Structural Isomerism Class 11 Notes | EduRev

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Class 11 : Structural Isomerism Class 11 Notes | EduRev

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Compounds having the same molecular formula (M.F.) but differ in their properties are known as isomers and this phenomenon is known as isomerism. 


Structural isomerism:

Compounds having the same M.F. and different in connectivity of atom (Structure is different).

Chain Isomerism:

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

e.g.1  CH3 - CH2 - CH2 - CH3   - butane (n - butane)


 - 2-methylpropane (Isobutane)


e.g.2  - butanoic acid and

 - 2-methylpropanoic acid


*  (iso group)
e.g. Isoheptane

* Isooctane (exception of 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


Alkyl Group :

CnH2n+2 Cn

Alkane                   Alkyl group


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



(Isopropyl alcohol)

Thus we can conclude that

C3H7 - → Two forms.

* C4H10

* C5H12 has it's three forms:

C - C - C - C - C  

* Consider n-pentane:


= 3 form 

Consider Isopentane,

Consider neopentane,

= only one form.

Total forms of C5H11 - = 8

Ex.1 Find all the structural isomers of C6H14 


C - C - C - C - C - C  


Position Isomerism

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

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)


*  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 IInd 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










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

c, d →chain isomers. d, e →position isomers.

* Monochlorination →Replace one H by Cl

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


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


(2, 2-dimethyl pentane)

Ex.5 Mono chlorinated products (Excluding Stereoisomers)? 




Ex.6 Find the total structural isomer of C5H10. 

Sol. For solving 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 double bond.

 = 2

 = 3

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

Total open chain structural isomers = 5

* To form cyclic structural 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 bond or (one ring + one double bond)

 (¯ indicate 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.8 C3H8 ? (Only structural in all)


Sol. For monochloroderivative,







Total dichloroderivatives = 4











* To find di or trichloroderivative

We place two or three chlorine at last carbon and after that rotate one Cl by keeping the other two at the same the place.

Ex.9 Find all the structural dichloroderivative of cyclopentane. 


Total structural isomer = 3

Functional Group Isomerism


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

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

* 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 ester are functional isomers to each other.


* Cyanide and Isocyanide are functional isomer to each other but HCN and HNC are tautomers to each other.

* 1°, 2° and 3° amine are functional isomers to each other.

Ex.10 How many primary, secondary and tertiary alcohol 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 - CH ⇒ 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 amide will be there which 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



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.



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


Total = 5

Ex.18 Find the possible dichloroderivatives of C6H4Cl(Only structural) 


Ex.19 C6H4Cl2 →C6H3Cl(Only structural)

 → → →

find the value of x, y, z.


(1, 2, 4)

There are two possibilities placing Cl in place of H.

y = 3 There are one possibilities 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 which have same no. of a-hydrogen.(Only structural) 

Sol. For ketones,


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) 


(1)  = 3


(2)  = 4


(3)  = 1


(4)  = 2


(5)  = 3

total = 13 isomers.


Ex.22 Find the total conjugated diene in C5H(Only structural)

 →one possibility

 →one possibility

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


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




 →no form cumulated diene

total = 3

⇒ Isolated dienes,

C - C - C - C - C →

 →no form isolated diene

total = 1


This type of isomerism is found in those types of compounds which have polyvalent atoms or polyvalent functional group, e.g. ether, 2º amine, ester etc.

Compound having same molecular formula but differ form the nature of alkyl group directly attached with polyvalent atom or polyvalent functional group.


* (a)


(b) CH3 - CH2 - CH2 - NH - CH2 - CH3

a & b are metamers


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

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

Keto-enol Tautomerism:

Mechanism :


*OH- acts as catalyst.

Base Catalysed Tautomerism :


Mechanism →

* enol is more acidic than keto.

* After removal of H+ from both form.

 → .......(i)

 → .......(ii)

In Ist -ve charge is on C and in IInd -ve charge is on O therefore (ii) is more stable than (i) hence enol form is more acidic than keto form.

Acid Catalysed Tautomerism:





* In case of base catalysed tautomerism the stability of carbanion is the deciding factor.
* For acid catalysed tautomerism the stability at the product will be the deciding factor.


 P1 + P2




                             (major)               (minor)

(3 α H) +      (7 α H)

(minor)                            (major)


Ex.26 Write the enol form:




* 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


* 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 = CH% of enol 70 - 80%

G = OC2H% of enol 5 - 10%

(one side)


* Cross conjugation restricts the Resonance.

Ex.27 Compare the enol percent.








Sol. g > f > e > d > c > b > a

* Incase of a and b

After forming 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 the given compound.



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

Ex.29 Compare the enol content






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





* Enol percent ∝ stability of carbanion

a > b > d > c

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

Ex.30 Find the enol form of

Nitro and Acinitro form 






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






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):



Ring-chain Isomerism

If one isomer has 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, (CnH2n)

CH3 - CH = CH2

(ii) Alkyne and cycloalkene, (CnH2n - 2)

C4H6 : CH3 - CH2 - C º CH

(iii) Alkenols and cyclic ethers, (CnH2nO)

C3H6O : CH2 = CH - CH2OH

Note : Ring-chain isomers are always functional isomers.

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