Q.Classify the following as primary, secondary and tertiary alcohols:

Showing the 12 most recent of 15 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.

   [!FORMULA] $\text { Identify the correct structure of o-ethyl anisole. }$

   (A) (B) (C) (D)

   ›Reveal solutionSolution

   The name “o-ethyl anisole” tells us the substituents are an ethyl group and a methoxy group in the ortho (1,2) position. The correct structure is therefore the one with –OCH₃ and –C₂H₅ on adjacent carbons — option (D).

   The key is to decode the name “o-ethyl anisole” systematically. Anisole is the common name for methoxybenzene (C₆H₅–OCH₃). The prefix “o-” (ortho) indicates that the ethyl group is attached to the ring carbon adjacent to the methoxy group. So the molecule is 1-methoxy-2-ethylbenzene.

   Let’s check each option:

   1. Option (A) has –OC₂H₅ (ethoxy) and –C₂H₅. That would be o-ethyl phenetole, not anisole.
   2. Option (B) has –COCH₃ (acetyl) and –C₂H₅. That’s an ortho-substituted acetophenone, not anisole.
   3. Option (C) has –OH (hydroxyl) and –C₂H₅. That’s o-ethyl phenol, not anisole.
   4. Option (D) has –OCH₃ (methoxy) and –C₂H₅. That matches exactly: anisole (methoxybenzene) with an ethyl group in the ortho position.

   Watch out

   A common mistake is confusing “ethoxy” (–OC₂H₅) with “methoxy” (–OCH₃). Anisole specifically means the methoxy group, not any other alkoxy group.

   Tip

   Remember: “anisole” = methoxybenzene. If you see “phenetole” it’s ethoxybenzene. The prefix tells you the alkoxy chain length.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\text { Which one of the following is not the correct IUPAC name of the compound? }$

   (A) $\text { 2-Bromo-3-methylbut-2-en-1-ol: }\left(\mathrm{CH}_3\right)_2-\mathrm{C}=\mathrm{C}(\mathrm{Br})-\mathrm{CH}_2 \mathrm{OH}$ (B) $\text { 2,5-Dimethylhexane-1,3-diol: }\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{CH} 2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}\left(\mathrm{CH}_2 \mathrm{OH}\right)-\mathrm{CH}_3$ (C) $\text { 3-Methylbutoxybenzene: } \mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\left(\mathrm{CH}_2\right)_2-\mathrm{CH}-\left(\mathrm{CH}_3\right)_2$ (D)

   ›Reveal solutionSolution

   [!TLDR]

   Options A, B and C are each valid IUPAC names for the structures shown, so the name that is NOT correct is option (D).

   Concept

   An IUPAC name is correct only when the parent chain is the longest chain that carries the principal characteristic group with the lowest set of locants, and every substituent is numbered correctly (CBSE/NCERT nomenclature rules).

   Solution

   • (A) $(\mathrm{CH_3})_2\mathrm{C{=}C(Br){-}CH_2OH}$: the chain is but-2-en-1-ol (OH at C1, double bond C2=C3) with Br on C2 and a methyl on C3 $\Rightarrow$ 2-bromo-3-methylbut-2-en-1-ol. Correct.
   • (B) $(\mathrm{CH_3})_2\mathrm{CH{-}CH_2{-}CH(OH){-}CH(CH_2OH){-}CH_3}$: the longest chain passing through both -OH groups is six carbons, with OH at 1 and 3 and methyls at 2 and 5 $\Rightarrow$ 2,5-dimethylhexane-1,3-diol. Correct.
   • (C) $\mathrm{C_6H_5{-}O{-}CH_2CH_2CH(CH_3)_2}$: the alkyl group $-\mathrm{CH_2CH_2CH(CH_3)_2}$ is 3-methylbutyl, so the ether is 3-methylbutoxybenzene. Correct.

   Since the names in A, B and C are all correct, the name that is not correct is the one in option (D).

   [!ANSWER]

   (D) option (D) is the incorrect IUPAC name

3. KCET 2026Set D31 markMCQ
   Q.The correct IUPAC name of $CH_3\text{-}C(CH_3)_2\text{-}O\text{-}C_2H_5$ is (A) Tertiary butoxy ethane (B) 1, 1-Dimethyl-1-ethoxyethane (C) 2-ethoxy-2-methyl propane (D) Ethoxy tertiary butane

   ›Reveal solutionSolution

   Ethers are named in the IUPAC substitutive system as an alkane parent chain carrying an "alkoxy" substituent, not by the older functional-class "alkyl alkyl ether" style.

   Step 1 — Identify the skeleton

   The compound is $(CH_3)_3C-O-C_2H_5$: a central carbon bearing three methyl groups (the tert-butyl-type carbon) connected through an oxygen to an ethyl group.

   Step 2 — Choose the parent chain

   Stripping away the ethoxy (-O-$C_2H_5$) substituent, the remaining carbon skeleton is $CH_3-C(CH_3)_2-CH_3$, a three-carbon chain (propane) with a methyl branch at C2. This gives the base name 2-methylpropane, and C2 also carries the ethoxy group.

   Step 3 — Name the ether as an alkoxy-substituted alkane

   The $-O-C_2H_5$ group becomes the substituent prefix ethoxy, attached at C2. Combining the substituents alphabetically on the 2-methylpropane parent gives 2-ethoxy-2-methylpropane.

   Step 4 — Why the other options fail

   "Tertiary butoxy ethane" and "Ethoxy tertiary butane" use non-IUPAC common-name fragments ("tert-butoxy", "tertiary butane"), and "1,1-Dimethyl-1-ethoxyethane" numbers the chain incorrectly (C1 cannot simultaneously be a chain terminus and carry two methyl substituents in a valid parent-chain sense here).

   ✓Final answer

   The correct option is (C) — 2-ethoxy-2-methylpropane.

4. KCET 2025Set D-41 markMCQ
   Q.Among the following, identify the compound that is not an isomer of hexane (A) $\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3}$ (B) $\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3}$ (C) (D) $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH_3}$

   ›Reveal solutionSolution

   Isomers must share the molecular formula $\mathrm{C_6H_{14}}$; the cyclopentane ring in option (C) is $\mathrm{C_5H_{10}}$, so it is not an isomer of hexane.

   Step 1 — What must be true of an isomer of hexane.

   Hexane is $\mathrm{C_6H_{14}}$ (a saturated, acyclic alkane, $\mathrm{C_nH_{2n+2}}$ with $n=6$). Any isomer of hexane must have exactly the same molecular formula, $\mathrm{C_6H_{14}}$: six carbons, fourteen hydrogens, zero degrees of unsaturation.

   Step 2 — Count the atoms in each option.

   (A) $\mathrm{CH_3-CH_2-CH(CH_3)-CH_2-CH_3}$ — 3-methylpentane.

   Carbons: $1+1+1+1+1$ in the pentane chain $= 5$, plus the $\mathrm{CH_3}$ branch $= 6$. Saturated and acyclic $\Rightarrow \mathrm{C_6H_{14}}$. Is an isomer.

   (B) $\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3}$ — n-hexane itself.

   Six carbons in a straight chain $\Rightarrow \mathrm{C_6H_{14}}$. It is hexane (the identity structure), so it certainly is not the odd one out.

   (D) $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH_3}$ — 2-methylpentane.

   Five-carbon chain + one methyl branch $= 6$ carbons; saturated, acyclic $\Rightarrow \mathrm{C_6H_{14}}$. Is an isomer.

   (C) the drawn five-membered ring — cyclopentane.

   Five carbons, each bearing two hydrogens:

   $\mathrm{C_5H_{10}} \quad (\mathrm{C_nH_{2n}},\; n=5)$

   This differs from $\mathrm{C_6H_{14}}$ on both counts — one carbon fewer, and it has one degree of unsaturation (the ring), so four hydrogens fewer than a $\mathrm{C_5}$ alkane would need to match. It is not an isomer of hexane.

   Step 3 — Sanity check with degrees of unsaturation.

   $\text{DoU} = \frac{2(5)+2-10}{2} = \frac{12-10}{2} = 1$

   One degree of unsaturation (the ring) — whereas hexane and all its isomers must have DoU $= 0$. Confirms (C).

   ✓Final answer

   The correct option is (C) — the five-membered ring, cyclopentane ($\mathrm{C_5H_{10}}$), which is not an isomer of hexane ($\mathrm{C_6H_{14}}$).

   ANSWER: C

5. KCET 2025Set D-41 markMCQ
   Q.The organic compound can be classified as ______________ (A) Allylic halide (B) Benzyl halide (C) Aryl halide (D) Alkyl halide

   ›Reveal solutionSolution

   Classify by which kind of carbon bears the halogen: here the C–Cl carbon is $sp^3$ and directly attached to an aromatic ring — the definition of a benzylic halide.

   Step 1 — Read the structure.

   From the figure: a benzene ring is bonded to a carbon that also carries two $\mathrm{CH_3}$ groups and one $\mathrm{Cl}$. So the compound is

   $\mathrm{C_6H_5{-}\overset{\mathrm{CH_3}}{\underset{\mathrm{CH_3}}{C}}{-}Cl} \;\equiv\; \text{2-chloro-2-phenylpropane}$

   The carbon bearing the Cl is $sp^3$ (four σ bonds: to the ring, to two methyls, to Cl) and it is directly attached to the aromatic ring.

   Step 2 — The classification rules for halides.

   Halides are classified by the hybridisation and environment of the carbon holding the halogen:

   Class	Where the halogen sits
   Alkyl halide	on an $sp^3$ carbon of a plain alkyl chain (no ring or double bond adjacent)
   Allylic halide	on an $sp^3$ carbon adjacent to a C=C double bond
   Benzylic halide	on an $sp^3$ carbon directly attached to an aromatic ring
   Aryl halide	on an $sp^2$ carbon of the ring itself (e.g. chlorobenzene)

   Step 3 — Apply the rules.

   • Is it aryl? No — the Cl is not on a ring carbon. In chlorobenzene the C–Cl carbon is one of the six aromatic $sp^2$ carbons. Here the C–Cl carbon is an extra, $sp^3$ carbon hanging off the ring. Rejected.
   • Is it allylic? No — there is no isolated C=C next to the C–Cl carbon; the unsaturation is an aromatic ring, which makes it benzylic, not allylic. Rejected.
   • Is it a plain alkyl halide? It is $sp^3$, but the aryl attachment is the defining structural feature and the more specific class always wins. Rejected.
   • Is it benzylic? The Cl-bearing $sp^3$ carbon is bonded straight to the benzene ring — yes. Specifically it is a tertiary benzylic halide.

   Step 4 — Why the distinction matters chemically.

   Benzylic halides ionise readily because the resulting benzylic carbocation is stabilised by resonance delocalisation into the ring, so they are exceptionally reactive by $\mathrm{S_N1}$. Aryl halides do the exact opposite — the C–Cl bond has partial double-bond character and is inert to ordinary nucleophilic substitution. Getting the class right therefore predicts the reactivity.

   ✓Final answer

   The correct option is (B) — Benzyl halide.

   ANSWER: B

6. KCET 2025Set D-41 markMCQ
   Q.$\mathrm{CH_3-\underset{\underset{\textstyle CH_3}{|}}{\overset{\overset{\textstyle CH_3}{|}}{C}}-OCH_3} + HI \longrightarrow A + B$ A and B respectively are (A) A = , B = (B) A = , B = (C) A = , B = (D) A = , B =

   ›Reveal solutionSolution

   A tertiary alkyl ether cleaved by HI proceeds via SN1: the bond that breaks is the one that forms the more stable (tertiary) carbocation, sending iodide to the tert-butyl side and leaving methanol as the other fragment.

   Step 1 — Protonation.

   The ether oxygen is protonated by HI, making it a good leaving group.

   Step 2 — C–O bond cleavage (SN1, tertiary substrate).

   With a tertiary carbon on one side, the bond breaks heterolytically to form the stable tertiary carbocation $(CH_3)_3C^+$, releasing methanol ($CH_3OH$) as the neutral leaving fragment.

   Step 3 — Nucleophilic attack.

   Iodide attacks the tertiary carbocation directly: $(CH_3)_3C-I$.

   ✓Final answer

   A = $CH_3OH$, B = $(CH_3)_3C-I$ — option (C).

7. COMEDK 2025Set 2025-A1 markMCQ
   Q.Which of the following is the correct name according to IUPAC rules? (A) (B) (C) (D)

   ›Reveal solutionSolution

   The key idea is to apply IUPAC nomenclature rules for alkenes, alkynes, alcohols, and ethers. Only option (A) correctly names its structure as 3‑bromoprop‑1‑ene; the others contain errors in numbering, suffix order, or functional group priority.

   Concept and Intuition

   IUPAC naming follows a strict hierarchy: the principal functional group determines the suffix, the longest carbon chain containing that group is chosen, and numbering gives the lowest locants to the principal group and then to multiple bonds. For compounds with multiple bonds, the suffix for the highest‑priority bond (alkene > alkyne) comes last, and locants are placed before the suffix. For alcohols, the –OH group takes precedence over multiple bonds, and the chain must include the carbon bearing the –OH. For ethers, the smaller alkyl group is named as an alkoxy substituent on the larger alkane. Each option must be checked against these rules.

   Step‑by‑Step Analysis

   1. Option (A): Br–CH₂–CH=CH₂ named “3‑bromoprop‑1‑ene”

      • The longest chain is three carbons (prop‑) with a double bond between C1 and C2.
      • The bromine is a substituent on C3.
      • Numbering starts from the end nearer the double bond, so the double bond gets locant 1.
      • The name “3‑bromoprop‑1‑ene” correctly places the bromine locant before the parent, and the double‑bond locant before “‑ene”.
      • Verdict: Correct.

   2. Option (B): HC≡C–CH=CH₂ named “3‑butene‑1‑yne”

      • The longest chain is four carbons (but‑). It contains both a triple bond and a double bond.
      • IUPAC rule: when both alkene and alkyne are present, the suffix “‑ene” comes before “‑yne” (alphabetical order of suffixes), and numbering gives the lowest locants to the multiple bonds as a set.
      • Numbering from the left: triple bond at C1, double bond at C3 → locants 1 and 3.
      • Numbering from the right: double bond at C1, triple bond at C3 → locants 1 and 3 (same).
      • The correct name is but‑1‑en‑3‑yne (or 1‑buten‑3‑yne), not “3‑butene‑1‑yne”. The locant for the double bond should be before “‑ene”, and the triple bond locant before “‑yne”. The given name reverses the order and misplaces locants.
      • Verdict: Incorrect.

   3. Option (C): H₃C–CH(OH)–CH₃ named “propan‑1‑ol”

      • The structure is propan‑2‑ol (the –OH is on the middle carbon).
      • The longest chain is three carbons, and the –OH group must get the lowest locant. Numbering from either end gives the –OH at C2.
      • The correct name is propan‑2‑ol (or isopropyl alcohol). “Propan‑1‑ol” would have the –OH on a terminal carbon, which is not the case.
      • Verdict: Incorrect.

   4. Option (D): H₃C–O–CH₂–CH₃ named “ethoxymethane”

      • This is an ether. IUPAC names ethers as alkoxyalkanes: the smaller alkyl group (methyl) becomes the alkoxy substituent (methoxy‑), and the larger alkyl group (ethyl) is the parent alkane (ethane).
      • The correct name is methoxyethane, not “ethoxymethane”. (The older common name “ethyl methyl ether” is also acceptable, but IUPAC prefers the alkoxyalkane form with the smaller group as the prefix.)
      • Verdict: Incorrect.

   Watch out

   A common mistake is to think that “ethoxymethane” is acceptable because it swaps the order of the alkyl groups. IUPAC rules specify that the smaller (or less complex) alkyl group is named as the alkoxy prefix, so “methoxyethane” is correct.

   Tip

   For option (B), remember the mnemonic: when both double and triple bonds are present, the suffix order is “‑ene” then “‑yne” (alphabetical), and the numbering is chosen to give the lowest locants to the multiple bonds as a whole, not just to the highest‑priority bond.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. COMEDK 2025Set 2025-E1 markMCQ
   Q.Match the IUPAC names in Column II with the correct structures given in Column I. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Structures of compounds IUPAC name A P 1-Bromomethyl-3-(2,2-dimethylpropyl)benzene B Q 1-Bromo-4-(1-methylethyl)benzene C R 1-Bromo-2-(1-methylpropyl)benzene D S 1-Bromo-4-(2-methylpropyl)benzene (A) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$ (B) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (C) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R}$ (D) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$

   ›Reveal solutionSolution

   Name each drawn structure yourself using IUPAC rules — identify the parent benzene, the substituents, their positions (ortho/meta/para) and the lowest locants — then compare with the given names P, Q, R, S. The matching is A→R, B→P, C→S, D→Q, which is option (A).

   ---

   Concept & Intuition

   This is a "match the structure to the name" problem. The reliable method is to name each structure yourself using IUPAC rules and then compare with the given names, rather than guessing from the drawings.

   • Benzene is the parent ring.
   • Substituents are named as prefixes (bromo, bromomethyl, alkyl groups).
   • Number the ring for the lowest set of locants; bromo comes first alphabetically here, so it takes position 1.
   • Know the common alkyl-group names: isopropyl = 1-methylethyl, isobutyl = 2-methylpropyl, sec-butyl = 1-methylpropyl, neopentyl = 2,2-dimethylpropyl.

   ---

   Step-by-step naming

   1. Structure A

      • Two substituents: bromine (Br) and a side chain –CH(CH₃)CH₂CH₃, which is a 1-methylpropyl group (sec-butyl).
      • They sit on adjacent carbons (ortho), so bromo is at position 1 and the side chain at position 2.
      • IUPAC name: 1-Bromo-2-(1-methylpropyl)benzene → matches R.

   2. Structure B

      • Two substituents: –CH₂Br (bromomethyl) and –CH₂–C(CH₃)₃, a CH₂ attached to a carbon carrying three methyls, i.e. 2,2-dimethylpropyl (neopentyl).
      • The only name in Column II built from a bromomethyl group together with a 2,2-dimethylpropyl group is P (1-Bromomethyl-3-(2,2-dimethylpropyl)benzene).
      • So B matches P.

   3. Structure C

      • A –CH₂–CH(CH₃)₂ side chain (2-methylpropyl, isobutyl) para to Br.
      • IUPAC name: 1-Bromo-4-(2-methylpropyl)benzene → matches S.

   4. Structure D

      • An isopropyl group, –CH(CH₃)₂ = 1-methylethyl, para to Br.
      • IUPAC name: 1-Bromo-4-(1-methylethyl)benzene → matches Q.

   ---

   Matching summary

   Structure	IUPAC name (derived)	Column II label
   A	1-Bromo-2-(1-methylpropyl)benzene	R
   B	1-Bromomethyl-3-(2,2-dimethylpropyl)benzene	P
   C	1-Bromo-4-(2-methylpropyl)benzene	S
   D	1-Bromo-4-(1-methylethyl)benzene	Q

   Thus A→R, B→P, C→S, D→Q.

   ---

   Watch out

   A common mistake is to misidentify the alkyl groups: "1-methylpropyl" is sec-butyl (a four-carbon chain branched at the first carbon), while "2-methylpropyl" is isobutyl (a three-carbon chain with a methyl on the second carbon), and "2,2-dimethylpropyl" is neopentyl. Draw the group out if you are unsure.

   ---

   Tip

   First identify the functional groups (bromo, bromomethyl), then the alkyl substituent, then the positions (ortho = 1,2; meta = 1,3; para = 1,4). This quickly narrows the options.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. COMEDK 2025Set 2025-M1 markMCQ
   Q.Match the structures in Column I with their correct IUPAC names given in Column II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Column I Column II A. P. 2,3-Dibromo-1-phenylpentane. B. Q. 2,3- Dibromohexanedial. C. R. 2- ( 4- isobutylphenyl) propanoic acid D. S. 2- Hydroxy-1,2,3- propanetricarboxylic acid. (A) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R}$ (B) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (C) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (D) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$

   ›Reveal solutionSolution

   The key is to systematically match each drawn structure to its IUPAC name by identifying the parent chain, functional groups, and substituents. The correct matches are: A → R, B → S, C → Q, D → P, which corresponds to option (B).

   Let’s walk through each structure step by step, naming them from scratch so the matching becomes clear.

   Concept & Intuition

   IUPAC nomenclature is like a recipe: first find the longest carbon chain that includes the principal functional group (the one with highest priority, e.g., carboxylic acid > aldehyde > alcohol > alkane). Number the chain to give the functional group the lowest possible number, then name substituents (like bromo, phenyl, alkyl) with their positions. For rings, treat the ring as a substituent if the chain is longer, or as the parent if the ring is the main feature. Here, we have four distinct cases: a substituted benzene (A), a polyfunctional acid (B), a dialdehyde with bromines (C), and a phenyl-substituted alkane with bromines (D).

   ---

   1. Structure A – The Ibuprofen-like molecule

   • What we see: A benzene ring with a para-substituted pattern. On one side (top) is a carbon that has a methyl group and a carboxylic acid group – that’s a -CH(CH₃)COOH group. On the opposite side (bottom) is a -CH₂-CH(CH₃)₂ group (isobutyl).
   • Parent chain: The longest chain attached to the ring is the isobutyl part? No – the carboxylic acid is the highest priority functional group. The ring is a substituent here because the acid-containing chain is only 2 carbons long (propanoic acid). So the parent is propanoic acid.
   • Numbering: The carbon of the acid is C1. The carbon next to it (C2) bears the phenyl ring. So the name is 2-(4-isobutylphenyl)propanoic acid.
   • Match: This is name R in Column II.

   Tip

   The “isobutyl” group is a common name; in IUPAC it’s (2-methylpropyl). But here the given name uses “isobutyl”, so we match directly.

   ---

   2. Structure B – Citric acid

   • What we see: A central carbon with an –OH and a –COOH. From that central carbon, two -CH₂- arms each end in a –COOH. That’s three carboxylic acid groups total.
   • Parent chain: The longest chain that includes all three carboxyls is a three-carbon chain (propane) with the central carbon bearing the –OH. The systematic name for citric acid is 2-hydroxy-1,2,3-propanetricarboxylic acid.
   • Match: This is name S.

   Watch out

   A common mistake is to think the central carbon is part of a longer chain, but the three carboxyls are on a three-carbon backbone. The numbering starts from the end nearest the first carboxyl, giving the –OH at position 2.

   ---

   3. Structure C – Dialdehyde with bromines

   • What we see: A six-carbon straight chain with an aldehyde (–CHO) at both ends. Two bromine atoms are on adjacent carbons near the right end.
   • Parent chain: The longest chain includes both aldehyde carbons, so it’s a hexane chain with two aldehyde groups. The suffix for two aldehydes is “dial”.
   • Numbering: Aldehyde carbons are always C1 and C6 (the ends). The bromines are on C2 and C3 (from the right end, they are the second and third carbons from the right-hand CHO). So the name is 2,3-dibromohexanedial.
   • Match: This is name Q.

   Tip

   When both ends are the same functional group (here aldehyde), numbering can start from either end; the bromines get the lowest set of locants, which is 2,3.

   ---

   4. Structure D – Phenyl-substituted dibromopentane

   • What we see: A benzene ring attached to a five-carbon chain. The chain: phenyl–CH₂–CHBr–CHBr–CH₂–CH₃. So it’s a pentane chain with a phenyl on C1 and bromines on C2 and C3.
   • Parent chain: The longest carbon chain is pentane (5 carbons). The benzene ring is a substituent (phenyl).
   • Numbering: Start from the end nearest the phenyl (to give it the lowest number). So C1 is the carbon attached to the ring. Then bromines are on C2 and C3. The name is 2,3-dibromo-1-phenylpentane.
   • Match: This is name P.

   Watch out

   Do not number from the other end – that would give the phenyl at C5 and bromines at C3 and C4, which is a higher set of locants. Always give the first substituent the lowest number.

   ---

   Final Matching

   • A → R
   • B → S
   • C → Q
   • D → P

   This corresponds to option (B).

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. COMEDK 2024Set 2024-A1 markMCQ
    Q.Names of some organic compounds are given. Which one is not in IUPAC system? (A) (B) (C) (D)

    ›Reveal solutionSolution

    The key idea is to check each name against IUPAC rules for numbering, substituent order, and functional group suffixes. Option (A) violates the rule that the carboxylic acid carbon must be number 1, making its numbering incorrect; the correct option is (A).

    The question asks which name is not in the IUPAC system. That means three names follow IUPAC rules, and one does not. To find the odd one out, we must recall the core principles of IUPAC nomenclature:

    • The principal functional group (here, the highest priority group) determines the suffix and gets the lowest possible locant.
    • For carboxylic acids, the carbon of the –COOH group is always number 1.
    • Double bonds and other substituents are numbered to give the lowest set of locants, but the acid carbon’s position is fixed.

    Let’s examine each option step by step.

    1. Option (A): “4-oxo-2,3-dimethylpent-2-en-1-oic acid”

       • The structure is: H₃C–C(=O)–C(CH₃)=C(CH₃)–C(=O)OH. This is a five-carbon chain with a ketone (oxo) on carbon 4, two methyl groups on carbons 2 and 3, a double bond between carbons 2 and 3, and a carboxylic acid at the end.
       • In IUPAC, the carboxylic acid carbon must be carbon 1. So the chain is numbered from the –COOH end: C1 is the acid carbon, C2 has a methyl and a double bond, C3 has a methyl and a double bond, C4 has the ketone, and C5 is the terminal methyl.
       • The correct name should be: 5-oxo-2,3-dimethylpent-3-enoic acid (because the double bond is between C3 and C4, not C2 and C3, when numbering from the acid). The given name says “pent-2-en-1-oic acid,” which incorrectly places the double bond at position 2 and uses “-1-oic” (redundant, since the acid carbon is always 1). This is a clear violation.
       • Conclusion: Option (A) is not in the IUPAC system.

    2. Option (B): “1,3,3-trimethylcyclohex-1-ene”

       • Structure: a cyclohexene ring with a double bond between C1 and C2, a methyl on C1, and two methyls on C3 (gem-dimethyl).
       • Numbering: The double bond gets priority, so carbons 1 and 2 are the double-bonded atoms. The methyl groups are then at positions 1 and 3 (with two at 3). The name “1,3,3-trimethylcyclohex-1-ene” follows IUPAC: the locant for the double bond is given as “-1-ene” (the lower-numbered carbon of the double bond), and substituents are listed alphabetically (trimethyl is fine). This is correct.

    3. Option (C): “but-3-enoic acid”

       • Structure: CH₂=CH–CH₂–COOH. This is a four-carbon chain with a double bond between C3 and C4 (if numbered from the acid carbon, C1 is –COOH, C2 is CH₂, C3 is CH, C4 is CH₂). The double bond is between C3 and C4, so the locant is 3. The name “but-3-enoic acid” is perfectly IUPAC: the suffix “-enoic” indicates the double bond, and the locant 3 is correct. (Note: The older name “vinylacetic acid” is not IUPAC, but the given name is IUPAC.)

    4. Option (D): “2-methyl-3-phenylpentane”

       • Structure: H₃C–CH(CH₃)–CH(C₆H₅)–CH₂–CH₃. This is a pentane chain with a methyl on C2 and a phenyl on C3. Numbering from the end nearest the first substituent gives the lowest locants: 2-methyl and 3-phenyl. The name “2-methyl-3-phenylpentane” follows IUPAC rules (substituents in alphabetical order: methyl before phenyl). This is correct.

    Watch out

    A common mistake is to think that “-1-oic” is acceptable for carboxylic acids. In IUPAC, the “1” is implied and never written; writing “pent-2-en-1-oic acid” is incorrect because the acid carbon is always position 1, so the locant is redundant and the double bond numbering is wrong.

    Tip

    For carboxylic acids, always number from the –COOH carbon as 1. Then assign locants to double bonds and other groups. If the name includes “-1-oic,” it’s a red flag — it’s either redundant or indicates a numbering error.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. COMEDK 2024Set 2024-M1 markMCQ
    Q.Which one of the following is the correct IUPAC name of the given compound? (A) 5-Bromo-2, 3-dimethylheptanoyl chloride (B) 5,6-Dimethyl-3-bromohexanoyl chloride (C) 1-Chloro-5-bromo-2, 3-dimethyl-1-oxoheptane (D) 3-Bromo-5, 6-dimethylhexanoyl chloride

    ›Reveal solutionSolution

    The compound is a seven-carbon acyl chloride with methyl groups on C2 and C3 and a bromine on C5; the correct IUPAC name is 5-Bromo-2,3-dimethylheptanoyl chloride, which corresponds to option (A).

    The key to naming this compound correctly is to identify the principal functional group and the longest carbon chain that includes it. Here, the functional group is an acyl chloride (–COCl), which takes priority over all other substituents. The chain must be numbered starting from the carbonyl carbon of the acyl chloride, giving it the lowest possible locant (C1). Once the chain is numbered, we name the substituents (methyl groups and bromine) with their positions, and then assemble the name in alphabetical order.

    Let’s work through it step by step.

    1. Identify the principal functional group and the parent chain.

       The structure shows a carbonyl (C=O) with a chlorine atom attached to the same carbon – that’s an acyl chloride group (–COCl). This group is the highest priority functional group, so the parent chain must include it. The longest continuous carbon chain that includes the carbonyl carbon has seven carbons (heptane). The acyl chloride suffix replaces the “-e” of heptane with “-oyl chloride”, giving the parent name heptanoyl chloride.

    2. Number the chain from the carbonyl carbon.

       In IUPAC nomenclature, the carbonyl carbon of an acyl halide is always assigned position 1. So we number the chain as:

       C1 = carbonyl carbon (part of –COCl),

       C2 = next carbon (has a methyl group up),

       C3 = next carbon (has a methyl group down),

       C4 = CH₂,

       C5 = carbon with Br (down),

       C6 and C7 = ethyl end.

    3. Locate and name the substituents.

       • At C2: a methyl group → 2-methyl
       • At C3: a methyl group → 3-methyl
       • At C5: a bromine atom → 5-bromo Since there are two methyl groups, we combine them as 2,3-dimethyl.

    4. Assemble the name in alphabetical order.

       Substituents are listed alphabetically (ignoring prefixes like di-). “Bromo” comes before “methyl”, so the order is:

       5-bromo-2,3-dimethylheptanoyl chloride.

    5. Check the other options for common mistakes.

       • Option (B): “5,6-Dimethyl-3-bromohexanoyl chloride” – wrong chain length (hexanoyl instead of heptanoyl) and wrong numbering.
       • Option (C): “1-Chloro-5-bromo-2,3-dimethyl-1-oxoheptane” – this treats the acyl chloride as a ketone and a chloro substituent, which is incorrect; the acyl chloride is the functional group, not a separate chloro and oxo.
       • Option (D): “3-Bromo-5,6-dimethylhexanoyl chloride” – again, wrong chain length (hexanoyl) and incorrect numbering.

    Watch out

    A common pitfall is to number the chain from the end farthest from the carbonyl, or to treat the acyl chloride as a chloro ketone. Always remember: the carbonyl carbon of an acyl halide is C1, and the suffix is “-oyl chloride”, not “-one” or “-oxo”.

    Tip

    To quickly verify, count the carbons in the longest chain including the carbonyl: here it’s seven, so the parent must be heptanoyl. Any option with “hexanoyl” is automatically wrong.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. KCET 2023Set D-21 markMCQ
    Q.IUPAC name of the compound is (A) 2, 3-dimethylbut-2-ene (B) 2, 3-dimethyl butyne (C) 1, 1, 2, 2-tetra methylethene (D) 2, 3-dimethyl butene

    ›Reveal solutionSolution

    The compound is a symmetrical alkene with four methyl groups on a two-carbon double bond; the longest chain is but-2-ene, and the correct IUPAC name is 2,3-dimethylbut-2-ene.

    The question asks for the IUPAC name of a compound, but the compound itself is not drawn in the text. From the options, it is clear we are dealing with a six-carbon alkene — specifically, the one where the double bond is between C2 and C3 of a butane chain, and each of those two carbons carries two methyl groups. That structure is $(CH_3)_2C=C(CH_3)_2$, commonly called tetramethylethene.

    The key to IUPAC naming is to identify the longest continuous carbon chain that includes the double bond. Here, the longest chain that contains the double bond has four carbons — that is the but-2-ene backbone. The two extra carbons (the methyl groups) are then treated as substituents.

    Let’s work through the naming step by step.

    1. Identify the parent chain. The longest chain containing the double bond is a four-carbon chain: C1–C2=C3–C4. The double bond is between C2 and C3, so the parent name is but-2-ene.

    2. Number the chain to give the double bond the lowest possible locant. Since the double bond is between C2 and C3, numbering from either end gives the same locant (2). So the parent is but-2-ene.

    3. Identify and locate substituents. On C2, there are two methyl groups; on C3, there are also two methyl groups. So we have four methyl substituents — two at position 2 and two at position 3.

    4. Assemble the name. The substituents are listed alphabetically (ignoring multiplying prefixes like di-, tri-). Here, all substituents are methyl, so we combine: 2,3-dimethyl (for the two methyls on C2 and C3) — but wait, there are four methyls total. The correct prefix is tetramethyl, with locants 2,2,3,3. So the full name is 2,2,3,3-tetramethylbut-2-ene? That would be wrong — let’s check.

    Watch out

    A common mistake is to write "2,2,3,3-tetramethylbut-2-ene". But that name implies the parent chain is butane with four methyl substituents, which is correct in terms of substituent count, but the IUPAC convention for alkenes requires the double bond to have the lowest possible locant, and here the double bond is already at position 2. However, the name "2,2,3,3-tetramethylbut-2-ene" is actually acceptable but not the simplest. The simpler name uses the fact that the two methyls on C2 and the two on C3 can be described as "2,3-dimethyl" if we consider that each of C2 and C3 already has one methyl from the parent chain? No — the parent chain is but-2-ene, which has no methyl groups on C2 or C3 in the parent. So each methyl is a substituent.

    Let’s re-evaluate: The structure $(CH_3)_2C=C(CH_3)_2$ has a four-carbon backbone: C1 (CH3), C2 (C with two CH3), C3 (C with two CH3), C4 (CH3). So the substituents are: at C2, two methyls; at C3, two methyls. That gives 2,2,3,3-tetramethylbut-2-ene. But option (A) is 2,3-dimethylbut-2-ene — that would correspond to only two methyl substituents, i.e., $(CH_3)_2C=CHCH_3$? No, that’s 2-methylbut-2-ene? Let’s check: 2,3-dimethylbut-2-ene means the parent is but-2-ene with a methyl at C2 and a methyl at C3 — that structure is $CH_3C(CH_3)=C(CH_3)CH_3$, which is exactly $(CH_3)_2C=C(CH_3)_2$! Because the parent chain already includes the four carbons: C1 (CH3), C2 (C with one CH3 substituent), C3 (C with one CH3 substituent), C4 (CH3). So the two methyl substituents are the ones attached to C2 and C3, giving the same molecule.

    Tip

    In IUPAC naming, the parent chain is the longest chain that includes the double bond. For $(CH_3)_2C=C(CH_3)_2$, the longest chain is indeed four carbons: CH3–C(=C)–C–CH3. The two "extra" methyls are the ones on C2 and C3. So the name is 2,3-dimethylbut-2-ene, not 2,2,3,3-tetramethylbut-2-ene. The latter would imply a butane parent (no double bond) with four methyls, which is a different compound (2,2,3,3-tetramethylbutane). Always ensure the parent chain includes the double bond.

    Thus, the correct IUPAC name is 2,3-dimethylbut-2-ene.

    ✓Final answer

    The correct option is (A) 2,3-dimethylbut-2-ene.