An aromatic compound 'A' (molecular formula $C_8H_8O$) gives a positive 2,4-DNP test. It gives a yellow precipitate of compound 'B' on treatment with iodine and sodium hydroxide solution. Compound 'A' does not give the Tollens or Fehling's test. On drastic oxidation with potassium permanganate it forms a carboxylic acid 'C' (molecular formula $C_7H_6O_2$), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.

An aromatic compound 'A' (molecular formula C₈H₈O) gives a positive 2,4-DNP test. It gives a yellow precipitat

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C_8H_8O

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Concept understanding — IUPAC Nomenclature

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

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.
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.
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 out

When 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.
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.
Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).

Worked Example

$CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3$ : 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 …

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The key is that compound A is an aromatic methyl ketone (acetophenone, $\ce{C6H5COCH3}$ ) — it gives a positive 2,4-DNP test (carbonyl), a positive iodoform test (methyl ketone), but fails Tollens/Fehling (not an aldehyde). Drastic oxidation cleaves the side chain to benzoic acid (C, $\ce{C7H6O2}$ ), which is also a byproduct of the iodoform reaction. So A = acetophenone, B = iodoform ( $\ce{CHI3}$ ), C = benzoic acid ( $\ce{C6H5COOH}$ ).

The problem is a classic organic identification puzzle from the IUPAC nomenclature and reactions chapter. The clues are all about functional group tests and oxidation behaviour. Let’s decode them one by one.

1. Molecular formula $\ce{C8H8O}$ and aromatic nature

The formula has 8 carbons, 8 hydrogens, and 1 oxygen. For an aromatic compound, the benzene ring itself accounts for $\ce{C6H5-}$ (6C, 5H). That leaves $\ce{C2H3O}$ for the side chain. The side chain must contain the oxygen — so it’s either a carbonyl group ( $\ce{C=O}$ ) or an alcohol/ether. But the tests will tell us which.

2. Positive 2,4-DNP test

This test (with 2,4-dinitrophenylhydrazine) gives an orange-red precipitate for any carbonyl compound — aldehyde or ketone. So A has a $\ce{C=O}$ group. The side chain is therefore an acyl group, not an alcohol or ether.

3. Positive iodoform test (iodine + NaOH gives yellow precipitate)

The iodoform test is specific for methyl ketones ( $\ce{R-CO-CH3}$ ) or compounds that can be oxidised to a methyl ketone (like ethanol or secondary alcohols with a $\ce{CH3CH(OH)-}$ group). The yellow precipitate is iodoform, $\ce{CHI3}$ . So A must contain the $\ce{COCH3}$ (acetyl) group. That fits the leftover $\ce{C2H3O}$ perfectly: $\ce{-COCH3}$ .

4. Negative Tollens and Fehling’s tests

These tests are positive for aldehydes (and some α-hydroxy ketones). A fails both, so it is not an aldehyde. This confirms A is a ketone — specifically an aromatic methyl ketone.

5. Drastic oxidation with $\ce{KMnO4}$ gives carboxylic acid C ( $\ce{C7H6O2}$ )

Drastic oxidation (hot, alkaline $\ce{KMnO4}$ ) cleaves alkyl side chains on benzene rings down to the ring, turning any carbon chain attached to the ring into a carboxyl group ( $\ce{-COOH}$ ). The product $\ce{C7H6O2}$ is benzoic acid ( $\ce{C6H5COOH}$ ). This tells us the benzene ring has a single carbon side chain that gets fully oxidised to $\ce{-COOH}$ . Since A is $\ce{C6H5COCH3}$ , oxidation removes the methyl carbon and converts the carbonyl carbon to carboxyl — exactly giving benzoic acid.

6. The iodoform reaction also produces C

In the iodoform reaction, a methyl ketone $\ce{R-COCH3}$ reacts with $\ce{I2/NaOH}$ to give $\ce{RCOONa}$ (the sodium salt of the carboxylic acid) and $\ce{CHI3}$ (yellow precipitate). For A, $\ce{R = C6H5-}$ , so the salt is sodium benzoate, which on acidification gives benzoic acid (C). This matches perfectly. …

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Method: Retro-synthesis with Functional Group Analysis

This method works backwards from the chemical tests and degradation products to deduce the structure.

Step 1 — Analyse the molecular formula and unsaturation

Formula of A: $\mathrm{C_8H_8O}$
Saturation formula for 8 carbons: $\mathrm{C_8H_{18}}$
Degree of unsaturation = $\frac{2(8) + 2 - 8}{2} = 5$

Five degrees of unsaturation strongly suggests a benzene ring (4 unsaturations) plus one more double bond or ring.

Step 2 — Interpret the chemical tests

Test	Result	What it tells us
2,4-DNP test	Positive	Contains a carbonyl group ( $\mathrm{C=O}$ ) — aldehyde or ketone
Tollens / Fehling’s test	Negative	Not an aldehyde → must be a ketone
Iodoform test (I₂ + NaOH)	Yellow precipitate (B)	Positive iodoform test → methyl ketone ( $\mathrm{CH_3CO-}$ ) or ethanol derivative

So A is an aromatic methyl ketone.

Step 3 — Identify compound B (iodoform reaction)

The iodoform reaction:

\mathrm{RCOCH_3 + 3I_2 + 4NaOH \rightarrow RCOONa + CHI_3 \downarrow + 3NaI + 3H_2O}

B is the yellow precipitate: iodoform ( $\mathrm{CHI_3}$ )

Step 4 — Identify compound C (oxidation product)

Drastic oxidation with $\mathrm{KMnO_4}$ gives C: $\mathrm{C_7H_6O_2}$
This formula matches benzoic acid ( $\mathrm{C_6H_5COOH}$ )
Also formed alongside iodoform in the iodoform test (as sodium benzoate, then acidified)

This means the aromatic ring has a methyl ketone side chain that gets oxidised to a carboxyl group.

Step 5 — Deduce compound A

A = aromatic methyl ketone = $\mathrm{C_6H_5COCH_3}$
Check formula: $\mathrm{C_8H_8O}$ ✓
This is acetophenone

Final Answer

| Compound | Name | Structure | …

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Step 1: Understanding the clues

Molecular formula of A: $C_8H_8O$

Degree of unsaturation = $\frac{2C + 2 - H}{2} = \frac{16+2-8}{2} = 5$ → aromatic ring (4) + one more double bond (likely C=O).
Positive 2,4-DNP test → A has a carbonyl group (aldehyde or ketone).
Yellow precipitate with $I_2/NaOH$ → Iodoform test positive → A must have a $CH_3CO-$ group (methyl ketone) or $CH_3CH(OH)-$ group (but here it’s a ketone, since no alcohol).
No Tollens or Fehling’s test → A is not an aldehyde → it’s a ketone.
Drastic oxidation with $KMnO_4$ gives $C_7H_6O_2$ → benzoic acid ( $C_6H_5COOH$ ).

So A has a benzene ring with a side chain that gets oxidised to $COOH$ .
Same carboxylic acid C is also formed along with the yellow compound → the iodoform reaction on A gives benzoic acid + iodoform (yellow).

Step 2: Identifying A, B, C

A must be acetophenone ( $C_6H_5COCH_3$ ).
- IUPAC name: 1-phenylethanone (or phenyl methyl ketone).
- Iodoform test: $C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow + 3NaI + 3H_2O$ Yellow precipitate B = iodoform ( $CHI_3$ ). Carboxylic acid C = benzoic acid ( $C_6H_5COOH$ ).
Oxidation with $KMnO_4$ :

$C_6H_5COCH_3 \xrightarrow[KMnO_4, \Delta]{} C_6H_5COOH + CO_2 + H_2O$

Step 3: Common mistakes & how to avoid them

Mistake	Why it happens	How to avoid
Thinking A is an aldehyde	Positive 2,4-DNP is common to both aldehydes and ketones.	Always check Tollens/Fehling’s — if negative, it’s a ketone.
Ignoring the iodoform clue	Students forget that only methyl ketones (or ethanol/ secondary alcohols with $CH_3CH(OH)-$ ) give iodoform.	Memorise: Iodoform test positive → $CH_3CO-$ or $CH_3CH(OH)-$ group.
Wrong oxidation product	Assuming the side chain becomes $CO_2$ or something else.	Count carbons: $C_8H_8O$ → $C_7H_6O_2$ means one carbon lost (as $CO_2$ ). That fits acetophenone → benzoic acid.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 35 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$\mathrm{CH_3-CO-C(CH_3)=C(CH_3)-COOH}$
›Reveal solutionSolution
Carboxylic acid outranks ketone in IUPAC seniority, so $-\mathrm{COOH}$ must be C1 and the ketone is named as a substituent prefix "oxo-". Answer: 2,3-dimethyl-4-oxopent-2-enoic acid.

Concept and Intuition

IUPAC nomenclature ranks functional groups by seniority for choosing the "parent" suffix: carboxylic acids outrank ketones. So whenever both a $-\mathrm{COOH}$ and a $\mathrm{C=O}$ (ketone) are present in the same chain, the compound is named as a "...oic acid" with the ketone expressed as an "oxo-" prefix, and numbering must start from the end nearest the $-\mathrm{COOH}$ group (giving it locant 1) regardless of where the ketone or double bond fall.

Step-by-Step Solution

Identify the five-carbon backbone: $\mathrm{COOH-C(CH_3)=C(CH_3)-CO-CH_3}$ .

Since $-\mathrm{COOH}$ has higher seniority than the ketone, it is named as the suffix "-oic acid" and must be numbered C1.

Number from the $\mathrm{COOH}$ end: C1 = $\mathrm{COOH}$ ; C2 = $=\mathrm{C(CH_3)}-$ (attached to COOH); C3 = $\mathrm{C(CH_3)}=$ ; C4 = $\mathrm{C(=O)}-$ ; C5 = $\mathrm{CH_3}$ .

The double bond lies between C2 and C3: so the parent name is "pent-2-enoic acid".

There is a methyl substituent on both C2 and C3: "2,3-dimethyl-".

The ketone oxygen sits on C4, expressed as the prefix "4-oxo-" (since the acid already claims the suffix). …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Q. The correct IUPAC names of the compounds X and Y given below are respectively [FIGURE: X is a branched heptane-type skeletal structure with an ethyl and a methyl substituent; Y is a phenyl group attached via CH2 to a CH(OH) carbon which continues to an ethyl group (i.e. a phenyl-substituted butan-2-ol)] (A) 3-Methyl-4-ethylheptane ; 1-Phenylbutan-2-ol (B) 4-Ethyl-3-Methylheptane ; 1-Phenylbutan-2-ol (C) 3-(sec-butyl)-hexane ; 2-Hydroxy-1-phenylbutane (D) 3-Methyl-4-Ethylheptane ; 2-Hydroxybutylbenzene$
›Reveal solutionSolution
Naming heptane with adjacent methyl/ethyl branches must alphabetise the substituent names (ethyl before methyl) even though the locants (3,4) are the same either way; the second structure is simply 1-phenylbutan-2-ol. Only option (B) gets both names right.

Concept and Intuition

IUPAC substitutive nomenclature has two separate rules that both matter here:

Lowest locants — number the main chain from whichever end gives the substituents the lowest possible position numbers. For two adjacent substituents on a 7-carbon chain, one numbering gives $\{3,4\}$ and the other gives $\{4,5\}$ ; since $3<4$ at the first point of difference, $\{3,4\}$ is chosen — this fixes which carbon is 3 and which is 4, but not yet how they're written in the name.

Alphabetical citation — once locants are fixed, the substituents must be listed in the name in alphabetical order of their names, regardless of which locant number each one carries. Since "ethyl" (e) precedes "methyl" (m) alphabetically, the name must read "...-ethyl-...-methyl-heptane", i.e. 4-ethyl-3-methylheptane — even though methyl sits at the numerically smaller position 3.

Options (A) and (D) both list "3-Methyl-4-ethyl..." — citing methyl before ethyl — which violates the alphabetisation rule, even though the locants are numerically correct. Option (B) lists "4-Ethyl-3-Methyl...", correctly alphabetised.

For structure Y: ring– $CH_2$ – $CH(OH)$ – $CH_2$ – $CH_3$ is a four-carbon chain (butane) bearing $OH$ on the second carbon (numbering from the phenyl-bearing end so the alcohol gets the lowest locant, and phenyl sits on C1): this is exactly 1-phenylbutan-2-ol. "2-Hydroxy-1-phenylbutane" and "2-Hydroxybutylbenzene" (options C, D) are both non-standard — the hydroxyl group, being the principal characteristic group here, must be expressed as the suffix "-ol", not as a "hydroxy-" prefix.

Step-by-Step Solution …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\mathrm{CH_2=CH-CH_2-CH_2-C(OH)(CH_3)(C_2H_5)}$
›Reveal solutionSolution
This tests IUPAC chain-selection (longest chain through the principal group) and numbering priority (lowest locant to the suffix -ol over the double bond). The answer is 3-methylhept-6-en-3-ol.

Concept and Intuition

When naming a polyfunctional/branched molecule, two rules matter here: (1) choose the longest continuous carbon chain that contains the principal characteristic group (here, the carbon bearing –OH); if there's a choice among chains of different lengths through that carbon, pick the longest one, even if it means treating a substituent as part of chain vs branch differently. (2) Once the chain is fixed, number it so the principal characteristic group (the one cited as suffix, i.e. -ol) gets the lowest locant — this takes priority over giving the double bond the lowest locant.

Step-by-Step Solution

Draw the structure: $CH_2{=}CH{-}CH_2{-}CH_2{-}C(OH)(CH_3)(CH_2CH_3)$ . The central carbon bearing OH is attached to: (a) a 4-carbon chain ending in the double bond ( $CH_2{=}CH{-}CH_2{-}CH_2{-}$ ), (b) a methyl group, (c) an ethyl group.

To maximize chain length while keeping the OH-bearing carbon in the chain, extend through the ethyl group rather than the methyl: this chain is $CH_2{=}CH{-}CH_2{-}CH_2{-}C(OH)(CH_3){-}CH_2{-}CH_3$ — count the carbons: 1,2,3,4,5,6,7 → seven carbons, i.e. a heptene skeleton, with a methyl branch left over at C5 (numbering from the vinyl end) or C3 (numbering from the ethyl end).

Now decide the numbering direction. IUPAC numbering priority: lowest locant to the principal characteristic group (suffix, here "-ol") beats lowest locant to unsaturation ("-ene"). …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$C_6H_5OH$
›Reveal solutionSolution
Matching each phenolic compound to its trivial/common name: phenol = carbolic acid, 2,4,6-trinitrophenol = picric acid, hydroquinone = quinol, p-cresol = p-cresol.

Concept and Intuition

Many simple phenolic compounds carry historic trivial names that are still in common industrial/medical use, independent of their IUPAC names. Recognising the structure (substitution pattern and functional groups) is the key to recalling the correct trivial name.

Step-by-Step Solution

A. $C_6H_5OH$ (phenol) — historically used as a disinfectant, its common name is Carbolic acid $\Rightarrow$ A–II.

B. 2,4,6-trinitrophenol — this compound's trivial name is literally Picric acid (used as an explosive/dye) $\Rightarrow$ B–IV.

C. 1,4-dihydroxybenzene (hydroquinone) — commonly called Quinol (used in photographic developers) $\Rightarrow$ C–I.

D. 4-methylphenol — this is literally p-Cresol $\Rightarrow$ D–III. …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3}$
›Reveal solutionSolution
The longest chain is an 8-carbon octane bearing three methyl branches; numbering to give the lowest locant set yields 2,5,6-trimethyloctane.

Concept and Intuition

IUPAC naming requires (1) finding the longest continuous carbon chain, (2) identifying substituents, and (3) numbering the chain from whichever end gives the lowest set of locants to the substituents (compared term-by-term at the first point of difference).

Step-by-Step Solution

Write out the structure and number the main chain left to right: $CH_3(1)-CH(CH_3)(2)-CH_2(3)-CH_2(4)-CH(CH_3)(5)-CH(CH_3)(6)-CH_2(7)-CH_3(8)$ .

This is an 8-carbon chain (octane) — check it's the longest: all branches are single methyl groups, so none of them can extend the main chain further; 8 carbons is indeed the longest possible chain.

Methyl substituents sit at carbons 2, 5, and 6 when numbered from the left end.

Now number from the right end instead: original C8→1, C7→2, C6→3, C5→4, C4→5, C3→6, C2→7, C1→8. The methyls (originally at C2, C5, C6) now sit at positions 7, 4, 3 — locant set {3,4,7}.

Compare the two locant sets at the first point of difference: {2,5,6} vs {3,4,7} → 2 < 3, so the left-to-right numbering ({2,5,6}) is preferred (lower locants rule). …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$Q. What is the IUPAC name of the compound formed when m-cresol is subjected to dinitration? (A) 3-methyl-4,6-dinitrophenol (B) 5-methyl-2,4-dinitrophenol (C) 2-methyl-4,6-dinitrophenol (D) 4-methyl-2,6-dinitrophenol$
›Reveal solutionSolution
Nitration of m-cresol is directed by both –OH and –CH3 to the two positions that avoid steric crowding; correctly renumbering for lowest locants gives 5-methyl-2,4-dinitrophenol.

Concept and Intuition

In m-cresol (–OH at C1, –CH3 at C3), both substituents are activating, ortho/para-directing groups. The ring position between them (C2) is doubly activated but sterically hindered, so electrophilic nitration preferentially occurs at the two positions that are activated by both groups without steric clash: C4 (para to OH, ortho to CH3) and C6 (ortho to OH, para to CH3). Once the substitution pattern is fixed on the ring, IUPAC numbering (with OH's carbon fixed at C1, since it carries the principal -ol suffix) is chosen in whichever direction gives the lowest set of locants to all substituents.

Step-by-Step Solution

Place OH at C1, CH3 at C3 (m-cresol).

Directing effects favour nitration at C4 and C6, avoiding sterically hindered C2 and electronically unfavoured C5.

Numbering one way (OH=1→C2→C3(CH3)→C4(NO2)→C5→C6(NO2)) gives substituent locants {3,4,6}.

Numbering the other way around the ring (OH=1→C2(NO2, was old C6)→C3→C4(NO2, was old C4)→C5(CH3, was old C3)→C6) gives locants {2,4,5}. …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$CH_3$
(1) a cyclohexene r](/figures/ap-eapcet/eng-2026-05-18-AN-chemistry-q154.svg)
›Reveal solutionSolution
Terpineol is p-menth-1-en-8-ol: a cyclohexene ring with a ring methyl on the double bond and a tertiary alcohol side-chain $-C(CH_3)_2OH$ hanging off the ring — matching structure (A), not the ring-bound-OH or aromatic alternatives.

Concept and Intuition

Terpineol is a monoterpenoid, biosynthetically related to the menthane (p-menthane) skeleton — a cyclohexane/cyclohexene ring with a methyl group at C1 and an isopropyl-derived group at C4. In $\alpha$ -terpineol specifically, the isopropyl-derived group has been oxidised to a tertiary alcohol, giving the side chain $-C(CH_3)_2OH$ instead of a simple isopropyl. The $-OH$ is on the side-chain carbon, NOT directly on the ring.

Step-by-Step Solution

Recall the terpineol skeleton: it is derived from limonene-type monoterpenes, i.e. a cyclohexene ring with a methyl group on the ring at the double bond, and (at the para-like ring position) a three-carbon side chain.

In $\alpha$ -terpineol, that three-carbon side chain is oxidised at its central carbon to a tertiary alcohol: $-C(CH_3)_2-OH$ , attached to the ring by a single bond (not part of the ring itself).

This matches structure (A): cyclohexene ring, $CH_3$ on the double-bond carbon (top), and the $-C(CH_3)_2OH$ group on the ring carbon on the other side (bottom) — a side-chain alcohol, not a ring-bound one.

Structure (B) instead places the $-OH$ directly ON the ring with an adjacent isopropyl group — that describes a different terpene alcohol (a cyclic/ring alcohol), not terpineol. …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$CH_3$
›Reveal solutionSolution
Option (A)'s name violates the lowest-locants rule — the correct IUPAC name is 3-ethyl-1,1-dimethylcyclohexane, not 1-ethyl-3,3-dimethylcyclohexane.

Concept and Intuition

When a ring or chain carries several substituents, IUPAC numbering is fixed by comparing the entire set of locants term-by-term at the first point of difference, and choosing whichever numbering gives the lowest set overall — this comes before any alphabetical tie-breaking (which only applies when two numbering choices give identical locant sets).

Step-by-Step Solution

The ring carries an ethyl group and a gem-dimethyl carbon, two positions apart around the ring.

Numbering with the ethyl carbon as C1 gives dimethyl at C3,C3 → locant set $\{1,3,3\}$ .

Numbering with the gem-dimethyl carbon as C1 instead gives ethyl at C3 → locant set $\{1,1,3\}$ .

Compare term-by-term: both start with 1; at the second position, $1 < 3$ , so $\{1,1,3\}$ is the lower (preferred) set.

Therefore the correct name is 3-Ethyl-1,1-dimethylcyclohexane, and the given name '1-Ethyl-3,3-dimethylcyclohexane' (option A) is NOT correctly IUPAC-named. …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$Q. The IUPAC name of the following compound is [FIGURE] (a branched skeletal structure: an ethyl group (CH3-CH2-) attached to a carbon that carries a C=C double bond going down to a CH which connects to a CH-Br carbon, which continues as a propyl chain -CH2-CH2-CH3; the same double-bond-bearing carbon also connects on the other side to a CH-OH carbon, then a CH2, then a CH bearing a methyl branch and terminating in a CH3 -- i.e. a line structure of 7-bromo-5-ethyl-2-methyldec-5-en-4-ol) (A) 6-Ethyl-9-methyl-4-bromodec-5-en-7-ol (B) 7-Bromo-2-methyl-5-ethyldec-5-en-4-ol (C) 7-Bromo-5-ethyl-2-methyldec-5-en-4-ol (D) 4-Bromo-6-ethyl-9-methyldec-5-en-7-ol$
›Reveal solutionSolution
This tests IUPAC numbering priority: the principal characteristic group (here, $-OH$ , suffixed as $-ol$ ) must receive the lowest possible locant, which fixes the numbering direction of the parent chain over the alternative that would look inverted.

Concept and Intuition

When naming a polyfunctional compound, IUPAC rules require choosing the numbering direction that gives the lowest locant to the principal characteristic group expressed as a suffix (here $-ol$ ), ahead of unsaturation (ene) and substituents (bromo, ethyl, methyl), which are only used as tie-breakers if the principal group's locant is the same either way.

Step-by-Step Solution

Identify the parent chain: a straight 10-carbon chain (dec-) containing one $C=C$ double bond, one $-OH$ , and three substituents (Br, ethyl, methyl).

Number from the end nearer the $-OH$ group so that it gets the lower possible locant (Rule: principal characteristic group gets priority for lowest locant).

Numbering from that end: methyl at C2, $-OH$ at C4, double bond C5=C6 ("5-ene"), ethyl at C5, bromo at C7.

Numbering the other way would push $-OH$ to C7 instead of C4 — a higher locant for the principal group — so it is rejected.

Arrange substituent prefixes alphabetically: bromo, ethyl, methyl → "7-Bromo-5-ethyl-2-methyldec-5-en-4-ol". …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$-\mathrm{CHO} > -\mathrm{OH} > -\mathrm{CONH_2} > -\mathrm{COCl}$
›Reveal solutionSolution
This tests the IUPAC seniority (priority) order of functional groups for choosing the principal characteristic group in nomenclature. Among acid halide, amide, aldehyde and alcohol, the correct decreasing priority is $-\mathrm{COCl} > -\mathrm{CONH_2} > -\mathrm{CHO} > -\mathrm{OH}$ .

Concept and Intuition

IUPAC nomenclature ranks characteristic groups by a fixed seniority order (roughly following decreasing oxidation state/reactivity of the carbon-based functional groups): cations > carboxylic acids > sulfonic acids > anhydrides > esters > acid halides > amides > nitriles > aldehydes > ketones > alcohols > amines > ethers. The group highest in this order is chosen as the principal characteristic group (suffix), and all others are cited as prefixes.

Step-by-Step Solution

Identify the four groups in the question: acid chloride (-COCl), amide (-CONH $_2$ ), aldehyde (-CHO), and alcohol (-OH).

Apply the known IUPAC seniority list: acid halides rank above amides, amides rank above aldehydes, and aldehydes rank above alcohols. …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$Q. IUPAC names of mesityl oxide and oxalic acid are respectively (A) 4-Methylpent-3-en-2-one; Ethanedioic acid (B) 4-Methylpent-3-en-2-one; Propanedioic acid (C) 3-Methylpent-3-en-2-one; Propanedioic acid (D) 3-Methylpent-3-en-2-one; Ethanedioic acid$
›Reveal solutionSolution
This tests IUPAC naming of two common named compounds: mesityl oxide is 4-methylpent-3-en-2-one, and oxalic acid (the two-carbon diacid) is ethanedioic acid.

Concept and Intuition

IUPAC naming of a ketone requires finding the longest carbon chain that includes the carbonyl carbon, numbering from the end that gives the carbonyl the lowest possible locant, and citing all substituents/unsaturations with their locants. For a simple dicarboxylic acid, the name is built directly off the number of carbons in the chain (including both -COOH carbons), using the suffix '-dioic acid'.

Step-by-Step Solution

Mesityl oxide: structure $(CH_3)_2C=CH-CO-CH_3$ . The longest chain through the carbonyl carbon is 5 carbons: $CH_3-CO-CH=C(CH_3)-CH_3$ . Numbering from the carbonyl end (to give it the lowest locant, position 2): $C1(CH_3)-C2(=O)-C3(H)=C4(CH_3)-C5(H_3)$ , i.e. a methyl branch sits on C4. This gives 4-methylpent-3-en-2-one. …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{HO-C_6H_4-OH}$
›Reveal solutionSolution
Matching common (trivial) names of aromatic compounds: 1,4-dihydroxybenzene = Quinol, phenetole = ethyl phenyl ether, catechol = 1,2-dihydroxybenzene, cumene = isopropylbenzene.

Concept and Intuition

Many simple aromatic compounds carry historical trivial names that are still in common industrial/chemical usage, and these names are frequently tested for recall in organic chemistry.

Step-by-Step Solution

A) $HO$ - $C_6H_4$ - $OH$ at the para (1,4) position is the classic photographic developer Quinol (also called hydroquinone) → matches IV.

B) $C_6H_5$ - $O$ - $CH_2$ - $CH_3$ is ethyl phenyl ether, whose trivial name is Phenetole → matches III.

C) The 1,2- (ortho) dihydroxybenzene isomer is Catechol → matches I.

D) $C_6H_5$ - $CH(CH_3)_2$ (isopropylbenzene) is Cumene, the industrial precursor to phenol via the cumene process → matches II. …
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