring strain and produce a secondary cycloheptyl carbocation. Loss of a proton then yields cycloheptene as the major rearranged product.

easy

8. In diethyl ether, the oxygen atom uses which type of hybrid orbitals to form σ-bonds with carbon atoms?

  1. sp³ hybrid orbitals
  2. sp² hybrid orbitals
  3. sp hybrid orbitals
  4. dsp² hybrid orbitals
Show answer and explanation

Answer: (A) sp³ hybrid orbitals

In ethers such as diethyl ether, the central oxygen atom is surrounded by two bond pairs and two lone pairs, adopting a steric number of 4. Hence, oxygen is sp³ hybridized and utilizes sp³ hybrid orbitals to form σ-bonds with carbon.

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9. What is the total number of constitutionally isomeric secondary (2°) alcohols possible for the molecular formula C₅H₁₂O?

  1. 2
  2. 3
  3. 4
  4. 5
Show answer and explanation

Answer: (B) 3

The secondary alcohols with formula C₅H₁₂O are: (1) pentan-2-ol, (2) pentan-3-ol, and (3) 3-methylbutan-2-ol. Thus, exactly 3 constitutional secondary isomers exist.

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10. The s-character of the hybrid orbital of the ring carbon atom directly bonded to the oxygen atom in phenetole (ethoxybenzene) is:

  1. 25%
  2. 33.3%
  3. 50%
  4. 75%
Show answer and explanation

Answer: (B) 33.3%

In phenetole (C₆H₅-O-C₂H₅), the aromatic carbon attached to oxygen is sp² hybridized. An sp² hybrid orbital contains one part s and two parts p, giving an s-character of 1/3 or approximately 33.3%.

easy

11. The correct decreasing order of acidity for the following substituted propanoic acids is: (I) 2-Fluoropropanoic acid (II) 3-Fluoropropanoic acid (III) 2-Chloropropanoic acid (IV) Propanoic acid

  1. I > III > II > IV
  2. I > II > III > IV
  3. III > I > II > IV
  4. II > I > III > IV
Show answer and explanation

Answer: (A) I > III > II > IV

Inductive effect decreases sharply with increasing distance from the carboxyl carbon. At carbon-2 (α-position), fluorine exerts a stronger -I effect than chlorine, making 2-fluoropropanoic acid stronger than 2-chloropropanoic acid. Fluorine at carbon-3 (β-position) has a weaker -I effect on the -COOH group than at carbon-2, but is still stronger than unsubstituted propanoic acid. Hence, the order is I > III > II > IV.

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12. Given below are two statements: Statement I: Primary alcohols undergo acid-catalysed dehydration predominantly via an E1 mechanism involving a free carbocation intermediate. Statement II: Secondary and tertiary alcohols undergo acid-catalysed dehydration via carbocation intermediates. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Primary alcohols typically undergo dehydration via an E2 pathway without forming an unstable primary carbocation intermediate (protonated alcohol loses water and a proton simultaneously). Secondary and tertiary alcohols proceed through carbocation intermediates via an E1 pathway. Hence, Statement I is incorrect and Statement II is correct.

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13. An organic compound 'A' with formula C₄H₈O gives a positive test with 2,4-dinitrophenylhydrazine and forms a white crystalline adduct with saturated NaHSO₃. However, it does not reduce ammoniacal AgNO₃ solution. Compound 'A' is:

  1. Butanal
  2. 2-Methylpropanal
  3. Cyclobutanol
  4. Butan-2-one
Show answer and explanation

Answer: (D) Butan-2-one

Reaction with 2,4-DNP and NaHSO₃ indicates the presence of a carbonyl group (aldehyde or methyl/aliphatic ketone). The compound does not reduce ammoniacal AgNO₃ (Tollens' reagent), confirming it is a ketone, not an aldehyde. Butan-2-one has the formula C₄H₈O and fulfills all these observations.

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14. 1 mol of ethane-1,2-diol reacts completely with excess sodium metal. The volume of hydrogen gas evolved at STP (22.4 L/mol) is:

  1. 11.2 L
  2. 33.6 L
  3. 22.4 L
  4. 44.8 L
Show answer and explanation

Answer: (C) 22.4 L

Each −OH gives ½ H₂. ethane-1,2-diol has 2 −OH, so moles H₂ = 1 × 2/2 = 1; volume = 22.4 L.

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15. A chemist is presented with three unlabelled bottles containing pentanal, pentan-2-one, and pentan-3-one. Which sequence of tests will definitively identify each compound?

  1. Schiff's test followed by Fehling's test
  2. Tollens' test followed by 2,4-DNP test
  3. Fehling's test followed by Iodoform test
  4. NaHSO₃ test followed by 2,4-DNP test
Show answer and explanation

Answer: (C) Fehling's test followed by Iodoform test

Fehling's test will yield a red precipitate with pentanal (aliphatic aldehyde), distinguishing it from the two ketones. Then, the iodoform test (I₂/NaOH) gives a yellow precipitate with pentan-2-one (a methyl ketone) but not with pentan-3-one, identifying all three compounds uniquely.

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