Definition
Alcohols, phenols, ethers
Alcohol: R-OH (alkyl-OH). Phenol: Ar-OH (aryl-OH). Ether: R-O-R'. Different chemical reactivity due to electronic environment of OH or O.
-- NCERT Class 12 Chemistry, Ch. 7, p. 193Alcohols Primary Secondary Tertiary
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 27 Sep 2026
Alcohols Primary Secondary Tertiary, explained for NEET
The classification trap: NEET regularly tests whether you can correctly identify which carbon the –OH is attached to — and the common error is counting substituents on the oxygen instead of on the carbinol carbon.
The rule (NCERT Class 12 Chemistry Chapter 7, page 194): Alcohols are classified by the nature of the carbon bearing the hydroxyl group.
- Primary (1°) alcohol: The –OH is on a carbon bonded to one other carbon (or none, in methanol's special case). Examples: CH₃OH, CH₃CH₂OH, (CH₃)₂CHCH₂OH (2-methylpropan-1-ol — the branch is on an adjacent carbon, not on the carbinol carbon).
- Secondary (2°) alcohol: The –OH is on a carbon bonded to two other carbons. Example: (CH₃)₂CHOH (propan-2-ol).
- Tertiary (3°) alcohol: The –OH is on a carbon bonded to three other carbons. Example: (CH₃)₃COH (2-methylpropan-2-ol).
The NEET-relevant confusion (NCERT Class 12 Chemistry Chapter 7, page 195): Branched-chain alcohols trick students. In 2-methylpropan-1-ol, students see three methyl groups near the –OH and mark it tertiary — but the carbinol carbon itself is bonded to only ONE other carbon (the branched one). Classification depends solely on the carbinol carbon's connectivity, not on how many carbons are "nearby."
Lucas test connection: The practical consequence is reaction rate with Lucas reagent (ZnCl₂/conc. HCl): 3° gives immediate turbidity, 2° takes 5–10 minutes, 1° shows no reaction at room temperature. This rate difference is a direct NEET application of the classification.
Watch-out: When a structure is written in condensed form, always explicitly identify the C–OH carbon and count its C–C bonds. Do not count hydrogens or the oxygen.
Can you answer these Alcohols Primary Secondary Tertiary MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
2-Methylpropan-1-ol is classified as a:
Show answer and why every option is right or wrong
Answer: D. The –OH group is attached to a –CH₂– carbon that is bonded to only one other carbon atom (the branched carbon bearing two methyl groups). One C–C bond on the carbinol carbon = primary alcohol. (NCERT Class 12 Chemistry Chapter 7, page 194)
Why A is wrong: Incorrect. Any alcohol with –OH on a carbon in a chain is classifiable as 1°, 2°, or 3°.
Why B is wrong: Incorrect. Secondary requires the carbinol carbon to be bonded to two other carbons. Here the –CH₂OH carbon has only one C–C bond — the branch is on the adjacent carbon, not on the carbinol carbon itself.
Why C is wrong: Incorrect. Tertiary requires three C–C bonds on the carbinol carbon. Students confuse the total number of methyl groups in the molecule with the carbinol carbon's connectivity.
MCQ 2Easy RecallPractice
Which of the following is a tertiary alcohol?
Show answer and why every option is right or wrong
Answer: A. In 2-methylbutan-2-ol, the –OH is on carbon-2 which is bonded to three other carbons (C-1, C-3, and the methyl branch). Three C–C bonds = tertiary. (NCERT Class 12 Chemistry Chapter 7, page 194)
Why B is wrong: Butan-2-ol: the carbinol carbon (C-2) is bonded to two other carbons (C-1 and C-3). This is secondary, not tertiary.
Why C is wrong: Butan-1-ol: the carbinol carbon (C-1) is bonded to only one other carbon (C-2). This is primary.
Why D is wrong: 3-Methylbutan-1-ol: the carbinol carbon (C-1) is bonded to one other carbon (C-2). The methyl branch at C-3 does not affect the carbinol carbon's classification. This is primary.
MCQ 3Easy RecallPractice
The compound (CH₃)₂CHOH is classified as:
Show answer and why every option is right or wrong
Answer: C. In (CH₃)₂CHOH (propan-2-ol/isopropanol), the carbinol carbon bears two C–C bonds (to two methyl groups). Two other carbons attached = secondary. (NCERT Class 12 Chemistry Chapter 7, page 194)
Why A is wrong: Primary requires only one C–C bond on the carbinol carbon. Here the –CHOH carbon is bonded to two methyl carbons.
Why B is wrong: Tertiary requires three C–C bonds. This carbon has only two.
Why D is wrong: Phenols have –OH directly bonded to an aromatic ring carbon. (CH₃)₂CHOH is aliphatic.
MCQ 4Direct ApplicationPractice
A student claims neopentyl alcohol [(CH₃)₃CCH₂OH] is tertiary because it contains a quaternary carbon. The correct classification is:
Show answer and why every option is right or wrong
Answer: B. The carbinol carbon is the –CH₂– bearing the –OH. That carbon is bonded to only one other carbon (the quaternary carbon). One C–C bond on the carbinol carbon = primary. The quaternary carbon is adjacent but is not the carbinol carbon. (NCERT Class 12 Chemistry Chapter 7, page 194)
Why A is wrong: The classification depends on the carbon bearing –OH, not on any other carbon in the molecule. The quaternary carbon does not bear the –OH.
Why C is wrong: The –CH₂– does connect to O and to C, but the classification counts only C–C bonds. The carbinol carbon has one C–C bond, not two.
Why D is wrong: Classification is purely structural — identify the carbinol carbon and count its C–C bonds. No spectral data needed.
MCQ 5Easy RecallPractice
With Lucas reagent (ZnCl₂/conc. HCl), which observation indicates a tertiary alcohol?
Show answer and why every option is right or wrong
Answer: D. Tertiary alcohols react immediately with Lucas reagent to form the alkyl chloride (turbidity) because the tertiary carbocation is the most stable. Secondary takes 5–10 min; primary shows no reaction at room temperature. (NCERT Class 12 Chemistry Chapter 7, page 195)
Why A is wrong: 5–10 minutes of standing before turbidity appears is the characteristic observation for a secondary alcohol, not tertiary.
Why B is wrong: This describes a forced reaction. Lucas test for tertiary is defined by immediate turbidity at room temperature without heating.
Why C is wrong: No turbidity at room temperature is the primary alcohol response — the primary carbocation is too unstable to form under these mild conditions.
MCQ 6CalculationPractice
3-Methylpentan-3-ol is treated with Lucas reagent (ZnCl₂/conc. HCl) at room temperature. Based on its structure, what observation should appear, and why?
Show answer and why every option is right or wrong
Answer: A. In 3-methylpentan-3-ol [CH₃CH₂–C(CH₃)(OH)–CH₂CH₃], the carbinol carbon (C-3) is bonded to three other carbons: the two ethyl-chain carbons (C-2 and C-4) and the methyl branch. Three C–C bonds makes it a tertiary alcohol (NCERT Class 12 Chemistry Chapter 7, page 194), and tertiary alcohols give immediate turbidity with Lucas reagent at room temperature because the tertiary carbocation they form is the most stable (NCERT Class 12 Chemistry Chapter 7, page 208).
Why B is wrong: Secondary alcohols have only two C–C bonds on the carbinol carbon. Here C-3 has three C–C bonds (to C-2, C-4, and the methyl branch), so the classification is tertiary, not secondary.
Why C is wrong: Primary alcohols have only one C–C bond on the carbinol carbon. C-3 in this molecule has three, so it cannot be primary — and primary alcohols show no turbidity, not tertiary ones.
Why D is wrong: The carbinol carbon in this molecule does have three C–C bonds, so a carbocation forms readily on treatment with Lucas reagent; the alcohol is tertiary, not carbon-isolated.
MCQ 7Direct ApplicationPractice
1-Methylcyclohexan-1-ol is a:
Show answer and why every option is right or wrong
Answer: C. The carbinol carbon (ring C-1) bears the –OH and is bonded to three other carbons: two ring carbons (C-2 and C-6) and the exocyclic methyl carbon. Three C–C bonds = tertiary. (NCERT Class 12 Chemistry Chapter 7, page 194)
Why A is wrong: Primary requires one C–C bond. This carbinol carbon has three.
Why B is wrong: Without the methyl group, cyclohexanol's C-1 would have two C–C bonds (secondary). But the methyl substituent adds a third C–C bond, making it tertiary.
Why D is wrong: This option incorrectly ignores the methyl group's contribution. The methyl adds a third C–C bond to the carbinol carbon, elevating it from secondary to tertiary.
MCQ 8Concept TrapPractice
Among the following, identify the correct order of reactivity with Lucas reagent:
Show answer and why every option is right or wrong
Answer: A. Lucas reagent (SN1 pathway): tertiary carbocation forms fastest (most stable), secondary slower, primary does not form under room temperature conditions. Reactivity order: 3° > 2° > 1°. (NCERT Class 12 Chemistry Chapter 7, page 195)
Why B is wrong: This is the reverse order. 1° alcohols are the LEAST reactive because primary carbocations are highly unstable.
Why C is wrong: 2° is not more reactive than 3°. Carbocation stability follows 3° > 2° > 1°.
Why D is wrong: 1° alcohols do not react before 2° alcohols. The order is strictly governed by carbocation stability.
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How do you solve a Alcohols Primary Secondary Tertiary question? A worked example
- 1
Given
Three unlabelled samples are known to be: butan-1-ol, butan-2-ol, and 2-methylpropan-2-ol. Each is treated with Lucas reagent (ZnCl₂/conc. HCl) at room temperature.
- 2
Required
Identify which sample is 1°, 2°, and 3° based on observed turbidity times.
- 3
Concept
Lucas test exploits the SN1 mechanism: the alcohol forms a carbocation intermediate whose stability determines reaction rate. More stable carbocation → faster alkyl chloride formation → earlier turbidity.
- 4
Principle
Carbocation stability: 3° > 2° > 1°. Therefore reaction rate with Lucas reagent: 3° (immediate) > 2° (5–10 min) > 1° (no reaction at RT).
- 5
Classification of given compounds
• Butan-1-ol: carbinol carbon bonded to one other C → 1°• Butan-2-ol: carbinol carbon bonded to two other C → 2°• 2-Methylpropan-2-ol: carbinol carbon bonded to three other C → 3°
- 6
Application
• Sample showing immediate turbidity = 2-methylpropan-2-ol (3°)• Sample showing turbidity after 5–10 min = butan-2-ol (2°)• Sample showing no turbidity at room temperature = butan-1-ol (1°)
- 7
Final answer
Immediate turbidity → tertiary (2-methylpropan-2-ol); delayed turbidity → secondary (butan-2-ol); no turbidity at RT → primary (butan-1-ol).
- 8
Common trap
Students confuse "no reaction" with "the test failed." For primary alcohols, no turbidity at room temperature IS the expected result — it confirms 1° classification. Do not heat the sample to force a reaction; the test is defined by room-temperature observations.
- 9
Similar NEET-style question
Three isomeric pentanols (pentan-1-ol, pentan-3-ol, 2-methylbutan-2-ol) are treated with Lucas reagent. Predict the order of turbidity appearance and classify each.
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What to remember before solving Alcohols Primary Secondary Tertiary questions
Group 15 includes nitrogen, phosphorus, arsenic, antimony, bismuth and moscovium. As we go down the group, there is a shift from non-metallic to metallic through metalloidic character. Nitrogen and phosphorus are non-metals, arsenic and antimony metalloids, bismuth and moscovium are typical metals. … 7.1.2 Electronic Configuration The valence shell electronic configuration of these elements is ns2np3. The s orbital in these elements is completely filled and p orbitals are half-filled, making their electronic configuration extra stable. 7.1.3 Atomic and Ionic Radii Covalent and ionic (in a particular state) radii increase in size down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and/or f orbitals in heavier members. 7.1.4 Ionisation Enthalpy Ionisation enthalpy decreases down the group due to gradual increase in atomic size. Because of the extra stable half-filled p orbitals electronic configuration and smaller size, the ionisation enthalpy of the group 15 elements is much greater than that of group 14 elements in the corresponding periods. The order of successive ionisation enthalpies, as expected is ∆iH1 < ∆iH2 < ∆iH3 (Table 7.1). 7.1.5 Electronegativity The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the difference is not that much pronounced. 7.1.6 Physical Properties All the elements of this group are polyatomic. Dinitrogen is a diatomic gas while all others are solids. Metallic character increases down the group. Nitrogen and phosphorus are non-metals, arsenic and antimony metalloids and bismuth is a metal. This is due to decrease in ionisation enthalpy and increase in atomic size. The boiling points, in general, increase from top to bottom in the group but the melting point increases upto arsenic and then decreases upto bismuth. Except nitrogen, all the elements show allotropy. Table 7.1 (N, P, As, Sb, Bi): Ionisation enthalpy I (kJ mol–1) 1402, 1012, 947, 834, 703; Electronegativity 3.0, 2.1, 2.0, 1.9, 1.9.
-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 170More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Alcohols Primary Secondary Tertiary questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys.
NEET 2022
1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
Sources
NCERT refs: Class 12 Chemistry Chapter 7, p.194 | Class 12 Chemistry Chapter 7, p.195
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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