Q.Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH3)2CHCH2CH3.
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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.
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 |
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 C4H10, the formula satisfies 4(4)+10(1)=26 valence electrons. But the carbon atoms can be arranged as:
- A straight chain: CH3−CH2−CH2−CH3 (n-butane)
- A branched chain: CH3−CH(CH3)−CH3 (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:
- C4H10: 2 structural isomers
- C5H12: 3 structural isomers
- C6H14: 5 structural isomers
- C10H22: 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 C5H12:
- 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 CnH2n+2O), the position of the -OH group creates isomers:
- CH3CH2CH2OH (propan-1-ol) — OH at end
- CH3CH(OH)CH3 (propan-2-ol) — OH in middle …
The key idea is structural isomerism — specifically, constitutional isomers formed by replacing different hydrogen atoms with chlorine.
Reasoning:
-
The given compound is 2-methylbutane. Its carbon skeleton is:
- C1 (primary, end CH3), C2 (tertiary, CH), C3 (secondary, CH2), C4 (primary, end CH3), plus a methyl branch (primary CH3) on C2.
-
Free radical chlorination substitutes H atoms at different carbon positions. Each distinct carbon type yields a different monochloro product.
-
The possible products are:
- 1-chloro-2-methylbutane (Cl on C1)
- 2-chloro-2-methylbutane (Cl on C2 — tertiary)
- 2-chloro-3-methylbutane (Cl on C3)
- 1-chloro-3-methylbutane (Cl on C4) …
Free radical monochlorination of 2-methylbutane yields four distinct monochloro structural isomers: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 1-chloro-3-methylbutane, and 2-chloro-3-methylbutane. The key is to identify all unique carbon environments where a hydrogen can be replaced, ignoring stereoisomers.
Why This Approach Works
Structural isomerism in monochlorination arises from replacing a hydrogen atom on different carbon atoms of the parent alkane. The free radical mechanism is unselective — chlorine radicals abstract hydrogens from all available positions, though with different rates depending on the type of carbon (primary, secondary, tertiary). But for structural isomers, we only care about which carbon gets the chlorine, not how fast. So the task reduces to: count the distinct carbon atoms in the molecule, then consider if chlorination at each gives a unique product.
The molecule given is (CH3)2CHCH2CH3 — that's 2-methylbutane (isopentane). Let's draw it properly.
Structure of 2-methylbutane:
CH3∣CH3−CH−CH2−CH3
Number the carbon chain for clarity:
- Carbon 1: the CH3 at the left end (attached to the branch point)
- Carbon 2: the CH (the branch point, with a methyl group attached)
- Carbon 3: the CH2 in the main chain
- Carbon 4: the CH3 at the right end
- Carbon 5: the CH3 branch on carbon 2
Now, each distinct carbon environment can yield a different monochlorinated product. But careful: some carbons are equivalent by symmetry.
Step-by-Step Reasoning
1. Identify all unique carbon environments.
Look at the molecule: carbon 1 is a primary carbon (attached to one other carbon). Carbon 4 is also primary, but is it the same as carbon 1? No — carbon 1 is attached to carbon 2 (a tertiary carbon), while carbon 4 is attached to carbon 3 (a secondary carbon). So they are in different chemical environments. What about carbon 5 (the branch methyl)? It is attached to carbon 2, exactly like carbon 1 — the two methyls of the (CH3)2CH− group are equivalent by symmetry. So there are only two distinct primary environments: the C1/C5 pair, and C4.
Carbon 2 is tertiary (attached to three other carbons). Carbon 3 is secondary (attached to two other carbons). So in total, there are four distinct carbon environments — and hence four different types of hydrogen atoms — in the molecule: the C1/C5 pair, C2, C3 and C4.
A common mistake is to think that all methyl groups are equivalent. They are not — the environment matters. For example, the methyl at the end of a chain is different from a methyl branch on a tertiary carbon.
2. Consider chlorination at each carbon atom.
When a hydrogen is replaced by chlorine on a given carbon, we get a structural isomer. Let's name each product systematically:
-
Chlorination at carbon 1 (the left-end CH3): gives CH2Cl−CH(CH3)−CH2−CH3. The IUPAC name is 1-chloro-2-methylbutane.
-
Chlorination at carbon 2 (the tertiary CH): gives (CH3)2CCl−CH2−CH3. That's 2-chloro-2-methylbutane.
-
Chlorination at carbon 3 (the CH2 in the chain): gives (CH3)2CH−CHCl−CH3. That's 2-chloro-3-methylbutane (note: numbering starts from the end nearer the chlorine, so the methyl is on carbon 3, chlorine on carbon 2).
-
Chlorination at carbon 4 (the right-end CH3): gives (CH3)2CH−CH2−CH2Cl. That's 1-chloro-3-methylbutane.
-
Chlorination at carbon 5 (the branch methyl): gives (CH3)(CH2Cl)CH−CH2−CH3, i.e. CH3−CH(CH2Cl)−CH2−CH3.
3. Check for duplicates. …
Method: Free Radical Substitution – Product Analysis via Carbon Skeleton & Symmetry
This method identifies all structural isomers by:
- Drawing the parent alkane’s carbon skeleton.
- Classifying each unique carbon environment (1°, 2°, 3°).
- Replacing one H on each distinct carbon with Cl — each gives a different structural isomer.
Step 1: Identify the parent alkane
The given compound is:
(CH3)2CHCH2CH3
This is 2-methylbutane (isopentane). Its carbon skeleton:
C
|
C-C-C-C
Number the chain for clarity:
C1
|
C2-C3-C4-C5
Step 2: Identify unique carbon environments
- C1 (methyl, attached to C3) → 1° carbon
- C2 (methyl, attached to C3) → 1° carbon (identical to C1 by symmetry)
- C3 (CH, attached to C1, C2, C4) → 3° carbon
- C4 (CH₂, attached to C3 and C5) → 2° carbon
- C5 (methyl, attached to C4) → 1° carbon (different from C1/C2 because it’s on a different branch)
So we have 4 distinct carbon types:
- 1° (type A): C1 and C2 (equivalent)
- 1° (type B): C5 (different)
- 2°: C4
- 3°: C3
Step 3: Replace one H on each distinct carbon with Cl
Each unique carbon gives one structural isomer of monochlorinated product:
| Carbon type | Position | Product name |
|-------------|----------|--------------| …
This is a classic exam trap in free radical halogenation and structural isomerism. Let’s break down the common mistakes and how to avoid them.
🧪 The Reaction
We are monochlorinating:
(CH3)2CHCH2CH3
This is 2-methylbutane (isopentane). Free radical chlorination replaces one H with Cl.
✗ Mistake #1: Forgetting that all non-equivalent hydrogens give different products
Why it happens
Students often count only the carbon skeleton and miss that hydrogens on the same carbon but in different chemical environments can yield different isomers.
✓ How to avoid
- Identify all unique types of hydrogen atoms in the molecule.
- Draw the structure and label each distinct H environment.
Structure of 2-methylbutane:
CH3
|
CH3 — C — CH2 — CH3
|
H
Number the carbons:
- C1: CH₃ (attached to C2)
- C2: CH (tertiary, attached to C1, C3, C4)
- C3: CH₂ (attached to C2 and C4)
- C4: CH₃ (attached to C3)
- C5: CH₃ (attached to C2, the branch)
Hydrogen types:
| Carbon | Type of H | Number of H | Product after Cl |
|---|---|---|---|
| C1 | Primary | 3 | 1-chloro-2-methylbutane |
| C2 | Tertiary | 1 | 2-chloro-2-methylbutane |
| C3 | Secondary | 2 | 2-chloro-3-methylbutane |
| C4 | Primary | 3 | 1-chloro-3-methylbutane |
| C5 | Primary | 3 | 1-chloro-2-methylbutane (same as C1) |
So 4 structural isomers are possible.
✗ Mistake #2: Counting the same product twice
Why it happens
C1 and C5 are both primary methyl groups attached to C2 — they are equivalent by symmetry.
✓ How to avoid
- Always check for symmetry in the molecule.
- If two carbons are in identical environments, they give the same product.
Here, C1 and C5 are equivalent → only one product from both.
✗ Mistake #3: Ignoring optical isomerism when asked only for structural isomers
Why it happens
Some students list optical isomers (chiral centres) as separate structural isomers.
✓ How to avoid
- Structural isomers = different connectivity (constitutional isomers).
- Stereoisomers (optical, geometrical) are not structural isomers.
- In this reaction, 2-chloro-3-methylbutane has a chiral centre, but its enantiomers are not counted as separate structural isomers.
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Showing the 12 most recent of 15 on this concept.
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.An isomer of C4H8 is X, which exhibits cis-trans isomerism. Y is the chain isomer of X. Products obtained from ozonolysis of Y are (A) CH3CHO+CH3CHO (B) CH3CH2CHO+HCOOH (C) CH3COCH3+HCHO (D) CH3CH2COOH+CO2
›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
C4H8 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 (CH3−CH=CH−CH3) 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, CH2=C(CH3)2 — a branched C4H8 alkene. So Y = isobutylene.
Step-by-Step Solution
- Identify X: but-2-ene, CH3−CH=CH−CH3 (cis/trans possible).
- Identify Y (chain isomer of X): 2-methylprop-1-ene, (CH3)2C=CH2.
- Ozonolysis of Y breaks the C=C bond and replaces it with C=O on each fragment: …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.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: CH3−CH(CH3)−CH2−CH3. 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(CH3)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. …
- AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQQ.Identify X in the following reaction sequence C4H9BrOHC4H10OCu573 KC4H8 (X) (A) (CH3)3CBr (tert-butyl bromide, i.e. 2-bromo-2-methylpropane) (B) (CH3)2CHCH2Br (isobutyl bromide, i.e. 1-bromo-2-methylpropane) (C) CH3CH2CHBrCH3 (sec-butyl bromide, i.e. 2-bromobutane) (D) CH3CH2CH2CH2Br (n-butyl bromide, i.e. 1-bromobutane)
›Reveal solutionSolution
Reading the final product (C4H8, 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 H2O) to give an alkene. The molecular formula given for the final product, C4H8, has lost an H2O relative to C4H10O (not just H2, which would give C4H8O), confirming dehydration, i.e. a tertiary alcohol.
Step-by-Step Solution
- Final step: C4H10O→C4H8 over Cu/573K. Formula change is loss of H2O (C4H10O−H2O=C4H8), 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 C4H10O intermediate is tert-butanol, (CH3)3C−OH, which indeed dehydrates over Cu/573K to isobutylene, (CH3)2C=CH2 (C4H8). …
- AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQQ.The functional isomer of Z formed in the given sequence of reactions is CH3CH2CH2OHConc. H2SO4443 KX(i) Br2 ∣ CCl4(ii) alc.KOH,Δ (iii) NaNH2YH2O, Hg2+H+ ∣ 333 KZ (A) CH3CH2COOH (propanoic acid) (B) CH2=CH−OH (vinyl alcohol / ethenol) (C) CH3CH2CHO (propanal) (D) CH3CH2−O−CH3 (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. H2SO4 at 443 K = dehydration (E1, alcohol → alkene); Br2/CCl4 = anti addition across a C=C (vicinal dibromide); alc. KOH/Δ = dehydrohalogenation (E2); NaNH2 (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; H2O/Hg2+/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
- CH3CH2CH2OHconc. H2SO4443KX: dehydration removes H2O to give propene, X=CH3−CH=CH2.
- X(i) Br2/CCl4 1,2-dibromopropane, CH3−CHBr−CH2Br (anti addition of Br across the double bond).
- (ii) alc. KOH,Δ removes one HBr (E2) to give a bromopropene.
- (iii) NaNH2 removes the second HBr, giving the alkyne; with only 3 carbons the triple bond can only sit terminally, so Y=HC≡C−CH3 (propyne).
- YH2O, Hg2+/H+333K: Markovnikov hydration of the terminal alkyne places OH on the more substituted alkyne carbon, giving an enol that tautomerises to the methyl ketone: Z=CH3−CO−CH3 (acetone, C3H6O). …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.The number of primary (1°), secondary (2°) and tertiary (3°) alcohols possible for the formula C5H12O 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 CnH2n+2O 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 CH3−CH2−CH2−CH2−CH2−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 (CH3)2CH−CH2−CH3: 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. …
- Straight (n-pentane) skeleton CH3−CH2−CH2−CH2−CH2−OH type: OH can go on C1, C2, or C3 (C4, C5 are equivalent to C2, C1 by the chain's symmetry).
- AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQQ.How many amines with molecular formula C3H9N can react with benzene sulphonyl chloride ? (A) 2 (B) 3 (C) 4 (D) 1
›Reveal solutionSolution
Tests enumerating the isomeric amines of C3H9N and applying the Hinsberg-test rule that only 1° and 2° amines react with benzenesulfonyl chloride.
Concept and Intuition
Benzenesulfonyl chloride (C6H5SO2Cl) 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 C3H9N:
- CH3CH2CH2NH2 — n-propylamine (1°)
- (CH3)2CHNH2 — isopropylamine (1°)
- CH3−NH−CH2CH3 — N-methylethanamine (2°)
- (CH3)3N — trimethylamine (3°)
- Both 1° amines have two N–H bonds and react with benzenesulfonyl chloride to give sulfonamides. …
- List all isomers of C3H9N:
- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.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: CH3−C(CH3)2−CH2−CH3
- The carbon skeleton is C1−C2(−CH3)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 CH3 groups on C2, (ii) the CH2 at C3, (iii) the terminal CH3 at C4.
- That's 3 distinct environments → 3 monochloro products.
2,3-Dimethylbutane: (CH3)2CH−CH(CH3)2 …
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.An isomer of C5H12 on reaction with Br2 / light gave only one isomer C5H11Br (X). Reaction of X with AgNO2 gave Y as major product. What is Y? (A) O2N−C(CH3)2−CH2CH3 (B) ONO−C(CH3)2−CH2CH3 (C) (CH3)3C−CH2−ONO (D) (CH3)3C−CH2−NO2
›Reveal solutionSolution
X is neopentyl bromide; AgNO2 gives mainly the nitroalkane, so Y is (CH3)3C-CH2-NO2 → (D).
Concept and Intuition
A C5H12 isomer that yields a single monobromo product under free-radical bromination must have all its hydrogens equivalent — that is neopentane, C(CH3)4. Silver nitrite (AgNO2) 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 (CH3)4C (12 equivalent H).
- Br2/light → X = neopentyl bromide (CH3)3C-CH2-Br.
- X + AgNO2 (silver salt) → attack via N → major product is the nitroalkane. …
- AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQQ.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: CH3−CH2−CHCl2 — both Cl on C1 → geminal.
- 1,2-Dichloropropane: CH3−CHCl−CH2Cl — Cl on C1 and C2 (adjacent carbons) → vicinal.
- 1,3-Dichloropropane: ClCH2−CH2−CH2Cl — Cl atoms separated by one carbon, not even vicinal. …
- AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQQ.The number of cyclic isomers possible for C4H6, with one double bond in the ring is (A) 2 (B) 3 (C) 4 (D) 5
›Reveal solutionSolution
C4H6 with a ring double bond has exactly 3 possible cyclic structures: cyclobutene, 1-methylcyclopropene, and 3-methylcyclopropene.
Concept and Intuition
Degree of unsaturation for C4H6 is 22(4)+2−6=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 C4H6 with no substituent needed. → 1 structure.
- 3-membered ring + methyl: cyclopropene is C3H4; adding a CH3 (replacing one H) gives C4H6.
- Cyclopropene's ring carbons: C1=C2 (each bearing one H) and C3 (sp³, bearing two H). Placing the methyl on C1 or C2 gives the same molecule by the ring's mirror symmetry → 1-methylcyclopropene (1 structure).
- Placing the methyl on C3 gives a distinct molecule → 3-methylcyclopropene (1 structure). …
- AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQQ.The number of possible aromatic benzenoid isomers for C6H4Cl2 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
- 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. …
- AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQQ.Alcohols with molecular formula CnH2n+2O are isomeric with ________ (A) Acids (B) Ethers (C) Esters (D) Aldehydes
›Reveal solutionSolution
Alcohols and ethers share the same general formula CnH2n+2O, 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 CnH2n+2O (e.g. ethanol C2H6O and dimethyl ether C2H6O).
Step-by-Step Solution
- General formula of saturated alcohols: CnH2n+1OH=CnH2n+2O.
- General formula of saturated ethers (CmH2m+1−O−CkH2k+1 with m+k=n): also simplifies to CnH2n+2O.
- Acids (CnH2nO2), esters (CnH2nO2), and aldehydes (CnH2nO) all have different general formulas (different O count or different H count), so they are ruled out. …
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