Q.Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
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IUPAC Nomenclature (Organic Compounds)
IUPAC nomenclature is a systematic way to name a compound so that its name alone tells you its exact structure, with no ambiguity. Every organic name follows the same underlying recipe, whatever the functional group.
The Recipe
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Identify the principal characteristic group. If the molecule has a functional group senior enough to be named as a suffix (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine, and so on down the seniority order), that group decides the suffix and must be included in the parent chain. A halogen is never senior enough to be a suffix — it is always named as a prefix ("halo-"), whatever else is present.
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Choose the parent chain. The parent is the longest continuous carbon chain that contains the principal characteristic group (if there is one). Among chains of the same length, the one with the most substituents wins.
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Number the chain. Number from whichever end gives the LOWEST LOCANT to the principal characteristic group first. If there is no principal group (e.g. a simple haloalkane, or an alkene with a halogen substituent), lowest locant goes to the site of unsaturation (double/triple bond) first, then to substituents as a set.
Watch outWhen two numbering directions give the SAME locant for the principal group/unsaturation (a genuine tie), the tie-break is the lowest locant SET for the substituents as a group — compare the two sets at their first point of difference. Only if the sets are themselves tied does the alphabetically-first substituent get the lower number.
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Name and cite the substituents as prefixes, in alphabetical order (ignoring multiplying prefixes like di-/tri- but not ignoring structural prefixes like iso-/cyclo-), each with its own locant.
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Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).
Worked Example
CH3−CH(Cl)−CH(CH3)−CH2−CH3: the longest chain is 5 carbons (pentane), no principal characteristic group (just a halogen substituent), so number for the lowest locant set. From the left: Cl at C2, methyl at C3 → set {2,3}. From the right: methyl at C3, Cl at C4 → set {3,4}. {2,3} is lower, so numbering from the left wins: 2-chloro-3-methylpentane. …
Why this formula?
IUPAC Nomenclature: Why the Rules Work the Way They Do
IUPAC nomenclature is not a single formula, but a system of rules designed to give every organic compound a unique, unambiguous name. The "why" behind these rules lies in clarity, consistency, and communication — ensuring that a chemist in Tokyo and one in Toronto draw the same structure from the same name.
1. The Core Principle: The Longest Carbon Chain
Rule: Identify the longest continuous chain of carbon atoms. This becomes the parent chain (e.g., pentane, hexane).
Why?
- The longest chain represents the backbone of the molecule.
- It gives the most stable, fundamental name — shorter chains would be branches, not the main structure.
- Example: In a molecule with 5 carbons in a row and a 2-carbon branch, calling it "pentane" (not "ethane") tells you the core skeleton is 5 carbons long.
Key idea: The parent chain is the maximum continuous path — not necessarily the one that looks "straight" on paper.
2. Numbering: Lowest Locants (The "First Point of Difference" Rule)
Rule: Number the parent chain so that substituents get the smallest possible numbers. When there's a tie, compare the first point of difference.
Why?
- This ensures reproducibility — two chemists will always number the same way.
- It avoids ambiguity: "2-methylpentane" is unambiguous; "3-methylpentane" would be a different compound.
- The first point of difference rule: If you have substituents at positions 2,4 and 3,5, choose 2,4 because 2 < 3 (the first number is smaller).
Example:
- For a methyl group on carbon 2 vs. carbon 4 of a 5-carbon chain:
- 2-methylpentane (correct)
- 4-methylpentane (wrong — higher number)
3. Alphabetical Order of Substituents
Rule: List substituents in alphabetical order (ignoring prefixes like di-, tri-, sec-, tert- but not iso-).
Why?
- Alphabetical order is a universal sorting convention — no need to remember priority based on size or complexity.
- It makes names searchable and predictable.
- Example: "3-ethyl-2-methylpentane" (e before m) — not "2-methyl-3-ethylpentane".
Exception: Prefixes like iso- and neo- are considered part of the name (e.g., isopropyl comes before methyl because "i" < "m").
4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes
Rule: Number the chain so that the double or triple bond gets the lowest possible number, even if it means giving a substituent a higher number.
Why?
- The functional group (alkene/alkyne) is more important than alkyl substituents.
- The bond position defines the compound's reactivity and geometry.
- Example: In pent-2-ene (not pent-3-ene), the double bond is between carbons 2 and 3 — the lower number (2) is used.
Priority order:
- Principal functional group (e.g., -OH, -COOH, C=C)
- Multiple bonds
- Substituents (alkyl, halo, etc.)
5. The "Suffix" and "Prefix" System …
Concept: IUPAC Nomenclature of Monohydric Phenols
A monohydric phenol has one –OH group directly attached to a benzene ring. For C7H8O, the benzene ring (C6H5) accounts for 6 carbons and 5 hydrogens, leaving one extra carbon and three hydrogens — that extra carbon must be a methyl substituent on the ring.
Reasoning steps:
- The –OH group is the principal functional group (suffix: -ol). The benzene ring is the parent chain (phenol).
- The methyl group can be placed at three positions relative to –OH: ortho (1,2), meta (1,3), or para (1,4).
- Name each isomer using locants, with –OH at position 1 (implied in "phenol").
Structures and IUPAC names: …
The molecular formula C7H8O with a phenolic –OH group means a benzene ring plus one methyl substituent and one –OH group — that is, a methylphenol (cresol). The three positional isomers are 2-methylphenol (o-cresol), 3-methylphenol (m-cresol), and 4-methylphenol (p-cresol).
The key to solving this is to first recognise what “monohydric phenol” means. A phenol has a hydroxyl group (–OH) directly attached to a benzene ring. “Monohydric” tells us there is exactly one such –OH group. So the core is a benzene ring (C6H5–) with one –OH, which accounts for C6H5O (that’s 6 carbons, 5 hydrogens, 1 oxygen). The molecular formula given is C7H8O. Subtract the phenol core: C7H8O−C6H5O=CH3. That leftover is exactly one methyl group (−CH3). So the molecule is a benzene ring with an –OH and a –CH₃ attached — a methylphenol, commonly called a cresol.
Now, the methyl group and the hydroxyl group can be placed in three different relative positions on the benzene ring. These are the ortho, meta, and para isomers. In IUPAC nomenclature, we number the ring so that the –OH gets the lowest possible number (since –OH is the principal functional group for phenols). The methyl group then takes the position number accordingly.
Let’s go through each isomer step by step.
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Ortho isomer — The methyl group is on carbon 2, adjacent to the –OH (which is on carbon 1).
Structure: a benzene ring with –OH at position 1 and –CH₃ at position 2.
IUPAC name: 2-methylphenol. Common name: o-cresol.
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Meta isomer — The methyl group is on carbon 3, one carbon away from the –OH.
Structure: –OH at position 1, –CH₃ at position 3.
IUPAC name: 3-methylphenol. Common name: m-cresol.
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Para isomer — The methyl group is on carbon 4, directly opposite the –OH.
Structure: –OH at position 1, –CH₃ at position 4.
IUPAC name: 4-methylphenol. Common name: p-cresol. …
Method: Systematic Isomer Enumeration + IUPAC Nomenclature
This method works by first listing all possible structural isomers (different carbon skeletons and positions of the –OH group), then naming each according to IUPAC rules.
Step 1: Determine the molecular formula and identify the functional group
- Formula: C7H8O
- “Monohydric phenol” means one –OH group directly attached to a benzene ring.
- So the benzene ring contributes C6H5 — the remaining part is CH3 (a methyl group).
Thus, all isomers are methylphenols (cresols) — one methyl group and one hydroxyl group on the benzene ring.
Step 2: Enumerate positional isomers
The –OH and –CH3 groups can be placed in three relative positions on the benzene ring:
| Position of –CH3 relative to –OH | IUPAC name prefix |
|---|---|
| Adjacent (1,2) | ortho- (or 2-) |
| One carbon apart (1,3) | meta- (or 3-) |
| Opposite (1,4) | para- (or 4-) |
Step 3: Draw structures and assign IUPAC names
1. 2-Methylphenol (ortho-cresol)
OH
|
— C₆H₄ — CH₃ (at position 2)
- IUPAC: 2-methylphenol
- Common name: o-cresol
2. 3-Methylphenol (meta-cresol)
OH
|
— C₆H₄ — CH₃ (at position 3)
- IUPAC: 3-methylphenol
- Common name: m-cresol …
Step 1: Understand the molecular formula
The formula is C7H8O.
For a monohydric phenol, we have:
- A benzene ring (C6H5–)
- One –OH group (phenolic)
- Remaining: C7H8O−C6H5OH=CH3
So the extra carbon is a methyl group attached to the ring.
Step 2: The correct structures and names
There are three positional isomers:
| Structure | IUPAC Name |
|---|---|
| 2-methylphenol | 2-methylphenol (or o-cresol) |
| 3-methylphenol | 3-methylphenol (or m-cresol) |
| 4-methylphenol | 4-methylphenol (or p-cresol) |
All are monohydric phenols with formula C7H8O.
Common Mistakes & How to Avoid Them
✗ Mistake 1: Including benzyl alcohol as a monohydric phenol
Why it's wrong:
Benzyl alcohol (C6H5CH2OH) has the –OH on a side chain, not directly on the benzene ring. That makes it an alcohol, not a phenol.
How to avoid:
Remember: phenol = –OH directly attached to an aromatic ring. If the –OH is on a carbon chain attached to the ring, it's an alcohol.
✗ Mistake 2: Forgetting the methyl group can be at positions 2, 3, or 4
Why it's wrong:
Some students only write o-cresol (2-methylphenol) and p-cresol (4-methylphenol), missing the m-cresol (3-methylphenol).
How to avoid:
Always systematically check all possible positions on the ring: 2, 3, and 4 (since position 1 is taken by –OH).
✗ Mistake 3: Using common names instead of IUPAC names
Why it's wrong:
"Cresol" is a common name. In exams, IUPAC names like 2-methylphenol are required.
How to avoid:
Learn the IUPAC naming rules:
- Phenol is the parent.
- Substituents are named as prefixes with locants.
- Number the ring so that –OH gets position 1.
✗ Mistake 4: Incorrect numbering (giving –OH a number > 1)
Why it's wrong:
In phenols, the –OH group always gets position 1 by priority. Some students number from the methyl group.
How to avoid: …
- KEAM 2026Set eng-2026-04174 marksMCQQ.The IUPAC name of mesityl oxide is (A) 2-Methylpent-2-en-3-one (B) 3-Methylpent-2-en-4-one (C) 4-Methylpent-2-en-3-one (D) 4-Methylpent-3-en-2-one (E) 2-Methylpent-3-en-4-one
›Reveal solutionSolution
The structure (CH3)2C=CH−CO−CH3 names as 4-methylpent-3-en-2-one.
Mesityl oxide has the structure (CH3)2C=CH−CO−CH3.
The longest chain containing the carbonyl is five carbons (pent-). Numbering to give the ketone the lowest locant, start from the methyl next to the C=O:
- C1: CH3
- C2: C=O (ketone → -2-one) …
- KEAM 2026Set eng-2026-04194 marksMCQQ.IUPAC name of (CH3)3C-CH2Br is (A) 1-Bromotrimethylpropane (B) neo-pentylbromide (C) 1-Bromo-2,2-dimethylpropane (D) 2,2-dimethylethylenediamine (E) 3-bromo-2,2-dimethylpropane
›Reveal solutionSolution
The five-carbon skeleton is propane with two methyls on C-2 and Br on C-1: 1-bromo-2,2-dimethylpropane.
Structure. (CH3)3C-CH2Br = a central carbon bearing three methyls and a CH2Br. The longest chain is propane (3 C); numbering to give Br the lowest locant puts CH2Br as C-1, the quaternary carbon as C-2 car …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.The IUPAC name of the following alkane is CH3−CH2−CH(C2H5)−CH2−CH(CH3)−CH2−CH3 (A) 3-methyl-5-ethylheptane (B) 3,5-diethylhexane (C) 4,6-diethylhexane (D) 3-ethyl-5-methylheptane (E) 3-ethyl-5,6-dimethylhexane
›Reveal solutionSolution
Longest chain = 7 C (heptane), ethyl at C3, methyl at C5.
The structure CH3−CH2−CH(C2H5)−CH2−CH(CH3)−CH2−CH3 has a 7-carbon parent chain. Numbering to give the lowest locants (tie {3,5} both ways) gives the lower number to the first-cited substituent alphabetically (ethyl before methyl), so ethyl = 3, methy …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.The IUPAC name of the following compound is (A) 2-Methylpent-2-en-2-one (B) 3-Methylpent-2-en-2-one (C) 4-Methylpent-2-en-3-one (D) 4-Methylpent-3-en-2-one (E) 1,1-Dimethylbuten-2-one
›Reveal solutionSolution
Numbering from the carbonyl end (mesityl oxide) gives 4-methylpent-3-en-2-one.
The structure is CH3−CO−CH=C(CH3)−CH3 (mesityl oxide). Choosing the longest chain containing the carbonyl (the principal group) and numbering to give the ketone the lowest locant:
- C1 = CH3, C2 = C=O (the 2-one), C3 = CH, C4 = C, C5 = CH3. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.The IUPAC name of phenyl isopentyl ether is (A) 3-Methtylbutoxybenzene (B) 2-Methylbutoxybenzene (C) 2-Methylphenoxybutane (D) 4-Methylbutoxybenzene (E) 1-Methylbutoxybenzene
›Reveal solutionSolution
Phenyl isopentyl ether is named 3-methylbutoxybenzene.
Concept and Intuition
Ethers are named as (alkoxy)benzene when one group is phenyl. Isopentyl (isoamyl) is the 3-methylbutyl group, (CH3)2CH-CH2-CH2-. Attaching it via oxygen to benzene gives 3-methylbutoxybenzene.
Step-by-Step Solution
- Isopentyl = isoamyl = 3-methylbutyl = (CH3)2CHCH2CH2-.
- As an -O- substituent it becomes 3-methylbutoxy.
- On benzene → 3-methylbutoxybenzene. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The IUPAC name of the compound HOCH2(CH2)3CH2COCH3 is (A) 7-Hydroxyheptan-2-one (B) 2-Oxoheptan-7-ol (C) 1-Hydroxyheptan-2-one (D) 5-Oxoheptan-2-ol (E) 6-Hydroxyheptan-3-one
›Reveal solutionSolution
The molecule is a seven-carbon chain bearing a ketone and an alcohol. The ketone (higher priority) gets the suffix '-one' with the lowest locant, and −OH becomes the 'hydroxy' prefix: 7-hydroxyheptan-2-one.
Expanding HOCH2(CH2)3CH2COCH3 gives a continuous chain of 7 carbons:
HO−CH2−CH2−CH2−CH2−CH2−CO−CH3
Priority: the ketone (C=O) outranks the alcohol, so it defines the suffix and gets the lower locant. Numbering from the methyl-ketone end: …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The IUPAC name of allylamine is (A) But-2-en-1-amine (B) But-1-en-2-amine (C) Prop-2-en-1-amine (D) Prop-1-en-2-amine (E) 2-Amino 1-propene
›Reveal solutionSolution
Allylamine is a 3-carbon chain with a C=C at position 2 and –NH2 at C1: prop-2-en-1-amine.
Allylamine is CH2=CH−CH2−NH2. Numbering to give the amine the lowest locant: C1 bears the –NH2, and the double bond starts at C2. The three-carbon parent is 'prop', the double …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.The hydrocarbon with molecular formula C20H42 is (A) Didodecane (B) Didecane (C) Dodidecane (D) Didocene (E) Eicosane
›Reveal solutionSolution
C20H42 fits the alkane formula CnH2n+2 with n=20; the straight-chain C20 alkane is named eicosane.
Derivation: Alkanes obey CnH2n+2. Setting 2n+2=42 gives n=20, so the molecule is a 20-carbon alkane. …
- KEAM 2025Set pha-2025-0424A4 marksMCQQ.Phenetole is (A) Ethoxybenzene (B) Methoxyethane (C) Methoxybenzene (D) 1-Methoxypropane (E) 2-Methoxypropane
›Reveal solutionSolution
Phenetole = ethyl phenyl ether =C6H5OC2H5= ethoxybenzene.
By analogy, anisole is methoxybenzene (C6H5OCH3); phenetole is its ethyl homologue, ethoxybenzene. Methoxyethane and 1-/2-methoxypropane are aliphatic eth …
- KEAM 2024Set eng-2024-06074 marksMCQQ.The IUPAC name of HOCH2(CH2)3CH2COCH3 (A) 2-oxo-heptan-7-ol (B) 7-hydroxyheptan-2-one (C) hydroxyheptan-6-one (D) 2-oxo-heptan-7-ol (E) hydroxy pentyl methyl ketone
›Reveal solutionSolution
[!TLDR]
The compound is a 7-carbon ketone with a terminal OH; naming the ketone as the senior group gives 7-hydroxyheptan-2-one.
Concept
When a molecule contains more than one functional group, the principal characteristic group (chosen by IUPAC seniority) takes the suffix and the lowest locant; others become prefixes. Ketones rank above alcohols in this order — a standard NCERT/CBSE nomenclature rule.
Solution
Expand the structure:
HO-CH2-CH2-CH2-CH2-CH2-CO-CH3
Counting carbons gives a chain of 7 (heptane skeleton). The functional groups are a ketone (C=O) and a hydroxyl (-OH).
Since a ketone is senior to an alcohol, the suffix is -one and the OH becomes a hydroxy prefix. Number the chain to give the ketone the lowest locant: …
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.Resorcinol is (A) Benzene-1, 3-diol (B) Benzene-1, 4-diol (C) Benzene-1, 2-diol (D) 3-Methylphenol (E) 4-Methylphenol
›Reveal solutionSolution
Resorcinol is benzene-1,3-diol.
Concept and Intuition
Resorcinol is a common dihydroxybenzene isomer; the three isomers are catechol (1,2), resorcinol (1,3) and hydroquinone (1,4).
Step-by-Step Solution
- Resorcinol has two -OH groups on a benzene ring.
- They occupy the meta (1,3) positions.
- Therefore resorcinol = benzene-1,3-diol. …
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.Which one of the following represents valeraldehyde? (A) CH3CH2CH2CH2CHO (B) CH3CH(CH3)CH2CHO (C) CH3CH(OCH3)CHO (D) (CH3)2CHCHO (E) CH3CH2CH(CH3)CHO
›Reveal solutionSolution
Valeraldehyde is pentanal, CH3CH2CH2CH2CHO.
Concept and Intuition
The common name valeraldehyde denotes the straight-chain five-carbon aldehyde, pentanal.
Step-by-Step Solution
- Valer- corresponds to a five-carbon (valeric acid, pentanoic acid) chain.
- The -aldehyde suffix places -CHO at the chain end.
- Straight-chain pentanal = CH3CH2CH2CH2CHO. …
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