The functional group is the atom or group of atoms within an organic molecule that determines its chemical behaviour. Two molecules may have identical carbon chain lengths but behave completely differently because they carry different functional groups. This is the basis for classifying organic compounds — not by molecular formula or carbon count, but by the reactive site.
NCERT Class 11 Chemistry Chapter 8, page 267 (Table 8.4), lists the major functional groups systematically: hydroxyl (–OH) for alcohols and phenols, carbonyl (C=O) for aldehydes and ketones, carboxyl (–COOH) for carboxylic acids, amino (–NH₂) for amines, nitro (–NO₂), halide (–X), and so on. Each functional group defines a homologous series — a family of compounds with the same functional group but differing chain lengths, sharing similar chemical properties.
What NEET actually tests here: The exam checks whether you can identify the functional group present in a given structure, classify the compound into the correct homologous series, and distinguish between compounds that look similar but belong to different classes. A common confusion is between aldehydes and ketones — both contain C=O, but the position differs (terminal in aldehydes, internal in ketones). Another frequent error is confusing alcohols (–OH on sp³ carbon) with phenols (–OH directly on benzene ring) — they are classified separately because their chemical reactivity differs substantially.
The classification hierarchy works as follows: first identify the functional group, then name the homologous series, then apply IUPAC naming rules specific to that class. Compounds with more than one functional group are classified by the principal (highest-priority) functional group according to the IUPAC priority order: –COOH > –SO₃H > –COOR > –COX > –CONH₂ > –CHO > >C=O > –OH > –NH₂ > –C≡C– > –C=C–.
Watch out: when a molecule has both –OH and –CHO, it is classified as an aldehyde (not an alcohol), because –CHO has higher priority.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which functional group is present in carboxylic acids?
Answer: B. Carboxylic acids are defined by the carboxyl group (–COOH), which is a combination of carbonyl (C=O) and hydroxyl (–OH) on the same carbon. NCERT Class 11 Chemistry Chapter 8, Part 2, page 262.
Why A is wrong: A is wrong because –CHO is the aldehyde functional group, not the carboxylic acid group. The key difference is that –COOH has an –OH attached to the carbonyl carbon, whereas –CHO has only a hydrogen.
Why C is wrong: C is wrong because –OH alone is the functional group of alcohols (on aliphatic sp³ carbon) or phenols (on aromatic ring). It is only part of –COOH, not the complete carboxylic acid functional group.
Why D is wrong: D is wrong because –CONH₂ is the amide functional group, formed when the –OH of –COOH is replaced by –NH₂. Different class entirely.
Which of the following pairs of compounds belong to the same homologous series?
Answer: D. CH₃OH (methanol) and C₂H₅OH (ethanol) are both aliphatic alcohols — same functional group (–OH on sp³ carbon), differing by one –CH₂– unit. This is the definition of a homologous series (NCERT Class 11 Chemistry Chapter 8, pages 262–263).
Why A is wrong: A is wrong because CH₃OH is an aliphatic alcohol (–OH on sp³ carbon) while C₆H₅OH is a phenol (–OH directly on the aromatic ring). Despite both containing –OH, they belong to different classes with different chemical reactivities.
Why B is wrong: B is wrong because CH₃CHO is an aldehyde (terminal C=O with H) and CH₃COCH₃ is a ketone (C=O flanked by two carbon groups). Both contain carbonyl but are classified as different homologous series.
Why C is wrong: C is wrong because CH₃COOH is a carboxylic acid (–COOH group) and CH₃CHO is an aldehyde (–CHO group). Different functional groups, different series.
The functional group present in an ester is:
Answer: C. Esters contain the –COOR functional group, where the hydrogen of –COOH has been replaced by an alkyl or aryl group R. NCERT Class 11 Chemistry Chapter 8, page 267 (Table 8.4), lists this as a distinct functional group class.
Why A is wrong: A is wrong because –COOH is the carboxylic acid functional group. An ester is derived from a carboxylic acid but the –OH hydrogen is replaced by –OR, making the functional group –COOR.
Why B is wrong: B is wrong because –CHO is the aldehyde functional group (terminal carbonyl with hydrogen). Esters have an –OR group attached to the carbonyl, not –H.
Why D is wrong: D is wrong because –COX (where X = halogen) is the acyl halide functional group, not the ester group. In esters, the group bonded to carbonyl is –OR (alkoxy), not a halogen.
A compound has the molecular formula C₃H₆O. It gives a positive silver mirror test. The functional group present is:
Answer: A. The silver mirror test (Tollens' test) is positive for aldehydes, which contain the –CHO group. C₃H₆O with –CHO is propanal (CH₃CH₂CHO). Ketones with the same formula (acetone, CH₃COCH₃) do not give a positive silver mirror test (NCERT Chemistry Lab Manual Class 12, Unit 8, page 94).
Why B is wrong: B is wrong because ketones (–COCH₃) do not reduce Tollens' reagent. Acetone (CH₃COCH₃) has the same molecular formula but gives a negative silver mirror test — this is precisely how aldehydes and ketones are distinguished.
Why C is wrong: C is wrong because alcohols (–OH) do not give a positive silver mirror test. An alcohol with C₃H₆O would be allyl alcohol, but it would not reduce Tollens' reagent.
Why D is wrong: D is wrong because carboxylic acids have the general formula CₙH₂ₙO₂ (for saturated monocarboxylic acids), requiring two oxygen atoms per carboxyl group. C₃H₆O has only one oxygen, ruling out –COOH.
A compound contains both –OH and –CHO functional groups. According to IUPAC priority rules, it is classified as:
Answer: D. When multiple functional groups are present, the compound is classified by the principal (highest-priority) functional group. IUPAC priority order places –CHO above –OH. Therefore, the compound is classified as a hydroxy aldehyde (priority order: NCERT Class 11 Chemistry Chapter 8, page 266).
Why A is wrong: A is wrong because –OH (alcohol) has lower priority than –CHO (aldehyde) in IUPAC classification. The –OH group becomes a substituent prefix (hydroxy-), not the basis of classification.
Why B is wrong: B is wrong because carboxylic acids require the –COOH group. Having –OH and –CHO separately does not make –COOH — the –OH must be directly bonded to the same carbon as C=O to form a carboxyl group.
Why C is wrong: C is wrong because ethers have the functional group R–O–R (two carbon groups joined by oxygen). The compound described has –OH and –CHO, neither of which is an ether linkage.
Which of the following compounds is correctly classified?
Answer: B. CH₃CH₂NH₂ (ethylamine) has the –NH₂ group bonded to one carbon, making it a primary amine. The classification depends on how many carbons are directly attached to nitrogen: one = primary, two = secondary, three = tertiary.
Why A is wrong: A is wrong because C₆H₅OH is phenol, not an aliphatic alcohol. The –OH is bonded directly to an aromatic ring carbon (sp² hybridised), which gives phenol distinctly different acidity and reactivity compared to aliphatic alcohols.
Why C is wrong: C is wrong because CH₃COCH₃ is acetone, a ketone. The carbonyl (C=O) is flanked by two carbon groups (both –CH₃). In aldehydes, the carbonyl carbon must be bonded to at least one hydrogen.
Why D is wrong: D is wrong because CH₃OCH₃ is dimethyl ether, not an alcohol. The oxygen is bonded to two carbon groups (R–O–R pattern). An alcohol requires –OH (oxygen bonded to one carbon and one hydrogen).
Two compounds have the molecular formula C₂H₆O. Compound X reacts with sodium metal to release hydrogen gas; compound Y does not. What are the functional groups in X and Y respectively?
Answer: C. C₂H₆O has two possible structures: ethanol (CH₃CH₂OH, containing –OH) and dimethyl ether (CH₃OCH₃, containing –O–). Sodium metal reacts with the acidic hydrogen in –OH to release H₂. Ethers lack an –OH hydrogen and do not react with sodium. Therefore X = alcohol (–OH) and Y = ether (–O–). (Alcohols liberating hydrogen with sodium: NCERT Class 11 Chemistry Chapter 8, page 266.)
Why A is wrong: A is wrong because –CHO (aldehyde) requires C₂H₄O as the minimum formula for an aldehyde with two carbons (acetaldehyde, CH₃CHO). C₂H₆O is saturated and cannot accommodate a carbonyl while retaining six hydrogens.
Why B is wrong: B is wrong because this inverts the reactivity. Ethers do NOT react with sodium metal (no acidic hydrogen). The compound that releases H₂ with Na must have –OH, not –O–. This is a direct test for the alcohol functional group.
Why D is wrong: D is wrong because a carboxylic acid (–COOH) with two carbons would be acetic acid (CH₃COOH), which has the formula C₂H₄O₂ — two oxygens, not one. C₂H₆O has only one oxygen.
Among the following functional groups, which has the highest priority in IUPAC nomenclature for determining the principal characteristic group?
Answer: A. The IUPAC priority order for principal characteristic groups is: –COOH > –CHO > –OH > –NH₂. Carboxylic acid (–COOH) has the highest priority among the listed options. When present alongside other groups, all others become substituent prefixes (NCERT Class 11 Chemistry Chapter 8, page 266).
Why B is wrong: B is wrong because –NH₂ (amino) has the lowest priority among these four groups. In the presence of any of the other three, –NH₂ becomes an 'amino-' prefix.
Why C is wrong: C is wrong because –CHO (aldehyde) ranks below –COOH but above –OH and –NH₂. In a compound with both –CHO and –COOH, the compound is named as a carboxylic acid with an 'oxo-' prefix for the aldehyde.
Why D is wrong: D is wrong because –OH (hydroxyl/alcohol) ranks below both –COOH and –CHO in IUPAC priority. When –OH and –COOH are both present, –OH is expressed as a 'hydroxy-' prefix.
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Given
• Molecular formula: C₄H₈O₂• Functional groups present: –OH and –CHO• Degree of unsaturation: (2×4 + 2 − 8) / 2 = 1 (consistent with one C=O, no ring or C=C)
Required
Classify the compound, identify the principal functional group, and write the IUPAC name.
Concept
When a molecule has multiple functional groups, IUPAC rules require classifying by the highest-priority group. Priority order (relevant subset): –COOH > –CHO > –OH > –NH₂. The lower-priority group becomes a prefix substituent.
Formula / Rule
IUPAC priority: –CHO > –OH. Therefore the compound is classified as a hydroxy aldehyde, not an "aldehydic alcohol."
Substitution / Setup
Structure: HOCH₂CH₂CH₂CHO (4-hydroxybutanal).• Longest chain including –CHO: 4 carbons → butanal.• –CHO carbon = C-1 (by rule, aldehyde always gets position 1).• –OH is on C-4.• Name: 4-hydroxybutanal.
Calculation
No arithmetic calculation needed. This is a classification and naming problem. The key operation is applying the priority hierarchy:• –CHO is principal group → suffix "-al" (butanal)• –OH is subordinate → prefix "hydroxy-" at position 4
Final answer
The compound is classified as an aldehyde (not an alcohol). IUPAC name: 4-hydroxybutanal. The –OH group appears as the prefix "hydroxy-" because it has lower priority than –CHO.
Common trap
Students sometimes classify this compound as an alcohol because –OH is the more "familiar" group, or they name it "4-oxobutan-1-ol" (treating –OH as principal). The IUPAC priority hierarchy is non-negotiable: –CHO always outranks –OH.
Similar NEET-style question
"A compound C₅H₁₀O₃ contains –OH and –COOH groups. What is the IUPAC classification and name?" (Answer: It is a carboxylic acid, not an alcohol — –COOH outranks –OH, so the acid is the suffix and the alcohol becomes the hydroxy- prefix. Name: 5-hydroxypentanoic acid, HOCH₂CH₂CH₂CH₂COOH = C₅H₁₀O₃. Check the oxygens before naming: –COOH accounts for two of the three and –OH for the third, so a C₅H₁₀O₂ compound could not carry both groups at once.)
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Group of atoms responsible for characteristic reactions of a compound. Common: -OH (alcohol), -CHO (aldehyde), >C=O (ketone), -COOH (carboxylic acid), -NH₂ (amine), -X (halide), -O- (ether), -NO₂ (nitro).
-- NCERT Class 11 Chemistry, Ch. 8, p. 265More in Organic Chemistry — Some Basic Principles: 9 exam traps and mistakes · 1 formula · 1 question pattern from its other lessons.
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