Q.Name the following compounds according to IUPAC system of nomenclature:

Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The IUPAC name of the following structure is $\mathrm{CH_3-CO-C(CH_3)=C(CH_3)-COOH}$ (a ketone $\mathrm{CH_3-C(=O)-}$ joined to a carbon bearing a methyl branch, double bonded to a second carbon also bearing a methyl branch, which is joined to a $-COOH$ group) (A) 2-Carboxy-3-methylpent-2-en-3-one (B) 4-Carboxy-3-methylpent-3-en-2-one (C) 4-oxo-2,3-dimethylpent-2-enoic acid (D) 2,3-Dimethyl-4-oxopent-2-enoic acid

   ›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

   1. Identify the five-carbon backbone: $\mathrm{COOH-C(CH_3)=C(CH_3)-CO-CH_3}$.
   2. Since $-\mathrm{COOH}$ has higher seniority than the ketone, it is named as the suffix "-oic acid" and must be numbered C1.
   3. 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 = $\mathrm{CH_3}$.
   4. The double bond lies between C2 and C3: so the parent name is "pent-2-enoic acid".
   5. There is a methyl substituent on both C2 and C3: "2,3-dimethyl-".
   6. The ketone oxygen sits on C4, expressed as the prefix "4-oxo-" (since the acid already claims the suffix).
   7. Assembling alphabetically/by rule: 2,3-dimethyl-4-oxopent-2-enoic acid.

   Common Mistakes

   • Naming the ketone as the parent suffix ("...-one") and the acid as a prefix ("carboxy-") — this reverses IUPAC seniority (acid > ketone), giving the wrong-looking names in options (A)/(B).
   • Numbering from the ketone end instead of the $-\mathrm{COOH}$ end, which would misplace all the locants.
   • Forgetting that the double bond locant (2-ene) is determined once numbering starts correctly from C1 at the acid carbon.

   ✓Final answer

   The correct option is (D) — 2,3-Dimethyl-4-oxopent-2-enoic acid.

   ANSWER: D

2. 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:

   1. 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.
   2. 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

   1. Number the heptane chain so the two substituents get the lowest locant set: $\{3,4\}$.
   2. Cite substituents alphabetically in the name: ethyl (at 4) before methyl (at 3) → "4-ethyl-3-methylheptane".
   3. Reject (A)/(D), which reverse this citation order.
   4. For Y, recognise ring–$CH_2$–$CH(OH)$–$CH_2CH_3$ as butan-2-ol substituted with phenyl at C1 → "1-phenylbutan-2-ol".
   5. Reject (C)/(D)'s non-standard "hydroxy-" based names for Y.
   6. Only (B) has both names correct.

   Common Mistakes

   • Citing substituents in the order they were "found" while drawing/numbering, rather than reordering alphabetically for the final name.
   • Naming an alcohol using a "hydroxy" prefix when the OH is the only/principal functional group — IUPAC requires the "-ol" suffix in that case.

   ✓Final answer

   The correct option is (B) — 4-Ethyl-3-Methylheptane ; 1-Phenylbutan-2-ol.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The IUPAC name of the given compound is [Structure: a hex-5-ene chain bearing a hydroxyl, methyl and ethyl group on the same carbon — $\mathrm{CH_2=CH-CH_2-CH_2-C(OH)(CH_3)(C_2H_5)}$] (A) 5-Ethylhex-1-en-5-ol (B) 2-Ethylhex-5-en-2-ol (C) 5-Methylhept-1-en-5-ol (D) 3-Methylhept-6-en-3-ol

   ›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

   1. 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.
   2. 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).
   3. Now decide the numbering direction. IUPAC numbering priority: lowest locant to the principal characteristic group (suffix, here "-ol") beats lowest locant to unsaturation ("-ene").

   • Numbering from the vinyl (CH2=) end: OH-bearing carbon is C5, double bond is C1=C2. This would be named 5-methylhept-1-en-5-ol.
   • Numbering from the ethyl-terminal end: OH-bearing carbon is C3, double bond is C6=C7, methyl branch at C3. This gives 3-methylhept-6-en-3-ol.

   4. Compare OH locants: 5 (first numbering) vs 3 (second numbering). Since OH (the suffix group) must get the lowest possible locant, we choose the numbering giving OH = 3.
   5. Final name: 3-methylhept-6-en-3-ol.

   Common Mistakes

   • Picking the shorter (hexene) chain through the methyl group instead of the longer heptene chain through the ethyl group.
   • Giving the double bond priority for low locants over the principal characteristic group (-ol) — the suffix functional group always gets lowest locant first.

   ✓Final answer

   The correct option is (D) — 3-Methylhept-6-en-3-ol.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Match the following List I (Compound) and List II (Common name): A. $C_6H_5OH$ ... I. Quinol B. [Structure: 2,4,6-trinitrophenol — a benzene ring with $OH$ and $NO_2$ groups at the 2, 4 and 6 positions, $\mathrm{C_6H_2(OH)(NO_2)_3}$] ... II. Carbolic acid C. [Structure: 1,4-dihydroxybenzene (hydroquinone) — a benzene ring with $OH$ groups para to each other, $\mathrm{C_6H_4(OH)_2}$] ... III. p-Cresol D. [Structure: 4-methylphenol (p-cresol) — a benzene ring with $OH$ and $CH_3$ groups para to each other, $\mathrm{CH_3-C_6H_4-OH}$] ... IV. Picric acid Correct answer is (A) A-IV, B-II, C-I, D-III (B) A-II, B-IV, C-I, D-III (C) A-II, B-IV, C-III, D-I (D) A-IV, B-I, C-III, D-II

   ›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

   1. A. $C_6H_5OH$ (phenol) — historically used as a disinfectant, its common name is Carbolic acid $\Rightarrow$ A–II.
   2. B. 2,4,6-trinitrophenol — this compound's trivial name is literally Picric acid (used as an explosive/dye) $\Rightarrow$ B–IV.
   3. C. 1,4-dihydroxybenzene (hydroquinone) — commonly called Quinol (used in photographic developers) $\Rightarrow$ C–I.
   4. D. 4-methylphenol — this is literally p-Cresol $\Rightarrow$ D–III.
   5. Assembling: A-II, B-IV, C-I, D-III, matching option (B).

   Common Mistakes

   • Swapping quinol and carbolic acid (both are simple, unsubstituted-looking phenols at first glance) — quinol specifically has the para-dihydroxy pattern.
   • Missing that picric acid's own common name is given directly in the list, no derivation needed.

   ✓Final answer

   The correct option is (B) — A-II, B-IV, C-I, D-III.

   ANSWER: B

5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The IUPAC name of the following hydrocarbon is $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3}$ (A) 2, 5, 6 – Trimethyloctane (B) 2 – Ethyl – 3, 6 – dimethylheptane (C) 3, 4, 7 – Trimethyloctane (D) 2 – Ethyl – 2, 6 – dimethylheptane

   ›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

   1. Write out the structure and number the main chain left to right: $CH_3(1)-CH(CH_3)(2)-CH_2(3)-CH_2(4)-CH(CH_3)(5)-CH(CH_3)(6)-CH_2(7)-CH_3(8)$.
   2. 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.
   3. Methyl substituents sit at carbons 2, 5, and 6 when numbered from the left end.
   4. 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}.
   5. 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).
   6. Name: 2,5,6-trimethyloctane.

   Common Mistakes

   • Numbering from the wrong end and reporting {3,4,7} instead of applying the "lowest locants at first point of difference" rule.
   • Mistakenly trying to construct an "ethyl" branch (as in the ethyl-dimethylheptane distractor options) — since every branch here is a single, separate $CH_3$ group (not two adjacent branch carbons that could be renamed as one longer substituent chain), no ethyl group actually exists in this structure.

   ✓Final answer

   The correct option is (A) — 2, 5, 6 – Trimethyloctane.

   ANSWER: A

6. 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

   1. Place OH at C1, CH3 at C3 (m-cresol).
   2. Directing effects favour nitration at C4 and C6, avoiding sterically hindered C2 and electronically unfavoured C5.
   3. Numbering one way (OH=1→C2→C3(CH3)→C4(NO2)→C5→C6(NO2)) gives substituent locants {3,4,6}.
   4. 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}.
   5. Compare {2,4,5} vs {3,4,6} term by term: 2 < 3, so {2,4,5} is the lower set and is the correct IUPAC numbering.
   6. Name: nitro groups at 2 and 4, methyl at 5 → 5-methyl-2,4-dinitrophenol.

   Common Mistakes

   • Keeping the 'original' numbering direction of the starting m-cresol instead of re-optimising for lowest locants in the final product name.
   • Nitrating at the sterically hindered position between OH and CH3 instead of the two open activated positions.

   ✓Final answer

   The correct option is (B) — 5-methyl-2,4-dinitrophenol.

   ANSWER: B

7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Which of the following represents the structure of 'Terpineol'? (A) [FIGURE] (a cyclohexene ring with a $CH_3$ group on the double-bond carbon at the top, and a $-C(CH_3)_2OH$ group attached at the bottom ring carbon) (B) [FIGURE] (a cyclohexene ring with $CH_3$ on the double-bond carbon at the top, and an isopropyl group plus an $-OH$ on two adjacent lower ring carbons) (C) [FIGURE] (a benzene ring with an $-OH$ at one position, two $CH_3$ groups, and a $Cl$ substituent — a chlorocresol-type aromatic structure) (D) [FIGURE] (a cyclohexadiene ring with $CH_3$ at the top, and $-OH$ plus an isopropyl group on lower ring carbons) ![Four candidate structures for 'Terpineol', numbered 1-4 top to bottom:

   (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

   1. 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.
   2. 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).
   3. 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.
   4. 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.
   5. Structure (C) is an aromatic (benzene-ring) chlorophenol derivative — has nothing to do with terpineol's terpenoid skeleton.
   6. Structure (D) has two ring double bonds (a diene) and the $-OH$ directly on the ring — also inconsistent with terpineol's actual structure.

   Common Mistakes

   • Confusing terpineol's side-chain tertiary alcohol with a ring-bound (cyclic secondary) alcohol.
   • Mistaking an aromatic phenolic structure for a terpenoid (terpineol is derived from isoprene units, not benzene).

   ✓Final answer

   The correct option is (A) — the cyclohexene ring with $CH_3$ on the double-bond carbon and $-C(CH_3)_2OH$ at the opposite ring carbon.

   ANSWER: A

8. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Identify the correct option in which the compound is not named as per IUPAC (A) 1-Ethyl-3, 3-dimethyl cyclohexane (structure: cyclohexane ring bearing an ethyl substituent and a gem-dimethyl substituent) (B) Cyclohex-2-en-1-ol (structure: cyclohexene ring with OH at C1 and the double bond between C2-C3) (C) 2-Chloro-1-methyl-4-nitrobenzene (structure: benzene ring with $CH_3$, $Cl$ adjacent to it, and $NO_2$ para to $CH_3$) (D) 4-Ethyl-1-fluoro-2-nitrobenzene (structure: benzene ring with $F$, $NO_2$ adjacent to it, and $C_2H_5$ substituent)

   ›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

   1. The ring carries an ethyl group and a gem-dimethyl carbon, two positions apart around the ring.
   2. Numbering with the ethyl carbon as C1 gives dimethyl at C3,C3 → locant set $\{1,3,3\}$.
   3. Numbering with the gem-dimethyl carbon as C1 instead gives ethyl at C3 → locant set $\{1,1,3\}$.
   4. Compare term-by-term: both start with 1; at the second position, $1 < 3$, so $\{1,1,3\}$ is the lower (preferred) set.
   5. 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.
   6. Checking the others: (B) cyclohex-2-en-1-ol correctly gives the -ol suffix the lowest locant (1); (C) and (D) benzene derivatives were checked and their given numbering already achieves the minimum locant set (forced, not a tie), so they are correctly named.

   Common Mistakes

   • Assuming the substituent named first alphabetically (ethyl, before methyl) automatically gets locant 1 — the lowest-locant-SET rule takes priority; alphabetical order is only a tie-breaker.
   • Not checking both possible numbering directions before concluding a name is correct.

   ✓Final answer

   The correct option is (A) — 1-Ethyl-3,3-dimethyl cyclohexane is NOT correctly named as per IUPAC.

   ANSWER: A

9. 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

   1. Identify the parent chain: a straight 10-carbon chain (dec-) containing one $C=C$ double bond, one $-OH$, and three substituents (Br, ethyl, methyl).
   2. 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).
   3. Numbering from that end: methyl at C2, $-OH$ at C4, double bond C5=C6 ("5-ene"), ethyl at C5, bromo at C7.
   4. Numbering the other way would push $-OH$ to C7 instead of C4 — a higher locant for the principal group — so it is rejected.
   5. Arrange substituent prefixes alphabetically: bromo, ethyl, methyl → "7-Bromo-5-ethyl-2-methyldec-5-en-4-ol".
   6. This matches option (C) exactly; options (A) and (D) use the rejected (reverse) numbering giving $-OH$ at C7, which violates the lowest-locant-to-principal-group rule; option (B) uses correct locants but wrong alphabetical prefix order (ethyl should come before methyl, not after).

   Common Mistakes

   • Numbering to give the double bond the lowest locant instead of the principal characteristic group — the $-OH$ suffix takes priority over "ene" locants.
   • Listing substituent prefixes out of alphabetical order in the final name (bromo, ethyl, methyl is the correct order).

   ✓Final answer

   The correct option is (C) — 7-Bromo-5-ethyl-2-methyldec-5-en-4-ol.

   ANSWER: C

10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The correct decreasing order of priority for the functional group of organic compounds in the IUPAC method of nomenclature is (A) $-\mathrm{CHO} > -\mathrm{OH} > -\mathrm{CONH_2} > -\mathrm{COCl}$ (B) $-\mathrm{CONH_2} > -\mathrm{CHO} > -\mathrm{COCl} > -\mathrm{OH}$ (C) $-\mathrm{COCl} > -\mathrm{CONH_2} > -\mathrm{CHO} > -\mathrm{OH}$ (D) $-\mathrm{CHO} > -\mathrm{CONH_2} > -\mathrm{COCl} > -\mathrm{OH}$

    ›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

    1. Identify the four groups in the question: acid chloride (-COCl), amide (-CONH$_2$), aldehyde (-CHO), and alcohol (-OH).
    2. Apply the known IUPAC seniority list: acid halides rank above amides, amides rank above aldehydes, and aldehydes rank above alcohols.
    3. So the decreasing order is: $-\mathrm{COCl} > -\mathrm{CONH_2} > -\mathrm{CHO} > -\mathrm{OH}$.
    4. This matches option (C).

    Common Mistakes

    • Misremembering aldehydes as senior to amides — amides actually outrank aldehydes in the standard IUPAC seniority table.
    • Placing -OH above -CHO, when alcohols are actually one of the lowest-ranked oxygen groups among these four.

    ✓Final answer

    The correct option is (C) — $-\mathrm{COCl} > -\mathrm{CONH_2} > -\mathrm{CHO} > -\mathrm{OH}$.

    ANSWER: C

11. 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

    1. 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.
    2. Oxalic acid: structure $HOOC-COOH$. This is a 2-carbon chain where both carbons carry a $-COOH$ group (i.e. it's the simplest dicarboxylic acid, ethane with two carboxylic acid groups at C1 and C2). Its IUPAC name is ethanedioic acid (not propanedioic acid, which would be malonic acid, $HOOC-CH_2-COOH$, a 3-carbon diacid).
    3. Combining both: 4-Methylpent-3-en-2-one; Ethanedioic acid.

    Common Mistakes

    • Numbering mesityl oxide's chain from the wrong end, which would incorrectly place the double bond/substituent locants and give '3-methylpent-3-en-2-one' or similar wrong names.
    • Confusing oxalic acid (2 carbons, ethanedioic acid) with malonic acid (3 carbons, propanedioic acid).

    ✓Final answer

    The correct option is (A) — 4-Methylpent-3-en-2-one; Ethanedioic acid.

    ANSWER: A

12. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.Match the following List-I (compound) — List-II (common name) A) $\mathrm{HO-C_6H_4-OH}$ (benzene ring with OH substituents at para positions, 1,4-dihydroxybenzene) — I) Catechol B) $\mathrm{C_6H_5-O-CH_2-CH_3}$ (benzene ring with an $\mathrm{O-CH_2-CH_3}$ substituent) — II) Cumene C) benzene ring with OH substituents at two adjacent (ortho) carbons, 1,2-dihydroxybenzene — III) Phenetole D) $\mathrm{C_6H_5-CH(CH_3)_2}$ (benzene ring with a $\mathrm{CH(CH_3)_2}$ substituent) — IV) Quinol The correct answer is (A) A-IV, B-III, C-I, D-II (B) A-IV, B-I, C-II, D-III (C) A-III, B-I, C-IV, D-II (D) A-II, B-III, C-IV, D-I

    ›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

    1. A) $HO$-$C_6H_4$-$OH$ at the para (1,4) position is the classic photographic developer Quinol (also called hydroquinone) → matches IV.
    2. B) $C_6H_5$-$O$-$CH_2$-$CH_3$ is ethyl phenyl ether, whose trivial name is Phenetole → matches III.
    3. C) The 1,2- (ortho) dihydroxybenzene isomer is Catechol → matches I.
    4. D) $C_6H_5$-$CH(CH_3)_2$ (isopropylbenzene) is Cumene, the industrial precursor to phenol via the cumene process → matches II.
    5. So the full match is A-IV, B-III, C-I, D-II.

    Common Mistakes

    • Confusing catechol (ortho) with quinol/hydroquinone (para) — the position of the two $-OH$ groups is the distinguishing feature.
    • Mixing up phenetole (an ether) with cumene (an alkylbenzene) — phenetole has an oxygen in the side chain, cumene does not.

    ✓Final answer

    The correct option is (A) — A-IV, B-III, C-I, D-II.

    ANSWER: A