Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Important Questions for CBSE Class 11 Chemistry Chapter 12 - Organic Chemistry - Some Basic Principles and Techniques

ffImage
Last updated date: 16th May 2024
Total views: 668.4k
Views today: 18.68k

CBSE Class 11 Chemistry Chapter-12 Important Questions - Free PDF Download

Important questions for Class 11 Chemistry Chapter 12, Organic Chemistry Some Basic Principles and Techniques, along with the answers have been prepared by the subject experts at Vedantu. These important questions cover all the topics given in this chapter. Students can download and refer to the PDF of these Organic Chemistry Class 11 Important Questions from Vedantu for free of cost. Class 11 organic chemistry important questions will help them to understand the type of questions asked in the competitive exams from this chapter.


The major reason for developing these significant questions was to assist students in organising their study materials ahead to tests. The most recent CBSE rules were used to develop these crucial questions. Students can also consult additional Class 11 Chemistry Chapter 12 study materials found on Vedantu for better exam preparation.


Download CBSE Class 11 Chemistry Important Questions 2024-25 PDF

Also, check CBSE Class 11 Chemistry Important Questions for other chapters:

CBSE Class 11 Chemistry Important Questions

Sl.No

Chapter No

Chapter Name

1

Chapter 1

Some Basic Concepts of Chemistry

2

Chapter 2

Structure of Atom

3

Chapter 3

Classification of Elements and Periodicity in Properties

4

Chapter 4

Chemical Bonding and Molecular Structure

5

Chapter 5

States of Matter

6

Chapter 6

Thermodynamics

7

Chapter 7

Equilibrium

8

Chapter 8

Redox Reactions

9

Chapter 9

Hydrogen

10

Chapter 10

The s-Block Elements

11

Chapter 11

The p-Block Elements

12

Chapter 12

Organic Chemistry - Some Basic Principles and Techniques

13

Chapter 13

Hydrocarbons

14

Chapter 14

Environmental Chemistry

Popular Vedantu Learning Centres Near You
centre-image
Mithanpura, Muzaffarpur
location-imgVedantu Learning Centre, 2nd Floor, Ugra Tara Complex, Club Rd, opposite Grand Mall, Mahammadpur Kazi, Mithanpura, Muzaffarpur, Bihar 842002
Visit Centre
centre-image
Anna Nagar, Chennai
location-imgVedantu Learning Centre, Plot No. Y - 217, Plot No 4617, 2nd Ave, Y Block, Anna Nagar, Chennai, Tamil Nadu 600040
Visit Centre
centre-image
Velachery, Chennai
location-imgVedantu Learning Centre, 3rd Floor, ASV Crown Plaza, No.391, Velachery - Tambaram Main Rd, Velachery, Chennai, Tamil Nadu 600042
Visit Centre
centre-image
Tambaram, Chennai
location-imgShree Gugans School CBSE, 54/5, School road, Selaiyur, Tambaram, Chennai, Tamil Nadu 600073
Visit Centre
centre-image
Avadi, Chennai
location-imgVedantu Learning Centre, Ayyappa Enterprises - No: 308 / A CTH Road Avadi, Chennai - 600054
Visit Centre
centre-image
Deeksha Vidyanagar, Bangalore
location-imgSri Venkateshwara Pre-University College, NH 7, Vidyanagar, Bengaluru International Airport Road, Bengaluru, Karnataka 562157
Visit Centre
View More
Competitive Exams after 12th Science

Study Important Questions for Class 11 Chemistry Chapter 12 – Organic Chemistry Some Basic Principles and Techniques

Very Short Answer Questions                                                                       1 Mark

1. How many $\sigma$ and $\pi $ bonds are present in each of the following molecules?

  1.  $HC\equiv C-C\equiv CC{{H}_{3}}$

Ans: $\sigma C-C:4,\sigma C-H:4,\pi C=C:4$

  1. $C{{H}_{2}}=C=CHC{{H}_{3}}$

Ans: $\sigma C-C:3,\sigma C-H:6,\pi C=C:2$


2. Why are electrons easily available to the attacking reagents in $\pi -$bonds? 

Ans: The electron charge cloud of the $\pi -$bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents.


3. Write the bond line formula for


Structure of Bond Line


Structure of CNCHOHCN


Ans:

Bond line structure


Bond line structure of CNCHOHCN


4. How are organic compounds classified?

Ans: (i) Acyclic or open chain compounds

(ii) Alicyclic or closed chain or ring compounds.

(iii) Aromatic compounds.


5.Define homologous series?

Ans: A group or a series of organic compounds each containing a characteristic functional group forms a homologous series and the members of the series are called homologous.


6. Write an example of a non-benzenoid compound.

Ans:

Structure of Tropolene


Structure of Tropolene


7. What is the cause of geometrical isomerism in alkenes?

Ans: Alkene has a $\pi -$bond and the restricted rotation around the $\pi -$ bond gives rise to geometrical isomerism.


8. Name the chain isomers of ${{C}_{5}}{{H}_{12}}$ which has a tertiary carbon atom.

Ans: 2-Methyl butane ${{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}-C{{H}_{3}}$ 


9. Define heterolytic cleavage.

Ans: In heterolytic cleavage the bond breaks in such a fashion that the shared pair of electrons remains with one of the fragments.


10. Define carbocation.

Ans: A species having a carbon atom possessing a sextet of electrons and a positive charge is called carbocation.


11. What are the nucleophiles?

Ans: The electron rich species are called nucleophiles. A nucleophile has affection for a positively charged centre.

Example- $O{{H}^{-}},{{I}^{-}},C{{N}^{-}}$


12. How can the mixture of kerosene oil and water be separated?

Ans: The mixture of kerosene oil and water can be separated by using a separating funnel.


13. Lassaigne’s test is not shown by diazonium salts. Why?

Ans: Diazonium salts usually leave ${{N}_{2}}$ on heating much before they have a chance to react with the fused sodium metal. Therefore, diazonium salts do not show positive Lassaigne’s test for nitrogen.


14. In which $C-C$ bond of $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Br$ , the inductive effect is expected to be the least?

Ans: Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence the effect is least in the third carbon $C-H$  bond.


15. Can you use potassium in place of sodium for fusing an organic compound in Lassaigne’s test?

Ans: No, because potassium is more reactive than sodium.


16. Give the reason for the fusion of an organic compound with sodium metal for testing nitrogen, sulphur and halogens.

Ans: The elements present in the compound are converted from covalent form into ionic form by fusing the compound with sodium metal.


17. Write the chemical composition of the compound formed when ferric chloride is added containing both N and S.

Ans:

$FeC{{l}_{3}}+NaSCN\to\underset{blood\text{ }red}{\mathop{{{[Fe(SCN)]}^{2+}}}}\,+3NaCl$


Short Answer Questions                                                                              2 Marks

1. Write the expanded form of the following condensed formulas into their complete structural formulas.

  1. $C{{H}_{3}}C{{H}_{2}}COC{{H}_{2}}C{{H}_{3}}$

Ans:

Expanded form


Expanded form of $C{{H}_{3}}C{{H}_{2}}COC{{H}_{2}}C{{H}_{3}}$


  1. $C{{H}_{3}}CH=CH{{(C{{H}_{2}})}_{3}}C{{H}_{3}}$

Ans: 

Expanded form


Expanded form of $C{{H}_{3}}CH=CH{{(C{{H}_{2}})}_{3}}C{{H}_{3}}$


2. How does hybridization affect electronegativity?

Ans: The greater the s-character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having a sp-hybrid orbital with $50$%  s-character is more electronegative than that possessing $s{{p}^{2}}$ or $s{{p}^{3}}$ hybridized orbitals.


3. Why is sp hybrid orbital more electronegative than $s{{p}^{2}}$ or $s{{p}^{3}}$ hybridized orbitals?

Ans: The greater the s-character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having a sp-hybrid orbital with $50$%  s-character is more electronegative than that possessing $s{{p}^{2}}$ or $s{{p}^{3}}$ hybridized orbitals.

Example: Hydroxyl group (-OH), Aldehyde group (-CHO), Carboxylic acid group (-COOH), etc.


4. Give two examples of aliphatic compounds.

Ans:

Aliphatic Compounds


Showing Examples of two Aliphatic Compounds


5. Write an example of an alicyclic compound.

Ans:

Examples of Alicyclic Compounds


Showing Examples of Alicyclic Compounds.


6. For each of the following compounds write a condensed formula and also their bond line formula.

  1. $HOC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}CH(C{{H}_{3}})CH(C{{H}_{3}})C{{H}_{3}}$

Ans: condensed formula - $HO{{(C{{H}_{2}})}_{5}}CHC{{H}_{3}}CH{{(C{{H}_{3}})}_{2}}$ 


Showing Bond line formula


Showing Bond line formula of $HOC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}CH(C{{H}_{3}})CH(C{{H}_{3}})C{{H}_{3}}$


Bond line formula - 

  1.                                   

Showing structure of Bond


Showing structure of CNCHOHCN


Ans: condensed formula - $HOCH{{(CN)}_{2}}$

Bond line formula - 


Bond line structure


Bond line structure of CNCHOHCN


7. Write the structural formula of 

  1. p-nitroaniline

  Ans:

Structural formula of p-nitroaniline


Structural formula of p-nitroaniline


  1. 2,3- Dibromo-1-phenylpentane

  Ans:

Structural formula of 2,3- Dibromo-1-phenylpentane


Structural formula of 2,3- Dibromo-1-phenylpentane


8. Derive the structure of 3-nitrocyclohexene

Ans:

Six membered ring containing a carbon – carbon double bond is implied by cyclohexene, which is numbered. The prefix 3-nitro means that a nitro group is parent on C-3. Thus, the complete structural formula of the compound is derived. Double bond is suffixed functional group whereas $-N{{O}_{2}}$ is prefixed functional group; therefore, double bond gets preference over $-N{{O}_{2}}$ group


Structure of cyclohexene and 3-nitrocyclohexene


Structure of cyclohexene and 3-nitrocyclohexene


9. Give the IUPAC name of the following:


2,5-dimethyl heptane

2,5-dimethyl heptane



  Ans: 2,5-dimethyl heptane

  1. $C{{l}_{2}}CHC{{H}_{2}}OH$ 

  Ans: 2,2-dichloro ethanol


10. Draw the two geometrical isomers of but-2-en-1,4 dioic acid. Which of the will have a higher dipole moment?

Ans:

Showing two geometrical isomers of but-2-en-1,4 dioic acid


Showing two geometrical isomers of but-2-en-1,4 dioic acid


11. How many structural isomers and geometrical isomers are possible for a cyclohexane derivative having the molecular formula ${{C}_{9}}{{H}_{16}}$ ?

Ans: five structural isomers


Showing structural isomers and geometrical isomers


Showing structural isomers and geometrical isomers are possible for a cyclohexane derivative having the molecular formula ${{C}_{9}}{{H}_{16}}$


Structure (a) has two geometrical isomers cis and trans


12. Alkynes does not exhibit geometrical isomers. Give a reason.

Ans: Because of linear geometry alkynes do not exhibit geometrical isomers. 


13. Which of the following shows geometrical isomerism?

  1. $CHCl=CHCl$

 Ans:    

Showing geometrical isomers


Showing geometrical isomers of $CHCl=CHCl$


  1. $C{{H}_{2}}=CC{{l}_{2}}$

Ans: this does not show geometric isomerism due to the same molecules on one side.

  1. $CC{{l}_{2}}=CHCl$

Ans: this does not show geometric isomerism


14. What is a functional group?

Ans: It may be defined as an atom or group of atoms joined in a specific manner which is responsible for the characteristic chemical properties of the organic compounds.


15. How many isomers are possible for monosubstituted and disubstituted benzene?

Ans: There is one, monosubstituted benzene as


Showing monosubstituted benzene


Showing monosubstituted benzene


There are three disubstituted benzene as


Showing disubstituted benzenes


Showing disubstituted benzenes


16. Identify electrophilic centre in the following: \[C{{H}_{3}}CH=O,C{{H}_{3}}C\equiv N,C{{H}_{3}}I\] 

Ans: The shared carbon atoms are electrophilic centers as they will have partial positive charge due to polarity of the bond. 

\[C{{H}_{3}}HC=O,{{H}_{3}}CC=N,{{H}_{3}}C-I\] 


17. For the following bond cleavages, use curved arrows to the electron flow and classify each as photolysis or heterolysis. Identify the reaction intermediates products as free radical carbocation or carbanion.

  1. $C{{H}_{3}}O-OC{{H}_{3}}\to C{{H}_{3}}{{O}^{-}}+{{O}^{-}}C{{H}_{3}}$

   Ans:

$C{{H}_{3}}O-OC{{H}_{3}}\to C{{H}_{3}}\overset{.}{\mathop{O}}\,+\overset{.}{\mathop{O}}\,C{{H}_{3}}$ 

  1.   

Attack of electrophile over benzene

Attack of electrophile over benzene


  Ans:

Attack of electrophile over benzene


Attack of electrophile over benzene following heteolysis


18. Write resonance structures of $C{{H}_{2}}=CH-CHO$ .Indicate relative stability of the contributing structure.

Ans:

Resonance structures


Resonance structures of $C{{H}_{2}}=CH-CHO$


Relative Stability $\text{I}>\text{II}>\text{III}$


19. Write the resonance structure of 

  1. $C{{H}_{3}}N{{O}_{2}}$ 

   Ans:

   

Resonance structures


Resonance structures of $C{{H}_{3}}N{{O}_{2}}$


  1. $C{{H}_{3}}CO{{O}^{-}}$ 

  Ans:

Resonance structures


Resonance structures of $C{{H}_{3}}CO{{O}^{-}}$


20. Explain why is ${{(C{{H}_{3}})}_{3}}{{C}^{+}}$ more stable than $C{{H}_{3}}C{{H}_{2}}^{+}$ and $C{{H}_{3}}^{+}$ is the least stable cation.

Ans: Hyper conjugation interaction in ${{(C{{H}_{3}})}_{3}}{{C}^{+}}$ is greater than in $C{{H}_{3}}C{{H}_{2}}^{+}$ as ${{(C{{H}_{3}})}_{3}}{{C}^{+}}$has nine C-H bonds. In $C{{H}_{3}}^{+}$ , the C-H bonds the nodal plane of the vacant 2p orbital and hence cannot overlap with it. Thus $C{{H}_{3}}^{+}$ locus conjugated stability.


21. Show how hyperconjugation occurs in propene molecules.

Ans:

Hyperconjugation in propene molecule


Hyperconjugation in propene molecule


22. Draw the orbital diagram showing hyperconjugation in ethyl cations.

Ans:

Orbital diagram


Orbital diagram showing hyperconjugation in ethyl cations


23. Name the common techniques used for purification of organic compounds.

Ans: (i) Sublimation (ii) Crystallization (iii) Distillation (iv) Differential extraction and (v) Chromatography.


24. Will $CC{{l}_{4}}$ give white precipitate of $AgCl$ on heating it with $AgN{{O}_{3}}$ ?

Ans: $CC{{l}_{4}}$ does not give white precipitate with silver nitrate solution.

\[CC{{l}_{4}}+AgN{{O}_{3}}\to no\text{ }reaction\] 

Carbon tetrachloride contains chlorine but it is bonded to carbon by a covalent bond.

Therefore, it is not in ionic form. Hence, it does not combine with silver nitrate  solution.


25. without using column chromatography, how will you separate a mixture of camphor and benzoic acid?

Ans: Sublimation cannot be used since both camphor and benzoic acid sublimate on heating. Therefore, a chemical method using $NaHC{{O}_{3}}$  solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dilute HCl, to get benzoic acid.


26. A liquid (0.1g) has three components. Which technique will you employ to separate them?

Ans: Column Chromatography is used to separate the three components which are present in the 0.1 g of the given liquid. 


27. Name two methods which can be safely used to purify aniline.

Ans: Vacuum distillation and Steam distillation method.


28. What is the basic principle of chromatography?

Ans: The method of chromatography is based on the difference in the rates at which the components of a mixture are adsorbed on a suitable adsorbent.


29. How will you separate a mixture of two organic compounds which have different solubilities in the same solvent?

Ans: By Fractional crystallization technique a mixture of two organic compounds which have different solubilities in the same solvent.


Long Answer Questions                                                                                  3 Marks

1. What is the shape of the following molecules:

  1. ${{H}_{2}}C=O$ 

  Ans: $s{{p}^{2}}$ hybridized carbon with trigonal planar

  1. $C{{H}_{3}}F$ 

  Ans: $s{{p}^{3}}$ hybridized carbon with tetrahedral

  1. $HC\equiv N$ 

Ans: sp hybridized carbon with linear structure


2. Giving justification, categories the following molecules or ions as nucleophile or electrophile : $H{{S}^{-}},B{{F}_{3}},{{C}_{2}}{{H}_{5}}{{O}^{-}},{{(C{{H}_{3}})}_{3}}N,C{{l}^{-}},C{{H}_{3}}{{C}^{+}}=O,{{H}_{2}}{{N}^{-}}$

Ans:

Nucleophiles: $H{{S}^{-}},{{C}_{2}}{{H}_{5}}{{O}^{-}},{{(C{{H}_{3}})}_{3}}N,{{H}_{2}}{{N}^{-}}$ have unshared pair of electrons which can be donated and shared with an electrophile

Electrophile: $B{{F}_{3}},C{{l}^{+}},C{{H}_{3}}{{C}^{+}}$ have only six electrons which can be accept electron from a nucleophile.


3. Using curved – arrow notation, show the formation of reactive intermediates when the following covalent bond undergo heterolytic cleavage.

  1. $C{{H}_{3}}-SC{{H}_{3}}$ 

Ans:

covalent bond


formation of reactive intermediates when the covalent bond undergo heterolytic cleavage.


  1. $C{{H}_{3}}-CN$ 

 Ans: 

covalent bond


formation of reactive intermediates when the covalent bond undergo heterolytic cleavage.


  1. $C{{H}_{3}}-Cu$ 

Ans: 

covalent bond


Formation of reactive intermediates when the covalent bond undergo heterolytic cleavage.


4. Benzyl carbocation is more stable than ethyl carbonation. Justify.              

Ans: In ethyl carbocation, there is only hyper conjugation of the three α – hydrogen atoms and as a result, the following contributing structures are feasible.


Hyper conjugation involved in ethyl carbocations


Hyper conjugation involved in ethyl carbocations.


But benzyl carbocation is more stable due to the presence of resonance and the following resonating structures are possible.


5. Which of the following pairs of structures do not constitute resonance structures?



Structure of nitromethane

Structure of nitromethane


 and $C{{H}_{3}}-O-N=O$ 

Ans: $C{{H}_{3}}-O-N=O$ 



Structure of prop-1-en-2-ol

Structure of prop-1-en-2-ol


and ${{(C{{H}_{3}})}_{2}}CO$

Ans: ${{(C{{H}_{3}})}_{2}}CO$  

  1. $C{{H}_{3}}CH=CHC{{H}_{3}}$ and $C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}$ 

 Ans: $C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}$ 


6. Write resonance structures of 

  1. $C{{H}_{3}}CO{{O}^{-}}$ 

  Ans: 

Resonance structures


Resonance structures of $C{{H}_{3}}CO{{O}^{-}}$


  1. ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$ 

 Ans:

Resonance structures


Resonance structures of ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$


7. Draw the resonance structures for the following compounds.

  1. ${{C}_{6}}{{H}_{5}}OH$ 

  Ans:

Resonance structures


Resonance structures of ${{C}_{6}}{{H}_{5}}OH$


  1. ${{C}_{6}}{{H}_{5}}-{{C}^{+}}{{H}_{2}}$ 

Ans:

Resonance structures


Resonance structures of ${{C}_{6}}{{H}_{5}}-{{C}^{+}}{{H}_{2}}$


8. 0.395g of an organic compound by Carius method for the estimation of sulphur gave 0.582g of $BaS{{O}_{4}}$. calculate the percentage of sculpture in the compound.

Ans: Mass of $BaS{{O}_{4}}$ =0.582g

Molecular weight of $BaS{{O}_{4}}$ = 233g/mol

233g of $BaS{{O}_{4}}$ contain sulphur = 32 g

0.582g of $BaS{{O}_{4}}$ contains sulphur = $\frac{32}{233}\times 0.582$ 

Percentage of sulphur = $\frac{weight\text{ }of\text{ }sulphur}{weight\text{ }of\text{ }compound}\times 100$ 

    = $\frac{32\times 0.582}{233\times 0.395}\times 100=20.24%$  


9. 0.40g of an organic compound gave 0.3g of AgBr by Carius method. Find the percentage of bromine in the compound.

Ans: Mass of the compound=0.40g

Now 188g of AgBr will contain Br = 80g

Therefore, 0.3g of AgBr will contain Br = $\frac{80}{188}\times 0.3=0.127g$ 

The percentage of Br in organic compound= $\frac{0.127}{0.40}\times 100=31.75%$ 


10. 0.12g of organic compound containing phosphorus gave 0.22g of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ by the usual analysis. Calculate the percentage of phosphorus in the compound.

 Ans: Here the mass of the compound taken= 0.12g

Mass of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ formed = 0.22g of atoms of P

Now 1 mole of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ formed = $(2\times 24+2\times 31+1687)$ = 222g of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ =62%

i.e, 222g of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ contain P = 62g

Therefore, 222g of $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ will contain Phosphorus = $\frac{62}{222}\times 0.22$ 

but this is the amount of phosphorus present in 0.12g of organic compound.

Hence, percentage of phosphorous = $\frac{62}{222}\times \frac{0.22}{0.12}\times 100=51.20$ 


11. Ammonia produced when 0.75g of a substance was kjeldahlized, neutralized $30c{{m}^{3}}$ of 0.25N ${{H}_{2}}S{{O}_{4}}$ . Calculate the percentage of nitrogen in the compound.                                                                                                    

Ans: Mass of the organic compound= $0.75g$

Volume of ${{H}_{2}}S{{O}_{4}}$ used  = $30c{{m}^{3}}$ 

Normality of ${{H}_{2}}S{{O}_{4}}$ = 0.25N

$30c{{m}^{3}}$ of ${{H}_{2}}S{{O}_{4}}$ of normality 0.25N $\equiv $ 30ml of $N{{H}_{3}}$ solution of normality 0.25N

But $1000c{{m}^{3}}$ of $N{{H}_{3}}$ of normality 1 contains 14g of nitrogen.

$\therefore 30c{{m}^{3}}$ of 0.25N $N{{H}_{3}}$contains nitrogen= $\frac{14}{1000}\times 30\times 0.25$ 

Percentage of nitrogen = $\frac{mass\text{ }of\text{ }nitrogen}{mass\text{ }of\text{ }subs\tan ce}\times 1000=\frac{14}{1000}\times \frac{30\times 0.25}{0.75}\times 100=14.00$ 


Download Organic Chemistry Class 11 Important Questions

Class 11 Chemistry Chapter 12 Important Questions

The topics covered under Class 11 Organic Chemistry important questions are as follows.

  • Classification of Organic Compounds.

  • Tetravalence of Carbon.

  • Shapes of Organic Compounds.

  • Nomenclature of Organic Compounds.

  • Isomerism

  • Fundamental Concepts in Organic Reaction Mechanism.

  • Methods of Purification of Organic Compounds.

  • Qualitative Analysis of Organic Compounds like Carbon, Hydrogen, Nitrogen, Halogens, Sulphur, Phosphorus, Oxygen.

Students have already learned some of these above-mentioned topics in Class X. After going through all these concepts, students will be able to understand the reasons for the tetravalence of carbon and the different shapes of organic molecules, naming of compounds according to the IUPAC system of nomenclature, and also derive their molecular formulas from the given IUPAC names. While going through these important questions they will be able to understand the techniques for the purification of organic compounds, and the three-dimensional representation of organic molecules. There are some important extra questions of Organic Chemistry Class 11, available on Vedantu, which explain the concepts of this chapter in an applicational approach.


Conclusion

Organic Chemistry is one of the most crucial areas in tests like the IIT or NEET because it is weighted between 20 and 30 percent. The majority of students have trouble comprehending the ideas in organic chemistry. Vedantu offers both online and offline access to the Class 11 Organic Chemistry Critical Questions course. One of the top websites for online learning, Vedantu, offers free PDFs to students so they may better understand the principles they are learning. Our team of professionals has created all of the study guides in accordance with the current CBSE board curriculum. In order to perform well in exams, students are recommended to practise more and more questions and answers.


Important Related Links for CBSE Class 11 Chemistry

FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 12 - Organic Chemistry - Some Basic Principles and Techniques

1. Is Chapter 12 of Class 11 Chemistry difficult?

Students will study different concepts of Chemistry in detail in Chapter 12 of Class 11 Chemistry. They study some basic concepts of Chemistry in Class 10. In Class 11, students study some similar topics in detail. Students should try to understand the basic concepts in Chemistry to get high marks in Chemistry. They can download Important Questions for Class 11 Chemistry from Vedantu free of cost. Important questions for all chapters of Class 11 are given on Vedantu that can help students to understand the concepts of Chemistry in Class 11. 

2. What are the main topics given in Chapter 12 of Class 11 Chemistry?

Chapter 12 of Class 11 Chemistry is mainly based on Organic Chemistry. Students learn basic concepts of Organic Chemistry in Class 10. In Class 11 Chemistry Chapter 12, students will study different concepts related to Organic Chemistry. They will study the tetravalence nature of carbon, the structure of organic compounds, different types of organic compounds and their names, what is isomerism, basic concepts in organic reaction, different methods used to purify organic compounds, etc.

3. Where can I find Important Questions of Chapter 12 of Class 11 Chemistry?

Students can download Important Questions of Chapter 12 of Class 11 Chemistry from Vedantu. All Important Questions of Chapter 12 of Class 11 Chemistry are available in simple language. Students can easily understand the basic concepts of Organic Chemistry in Class 11 from the Important Questions. These important questions are made by expert Chemistry teachers. 

4. Where can I read Chapter 12 of Class 11 Chemistry online?

Students of Class 11 can read complete Chapter 12 of Class 11 Chemistry on Vedantu. They can find Important Questions for Chapter 12 of Class 11 Chemistry along with solutions on Vedantu. Students can also find NCERT Solutions on Vedantu to prepare for their Chemistry exam. Students can prepare well for Chemistry from the information and notes available for Class 11 Chemistry on Vedantu. All concepts of the NCERT textbook for Class 11 Chemistry are explained thoroughly for easy understanding. 

5. Is Chapter 12 of Class 11 Chemistryimportant for entrance exams?

Chapter 12 of Class 11 Chemistry includes important topics based on Organic Chemistry. It is a part of Organic Chemistry therefore it is an important chapter for entrance exams. Students have to read Chapter 12 of Class 11 Chemistry carefully to understand the chemical reactions and other concepts related to Organic Chemistry. Students can get one mark or two mark questions from this Chapter in their exams. Therefore, they should read Important Questions for Chapter 12 of Class 11 Chemistry to prepare for their exams.