Structural Isomerism - Chemistry Class 11 - NEET PDF Download

What is Isomerism?

Compounds having the same molecular formula but different properties are called isomers. The phenomenon of existence of such compounds is called isomerism. The word isomer is derived from the Greek words isos (equal) and meros (parts). The term was introduced by the Swedish chemist Jacob Berzelius in 1830.

Classification of Isomerism

Isomerism is broadly classified into two main types:

• Structural (constitutional) isomerism: Isomers that differ in the connectivity or sequence in which atoms are bonded to one another. Differences may include changes in the carbon chain, position of functional groups, or the functional group itself.
• Stereoisomerism: Isomers that have the same connectivity but differ in the spatial arrangement of atoms. This includes geometrical isomerism (cis/trans) and optical isomerism (enantiomers and diastereomers).

Classification of Isomerism

What is Structural Isomerism?

Structural isomerism occurs when two or more compounds have the same molecular formula but different connectivity of atoms (different structures). A simple example is the alkane with formula C4H10, which shows two different connectivity patterns (n-butane and isobutane). As the number of carbon atoms in an alkane increases, the number of possible structural isomers increases rapidly.

Structural Isomers of Butane

Classification of Structural Isomers

Structural isomerism can be divided into the following types:

• Chain isomers: Same molecular formula but different arrangements of the carbon skeleton (straight chain vs branched chains).
• Position isomers: Same molecular formula and same principal chain but differ in the position of a functional group, substituent or multiple bond.
• Functional group isomers: Same molecular formula but different functional groups (for example, alcohol vs ether, aldehyde vs ketone).
• Tautomers: Isomers that are interconvertible by movement (migration) of an atom (usually hydrogen) and a shift of bonding electrons; they are in rapid equilibrium (e.g., keto-enol tautomerism).
• Metamers: Isomers with the same molecular formula where a polyvalent atom or group (such as oxygen in ethers) bears different alkyl groups on either side (R-O-R').
• Ring isomers: Compounds that differ by formation of different cyclic structures while retaining the same molecular formula (different ring size or ring connectivity).

Classification of Structural Isomers

Chain Isomerism

Definition: Chain isomers are compounds having the same molecular formula but differing in the length or branching of the principal carbon chain.

How to form Chain Isomers?

The general formula for alkanes is C_nH_2n+2. Removing one H gives an alkyl group C_nH_2n+1. To form chain isomers, change which carbon atoms are taken as the main chain and which as branches (iso-, neo- prefixes describe common branching patterns).

C_nH_2n+2 C_nH_2n+1

Alkane                   Alkyl group

The dark line (-) in structural sketches represents a vacant valency where any group (or bond) can be attached.

Iso-group

e.g. Isoheptane

e.g. Isooctane (note: some products are exceptions to simple iso-group rules)

Neo-group

To prepare a neo compound, write the central compact group first and then extend the straight chain as required.

e.g. Neopentane

e.g. Neoheptane

Examples of Chain Isomerism

1. C₄H₁₀ has two structural isomers (n-butane and isobutane). The input sketches show 4 forms-these correspond to distinct connectivity drawings and conformational depictions; the two unique connectivity isomers are:

2. C₅H₁₂ has three structural isomers:

n-pentane: C-C-C-C-C (straight chain)

isopentane (methylbutane)

neopentane (dimethylpropane)

Counting distinct alkyl radicals, the total distinct C₅H₁₁ fragments shown equal 8 in the worked example.

3. Butane examples:

CH₃-CH₂-CH₂-CH₃ = n-butane

2-methylpropane = isobutane

4. Carboxylic acids showing chain isomerism:

butanoic acid and 2-methylpropanoic acid:

Ex.1 Find all the structural isomers of C₆H₁₄

Sol.

Structural Isomers of C₆H₁₄

Position Isomerism

Definition: Position isomers have the same molecular formula and the same principal carbon chain, but differ in the position of a functional group, substituent or a multiple bond.

Position Isomerism

e.g. 1-butene (C-C-C= C) and 2-butene (C-C= C-C) are position isomers.

e.g. 1-chloropropane and 2-chloropropane:

 (1-chloropropane)  and  (2-chloropropane)  are position isomers.

e.g. 

 (1-propanol)  and  (2-propanol)

Note: Some pairs that look like a simple movement of a substituent actually change the principal chain and are therefore chain isomers, not position isomers. For example, shifting a terminal carbon into another position may change which chain is considered the principal chain.

Ex.2 Find the relation between the given compounds:

(A) C-C-C-C-C-C

(B)

(C)

(D)

(E)

Sol. a, b → chain isomers. b, c → position isomers. c, d → chain isomers. d, e → position isomers.

Ex.3 How many monochloro derivatives will be of C₄H₁₀ (Only structural)

Sol.

Monochloro Derivative of C₄H₁₀

Important: Monochlorination means replace one H by Cl; count unique positions considering symmetry and different types of hydrogen (primary, secondary, tertiary).

Ex.4 An alkane having molecular formula C₅H₁₂ can give only one product on monochlorination. Find the IUPAC name of the alkane.

Sol.

(2,2-dimethylpropane is the structure implied; IUPAC: 2,2-dimethylpropane is not a correct name for C5H12 - the intended structure in the input corresponds to neopentane derivative; accept the original figure as authoritative.)

Ex.5

 Mono chlorinated products (Excluding Stereoisomers)?

Sol.

Ex.6 Find the total structural isomer of C₅H₁₀.

Sol. To solve these problems, draw all possible corresponding alkanes and then check where a double bond can be placed without exceeding carbon valency.

= 2

= 3

no double bond can be placed in this compound as the valency of C would exceed 4.

Total open chain structural isomers = 5

To form cyclic structures, start with a 3-carbon ring and consider all placements of the remaining atoms:

Total structural isomers = 5 (open chain) + 5 (cyclic) = 10

Ex.7 Find the total structural isomers of C₄H₆.

Sol. Total degree of unsaturation for C₄H₆ is 2. Possibilities: one triple bond, two double bonds, or one ring + one double bond.

(positions for triple bond)

→ (no triple bond)

For alkenes:

Total open chain = 2 + 2 = 4

Cyclic possibilities:

Total structural isomers (cyclic + acyclic) = 9

Ex.9 Find all the structural dichloro derivatives of cyclopentane.

Sol.

Total structural isomers = 3

Functional Group Isomerism

Definition: Functional group isomers have the same molecular formula but different functional groups and therefore different chemical properties.

Important:

• Aldehydes, ketones, cyclic ethers, cyclic alcohols, and unsaturated alcohols can be functional isomers of one another.
• Alcohols and ethers are functional isomers (e.g., CH₃CH₂OH and CH₃OCH₃).
• Carboxylic acids and esters are functional isomers (examples shown in figures).
• Cyanides (nitriles) and isocyanides are functional isomers; note HCN and HNC are tautomers rather than classical functional isomers.
• Primary (1°), secondary (2°) and tertiary (3°) amines represent different functional classes and, for a given molecular formula, can be seen as functional isomers.

Example: C₃H₆O can correspond to propanal (an aldehyde), acetone (a ketone), or other structures such as unsaturated alcohols/ethers depending on connectivity:

CH₃-CH₂-CHO and corresponding enol/ether representations are shown in the figures.

Ex.10 How many primary, secondary, and tertiary alcohols are possible for C₅H₁₃O? (Only structural)

Sol. For 1° alcohol (-CH₂OH)

C₅H₁₂OH ⇒ C₄H₉ - CH₂ - OH (4 form)

for 2° alcohol  →  

Replace one C (1 form)

total = 3

for 3° alcohol  

total = 4 + 3 + 1 = 8

or C₅H₁₃O ⇒ C₅H₁₂ - OH (8 form)

Ex.11 How many ethers are possible in C₅H₁₂O? (Only structural)

Sol. C₄H₉OCH₃   ⇒    C₃H₇OC₂H₅
        (4 form)            (2 form)

total = 6

Ex.12 How many 1º, 2º, and 3º amines are possible for C₅H₁₃N (Only structural)

Sol. For 1º amine (- NH₂)

C₅H₁₁ NH₂ ⇒ (8 form) = 8

for 2º Amine, (- NH -) C₄H₉ - NH - CH₃  ⇒ 4 forms

Also, C₃H₇ - NH - C₂H₅

(2 form)

total form at 2º = 4 + 2 = 6

for 3º

C₃H₇ -  (2 form)

Also,  (1 form)

Total form of 3º = 2 + 1 = 3

Total no. of amines = 8 + 6 + 3 = 17

Ex.13 For molecular formula C₄H₉NO, how many amides will be there that will not form H-bond? (Only structural)

Sol. (1º amide)  (2º amide)  (3º amide)

for 1º amide

 (2 form)

for 2º amide,  (2 form)

total 2º amide = 4

for 3º amide

3º amide will not form H-bond hence there will be 2 amides which will not form H-bond.

Ex.14 Find all 1º, 2º and 3º amides for C₃H₇NO (Only structural)

Sol. For 1º amides

 (1 form)   , total 1º amide = 1

for 2º amide

 = 2 form

for 3º amide

 total 3º amide = 1

total amides (1º 2º 3º) = 1 + 2 + 1 = 4

Ex.15 Find the total no. of acids and esters from C₄H₈O₂ (Only structural)

Sol. For acid, (- COOH)    C₃H₇ - COOH (2 form)

for ester,  → alcohol part

CH₃COOH C₂H₅OH

for ester,  (2 form)

total esters = 4

Ex.16 C₄H₄O₄ may be (Only structural):

(i) Saturated dicarboxylic acid (ii) Unsaturated dicarboxylic acid (iii) Cyclic diester (iv) Saturated dialdehyde

Sol. Three unsaturation.

(ii) and (iii) is the Ans.

 and

Ex.17 How many aromatic isomers will be possible for C₇H₈O (Only structural)

Sol.

Total = 5

Ex.18 Find the possible dichloro derivatives of C₆H₄Cl₂ (Only structural)

Sol.

Ex.19 C₆H₄Cl₂ → C₆H₃Cl₃ (Only structural)

→→→

find the value of x, y, z.

Sol.

(1, 2, 4)

Two possibilities are there for placing Cl in place of H (figures show choices):

y = 3 (three possible placements), therefore z = 1

x = 2, y = 3, z = 1

Ex.20 Find the total carbonyl compounds (aldehydes and ketones) formed by C₅H₁₀O and also find the relation between carbonyl compounds that have the same number of α-hydrogen. (Only structural)

Sol. For ketones,

  and  

Isomerism - 

(2 form)

total ketones = 3

for aldehydes   
                          (4 forms)

total aldehydes = 4

total carbonyl compounds = 4 + 3 = 7

CH₃ - CH₂ - CH₂ - CH₂ - CHO (2 α H)  (one α H)

 (2 α H)  (no α hydrogen)

 ketone,  (5 α H)

Ex.21 Find total acyclic structural isomer of C₆H₁₂ (Only structural)

Sol.

(1)  = 3

(2)  = 4

(3)  = 1

(4)  = 2

(5)  = 3

total = 13 isomers.

Ex.22 Find the total conjugated diene in C₅H₈ (Only structural)

Sol.

→ one possibility

→ one possibility

→ no possibility (double bond placement exceeds valency)

Total conjugated diene possibilities listed in figures.

Ex.23 Find the total cumulated diene in C₅H₈ (Only structural)

Sol.

 →

 →no form cumulated diene

total = 3

⇒ Isolated dienes,

C - C - C - C - C →

 →no form isolated diene

total = 1

Metamerism

• Metamers are structural isomers having the same molecular formula but different alkyl groups attached to a polyvalent atom or functional group (for example, ethers R-O-R', secondary amines R-NH-R', esters R-COO-R').
• This type of isomerism appears in compounds containing a polyvalent atom or group which splits two alkyl moieties across it.

Examples of metamers:

1.

2.

(a)

(b) CH₃-CH₂-CH₂-NH-CH₂-CH₃

a & b are metamers.

Tautomerism

Definition: Tautomers are isomers which readily interconvert by migration of an atom (usually hydrogen) accompanied by a shift of a double bond; the equilibrium between them is called a tautomeric equilibrium. They have the same molecular formula but different bonding arrangements.

The conjugate anion formed after H+ removal is often resonance stabilised, which is why tautomerism is common for carbonyl compounds.

Keto-enol Tautomerism

Keto-Enol Tautomerism

Mechanism :

* OH^- acts as a catalyst in base-catalysed enolisation.

Base Catalysed Tautomerism

Mechanism:

Enols are generally more acidic than their keto forms at the α-C-H position because deprotonation leads to resonance-stabilised anions. The anion with negative charge on oxygen is usually more stable than that with negative charge on carbon, making some routes more favourable:

 → .......(i)

 → .......(ii)

In (i) the negative charge is mainly on carbon; in (ii) it is on oxygen; (ii) is more stable, so enol is comparatively acidic.

Acid Catalysed Tautomerism

Ex.24

• In base-catalysed tautomerism, the stability of the intermediate carbanion determines the pathway and product distribution.
• In acid-catalysed tautomerism, the stability of the protonated intermediates and the final product determines the major product.

Ex.25

P₁ + P₂

Sol.

+  

                             (major)               (minor)

(3 α H) +      (7 α H)

(minor)                            (major)

Ex.26 Write the enol form:

Sol.

Although the keto form is generally more stable, enol form may be more stable in special cases due to:

• Intramolecular hydrogen bonding
• Aromatic character gained on enolisation
• Extended conjugation (stabilising resonance)
• Steric effects that destabilise the keto form

Example shown in figure shows a 6-membered intramolecular H-bonded enol, which can make the enol percentage high.

General trends (approximate enol percentages depend on substituent G):

If G = H, % of enol: 80-90%

If G = Ph, % of enol: 90-99%

If G = CH₃, % of enol: 70-80%

If G = OC₂H₅, % of enol: 5-10%

Cross conjugation restricts resonance and generally reduces enol stability:

Ex.27 Compare the enol percent.

(A) CH₃CHO

(B)

(C)

(D)

(E)

(F)

(G)  

Sol. Ordering of enol content (from highest to lowest): g > f > e > d > c > b > a.

For (a) and (b), forming the enol leads to CH₂=CH-OH (no α H); structures with greater resonance/stabilisation give higher % enol.

Cases where intramolecular H-bonding stabilises enol, enol % > keto %; otherwise keto tends to be dominant.

 enol % > keto %

 (Keto < enol due to Intramolecular H-bond)

keto > enol

Ex.28 Find the enol form of the given compound.

Sol.

This example is a case of para-tautomerism where a hydrogen from a para position participates in tautomerism.

Ex.29 Compare the enol content

(A)

(B)

(C)

(D)

Sol.

After removing the acidic H (keto → enol pathway) the stability of the resulting carbanion determines enol percentage:

(A)

(B)

(C)

(D)

Enol percent ∝ stability of carbanion. Therefore ordering: a > b > d > c.

Formation of the carbanion is a key step in the keto → enol conversion.

Ex.30 Find the enol form of

Nitro and Acinitro form

 →

(acinitro)

 (acinitro)

Note: tertiary alkyl nitro compounds like (CH₃)₃C-NO₂ cannot show nitro↔acinitro tautomerism because there is no hydrogen on the carbon adjacent to NO₂.

Imine and Enamine

Tautomerism between imine and enamine forms requires a hydrogen on the carbon adjacent to the imine (-CH=NH type systems).

Amide and Amidol

Nitroso and Oxime form

Form II > Form I in stability due to extended conjugation in II (shown in figures).

Hydrazone and Azo form

Hydrazine (NH₂-NH₂) reacts with carbonyls to give hydrazones; azo compounds (R-N=N-R') gain extra conjugation and are often more stable:

Example: CH₃-CH=N-NH-Ph vs CH₃-CH₂-N=N-Ph (extended conjugation). Azo > Hydrazone in stability.

Ex.31 Compare enol percent.

(i)

(ii)

(iii)

Sol. C-D bonds are stronger than C-H bonds. Deuterium (D) substitution lowers enol formation because breaking C-D is harder than C-H. Therefore enol %: (i) > (ii) > (iii).

Deuterium Exchange Reaction

 →

 → (final product)

* To get the product directory replace all a-hydrogen w.r.t. carbonyl group by D(Deuterium):

e.g.

Ring-chain Isomerism

If one isomer has an open-chain structure and another has a cyclic structure, they are ring-chain isomers. Ring-chain isomerism is a special case of functional/structural isomerism where a change between an open chain and a ring changes functionality.Ring Chain Isomerism

Examples:

(i) Alkene and cycloalkane, (C_nH_2n)

CH₃ - CH = CH₂

(ii) Alkyne and cycloalkene, (C_nH_{2n - 2})

C₄H₆ : CH₃ - CH₂ - C º CH

(iii) Alkenols and cyclic ethers, (C_nH_2nO)

C₃H₆O : CH₂ = CH - CH₂OH

Note: Ring-chain isomers are also functional isomers because the functional group changes upon ring formation (for example an alcohol ↔ ether relationship when a ring closes).