Organic Compounds Containing Oxygen — NEET Chemistry MCQs with answers

Organic Compounds Containing Oxygen is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 1,107 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Alcohols · Phenols · Ethers · Aldehydes and ketones · Carboxylic acids and derivatives · Carboxylic acids · Carboxylic acids and derivatives-in-syllabus · Aldehydes · Ketones · Aldehydes, Ketones & Carboxylic Acids · Alcohols, Phenols & Ethers · Alcohols · Phenols · Ethers · Aldehydes and Ketones · Carboxylic acids · Carbonyl Compounds · Acidity comparison · Distinguishing tests · Preparation · Reimer-Tiemann reaction · Boiling point trends · Ether cleavage · Effect of substituents on phenol acidity · Dehydration of alcohols · Preparation of phenols · Nucleophilic addition · Aldol condensation · Cannizzaro reaction · Acidity of carboxylic acids · Haloform reaction · Distinguishing tests · HVZ reaction · Clemmensen and Wolff-Kishner reduction · Decarboxylation · Carboxylic acids · Carboxylic acid derivatives · Esters · Phenols and ethers · Alcohols and esters

8 free Organic Compounds Containing Oxygen practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Alcohols

   Which functional-group feature distinguishes a tertiary alcohol from primary and secondary alcohols?

   • A.The carbon bearing -OH is attached to three other carbon atoms
   • B.The carbon bearing -OH is attached to exactly two hydrogen atoms
   • C.It always has a lower boiling point than the corresponding primary alcohol due to weaker H-bonding, which is used to define it
   • D.It contains two -OH groups on adjacent carbons
   Show answer and explanation

   Answer: A. The carbon bearing -OH is attached to three other carbon atoms

   Alcohols are classified 1°, 2°, 3° by the number of carbon atoms directly bonded to the carbinol carbon (the C-OH carbon): one, two, and three respectively.

   The carbinol carbon in a tertiary alcohol is bonded to three other carbon atoms (and no H). In a primary alcohol it is bonded to one carbon (and two H), in a secondary alcohol to two carbons (and one H). This classification governs reactivity in Lucas test, oxidation, and dehydration ease.

   Common mistake: Confusing the number of H atoms on the carbinol carbon with the number of carbon substituents

2. Question 2 · medium · Alcohols

   Why is phenol considerably more acidic than cyclohexanol, even though both have an -OH group attached to a six-membered carbon ring?

   • A.Phenol has a lower molecular weight, and lighter molecules are generally more acidic
   • B.The aromatic ring in phenol donates electron density to oxygen through the sigma framework, weakening the O-H bond
   • C.Cyclohexanol's -OH is more sterically hindered, making its O-H bond harder to break
   • D.The phenoxide ion formed from phenol is stabilised by resonance delocalisation of the negative charge into the aromatic ring, while the cyclohexoxide ion has no such delocalisation
   Show answer and explanation

   Answer: D. The phenoxide ion formed from phenol is stabilised by resonance delocalisation of the negative charge into the aromatic ring, while the cyclohexoxide ion has no such delocalisation

   Phenol's conjugate base, phenoxide, spreads its negative charge onto the ortho/para ring carbons via resonance, stabilising it; cyclohexanol's alkoxide has no such delocalisation, so phenol is far more acidic (pKa ~10) than cyclohexanol (pKa ~16-18).

   In phenoxide, the oxygen lone pair (or negative charge) is in conjugation with the aromatic pi system, giving resonance structures with negative charge at ortho and para ring carbons; this delocalisation lowers the energy of the conjugate base substantially. Cyclohexanol's -OH is on an sp3 carbon with no adjacent pi system, so its alkoxide ion carries the full negative charge localised on oxygen with no resonance stabilisation. Consequently phenol (pKa ~10) is far more acidic than cyclohexanol (pKa ~16-18), and even more acidic than water in some contexts, despite cyclohexanol also being a six-membered ring alcohol.

   Common mistake: Treating phenol as 'just another cyclic alcohol' and ignoring the resonance available specifically because the ring is aromatic

3. Question 3 · medium · Alcohols

   A student predicts that treating propan-2-ol with PCC (pyridinium chlorochromate) will give no reaction, "because PCC only works on primary alcohols to stop at the aldehyde stage." Evaluate this claim.

   • A.Incorrect — PCC also oxidises secondary alcohols, converting propan-2-ol to propanone (acetone)
   • B.Correct — PCC is unreactive toward secondary alcohols and only oxidises primary alcohols
   • C.Incorrect — PCC would instead over-oxidise propan-2-ol all the way to propanoic acid via C-C cleavage
   • D.Correct, and no oxidation is possible with any Cr(VI) reagent since propan-2-ol lacks an alpha-hydrogen
   Show answer and explanation

   Answer: A. Incorrect — PCC also oxidises secondary alcohols, converting propan-2-ol to propanone (acetone)

   PCC oxidises both primary alcohols (to aldehydes) and secondary alcohols (to ketones); it is mild and stops at the carbonyl stage in both cases, so propan-2-ol is readily converted to propanone.

   PCC (pyridinium chlorochromate) is a widely used mild, anhydrous Cr(VI) oxidant. Its defining feature is that it avoids over-oxidation of primary alcohols to carboxylic acids (stopping at the aldehyde) because no water is present for the aldehyde-hydrate pathway; it is not restricted to primary alcohols. Secondary alcohols like propan-2-ol are oxidised normally to ketones (propanone here), since ketones are already the natural stopping point (no further oxidation is possible without C-C cleavage). The student's claim incorrectly narrows PCC's substrate scope.

   Common mistake: Over-restricting PCC's role to only primary-alcohol-to-aldehyde conversions

4. Question 4 · medium · Alcohols

   Which statement correctly explains why the acid-catalysed hydration of alkenes (Markovnikov addition of water) is generally NOT a good laboratory method for preparing primary alcohols from simple terminal alkenes?

   • A.The mechanism proceeds through the more stable (more substituted) carbocation, so -OH is delivered preferentially to the more substituted carbon, giving secondary/tertiary alcohols instead of primary
   • B.Acid-catalysed hydration only works with aromatic alkenes, not simple terminal alkenes
   • C.Water cannot act as a nucleophile toward carbocations, so no addition occurs at all
   • D.Terminal alkenes are too unreactive toward protonation to undergo this reaction
   Show answer and explanation

   Answer: A. The mechanism proceeds through the more stable (more substituted) carbocation, so -OH is delivered preferentially to the more substituted carbon, giving secondary/tertiary alcohols instead of primary

   Acid-catalysed hydration proceeds via the more stable carbocation (Markovnikov), so it delivers -OH to the more substituted carbon; a terminal alkene like propene thus gives the secondary alcohol, not the primary one, unlike anti-Markovnikov hydroboration-oxidation.

   In acid-catalysed hydration, protonation of the alkene generates the more stable carbocation (secondary or tertiary preferred over primary), and water then attacks this carbocation. For a terminal alkene like propene, this means the secondary carbocation forms preferentially, so water adds to the internal carbon, giving propan-2-ol (secondary), not propan-1-ol (primary). To obtain primary (anti-Markovnikov) alcohols from terminal alkenes, hydroboration-oxidation is used instead, which proceeds via a concerted, non-carbocation mechanism placing -OH on the less substituted carbon.

   Common mistake: Assuming acid-catalysed hydration of any terminal alkene gives the primary alcohol by analogy with simple addition across the double bond

5. Question 5 · easy · Alcohols

   In 2-methylpropan-2-ol (tert-butyl alcohol), the carbon bearing the –OH group is directly bonded to how many other carbon atoms, and how is the alcohol classified?

   • A.3 carbons; tertiary alcohol
   • B.2 carbons; secondary alcohol
   • C.1 carbon; primary alcohol
   • D.0 carbons; primary alcohol
   Show answer and explanation

   Answer: A. 3 carbons; tertiary alcohol

   Classification of alcohols (1°/2°/3°) is based purely on the number of carbon atoms directly bonded to the carbinol carbon. In tert-butyl alcohol, the OH-bearing carbon is joined to three methyl carbons, making it tertiary.

   The structure of 2-methylpropan-2-ol is (CH3)3C-OH. The central carbon carries the -OH group and is bonded to three separate methyl carbons plus the oxygen. Since three carbon atoms are directly attached to the carbinol carbon, the alcohol is tertiary (3°). This classification governs reactivity: tertiary alcohols undergo SN1/E1-type reactions readily (fast Lucas test) and resist oxidation to carbonyl compounds because there is no H atom on the carbinol carbon to remove.

   Common mistake: Counting the total methyl groups in the molecule instead of only those bonded to the carbinol carbon.

6. Question 6 · medium · Alcohols

   Consider the following statements about the physical properties of alcohols: I. Alcohols have higher boiling points than alkanes of comparable molecular mass because of intermolecular hydrogen bonding. II. Among isomeric alcohols, boiling point generally decreases with increasing branching of the carbon chain. III. Alcohols are more volatile (lower boiling) than ethers of comparable molar mass. Which of the statements is/are correct?

   • A.I and II only
   • B.I and III only
   • C.II and III only
   • D.I, II and III
   Show answer and explanation

   Answer: A. I and II only

   Alcohols hydrogen-bond extensively, so they boil higher than alkanes (I, true) and higher than isomeric ethers (III is false — alcohols are LESS volatile). Increased branching reduces surface area for intermolecular contact, lowering boiling point among isomers (II, true).

   Statement I: The O-H bond in alcohols allows hydrogen bonding between molecules, requiring extra energy to separate them, so alcohols boil much higher than alkanes of similar molar mass (e.g., ethanol 78°C vs ethane -89°C) — correct. Statement II: For isomeric alcohols (e.g., butan-1-ol vs 2-methylpropan-2-ol), increased branching decreases the surface area available for van der Waals contact between molecules, lowering the boiling point despite similar hydrogen-bonding capability — correct. Statement III: Ethers cannot hydrogen-bond with each other (no O-H), so they have much lower boiling points than isomeric alcohols, not higher; alcohols are LESS volatile (higher bp) than ethers of comparable molar mass, so statement III as written is false.

   Common mistake: Assuming alcohols are more volatile than ethers, overlooking that only alcohols can hydrogen-bond.

7. Question 7 · easy · Alcohols

   A student assumes that because methanol and octan-1-ol are both 'alcohols with one -OH group,' they must have essentially similar solubility in water. What is wrong with this reasoning?

   • A.Nothing is wrong; both are completely miscible with water in all proportions regardless of chain length
   • B.Water solubility of alcohols decreases as the hydrocarbon chain length increases, because the increasingly large non-polar alkyl portion cannot be accommodated by water's hydrogen-bonded network, even though the -OH group itself can still hydrogen-bond; methanol is fully miscible while octan-1-ol is only sparingly soluble
   • C.Water solubility of alcohols increases with chain length because longer chains provide more sites for hydrogen bonding with water
   • D.Solubility in water is unrelated to chain length and depends only on whether the alcohol is primary, secondary, or tertiary
   Show answer and explanation

   Answer: B. Water solubility of alcohols decreases as the hydrocarbon chain length increases, because the increasingly large non-polar alkyl portion cannot be accommodated by water's hydrogen-bonded network, even though the -OH group itself can still hydrogen-bond; methanol is fully miscible while octan-1-ol is only sparingly soluble

   Only the -OH group of an alcohol can hydrogen-bond with water; as the non-polar hydrocarbon chain grows longer, this hydrophobic portion increasingly dominates the molecule's interaction with water, so solubility drops sharply with increasing chain length — methanol is miscible in all proportions, but octan-1-ol is only sparingly soluble.

   Alcohol solubility in water arises from hydrogen bonding between the -OH group and water molecules, which can overcome the disruption of water's own hydrogen-bonded network. However, each alcohol molecule has only one -OH group regardless of chain length, while the hydrocarbon (alkyl) portion is entirely non-polar and cannot hydrogen-bond; it merely takes up space and disrupts water structure without compensating attraction. As the alkyl chain lengthens (methanol → octan-1-ol), the non-polar portion of the molecule grows much larger relative to the single polar -OH group, so the hydrophobic effect dominates increasingly, and solubility drops off sharply. Methanol, ethanol, propan-1-ol, and to a lesser extent butan-1-ol are miscible with water in all proportions, but from about pentan-1-ol onward, water solubility falls off rapidly, and octan-1-ol is only sparingly soluble in water despite still technically being an 'alcohol with one -OH group.'

   Common mistake: Assuming all alcohols with a single -OH group have comparable water solubility regardless of the size of the attached hydrocarbon chain.

8. Question 8 · hard · Aldehydes and ketones

   A student performs the Cannizzaro reaction on benzaldehyde with concentrated NaOH and obtains benzyl alcohol and sodium benzoate in a 1:1 mole ratio. The student then claims that doubling the initial concentration of benzaldehyde (while keeping excess NaOH) will shift the ratio to favour more benzyl alcohol, since "more starting material should push the equilibrium towards more reduction product." Is this claim correct?

   • A.Yes; doubling the benzaldehyde concentration will proportionally increase the fraction converted to benzyl alcohol relative to sodium benzoate
   • B.Yes, but only if the reaction is heated above 100°C, at which point the ratio becomes concentration-dependent
   • C.No, because increasing concentration completely stops the Cannizzaro reaction from occurring at all
   • D.No; the Cannizzaro reaction is a stoichiometric disproportionation (not a reversible equilibrium), so each pair of benzaldehyde molecules always yields one molecule of benzyl alcohol and one of sodium benzoate regardless of the initial concentration used
   Show answer and explanation

   Answer: D. No; the Cannizzaro reaction is a stoichiometric disproportionation (not a reversible equilibrium), so each pair of benzaldehyde molecules always yields one molecule of benzyl alcohol and one of sodium benzoate regardless of the initial concentration used

   The Cannizzaro reaction is a stoichiometric disproportionation in which NaOH is consumed (not catalytic) and each two molecules of aldehyde always give one molecule of alcohol and one of carboxylate salt; this intrinsic 1:1 ratio is fixed by the reaction mechanism and cannot be shifted by simply changing the initial concentration of the aldehyde.

   The Cannizzaro reaction is not a reversible equilibrium process whose product distribution can be shifted by Le Chatelier-type concentration effects; it is a stoichiometric disproportionation reaction. Mechanistically, hydroxide adds to one aldehyde molecule to form a tetrahedral alkoxide intermediate, which then transfers a hydride ion irreversibly to a second aldehyde molecule; the first molecule is oxidised to the carboxylate while the second is reduced to the alkoxide (later protonated to the alcohol). This 2:1:1 stoichiometry (2 aldehyde : 1 alcohol : 1 carboxylate) is intrinsic to the mechanism and does not change based on how much starting aldehyde is used — doubling the benzaldehyde simply doubles the total amounts of both products formed, keeping their 1:1 ratio to each other unchanged. The student's reasoning incorrectly imports equilibrium-shifting logic (appropriate for reversible reactions) onto a reaction that proceeds via an essentially irreversible hydride-transfer step under the stated conditions.

   Common mistake: Applying Le Chatelier-style equilibrium-shifting reasoning to a stoichiometric, essentially irreversible disproportionation reaction

Questions about Organic Compounds Containing Oxygen for NEET

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