Structural Isomerism | Chemistry Class 11 - NEET PDF Download

What is Isomerism?

Compounds having the same molecular formula but different properties are known as isomers and this phenomenon is known as isomerism. 

The term "isomer" comes from the Greek words "isos" and "meros," which together mean "equal parts." A Swedish chemist named Jacob Berzelius created this term in 1830.

Classification of Isomerism

Two common types of isomerism are structural isomerism and stereoisomerism. 

1. Structural Isomerism: In structural isomerism, molecules with identical molecular formulas have different arrangements of atoms. This can manifest as variations in the branching of carbon chains, positions of functional groups, or the presence of different functional groups.
2. Stereoisomerism: Stereoisomerism, on the other hand, involves molecules with the same connectivity of atoms but differ in their spatial arrangement. This can occur as geometric isomerism, where groups around a double bond have different spatial orientations, or optical isomerism, where molecules are mirror images of each other but cannot be superimposed. These isomers can exhibit unique properties due to their distinct spatial structures.

Classification of Isomerism

What is Structural isomerism? 

• Compounds having the same molecular formula and different connectivity of atoms (Structure is different) are called Structural Isomers.
• A straightforward example is the alkane with a molecular formula of C4H10, showcasing structural isomers with varied arrangements. As the number of carbon atoms in the alkane molecule increases, the number of structural isomers also increases.

Structural Isomers of Butane

Classification of Structural Isomers

There are six types of Structural Isomers: 

• Chain Isomers: These isomers have the same molecular formula but differ in the arrangement of the carbon skeleton, creating distinct chains or branches.
• Position Isomers: In this type, atoms are attached at different positions on the carbon chain, leading to unique compounds with distinct physical and chemical properties.
• Functional Group Isomers: Molecules with the same molecular formula but different functional groups fall under this category. For example, an alcohol and an ether can be functional group isomers.
• Tautomers: Tautomeric isomers exist in equilibrium with each other due to the migration of a hydrogen atom or a proton. They rapidly interconvert, creating dynamic structures.
• Metamers: Metamers have the same alkyl groups on either side of a functional group, such as an ether or a secondary amine, but with a different arrangement of alkyl groups.
• Ring Isomers: These isomers involve variations in the arrangement of atoms to form different cyclic structures while maintaining the same molecular formula. The shift from one ring structure to another characterizes this type of isomerism.

Classification of Structural Isomers

Chain Isomerism

Compounds having the same molecular formula but differ in the length of the principal chain.

How to Form Chain Isomers?

C_nH_2n+2 C_nH_2n+1

Alkane                   Alkyl group

dark line (-) represents vacant valency where any group can be attached.

- Iso-Group
e.g.Isoheptane

e.g. Isooctane (exception for the Iso group)

- Neo Group 

To prepare the neo compound firstly the above group  is written. After that required no. of carbon is added in the straight chain.

e.g. neopentane

e.g. neo heptane

Examples of Chain Isomerism:

1. C₄H₁₀ has 4 forms

2. C₅H₁₂ has it's three forms:

C - C - C - C - C  
(n-pentane)

Consider n-pentane:

= 3 form 

Consider Isopentane,

Consider neopentane,

= only one form.

Total forms of C₅H₁₁ - = 8

3. CH₃ - CH₂ - CH₂ - CH₃   - butane (n-butane)

and

 - 2-methylpropane (Isobutane)

4.  - butanoic acid and

 - 2-methyl propanoic acid

Ex.1 Find all the structural isomers of C₆H₁₄ 

Sol. 

Structural Isomers of C₆H₁₄

Position Isomerism

Compounds having the same molecular formula and same principal chain but differ in the position of functional group, multiple bonds, and substitution group are known as position isomers.

Position Isomerism

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

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

e.g.  (1-propanol)  and  (2-propanol)

Note:  and  are chain isomers, not position isomers.

The above example can be best understood taking the following example

C - C - C - C and

In this, the last carbon has been placed to II^nd position to form chain isomer the same has happened with above example and hence they are chain isomer to each other.

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 →Replace one H by Cl

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-dimethyl pentane)

Ex.5 Mono chlorinated products (Excluding Stereoisomers)? 

Sol.

Ex.6 Find the total structural isomer of C₅H₁₀. 

Sol. To solve these kinds of problems we should at once draw all possible structures of corresponding alkanes and then we should check how many possibilities are there to put a double bond.

 = 2

 = 3

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

Total open chain structural isomers = 5

* To form a cyclic structure we should always start with 3 carbon ring.

total structural isomers = 5 + 5 = 10

Ex.7 Find the total structural isomers of C₄H₆. 

Sol. Total unsaturation of C₄H₆ = 2

i.e. possibility = one triple bond, or 2 double bonds, or (one ring + one double bond)

 ( indicates the possible position of triple bond)

 → (No triple bond)

for alkene,

total open chain = 2 + 2 = 4

for cyclic

total structural isomers (cyclic, acyclic) = 9

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

Sol.

Total structural isomer = 3

Functional Group Isomerism

Compounds having same molecular formula but different in functional group are known as functional isomers.

Important:

• Aldehyde, Ketone, cyclic ethers, cyclic alcohol, unsaturated alcohol etc. are functional isomers to each other.
• Alcohol and ether are functional isomers to each other. e.g. CH₃CH₂OH, CH₃OCH₃
• Acids and esters are functional isomers to each other. eg.
• Cyanide and Isocyanide are functional isomers to each other but HCN and HNC are tautomers to each other.
• * 1°, 2° and 3° amine are functional isomers to each other.

e.g. C₃H₆O          CH₃ - CH₂ CHO        CH₂ = CH - CH₂OH      

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 amine 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 acid 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 di aldehyde 

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.

y = 3 There is one possibility of placing Cl, therefore z = 1

x = 2, y = 3, z = 1

Ex.20 Find the total carbonyl compound (aldehydes and ketones) formed by C₅H₁₀O and also find the relation between carbonyl compounds that have the same no. of a-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)

 →one possibility

 →one possibility

 no possibility of placing a double bond the valency of C will be more than four.

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

• Compounds having the same molecular formula but differ from the nature of the alkyl group directly attached to a polyvalent atom or polyvalent functional group.
• This type of isomerism is found in those types of compounds that have polyvalent atoms or polyvalent functional groups, e.g. ether, 2º amine, ester, etc.

Examples of metamers:

1.

 2.

(a)

(b) CH₃ - CH₂ - CH₂ - NH - CH₂ - CH₃

a & b are metamers.

Tautomerism

Compounds having the same molecular formula but different due to the oscillation of an atom (usually H ) are known as tautomers.

as after removal of H⁺, the anion formed is resonance stabilized.

Keto-enol Tautomerism: 

Keto-Enol Tautomerism

Mechanism :

*OH^- acts as a catalyst.

Base Catalysed Tautomerism :

Mechanism:

* enol is more acidic than keto.

* After removal of H⁺ from both form.

 → .......(i)

 → .......(ii)

In I^st -ve charge is on C and in II^nd -ve charge is on O therefore (ii) is more stable than (i) hence enol form is more acidic than the keto form.

Acid Catalysed Tautomerism:

Ex.24

• In the case of base-catalyzed tautomerism, the stability of carbanion is the deciding factor.
• For acid-catalyzed tautomerism, the stability of the product will be the deciding factor.

Ex.25

 P₁ + P₂

Sol.

+  

                             (major)               (minor)

(3 α H) +      (7 α H)

(minor)                            (major)

Ex.26 Write the enol form:

Sol.

Generally keto form is more stable than enol but in some cases, the stability of enol form is greater than keto. This is due to

(i) Intramolecular H-bonding

(ii) Aromatic character

(iii) Extended conjugation

(iv) Steric factor

e.g. 

* Due to intramolecular H-bonding formation of 6-member ring takes place which is the cause of stability.

* This can be also summarized as:

If G = H % of enol 80 - 90%

G = Ph % of enol 90 - 99%

G = CH₃ % of enol 70 - 80%

G = OC₂H₅ % of enol 5 - 10%

(one side)

* 

* Cross conjugation restricts the Resonance.

Ex.27 Compare the enol percent.

(A) CH₃CHO

(B)

(C)

(D)

(E)

(F)

(G)  

Sol. g > f > e > d > c > b > a

* In the case of a and b

After forming the enol form

CH₂ = CH - OH (no ∝ H)

 more stable = higher %

*

*

*  enol % > keto %

*  (Keto < enol due to Intramolecular H-bond)

*

keto > enol

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

Sol.

* The above case is called para tautomerism. Here g-Hydrogen participate in tautomerism.

Ex.29 Compare the enol content

(A)

(B)

(C)

(D)

Sol. After removing H  (acidic-H) from the compounds

(A)

(B)

(C)

(D)

* Enol percent ∝ stability of carbanion

a > b > d > c

* Formation of carbanion is one of the steps from keto to enol. Therefore can be calculated as a percent of the stability of the carbanion.

Ex.30 Find the enol form of

Nitro and Acinitro form 

 →

(acinitro)

 (acinitro)

* (CH₃)₃ C - NO₂ will not show nitro and acinitro form it has no H w.r.t to the NO₂ group.

Imine and Enamine 

For this type of tautomerism, there must a H w.r.t. (-CH = NH) group.

Amide and Amidol 

Nitroso and Oxime form 

II > I (stability)

due to extended conjugation in (II)

Hydrazone and Azoform 

NH₂ NH₂ (Hydrazine)

CH₃ - CH = N - NH - Ph  CH₃ - CH₂N = N - Ph

(extended conjugation)

Azo > Hydrazone (stability)

Ex.31 Compare enol percent.

(i) 

(ii) 

(iii) 

Sol. As we know C - D > C - H (Bond strength)

⇒ C - D will not break easily.

⇒ Compound will have less tendency to come into in enol form as C - D bond breaking is one of the step for conversion of keto into end.

⇒ enol percent will be less.

⇒ (i) > (ii) > (iii) (enol content)

Deuterium Exchange Reaction 

(Deuterium Exchange Tautomerism)

 →

 → (final product)

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

e.g.

Ring-chain Isomerism

Ring Chain IsomerismIf one isomer has an open chain structure and the other has cyclic structure then isomers are known as ring-chain isomers and isomerism between them is known as ring-chain isomerism.

For 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 always functional isomers.