Q.Primary alkyl halide C4H9Br
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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 …
Concept: Structural Isomerism — different compounds with the same molecular formula but different arrangements of atoms.
Reasoning:
- (a) is a primary alkyl halide CX4HX9Br. Possible isomers: 1-bromobutane or 1-bromo-2-methylpropane.
- (a) with alcoholic KOH gives alkene (b) via dehydrohalogenation. (b) with HBr gives (c), an isomer of (a). This means (c) must be a secondary or tertiary bromide, implying (b) is an unsymmetrical alkene.
- (a) with Na (Wurtz reaction) gives CX8HX18 (d), different from the product of n-butyl bromide with Na. n-Butyl bromide gives n-octane. So (d) must be a branched octane, confirming (a) is 1-bromo-2-methylpropane.
Reactions:
- (CHX3)X2CHCHX2Br+KOH (alc⋅)(CHX3)X2C=CHX2+KBr+HX2O …
The key is that (a) is a primary alkyl halide with formula C4H9Br, but its reaction with sodium gives a C8H18 product different from that of n-butyl bromide — this forces (a) to be isobutyl bromide. The sequence then proceeds through elimination, addition, and Wurtz coupling.
Let’s unpack this step by step. The problem is about structural isomerism and how different isomers behave differently in reactions. The formula C4H9Br has four isomers, of which only two are primary — and only one fits the entire puzzle.
1. Identify the possible primary alkyl halides with formula C4H9Br
A primary alkyl halide has the bromine attached to a terminal carbon. The four isomers of C4H9Br are:
- n-Butyl bromide: CH3CH2CH2CH2Br
- Isobutyl bromide: (CH3)2CHCH2Br
- sec-Butyl bromide: CH3CH2CHBrCH3 — but this is secondary, not primary, so it’s out.
- tert-Butyl bromide: (CH3)3CBr — tertiary, also out.
So only two primary isomers exist: n-butyl bromide and isobutyl bromide. The problem says (a) is primary, so (a) must be one of these two.
2. Reaction of (a) with alcoholic KOH gives (b) — an elimination
Alcoholic KOH favours elimination (dehydrohalogenation) over substitution. For a primary halide, the major product is the more substituted alkene (Saytzeff’s rule), but with only one possible alkene from each:
- n-Butyl bromide gives 1-butene: CH3CH2CH=CH2
- Isobutyl bromide gives 2-methylpropene: (CH3)2C=CH2
So (b) is either 1-butene or 2-methylpropene.
3. Compound (b) reacts with HBr to give (c), an isomer of (a)
Addition of HBr to an alkene follows Markovnikov’s rule — the hydrogen adds to the less substituted carbon, bromine to the more substituted.
- If (b) is 1-butene: HBr adds to give 2-bromobutane (secondary), which has formula C4H9Br but is not a primary halide — it’s an isomer of (a), but (a) is primary. That’s fine: (c) just needs to be an isomer, not necessarily primary.
- If (b) is 2-methylpropene: HBr adds to give tert-butyl bromide (tertiary), also an isomer of (a).
Both possibilities give an isomer of (a). So this step alone doesn’t decide.
4. Reaction of (a) with sodium metal gives (d), C8H18 — the Wurtz reaction
The Wurtz reaction couples two alkyl halides with sodium:
2RBr+2Na→R−R+2NaBr
For n-butyl bromide, the product is n-octane: CH3(CH2)6CH3
For isobutyl bromide, the product is 2,5-dimethylhexane: (CH3)2CHCH2CH2CH(CH3)2
The problem states that (d) is different from the compound formed when n-butyl bromide is reacted with sodium. That means (a) cannot be n-butyl bromide — because if it were, (d) would be n-octane, which is exactly what n-butyl bromide gives. So (a) must be isobutyl bromide.
A common mistake is to assume that the Wurtz product from isobutyl bromide is the same as from n-butyl bromide — but they are structural isomers. n-Octane is a straight chain; 2,5-dimethylhexane is branched. They are different compounds.
--- …
Method: Retrograde Analysis with Reaction Pathway Mapping
This method works backwards from the given products and constraints to deduce the unknown starting structure.
Step 1 — Identify the key constraint
We are told:
- (a) is a primary alkyl halide with formula CX4HX9Br.
- (c) is an isomer of (a) — meaning same formula but different structure.
- (d) is CX8HX18 from Wurtz reaction of (a), and it is different from the product of n-butyl bromide under the same conditions.
Key inference: If n-butyl bromide gives n-octane (CHX3(CHX2)X6CHX3), then (d) must be a branched CX8HX18 isomer. This means ** (a)** must be a branched primary alkyl bromide.
Step 2 — List possible primary CX4HX9Br isomers
| Name | Structure |
|---|---|
| n-Butyl bromide | CHX3CHX2CHX2CHX2Br |
| Isobutyl bromide | (CHX3)X2CHCHX2Br |
Only isobutyl bromide is primary and branched.
Step 3 — Verify with the reaction sequence
Reaction 1: Dehydrohalogenation with alcoholic KOH
(CHX3)X2CHCHX2Br+alc⋅KOH(CHX3)X2C=CHX2+KBr+HX2O
- (b) is isobutylene (2-methylpropene).
Reaction 2: Addition of HBr to (b)
(CHX3)X2C=CHX2+HBr(CHX3)X3CBr
- (c) is tert-butyl bromide, an isomer of ** (a)** (both CX4HX9Br). …
Here are the common mistakes students make on this exact problem, along with how to avoid each one.
Mistake 1: Assuming the primary alkyl halide is n-butyl bromide
Why it’s wrong:
The problem explicitly says that when (a) reacts with sodium (Wurtz reaction), the product C8H18 is different from the product formed when n-butyl bromide reacts with sodium. If (a) were n-butyl bromide, the Wurtz product would be n-octane — the same as the reference. So (a) cannot be n-butyl bromide.
How to avoid:
Read the “different from” condition carefully. It forces (a) to be a branched primary alkyl halide. The only primary C4H9Br that is not n-butyl bromide is isobutyl bromide:
- n-butyl bromide: CH3CH2CH2CH2Br
- Isobutyl bromide: (CH3)2CHCH2Br
So (a) must be isobutyl bromide.
Mistake 2: Forgetting that alcoholic KOH causes elimination, not substitution
Why it’s wrong:
Many students write substitution products (like an alcohol) when they see “KOH”. But alcoholic KOH favours elimination (dehydrohalogenation), especially with a primary halide and a strong base.
How to avoid:
Remember the rule:
- Aqueous KOH → substitution (alcohol)
- Alcoholic KOH → elimination (alkene)
Here, (b) is an alkene. For isobutyl bromide, elimination gives isobutylene (2-methylpropene):
(CH3)2CHCH2Br+KOH (alc.)Δ(CH3)2C=CH2+KBr+H2O
Mistake 3: Adding HBr to the alkene and getting the wrong isomer
Why it’s wrong:
When HBr adds to an unsymmetrical alkene, Markovnikov’s rule applies — the hydrogen goes to the carbon with more hydrogens. For isobutylene:
(CH3)2C=CH2+HBr→(CH3)3CBr
This gives tert-butyl bromide, which is an isomer of (a) (both are C4H9Br). Some students incorrectly add HBr the other way and get back isobutyl bromide — but that would mean (c) = (a), which contradicts “isomer of (a)”.
How to avoid:
Always apply Markovnikov’s rule for HX addition. Check that the product is different from (a) — here, (c) is a tertiary halide, while (a) is primary.
Mistake 4: Writing the Wurtz reaction incorrectly
Why it’s wrong:
The Wurtz reaction couples two alkyl halides with sodium:
2RBr+2Na→R−R+2NaBr
For isobutyl bromide: …
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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