Q.Give the IUPAC names of the following compounds:

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.An isomer of $\mathrm{C_4H_8}$ is X, which exhibits cis-trans isomerism. Y is the chain isomer of X. Products obtained from ozonolysis of Y are (A) $\mathrm{CH_3CHO + CH_3CHO}$ (B) $\mathrm{CH_3CH_2CHO + HCOOH}$ (C) $\mathrm{CH_3COCH_3 + HCHO}$ (D) $\mathrm{CH_3CH_2COOH + CO_2}$

   ›Reveal solutionSolution

   X (showing cis-trans isomerism) is but-2-ene; its chain isomer Y is isobutylene, whose ozonolysis gives acetone + formaldehyde.

   Concept and Intuition

   $\mathrm{C_4H_8}$ has several isomers: but-1-ene, cis/trans-but-2-ene, 2-methylprop-1-ene (isobutylene), plus the cyclic ones (cyclobutane, methylcyclopropane). Cis-trans (geometrical) isomerism needs each double-bond carbon to carry two different substituents — only but-2-ene ($\mathrm{CH_3-CH=CH-CH_3}$) among the open-chain alkenes satisfies this. So X = but-2-ene.

   A chain isomer must differ in the carbon skeleton itself (straight vs branched), not merely in the position of the double bond. But-1-ene is only a position isomer of but-2-ene (both are straight-chain butenes). The genuine chain (skeletal) isomer of but-2-ene is 2-methylprop-1-ene, $\mathrm{CH_2=C(CH_3)_2}$ — a branched $\mathrm{C_4H_8}$ alkene. So Y = isobutylene.

   Step-by-Step Solution

   1. Identify X: but-2-ene, $\mathrm{CH_3-CH=CH-CH_3}$ (cis/trans possible).
   2. Identify Y (chain isomer of X): 2-methylprop-1-ene, $\mathrm{(CH_3)_2C=CH_2}$.
   3. Ozonolysis of Y breaks the $\mathrm{C=C}$ bond and replaces it with $\mathrm{C=O}$ on each fragment:
      • The $\mathrm{=CH_2}$ end becomes $\mathrm{HCHO}$ (formaldehyde).
      • The $\mathrm{=C(CH_3)_2}$ end becomes $\mathrm{(CH_3)_2C=O}$ (acetone, $\mathrm{CH_3COCH_3}$).
   4. Products: $\mathrm{CH_3COCH_3 + HCHO}$.

   Common Mistakes

   • Confusing but-1-ene (a position isomer) with the true chain isomer of but-2-ene.
   • Forgetting that ozonolysis products depend on the substitution pattern at each alkene carbon, not just the molecular formula.

   ✓Final answer

   The correct option is (C) — $\mathrm{CH_3COCH_3 + HCHO}$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The number of monochloro derivatives possible for

   (i) Isopentane

   (ii) neopentane and

   (iii) 2,3-dimethylbutane are $x, y$ and $z$ respectively. The sum of $x, y$ and $z$ is (A) $6$ (B) $7$ (C) $8$ (D) $5$

   ›Reveal solutionSolution

   This tests counting distinct monochlorination products via symmetry analysis of hydrogens. The individual counts are 4, 1, 2, summing to 7.

   Concept and Intuition

   The number of distinct monochloro derivatives of an alkane equals the number of chemically non-equivalent (by molecular symmetry) sets of hydrogen atoms — because replacing any H within an equivalent set by Cl gives the identical product.

   Step-by-Step Solution

   1. Isopentane = 2-methylbutane: $CH_3{-}CH(CH_3){-}CH_2{-}CH_3$. Label the central CH carbon as C2, which bears two methyl groups (C1 and the branch methyl) that are equivalent to each other by symmetry. Distinct H-environments: (i) the two equivalent terminal methyls attached directly to C2 (C1 and branch-CH3), (ii) the tertiary H on C2, (iii) the CH2 hydrogens (C3), (iv) the terminal CH3 on C4 (different from group (i) since it's attached to a CH2, not to the tertiary C). That's 4 distinct types → $x = 4$.
   2. Neopentane = 2,2-dimethylpropane, $C(CH_3)_4$. All four methyl groups are equivalent by the high (tetrahedral) symmetry of the molecule, so there is only one kind of hydrogen → $y = 1$.
   3. 2,3-Dimethylbutane = $(CH_3)_2CH{-}CH(CH_3)_2$. The molecule has a center/axis of symmetry: C2 and C3 are equivalent to each other, and on each of C2/C3 the two attached methyl groups (chain-methyl and branch-methyl) are locally equivalent (free rotation, no stereocenter). This makes all four methyl groups in the molecule equivalent (one type), and the two methine H's (on C2, C3) equivalent to each other (a second type). So there are 2 distinct H-environments → $z = 2$.
   4. Sum: $x + y + z = 4 + 1 + 2 = 7$.

   Common Mistakes

   • Forgetting that the two methyls on the central carbon of isopentane are equivalent, and miscounting more or fewer than 4 types.
   • Assuming 2,3-dimethylbutane's four methyls are pairwise distinct rather than realizing the whole-molecule symmetry makes them all equivalent.

   ✓Final answer

   The correct option is (B) — $7$.

   ANSWER: B

3. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Identify X in the following reaction sequence $C_4H_9Br \xrightarrow{OH} C_4H_{10}O \xrightarrow[573\ K]{Cu} C_4H_8$ (X) (A) $(CH_3)_3CBr$ (tert-butyl bromide, i.e. 2-bromo-2-methylpropane) (B) $(CH_3)_2CHCH_2Br$ (isobutyl bromide, i.e. 1-bromo-2-methylpropane) (C) $CH_3CH_2CHBrCH_3$ (sec-butyl bromide, i.e. 2-bromobutane) (D) $CH_3CH_2CH_2CH_2Br$ (n-butyl bromide, i.e. 1-bromobutane)

   ›Reveal solutionSolution

   Reading the final product ($C_4H_8$, an alkene, not a carbonyl) backward reveals the alcohol must be tertiary (since only tertiary alcohols dehydrate rather than dehydrogenate over Cu/573K); hence X is tert-butyl bromide.

   Concept and Intuition

   Copper at ~573 K catalyses two different reactions on alcohols depending on how many H atoms sit on the carbinol (C–OH) carbon:

   • Primary alcohol (2 H's on that carbon) → dehydrogenates to an aldehyde.
   • Secondary alcohol (1 H) → dehydrogenates to a ketone.
   • Tertiary alcohol (0 H's on that carbon) → CANNOT dehydrogenate (no H to remove along with the O–H), so it instead undergoes dehydration (loses $H_2O$) to give an alkene. The molecular formula given for the final product, $C_4H_8$, has lost an $H_2O$ relative to $C_4H_{10}O$ (not just $H_2$, which would give $C_4H_8O$), confirming dehydration, i.e. a tertiary alcohol.

   Step-by-Step Solution

   1. Final step: $C_4H_{10}O \to C_4H_8$ over Cu/573K. Formula change is loss of $H_2O$ ($C_4H_{10}O - H_2O = C_4H_8$), which is dehydration — only possible (via this heterogeneous Cu route) for a tertiary alcohol, since it lacks the α-H needed for dehydrogenation.
   2. So the $C_4H_{10}O$ intermediate is tert-butanol, $(CH_3)_3C{-}OH$, which indeed dehydrates over Cu/573K to isobutylene, $(CH_3)_2C{=}CH_2$ ($C_4H_8$).
   3. First step: $C_4H_9Br \xrightarrow{OH^-} C_4H_{10}O$ is simple hydroxide substitution of the halide. For the product to be tert-butanol, the starting halide must already have the same carbon skeleton with Br in place of OH: tert-butyl bromide, $(CH_3)_3CBr$.
   4. Tertiary bromides indeed react readily with $OH^-$ (largely $S_N1$) to give the tertiary alcohol as the substitution product.

   Common Mistakes

   • Forgetting that Cu/573K distinguishes alcohols by dehydrogenation (1°/2°) vs. dehydration (3°) — treating all alcohols as giving a carbonyl product regardless of class.
   • Picking a primary or secondary bromide (options B, C, D) without checking that their corresponding alcohols would give a carbonyl ($C_4H_8O$), not $C_4H_8$.

   ✓Final answer

   The correct option is (A) — $(CH_3)_3CBr$ (tert-butyl bromide, i.e. 2-bromo-2-methylpropane).

   ANSWER: A

4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.The functional isomer of $\underline{Z}$ formed in the given sequence of reactions is $CH_3CH_2CH_2OH \xrightarrow[443\ K]{Conc.\ H_2SO_4} X \xrightarrow[(ii)\ alc.KOH,\Delta\ \ (iii)\ NaNH_2]{(i)\ Br_2\ |\ CCl_4} Y \xrightarrow[H^+\ |\ 333\ K]{H_2O,\ Hg^{2+}} Z$ (A) $CH_3CH_2COOH$ (propanoic acid) (B) $CH_2{=}CH{-}OH$ (vinyl alcohol / ethenol) (C) $CH_3CH_2CHO$ (propanal) (D) $CH_3CH_2{-}O{-}CH_3$ (ethyl methyl ether)

   ›Reveal solutionSolution

   Traces propan-1-ol through dehydration → dihalogenation/double-dehydrohalogenation → Markovnikov alkyne hydration to reach acetone (Z); its functional isomer (same formula, aldehyde instead of ketone) is propanal.

   Concept and Intuition

   This is a classic hydrocarbon-interconversion chain. Each named condition is a fixed textbook reagent-to-transformation rule: conc. $H_2SO_4$ at 443 K = dehydration (E1, alcohol → alkene); $Br_2/CCl_4$ = anti addition across a C=C (vicinal dibromide); alc. $KOH/\Delta$ = dehydrohalogenation (E2); $NaNH_2$ (excess, strong base/strong nucleophile) = a second dehydrohalogenation to reach a triple bond, and (for longer chains) also isomerises an internal alkyne to the terminal one via the acetylide anion; $H_2O/Hg^{2+}/H^+$ = Markovnikov hydration of an alkyne to a carbonyl (via an unstable enol that tautomerises). Two carbonyl compounds sharing the same molecular formula but differing in functional group (aldehyde vs ketone here) are called functional isomers.

   Step-by-Step Solution

   1. $CH_3CH_2CH_2OH \xrightarrow[443K]{conc.\ H_2SO_4} X$: dehydration removes $H_2O$ to give propene, $X = CH_3{-}CH{=}CH_2$.
   2. $X \xrightarrow{(i)\ Br_2/CCl_4}$ 1,2-dibromopropane, $CH_3{-}CHBr{-}CH_2Br$ (anti addition of Br across the double bond).
   3. $\xrightarrow{(ii)\ alc.\ KOH,\Delta}$ removes one HBr (E2) to give a bromopropene.
   4. $\xrightarrow{(iii)\ NaNH_2}$ removes the second HBr, giving the alkyne; with only 3 carbons the triple bond can only sit terminally, so $Y = HC{\equiv}C{-}CH_3$ (propyne).
   5. $Y \xrightarrow[333K]{H_2O,\ Hg^{2+}/H^+}$: Markovnikov hydration of the terminal alkyne places OH on the more substituted alkyne carbon, giving an enol that tautomerises to the methyl ketone: $Z = CH_3{-}CO{-}CH_3$ (acetone, $C_3H_6O$).
   6. A functional isomer of Z must share its molecular formula ($C_3H_6O$) but carry a different functional group: propanal, $CH_3CH_2CHO$, is the aldehyde isomer of acetone — same formula, different (aldehyde) group.
   7. Check the distractors: propanoic acid is $C_3H_6O_2$ (wrong formula); the "vinyl alcohol" option as drawn is only $C_2$ (wrong carbon count / formula); ethyl methyl ether is $C_3H_8O$ (wrong formula, no unsaturation) — none of these are isomers of Z.

   Common Mistakes

   • Forgetting that with only 3 carbons, an "alkyne" can only be terminal — no internal-vs-terminal ambiguity to resolve.
   • Confusing Markovnikov (Hg²⁺-catalysed) alkyne hydration, which gives a ketone from a terminal alkyne, with anti-Markovnikov hydroboration-oxidation, which would give an aldehyde.
   • Picking an option that "looks like" an isomer without actually checking its molecular formula matches Z's ($C_3H_6O$).

   ✓Final answer

   The correct option is (C) — $CH_3CH_2CHO$ (propanal).

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The number of primary ($1°$), secondary ($2°$) and tertiary ($3°$) alcohols possible for the formula $\mathrm{C_5H_{12}O}$ respectively are (A) 3, 3, 2 (B) 4, 2, 2 (C) 4, 3, 1 (D) 3, 4, 1

   ›Reveal solutionSolution

   Listing all 8 structural isomers of pentanol (C₅H₁₂O) and classifying each by the number of carbons attached to the carbinol carbon gives 4 primary, 3 secondary, and 1 tertiary alcohol. Answer: (C).

   Concept and Intuition

   A saturated monohydric alcohol $\mathrm{C_nH_{2n+2}O}$ is classified 1°/2°/3° by how many carbon groups are attached to the carbon bearing the -OH. For C₅, we must enumerate every distinct carbon skeleton (n-pentane and 2-methylbutane; 2,2-dimethylpropane skeleton also, once we allow the OH to sit on any carbon of any skeleton) and every distinct OH position on each, discarding symmetry-equivalent positions.

   Step-by-Step Solution

   1. Straight (n-pentane) skeleton $CH_3-CH_2-CH_2-CH_2-CH_2-OH$ type: OH can go on C1, C2, or C3 (C4, C5 are equivalent to C2, C1 by the chain's symmetry).
      • 1-pentanol (OH on C1): primary.
      • 2-pentanol (OH on C2): secondary.
      • 3-pentanol (OH on C3): secondary.
   2. 2-methylbutane skeleton $(CH_3)_2CH-CH_2-CH_3$: OH can go on the terminal methyl of the branch (giving 2-methyl-1-butanol), on the branch/tertiary CH carbon (2-methyl-2-butanol), on the CH₂ (2-methyl-3-butanol = 3-methyl-2-butanol by renumbering), or the far methyl (3-methyl-1-butanol, i.e., isoamyl alcohol).
      • 2-methyl-1-butanol: primary.
      • 3-methyl-1-butanol (isoamyl alcohol): primary.
      • 2-methyl-2-butanol (tert-amyl alcohol): tertiary (the carbinol carbon has 3 alkyl groups).
      • 3-methyl-2-butanol: secondary.
   3. 2,2-dimethylpropane (neopentane) skeleton $(CH_3)_3C-CH_2-OH$: OH must sit on the single exocyclic CH₂ — neopentyl alcohol: primary (carbinol carbon attached to only one carbon, the quaternary C, but bonded to 1 carbon group so classified 1°... more precisely the carbinol carbon $CH_2OH$ is attached to just ONE other carbon, so it's primary).
   4. Total count: 8 isomers.
      • Primary (1°): 1-pentanol, 2-methyl-1-butanol, 3-methyl-1-butanol, neopentyl alcohol → 4.
      • Secondary (2°): 2-pentanol, 3-pentanol, 3-methyl-2-butanol → 3.
      • Tertiary (3°): 2-methyl-2-butanol → 1.
   5. So the counts (1°, 2°, 3°) = (4, 3, 1), matching option (C).

   Common Mistakes

   • Missing the neopentyl alcohol isomer (easy to overlook the fully-branched skeleton).
   • Double counting symmetry-equivalent OH positions on the straight chain (e.g., counting both "C4" and "C2" as distinct positions).

   ✓Final answer

   The correct option is (C) — 4, 3, 1.

   ANSWER: C

6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.How many amines with molecular formula $\mathrm{C_3H_9N}$ can react with benzene sulphonyl chloride ? (A) 2 (B) 3 (C) 4 (D) 1

   ›Reveal solutionSolution

   Tests enumerating the isomeric amines of $\mathrm{C_3H_9N}$ and applying the Hinsberg-test rule that only 1° and 2° amines react with benzenesulfonyl chloride.

   Concept and Intuition

   Benzenesulfonyl chloride ($\mathrm{C_6H_5SO_2Cl}$) reacts with an amine's N–H bond(s) to form a sulfonamide. A primary amine (two N–H) forms an N,N-disubstituted-looking, acidic (soluble in alkali) sulfonamide; a secondary amine (one N–H) forms a neutral (alkali-insoluble) sulfonamide. A tertiary amine has NO N–H bond at all, so it cannot form a stable sulfonamide this way — it does not react (any salt formed simply hydrolyses back).

   Step-by-Step Solution

   1. List all isomers of $\mathrm{C_3H_9N}$:
      • $\mathrm{CH_3CH_2CH_2NH_2}$ — n-propylamine (1°)
      • $(\mathrm{CH_3})_2\mathrm{CHNH_2}$ — isopropylamine (1°)
      • $\mathrm{CH_3-NH-CH_2CH_3}$ — N-methylethanamine (2°)
      • $(\mathrm{CH_3})_3\mathrm{N}$ — trimethylamine (3°)
   2. Both 1° amines have two N–H bonds and react with benzenesulfonyl chloride to give sulfonamides.
   3. The 2° amine has one N–H bond and also reacts, giving a sulfonamide.
   4. The 3° amine, trimethylamine, has no N–H bond, so it cannot react to form a sulfonamide.
   5. Count of amines that react: n-propylamine, isopropylamine, N-methylethanamine = 3.

   Common Mistakes

   • Forgetting one of the four structural isomers (especially the branched 1° amine, isopropylamine).
   • Assuming the tertiary amine still reacts (perhaps by simple acid–base salt formation) and counting all four.

   ✓Final answer

   The correct option is (B) — 3.

   ANSWER: B

7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.The number of monochloro derivatives possible for 2,2-Dimethylbutane and 2,3-Dimethylbutane are respectively (A) 3, 2 (B) 2, 3 (C) 4, 2 (D) 2, 4

   ›Reveal solutionSolution

   Counting distinct (non-equivalent) hydrogen environments in each alkane gives the number of possible monochloro derivatives: 3 for 2,2-dimethylbutane and 2 for 2,3-dimethylbutane.

   Concept and Intuition

   In free-radical monochlorination, each type of chemically distinct hydrogen (by symmetry) gives, in principle, one distinct monochloro product (ignoring stereochemistry/enantiomers, which is the usual convention in these counting questions). So the count of products equals the count of symmetry-distinct C–H environments in the molecule.

   Step-by-Step Solution

   2,2-Dimethylbutane: $\mathrm{CH_3-C(CH_3)_2-CH_2-CH_3}$

   1. The carbon skeleton is $\mathrm{C1-C2(-CH_3)_2-C3-C4}$, where C2 is quaternary, bearing three methyl groups (C1 and its two substituents) that are all equivalent by the local symmetry around C2.
   2. Distinct H-types: (i) the three equivalent $\mathrm{CH_3}$ groups on C2, (ii) the $\mathrm{CH_2}$ at C3, (iii) the terminal $\mathrm{CH_3}$ at C4.
   3. That's 3 distinct environments → 3 monochloro products.

   2,3-Dimethylbutane: $\mathrm{(CH_3)_2CH-CH(CH_3)_2}$

   1. This molecule has a center of symmetry: C2 and C3 (each a tertiary CH) are equivalent to each other, and all four methyl groups (one on C1, one substituent on C2, one substituent on C3, one on C4) are equivalent to each other.
   2. Distinct H-types: (i) the four equivalent methyl groups, (ii) the two equivalent tertiary C–H's.
   3. That's 2 distinct environments → 2 monochloro products.

   Common Mistakes

   • Not recognizing the symmetry that makes several methyl groups equivalent, and over-counting products.
   • Confusing "number of distinct H atoms" with "number of substituent groups" without checking equivalence.

   ✓Final answer

   The correct option is (A) — 3, 2.

   ANSWER: A

8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.An isomer of $\mathrm{C_5H_{12}}$ on reaction with $\mathrm{Br_2}$ / light gave only one isomer $\mathrm{C_5H_{11}Br}$ (X). Reaction of X with $\mathrm{AgNO_2}$ gave Y as major product. What is Y? (A) $\mathrm{O_2N-C(CH_3)_2-CH_2CH_3}$ (B) $\mathrm{ONO-C(CH_3)_2-CH_2CH_3}$ (C) $\mathrm{(CH_3)_3C-CH_2-ONO}$ (D) $\mathrm{(CH_3)_3C-CH_2-NO_2}$

   ›Reveal solutionSolution

   X is neopentyl bromide; $\mathrm{AgNO_2}$ gives mainly the nitroalkane, so Y is $(CH_3)_3C\text{-}CH_2\text{-}NO_2$ → (D).

   Concept and Intuition

   A $\mathrm{C_5H_{12}}$ isomer that yields a single monobromo product under free-radical bromination must have all its hydrogens equivalent — that is neopentane, $\mathrm{C(CH_3)_4}$. Silver nitrite ($\mathrm{AgNO_2}$) reacts with alkyl halides through the more electronegative nitrogen (ambident nucleophile with a covalent Ag salt), giving the nitroalkane as the major product and the alkyl nitrite as minor.

   Step-by-Step Solution

   1. Only-one-product test → neopentane $\mathrm{(CH_3)_4C}$ (12 equivalent H).
   2. $\mathrm{Br_2}$/light → X = neopentyl bromide $\mathrm{(CH_3)_3C\text{-}CH_2\text{-}Br}$.
   3. X + $\mathrm{AgNO_2}$ (silver salt) → attack via N → major product is the nitroalkane.
   4. Retaining the neopentyl skeleton → Y $= \mathrm{(CH_3)_3C\text{-}CH_2\text{-}NO_2}$ → option (D).

   Common Mistakes

   • Choosing the alkyl nitrite (R–ONO): that is the minor product with $\mathrm{AgNO_2}$ (it is major only with $\mathrm{KNO_2}$).
   • Rearranging the carbon skeleton — the neopentyl framework is retained.

   ✓Final answer

   The correct option is (D) — $\mathrm{(CH_3)_3C\text{-}CH_2\text{-}NO_2}$.

   ANSWER: D

9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Which of the following is the geminal dichloride? (A) 1,1-Dichloropropane (B) 1,2-Dichloropropane (C) 1,3-Dichloropropane (D) 2,3-Dichloropropane

   ›Reveal solutionSolution

   Tests the definition of geminal vs. vicinal dihalides. Answer: 1,1-dichloropropane (option A), since both Cl atoms sit on the same carbon.

   Concept and Intuition

   "Geminal" (from Latin gemini, twins) means both substituents are on the same carbon atom, while "vicinal" (from vicinus, neighbouring) means the substituents are on adjacent carbons. This distinction matters chemically: geminal dihalides give aldehydes/ketones on hydrolysis, while vicinal dihalides give alkynes on double dehydrohalogenation.

   Step-by-Step Solution

   1. 1,1-Dichloropropane: $\mathrm{CH_3-CH_2-CHCl_2}$ — both Cl on C1 → geminal.
   2. 1,2-Dichloropropane: $\mathrm{CH_3-CHCl-CH_2Cl}$ — Cl on C1 and C2 (adjacent carbons) → vicinal.
   3. 1,3-Dichloropropane: $\mathrm{ClCH_2-CH_2-CH_2Cl}$ — Cl atoms separated by one carbon, not even vicinal.
   4. 2,3-Dichloropropane doesn't correspond to a valid propane numbering issue aside — Cl on C2/C3 would be vicinal, not geminal.
   5. Only option (A) has both chlorines on one carbon, so it is the geminal dichloride.

   Common Mistakes

   • Confusing geminal (same carbon) with vicinal (adjacent carbons) — a very common mix-up in nomenclature questions.

   ✓Final answer

   The correct option is (A) — 1,1-Dichloropropane.

   ANSWER: A

10. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.The number of cyclic isomers possible for $C_4H_6$, with one double bond in the ring is (A) 2 (B) 3 (C) 4 (D) 5

    ›Reveal solutionSolution

    $C_4H_6$ with a ring double bond has exactly 3 possible cyclic structures: cyclobutene, 1-methylcyclopropene, and 3-methylcyclopropene.

    Concept and Intuition

    Degree of unsaturation for $C_4H_6$ is $\dfrac{2(4)+2-6}{2}=2$. If the molecule is cyclic AND has one ring C=C, that accounts for both degrees (1 ring + 1 π bond), so there can be no other ring or double bond anywhere else. With only four carbons available, the ring must be either a 4-membered ring (using all 4 carbons in the ring, no substituents) or a 3-membered ring (using 3 carbons in the ring, with the 4th carbon as a methyl substituent).

    Step-by-Step Solution

    1. 4-membered ring: cyclobutane skeleton with one C=C = cyclobutene, molecular formula already $C_4H_6$ with no substituent needed. → 1 structure.
    2. 3-membered ring + methyl: cyclopropene is $C_3H_4$; adding a $CH_3$ (replacing one H) gives $C_4H_6$.
    3. Cyclopropene's ring carbons: $C_1=C_2$ (each bearing one H) and $C_3$ (sp³, bearing two H). Placing the methyl on $C_1$ or $C_2$ gives the same molecule by the ring's mirror symmetry → 1-methylcyclopropene (1 structure).
    4. Placing the methyl on $C_3$ gives a distinct molecule → 3-methylcyclopropene (1 structure).
    5. No cis/trans isomerism arises in either 3- or 4-membered ring alkenes (too strained to have a distinct geometric isomer).
    6. Total distinct structures = 1 (cyclobutene) + 1 (1-methylcyclopropene) + 1 (3-methylcyclopropene) = 3.

    Common Mistakes

    • Forgetting that placing the methyl on either double-bond carbon of cyclopropene gives the same compound (symmetry), which would over-count to 4 if not careful.
    • Missing the 3-membered-ring possibilities entirely and only counting cyclobutene.

    ✓Final answer

    The correct option is (B) — 3.

    ANSWER: B

11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.The number of possible aromatic benzenoid isomers for $\mathrm{C_6H_4Cl_2}$ are (A) 2 (B) 3 (C) 4 (D) 5

    ›Reveal solutionSolution

    Tests counting positional (structural) isomers for a disubstituted benzene with two identical substituents — ortho, meta, and para are the only 3 distinct arrangements.

    Concept and Intuition

    Benzene's six carbons are all equivalent by symmetry. When two identical substituents (here, two Cl atoms) are placed on the ring, the relative position between them can only take 3 distinct values due to the ring's symmetry: adjacent (1,2 - ortho), separated by one carbon (1,3 - meta), or directly opposite (1,4 - para). Any other numbering (e.g., 1,5 or 1,6) is just a renamed/rotated version of one of these three because of the ring's six-fold symmetry.

    Step-by-Step Solution

    1. Label ring positions 1 through 6. Fix one Cl at position 1 (by ring symmetry, this loses no generality).
    2. The second Cl can be at position 2 (ortho), 3 (meta), or 4 (para) — relative to position 1.
    3. Position 5 relative to 1 is equivalent to position 3 (meta) by symmetry (counting the other way around the ring); position 6 is equivalent to position 2 (ortho).
    4. So there are exactly 3 distinct isomers: 1,2-; 1,3-; and 1,4-dichlorobenzene.

    Common Mistakes

    • Counting positions 1–6 naively as 5 distinct pairings without recognizing the ring's mirror symmetry collapses them to 3 unique isomers.
    • Forgetting these must all be "benzenoid" (retaining the aromatic ring intact), which the question specifies, ruling out non-aromatic tautomers/isomers.

    ✓Final answer

    The correct option is (B) — 3.

    ANSWER: B

12. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.Alcohols with molecular formula $C_nH_{2n+2}O$ are isomeric with ________ (A) Acids (B) Ethers (C) Esters (D) Aldehydes

    ›Reveal solutionSolution

    Alcohols and ethers share the same general formula $C_nH_{2n+2}O$, so they are functional isomers of each other.

    Concept and Intuition

    Isomerism requires identical molecular formula but different structural arrangement/functional group. Alcohols ($R-OH$) and ethers ($R-O-R'$) are the textbook example of functional isomerism because both have exactly one oxygen and the same degree of saturation, giving the identical general formula $C_nH_{2n+2}O$ (e.g. ethanol $C_2H_6O$ and dimethyl ether $C_2H_6O$).

    Step-by-Step Solution

    1. General formula of saturated alcohols: $C_nH_{2n+1}OH = C_nH_{2n+2}O$.
    2. General formula of saturated ethers ($C_mH_{2m+1}-O-C_kH_{2k+1}$ with $m+k=n$): also simplifies to $C_nH_{2n+2}O$.
    3. Acids ($C_nH_{2n}O_2$), esters ($C_nH_{2n}O_2$), and aldehydes ($C_nH_{2n}O$) all have different general formulas (different O count or different H count), so they are ruled out.
    4. Hence alcohols and ethers are isomeric with each other.

    Common Mistakes

    • Mixing up the general formula of aldehydes ($C_nH_{2n}O$) with alcohols ($C_nH_{2n+2}O$) — they differ by 2 hydrogens (different degree of unsaturation), so they are NOT isomers.
    • Forgetting that acids/esters carry two oxygens, immediately disqualifying them.

    ✓Final answer

    The correct option is (B) — Ethers.

    ANSWER: B