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, p. 2The 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 2): Alcohols are classified by the nature of the carbon bearing the hydroxyl group.
The NEET-relevant confusion (NCERT Class 12 Chemistry Chapter 7, page 4): 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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
2-Methylpropan-1-ol is classified as a:
Which of the following is a tertiary alcohol?
The compound (CH₃)₂CHOH is classified as:
A student claims neopentyl alcohol [(CH₃)₃CCH₂OH] is tertiary because it contains a quaternary carbon. The correct classification is:
With Lucas reagent (ZnCl₂/conc. HCl), which observation indicates a tertiary alcohol?
Cyclohexanol is classified as:
1-Methylcyclohexan-1-ol is a:
Among the following, identify the correct order of reactivity with Lucas reagent:
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.
Required
Identify which sample is 1°, 2°, and 3° based on observed turbidity times.
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.
Principle
Carbocation stability: 3° > 2° > 1°. Therefore reaction rate with Lucas reagent: 3° (immediate) > 2° (5–10 min) > 1° (no reaction at RT).
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°
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°)
Final answer
Immediate turbidity → tertiary (2-methylpropan-2-ol); delayed turbidity → secondary (butan-2-ol); no turbidity at RT → primary (butan-1-ol).
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.
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. ---
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, p. 2Group 15 (N, P, As, Sb, Bi) have ns²np³ configuration with stable half-filled p subshell. Metallic character increases down the group: N and P are non-metals, As and Sb metalloids, Bi a metal. Ionisation enthalpy decreases down the group. IE of Group 15 > Group 16 in same period because of extra-stable half-filled p³ configuration. All are polyatomic; N₂ is diatomic gas, others are solids.
-- NCERT Class 12 Chemistry, Ch. 7, p. 1Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.
Question gives 1° alcohol oxidation with specified reagent.
PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.
Root cause: concept gap
Aldehydes more reactive: less steric hindrance, less +I from one R group. Order: HCHO > RCHO > R₂CO.
Root cause: concept gap
Iodoform test ONLY positive for: methyl ketones (CH₃-CO-R), ethanol, secondary alcohols of form CH₃-CH(OH)-R, ethanal.
20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
Identify the suitable reagent for the following conversion.
The correct order of decreasing acidity of the following aliphatic acids is
Identify the correct reagents that would bring about the following transformation.
In which of the following equilibria, K and K are NOT equal? p c
Taking stability as the factor, which one of the following represents correct relationship?
The right option for the statement "Tyndall effect is exhibited by", is :
Which of the following reactions is the metal displacement reaction? Choose the right option. → ↑
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
inverts aldehyde ketone reactivity
Believes ketones more reactive
includes non methyl ketones
Treats all ketones as positive
ignores resonance vs induction balance
Uses inductive only without considering resonance
Test yourself on this topic with real past-paper questions:
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