An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br2 to give a c

NCERT Exemplar · Q48

Q. An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br2 to give a compound 'C', which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' gives one mole of the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the reactions involved.

Andhra Pradesh BieapLong· 5mImportance★★★★★

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Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

Imagine you have a box of identical Lego bricks — four red, ten blue, and six yellow. You build two different models: a car and a house. Both use exactly the same number of each colour brick, but the structures are completely different. That is the core idea of isomerism: same atoms, different arrangement.

In chemistry, molecules are not just a list of atoms. How those atoms are connected matters enormously. Two molecules can have the exact same molecular formula (same number of each atom) but be connected in different ways. Those are structural isomers (also called constitutional isomers).

The Precise Statement

Structural isomers are compounds that have the same molecular formula but different connectivity of atoms — that is, different structural formulas.

The key word is connectivity. Which atom is bonded to which? If you change that, you get a different substance with different physical and chemical properties.

A Concrete Example: C₄H₁₀

Take butane, C₄H₁₀. There are exactly two ways to connect four carbon atoms and ten hydrogen atoms:

n-Butane — a straight chain: C–C–C–C
Isobutane (2-methylpropane) — a branched chain: a central carbon bonded to three methyl groups

Both have formula C₄H₁₀. But n-butane boils at –0.5 °C, while isobutane boils at –11.7 °C. Same atoms, different connectivity → different substance.

Note

Structural isomers are not the same molecule. They are distinct compounds that happen to share a molecular formula. You cannot rotate or flip one to get the other — you must break and reform bonds.

The Three Main Types

Structural isomerism comes in three flavours:

Type	What changes	Example (C₃H₆O)
Chain isomerism	The carbon skeleton (straight vs. branched)	Butane vs. isobutane
Position isomerism	The location of a functional group or substituent	Propan-1-ol vs. propan-2-ol (OH on carbon 1 vs. carbon 2)
Functional group isomerism	The atoms are rearranged into a different functional group	Propanal (aldehyde) vs. propanone (ketone) — both C₃H₆O

Watch out

Do not confuse structural isomers with stereoisomers. Stereoisomers have the same connectivity but differ in spatial arrangement (like left and right hands). That is a completely different chapter. For now: structural isomers = different bond connections.

Why This Matters …

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

Structural isomerism arises when molecules share the same molecular formula but differ in the connectivity of atoms. There is no single "formula" for structural isomerism — instead, the key is understanding why different arrangements are possible.

The Core Principle: Connectivity ≠ Composition

A molecular formula tells you how many of each atom are present, but not how they are joined. Structural isomers exist because atoms can form bonds in multiple distinct sequences while satisfying valency rules.

Why This Happens: The Valency Constraint

Each atom has a fixed bonding capacity (valency):

Carbon: 4 bonds
Hydrogen: 1 bond
Oxygen: 2 bonds
Nitrogen: 3 bonds

Example: For $C_4H_{10}$ , the formula satisfies $4(4) + 10(1) = 26$ valence electrons. But the carbon atoms can be arranged as:

A straight chain: $CH_3-CH_2-CH_2-CH_3$ (n-butane)
A branched chain: $CH_3-CH(CH_3)-CH_3$ (isobutane)

Both satisfy valency, but the connectivity differs.

The "Formula" for Counting Isomers: Why It's Not Simple

There is no closed-form formula to count structural isomers for a given molecular formula. The number grows rapidly and depends on:

Carbon skeleton branching possibilities
Functional group positions
Ring formation possibilities

Why No Simple Formula Exists

The problem is combinatorial — the number of possible trees (acyclic graphs) with $n$ carbon atoms grows exponentially. For example:

$C_4H_{10}$ : 2 structural isomers
$C_5H_{12}$ : 3 structural isomers
$C_6H_{14}$ : 5 structural isomers
$C_{10}H_{22}$ : 75 structural isomers

The pattern follows Cayley's formula for trees, but even that counts only carbon skeletons — not functional group positions.

Key Reasoning: The Branching Principle

The fundamental reason structural isomers exist is that carbon chains can branch. Consider $C_5H_{12}$ :

Straight chain: $C-C-C-C-C$ (n-pentane)
One branch: $C-C-C(C)-C$ (isopentane) — the branch can be at position 2 or 3, but these are identical due to symmetry
Two branches: $C-C(C)(C)-C$ (neopentane) — a quaternary carbon

Why position matters: The branch location changes the carbon's environment, altering physical and chemical properties.

The Functional Group Position Rule

For compounds with functional groups (e.g., alcohols $C_nH_{2n+2}O$ ), the position of the -OH group creates isomers:

$CH_3CH_2CH_2OH$ (propan-1-ol) — OH at end
$CH_3CH(OH)CH_3$ (propan-2-ol) — OH in middle …

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The alkyl halide A is 1-bromopentane; elimination gives 1-pentene (B), which adds Br₂ to form 1,2-dibromopentane (C). Dehydrobromination yields 1-pentyne (D), a terminal alkyne that reacts with sodium in liquid ammonia to form a sodium acetylide and liberates half a mole of H₂. Hydrogenation of D gives straight-chain pentane.

The heart of this problem lies in recognising structural isomerism and the characteristic reactions of alkynes, particularly terminal alkynes. The clue that "one mole of D gives one mole of sodium salt and half a mole of H₂" tells us D is a terminal alkyne—only terminal alkynes have an acidic hydrogen that reacts with strong bases like sodium in liquid ammonia. The requirement that complete hydrogenation yields a straight-chain alkane locks in the carbon skeleton.

Let's trace the reaction sequence backward and forward to identify each compound.

Understanding the Key Clue

The reaction of D with sodium metal in liquid ammonia is diagnostic:

\ce{D + Na ->[liq. NH3] D^- Na^+ + 1/2 H2}

This stoichiometry—one mole of alkyne producing half a mole of hydrogen—means D has exactly one acidic hydrogen. Terminal alkynes ( $\ce{RC≡CH}$ ) have a hydrogen on the sp-hybridised carbon with $\mathrm{p}K_\mathrm{a} \approx 25$ , acidic enough to be deprotonated by sodium metal. Internal alkynes ( $\ce{RC≡CR'}$ ) have no such hydrogen and do not react this way.

Therefore, D is a terminal alkyne.

Working Backward from D

Since complete hydrogenation of D gives a straight-chain alkane and D is $\ce{C5H8}$ (an alkyne with five carbons), the only straight-chain terminal alkyne with five carbons is:

\text{D: } \ce{HC≡C-CH2-CH2-CH3} \quad \text{(1-pentyne)}

Hydrogenation adds two moles of H₂:

\ce{HC≡C-CH2-CH2-CH3 + 2 H2 -> CH3-CH2-CH2-CH2-CH3} \quad \text{(pentane)}

Identifying C

C is formed by adding Br₂ to alkene B, and dehydrobromination of C gives D (1-pentyne). Dehydrobromination removes two HBr molecules from a vicinal dibromide to form a triple bond. Working backward, C must be a 1,2-dibromide:

\text{C: } \ce{CH2Br-CHBr-CH2-CH2-CH3} \quad \text{(1,2-dibromopentane)}

The dehydrobromination proceeds:

\ce{CH2Br-CHBr-CH2-CH2-CH3 + 2 KOH_{(alc)} -> HC≡C-CH2-CH2-CH3 + 2 KBr + 2 H2O}

Identifying B

B is the alkene formed by adding Br₂ to give C. Since C is 1,2-dibromopentane, B must be:

\text{B: } \ce{CH2=CH-CH2-CH2-CH3} \quad \text{(1-pentene)}

Bromination of the double bond:

\ce{CH2=CH-CH2-CH2-CH3 + Br2 -> CH2Br-CHBr-CH2-CH2-CH3}

Identifying A …

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

Showing the 12 most recent of 15 on this concept.

AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$\mathrm{C_4H_8}$
›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

Identify X: but-2-ene, $\mathrm{CH_3-CH=CH-CH_3}$ (cis/trans possible).

Identify Y (chain isomer of X): 2-methylprop-1-ene, $\mathrm{(CH_3)_2C=CH_2}$ .

Ozonolysis of Y breaks the $\mathrm{C=C}$ bond and replaces it with $\mathrm{C=O}$ on each fragment: …
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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

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 → $x = 4$ .

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 → $y = 1$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$C_4H_9Br \xrightarrow{OH} C_4H_{10}O \xrightarrow[573\ K]{Cu} C_4H_8$
›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

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.

So the $C_4H_{10}O$ intermediate is tert-butanol, $(CH_3)_3C{-}OH$ , which indeed dehydrates over Cu/573K to isobutylene, $(CH_3)_2C{=}CH_2$ ( $C_4H_8$ ). …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$\underline{Z}$
›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

$CH_3CH_2CH_2OH \xrightarrow[443K]{conc.\ H_2SO_4} X$ : dehydration removes $H_2O$ to give propene, $X = CH_3{-}CH{=}CH_2$ .

$X \xrightarrow{(i)\ Br_2/CCl_4}$ 1,2-dibromopropane, $CH_3{-}CHBr{-}CH_2Br$ (anti addition of Br across the double bond).

$\xrightarrow{(ii)\ alc.\ KOH,\Delta}$ removes one HBr (E2) to give a bromopropene.

$\xrightarrow{(iii)\ NaNH_2}$ 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).

$Y \xrightarrow[333K]{H_2O,\ Hg^{2+}/H^+}$ : 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, $C_3H_6O$ ). …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$1°$
›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

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-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. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{C_3H_9N}$
›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

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°)

Both 1° amines have two N–H bonds and react with benzenesulfonyl chloride to give sulfonamides. …
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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: $\mathrm{CH_3-C(CH_3)_2-CH_2-CH_3}$

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.

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.

That's 3 distinct environments → 3 monochloro products.

2,3-Dimethylbutane: $\mathrm{(CH_3)_2CH-CH(CH_3)_2}$ …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\mathrm{C_5H_{12}}$
›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

Only-one-product test → neopentane $\mathrm{(CH_3)_4C}$ (12 equivalent H).

$\mathrm{Br_2}$ /light → X = neopentyl bromide $\mathrm{(CH_3)_3C\text{-}CH_2\text{-}Br}$ .

X + $\mathrm{AgNO_2}$ (silver salt) → attack via N → major product is the nitroalkane. …
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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-Dichloropropane: $\mathrm{CH_3-CH_2-CHCl_2}$ — both Cl on C1 → geminal.

1,2-Dichloropropane: $\mathrm{CH_3-CHCl-CH_2Cl}$ — Cl on C1 and C2 (adjacent carbons) → vicinal.

1,3-Dichloropropane: $\mathrm{ClCH_2-CH_2-CH_2Cl}$ — Cl atoms separated by one carbon, not even vicinal. …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$C_4H_6$
›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

4-membered ring: cyclobutane skeleton with one C=C = cyclobutene, molecular formula already $C_4H_6$ with no substituent needed. → 1 structure.

3-membered ring + methyl: cyclopropene is $C_3H_4$ ; adding a $CH_3$ (replacing one H) gives $C_4H_6$ .

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).

Placing the methyl on $C_3$ gives a distinct molecule → 3-methylcyclopropene (1 structure). …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$\mathrm{C_6H_4Cl_2}$
›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

Label ring positions 1 through 6. Fix one Cl at position 1 (by ring symmetry, this loses no generality).

The second Cl can be at position 2 (ortho), 3 (meta), or 4 (para) — relative to position 1. …
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AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$C_nH_{2n+2}O$
›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

General formula of saturated alcohols: $C_nH_{2n+1}OH = C_nH_{2n+2}O$ .

General formula of saturated ethers ( $C_mH_{2m+1}-O-C_kH_{2k+1}$ with $m+k=n$ ): also simplifies to $C_nH_{2n+2}O$ .

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. …
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Understanding the Key Clue

Working Backward from D

Identifying C

Identifying B

Identifying A …

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