The correct IUPAC name for $CH_2{=}CHCH_2NHCH_3$ is ____. (i) Allylmethylamine (ii) 2-amino-4-pentene (iii) 4-aminopent-1-ene (iv) N-methylprop-2-en-1-amine

> [!TLDR] > The compound is a secondary amine with a three-carbon chain and a double bond at the 2-position. The correct IUPAC name is N-methylprop-2-en-1-amine, option (iv). The key to naming this compound is recognising it as an amine, not an alkene with an amino substituent - the amine is the principal characteristic group, so the parent chain takes the suffix "-amine" (with the double bond shown as an infix "-en-"), not the reverse. 1. Identify the functional group. The structure is CH2=CHCH2NHCH3. The -NHCH3 part is a secondary amine - nitrogen bonded to a methyl and to a three-carbon chain, CH2=CH-CH2-. No higher-priority group (e.g. carboxylic acid, aldehyde) is present, so the amine is the parent. 2. Determine the parent chain. The chain attached to nitrogen is CH2=CH-CH2-: three carbons with a double bond between C1 and C2 (numbering from the amine carbon). The parent is named as an amine with an "-en-" infix for the double bond: prop-2-en-1-amine. 3. Number so the amine carbon gets the lowest locant. The carbon directly bonded to N is position 1; the double bond then starts at C2. 4. Name the N-substituent. The methyl group on nitrogen is indicated by the prefix N-methyl. 5. Combine: N-methylprop-2-en-1-amine. > [!WARNING] > A common mistake is to treat the amine as a substituent ("amino") and name the compound as an alkene derivative - options (ii) and (iii) do exactly that, and are wrong because the amine has priority over the double bond as the parent suffix. Option (i), "Allylmethylamine," is a common name, not the systematic IUPAC name. > [!TIP] > When an amine is the principal group, the suffix is "-amine," with the nitrogen's position given a locant and the double bond shown by "-en-" with its own locant. > [!ANSWER] > The correct option is (iv) N-methylprop-2-en-1-amine.

The correct IUPAC name for CH₂=CHCH₂NHCH₃ is __. (i) Allylmethylamine (ii) 2-amino-4-pentene (iii) 4-aminopent

NCERT Exemplar · Q2

CH_2{=}CHCH_2NHCH_3

(i) Allylmethylamine

(ii) 2-amino-4-pentene

(iii) 4-aminopent-1-ene

(iv) N-methylprop-2-en-1-amine

Gujarat GsebMCQ· 1mImportance★★★★★

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✓ Free question

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

Structural isomers can have wildly different properties. Ethanol (C₂H₆O) is a drinkable alcohol; its isomer dimethyl ether is a gas used as a refrigerant. Same atoms, but one is a liquid you can consume, the other is a gas that would kill you. That is why chemists care so much about connectivity — it determines everything.

Quick Check

Question: Are these structural isomers?

Molecule A: CH₃–CH₂–CH₂–CH₃

Molecule B: CH₃–CH(CH₃)–CH₃

Answer: Yes. Both are C₄H₁₀. A is n-butane (straight chain), B is isobutane (branched). Different connectivity → structural isomers.

Important

The molecular formula must be identical. If the formulas differ, they are not isomers at all — just different compounds.

Structural isomerism is introduced in the NCERT/CBSE Class 11 Chemistry chapter on Organic Chemistry: Some Basic Principles and Techniques, and ‘structural isomerism examples class 11’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Correctly distinguishing structural isomers by connectivity, rather than just matching molecular formulas, is a skill tested throughout competitive organic chemistry exams.

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

Why these are distinct: The OH group's position changes the carbon's hybridization environment and the molecule's polarity.

The Ring-Chain Isomerism Reason

For unsaturated formulas like $C_4H_8$ , the same formula can represent:

A straight alkene: $CH_2=CH-CH_2-CH_3$
A branched alkene: $CH_3-C(=CH_2)-CH_3$
A cycloalkane: cyclobutane (ring)

Why rings form: Carbon atoms can bond to form closed loops, reducing the number of hydrogen atoms needed. The formula $C_nH_{2n}$ can be either an alkene (one double bond) or a cycloalkane (one ring).

Summary: The Takeaway

Aspect	Why It Holds
Different connectivity	Atoms can bond in multiple sequences while satisfying valency
No simple counting formula	The number of possible trees grows combinatorially
Branching creates isomers	Carbon chains can have branches at different positions
Position matters	Functional groups at different locations change properties
Rings vs. chains	Same formula can represent open chains or closed rings

The key insight: Structural isomerism exists because molecular formula is a constraint, not a blueprint — it tells you the ingredients, not the recipe.

The compound is a secondary amine with a three-carbon chain and a double bond at the 2-position. The correct IUPAC name is N-methylprop-2-en-1-amine, option (iv).

The key to naming this compound is recognising it as an amine, not an alkene with an amino substituent - the amine is the principal characteristic group, so the parent chain takes the suffix "-amine" (with the double bond shown as an infix "-en-"), not the reverse.

Identify the functional group. The structure is CH2=CHCH2NHCH3. The -NHCH3 part is a secondary amine - nitrogen bonded to a methyl and to a three-carbon chain, CH2=CH-CH2-. No higher-priority group (e.g. carboxylic acid, aldehyde) is present, so the amine is the parent.
Determine the parent chain. The chain attached to nitrogen is CH2=CH-CH2-: three carbons with a double bond between C1 and C2 (numbering from the amine carbon). The parent is named as an amine with an "-en-" infix for the double bond: prop-2-en-1-amine.
Number so the amine carbon gets the lowest locant. The carbon directly bonded to N is position 1; the double bond then starts at C2.
Name the N-substituent. The methyl group on nitrogen is indicated by the prefix N-methyl.
Combine: N-methylprop-2-en-1-amine.

Watch out

A common mistake is to treat the amine as a substituent ("amino") and name the compound as an alkene derivative - options (ii) and (iii) do exactly that, and are wrong because the amine has priority over the double bond as the parent suffix. Option (i), "Allylmethylamine," is a common name, not the systematic IUPAC name.

Tip

When an amine is the principal group, the suffix is "-amine," with the nitrogen's position given a locant and the double bond shown by "-en-" with its own locant.

✓Final answer

The correct option is (iv) N-methylprop-2-en-1-amine.

✗ Mistake 1: Choosing a common name instead of IUPAC

Example: Selecting (A) Allylmethylamine.

Why it happens:

Students see the allyl group ( $CH_2{=}CHCH_2-$ ) and a methyl group on nitrogen, and recall the common naming pattern: "allylmethylamine". This is a valid common name, but the question asks for the IUPAC name.

How to avoid:

Always check if the question explicitly asks for the IUPAC name.
Common names (like allyl, vinyl, phenyl) are not accepted in IUPAC unless specified.
For IUPAC, treat the amine as a substituted alkylamine — the longest carbon chain containing the $NH$ group gets the parent name.

✗ Mistake 2: Misidentifying the principal functional group

Example: Choosing (B) 2-amino-4-pentene or (C) 4-aminopent-1-ene.

Why it happens:

Students see the double bond and the $NH$ group and think "amine is a substituent, alkene is the parent". They then number the chain to give the double bond the lowest number.

How to avoid:

In IUPAC, amine ( $-NH_2$ or substituted $-NHR$ ) is the principal functional group when no higher priority group (like acid, aldehyde, etc.) is present.
The suffix for amine is -amine, not "amino" as a prefix (unless it's a substituent on a higher-priority group).
The parent chain must include the nitrogen atom and be numbered to give the amine the lowest locant.

Correct logic:

The longest chain containing the $NH$ is propane (3 carbons: $CH_2{=}CHCH_2-$ ).
The double bond is on carbon 1 (prop-2-en because numbering starts from the end nearer the $NH$ ).
The $NH$ has a methyl substituent → N-methyl.
So the name is N-methylprop-2-en-1-amine → option (D).

✗ Mistake 3: Wrong numbering priority

Example: Numbering from the double bond end instead of the amine end.

Why it happens:

Students apply the "lowest locant for the double bond" rule without checking functional group priority.

How to avoid:

Functional group priority order (for IUPAC nomenclature): Amine > Alkene > Alkyne > Alkane
Always number the parent chain to give the principal group (amine) the lowest number, even if it makes the double bond number higher.

Check:

If numbered from the $NH$ end: $NH$ on C1, double bond on C2 → prop-2-en-1-amine
If numbered from the double bond end: double bond on C1, $NH$ on C3 → prop-1-en-3-amine (incorrect because amine gets higher locant)

✓ Correct Answer: (D) N-methylprop-2-en-1-amine

📌 Quick Summary Table

Mistake	Why it happens	How to avoid
Choosing common name	Familiarity with "allyl"	Use IUPAC rules only
Treating amine as substituent	Double bond seems more important	Amine is principal group
Wrong numbering	Double bond priority confusion	Number to give amine lowest locant

Final tip: For any amine with a double bond, always:

Find the longest chain containing the $N$ atom.
Number from the end nearest the $N$ .
Name as N-substituted-alken-amine.

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

GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Total number of all the possible monochloro structural isomers expected to be formed on free radical Monochlorination of 2-Methyl butane$
(a) 2
(b) 5
(c) 3
(d) 4
›Reveal solutionSolution
2-Methylbutane has 4 chemically distinct sets of hydrogen atoms, so free-radical monochlorination gives 4 structurally different monochloro products.

2-Methylbutane: CH3-CH(CH3)-CH2-CH3

Identify the distinct (non-equivalent) carbon positions where a H can be replaced by Cl:

The two equivalent CH3 groups attached to C2 (C1 and the branch methyl - these are chemically identical by symmetry) -> substitution here gives 1-chloro-2-methylbutane.

The C2-H (the single tertiary hydrogen) -> substitution gives 2-chloro-2-methylbutane.

The C3-H2 (the CH2 group) -> substitution gives 2-chloro-3-methylbutane.

The C4-H3 (terminal methyl of the ethyl arm) -> substitution gives 1-chloro-3-methylbutane.

That gives 4 distinct structurally different monochloro products (the two equivalent methyls on C2 only count once, since substituting either gives the identical structure).

✓Final answer
$(d) 4 - free-radical monochlorination of 2-methylbutane gives 4 structural isomers (1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, 1-chloro-3-methylbutane).$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. How many monochloro structural isomers expected to be formed on free radical monochlorination of iso-pentane?$
(a) 5
(b) 4
(c) 2
(d) 3
›Reveal solutionSolution
Iso-pentane has 4 kinds of H, so 4 monochloro structural isomers form.

Iso-pentane = 2-methylbutane: (CH3)2CH-CH2-CH3. Its hydrogen environments:

The two equivalent CH3 on C2 (branch) -> gives 1-chloro-2-methylbutane.

The single tertiary H on C2 -> gives 2-chloro-2-methylbutane.

The CH2 (C3) hydrogens -> gives 2-chloro-3-methylbutane.

The terminal CH3 (C4) hydrogens -> gives 1-chloro-3-methylbutane.

Four distinct types of H -> four monochloro structural isomers.

✓Final answer
$(b) 4.$
GUJCET 2022Set 171 markMCQ
$C_3H_9N$
›Reveal solutionSolution
$C_3H_9N$ → 4 isomeric amines.

Concept. Enumerate all amines (1°, 2°, 3°) with three carbons and the saturated formula $C_3H_9N$ :

Propan-1-amine, $CH_3CH_2CH_2NH_2$ (1°)

Propan-2-amine, $(CH_3)_2CHNH_2$ (1°)

N-methylethanamine, $CH_3CH_2NHCH_3$ (2°)

Trimethylamine, $(CH_3)_3N$ (3°)

Total = 4.

✓Final answer
$(C) 4 isomers.$

ANSWER: (C)
GUJCET 2020Set 071 markMCQ
$C_4H_9Br$
›Reveal solutionSolution

Among the C $_4$ H $_9$ Br isomers only sec-butyl bromide (2-bromobutane) is chiral, giving 2 optically active forms.

Concept. Optical activity requires a stereocentre. The isomers of C $_4$ H $_9$ Br are 1-bromobutane, 2-bromobutane, isobutyl bromide, and tert-butyl bromide.

Steps. Only 2-bromobutane, CH $_3$ CH $_2$ CHBrCH $_3$ , has an asymmetric carbon, so it exists as two enantiomers (R and S) — 2 optically active isomers.

✓Final answer
$(B) 2$

ANSWER: (B)
GUJCET 2020Set 071 markMCQ
$R' - Cl \xrightarrow{Na/ether} 2,3\text{-dimethyl butane}$
›Reveal solutionSolution

Wurtz couples $R'-R'$ ; two isopropyl groups give $(\text{CH}_3)_2\text{CH–CH(CH}_3)_2$ = 2,3-dimethylbutane.

Concept — Wurtz reaction. $2R'-\text{Cl}\xrightarrow{\text{Na/ether}}R'-R'$ . For the product 2,3-dimethylbutane $(\text{CH}_3)_2\text{CH–CH(CH}_3)_2$ , each half is $(\text{CH}_3)_2\text{CH}-$ , i.e. isopropyl.

✓Final answer
$(C) isopropyl$

ANSWER: (C)
GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Possible isomers of monohydric phenol having molecular formula C7H8O are ______.$
(a) 1
(b) 4
(c) 3
(d) 2
›Reveal solutionSolution

A monohydric phenol of formula C7H8O must be a methyl-substituted phenol (cresol); the methyl group can sit at three distinct ring positions relative to -OH.

C7H8O with the -OH directly attached to the aromatic ring (a phenol, not an ether like anisole, C6H5-OCH3, which is also C7H8O but is NOT a phenol) must be a cresol: a benzene ring bearing one -OH and one -CH3 group. By the symmetry of the benzene ring, the -CH3 can be ortho, meta, or para to the -OH, giving exactly 3 distinct constitutional isomers: o-cresol, m-cresol, and p-cresol.

✓Final answer
$(c) 3.$

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Q.The correct IUPAC name for $CH_2{=}CHCH_2NHCH_3$ is ____.

Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

The Precise Statement

A Concrete Example: C₄H₁₀

The Three Main Types

Why This Matters

Quick Check

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

The Core Principle: Connectivity ≠ Composition

Why This Happens: The Valency Constraint

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

Why No Simple Formula Exists

Key Reasoning: The Branching Principle

The Functional Group Position Rule

The Ring-Chain Isomerism Reason

Summary: The Takeaway

Method: IUPAC Naming of Secondary Amines (Substitutive Nomenclature)

Why this method?

Steps

Why other options are wrong

✗ Mistake 1: Choosing a common name instead of IUPAC

✗ Mistake 2: Misidentifying the principal functional group

✗ Mistake 3: Wrong numbering priority

✓ Correct Answer: (D) N-methylprop-2-en-1-amine

📌 Quick Summary Table

Q.The correct IUPAC name for CH2=CHCH2NHCH3CH_2{=}CHCH_2NHCH_3CH2​=CHCH2​NHCH3​ is ____.

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Q.The correct IUPAC name for $CH_2{=}CHCH_2NHCH_3$ is ____.

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