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

Students may find Chemistry an exciting subject if they receive proper study material, guidance and step-by-step solutions to the various problems covered in the NCERT textbook. The Extramarks team has created the most precise and reliable NCERT based solutions for important questions in Class 11 Chemistry Chapter 12. Since these questions and their answers are based on the CBSE syllabus, they can help you achieve good marks in Chemistry exams. So our Chemistry Class 11 Chapter 12 important questions and their solutions have proven to be very helpful from an exam perspective.

In this chapter, students will learn how organic compounds are crucial for maintaining life on earth and contain complex molecules like DNA, which carries genetic information, and proteins, which are constituents of our blood, muscles, and skin. Materials like textiles, fuels, polymers, dyes, and medications all contain organic molecules. These are a few crucial fields in which these chemicals are used. The special ability of the element carbon to make covalent bonds with other carbon atoms is known as catenation, as you learned in the last chapter. This chapter includes some fundamental analytical ideas and methods needed to comprehend the synthesis and characteristics of organic molecules.

.At Extramarks, we highlight key ideas and exam-oriented questions from every chapter to assist students with their studies before exams. Students will be familiar with the questions likely to come in final exams by completing  Class 11 Chemistry Chapter 12 important questions. Since Chemistry is a subject that demands in-depth conceptual understanding, only memorising the structures and formulas will not be enough.

Students must therefore understand the core concepts and theories to answer the questions. These questions are a combination of the NCERT textbook, NCERT exemplar, and other reference books. To create a solid basis for the chapter, students should try to solve all questions from our important questions Class  11 Chemistry Chapter 12. To study the fundamentals of Organic Chemistry and some frequent terms used in this area of Chemistry, students are recommended to regularly refer to these Class 11 Chemistry Chapter 12 important questions.

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Organic Chemistry Class 11 Important Questions and Answers

The Important Questions Class 11 Chemistry Chapter 12 have been compiled by the Extramarks team from various sources. The problems span many topics, including classifying organic compounds, homologous series, and reactivity series. Students can better grasp IUPAC naming using these real-world questions and their solutions. Students may enjoy the clarity of concepts which in turn is useful in answering the most difficult questions. 

The NCERT books are well designed such that it covers all the concepts along with their clarity up to the student irrespective of their intelligence level and it is also meant to clear their doubts and polish their basics as well.

A few Important Questions Class 11 Chemistry Chapter 12 are provided here, along with their answers:

Question 1. The reaction:

CH3CH2I + KOH(aq) → CH3CH2OH + KI

is classified as :

(a) electrophilic substitution 

(b) nucleophilic substitution

(c) elimination 

(d) addition

Answer 1. (b) Nucleophilic substitution

Explanation:

It is a nucleophilic substitution reaction. As a nucleophile, the hydroxyl group of KOH with a lone pair of itself replaces the iodide ion in CH3CH2I to produce ethanol.

Question 2. Which of the following is the correct IUPAC name?

(i) 3-Ethyl-4, 4-dimethylheptane

(ii) 4,4-Dimethyl-3-ethylheptane

(iii) 5-Ethyl-4, 4-dimethylheptane

(iv) 4,4-Bis(methyl)-3-ethylheptane

Answer 2. (i) 3-Ethyl-4, 4-dimethylheptane

Explanation:

The various alkyl groups present in a compound are listed in the IUPAC name in alphabetical order. Di, tri, and tetra are left out of the alphabetical list. Therefore, the ethyl group (e) comes before the methyl group (m). As a result, the ethyl group is assigned the number 3 and is written first.

Question 3. Benzene hexachloride is prepared from benzene and chlorine in sunlight by

(a) Substitution reaction

(b) Elimination reaction

(c) Addition reaction

(d) Rearrangement reaction

Answer 3. (c) Addition reaction

Explanation: In addition to the reaction, benzene hexachloride is prepared from benzene and chlorine in sunlight. It occurs in the absence of oxygen, and a catalyst may be involved to increase the reaction rate.

C6H6    +    3Cl2     →   C6H6Cl6

Question 4. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue

colour is obtained due to the formation of:

(a) Na4[Fe(CN)6] 

(b) Fe4[Fe(CN)6]3

(c) Fe2[Fe(CN)6] 

(d) Fe3[Fe(CN)6]4

Answer 4. (b) Fe4[Fe(CN)6]3

Explanation:

In Lassaigne’s test for nitrogen in an organic molecule, the sodium fusion extract is heated with iron (II) sulphate. It is then acidified with sulphuric acid. Iron (II) sulphate and sodium cyanide first combine to create sodium hexacyanoferrate (II). Sulphuric acid is subsequently used to oxidise the iron (II), resulting in the production of iron (III)hexacyanoferrate (II) or Prussian blue.

The following are the chemical reactions:

6CN–  +  Fe2+ ⟶ [Fe (CN)6]4-

3[Fe(CN)6]4-      +    4Fe3+    xH2O⟶      Fe4[Fe(CN)6]3.xH2O

                                                             (Prussian blue)

Question 5. The fragrance of flowers is due to the presence of some steam volatile organic

compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in the vapour phase. A suitable method for the extraction of these oils from the flowers is:

(i) Distillation

(ii) Crystallisation

(iii) Distillation under reduced pressure

(iv) Steam distillation

Answer 5. (iv) Steam distillation

Explanation:

Steam distillation is used to distinguish between compounds that are immiscible with water and are steam volatile. In steam distillation, heated flasks carrying the liquid to be distilled are passed through the steam produced by a generator. The steam and the volatile organic compound mixture are then condensed and collected. The separating funnel is afterwards used to separate the chemical from the water.

Question 6. Which of the following compounds will not exist as a resonance hybrid? Give reasons for your answer.

(i) CH3OH

(ii) R−CONH2

(iii) CH3CH=CHCH2NH2

Answer 6. (i) CH3OH   and  (iii) CH3CH=CHCH2NH2

Explanation:

(i) CH3OH does not contain pi-electrons and thus cannot form a resonance hybrid. 

(iii) In CH3CH=CHCH2NH2, the lone pair of electrons on the nitrogen atom is not coupled to the pi -electrons. Thus, the formation of a resonance hybrid is not possible.

Question 7. Which of the two:

 O2NCH2CH2O–    or CH3CH2O–     is expected to be more stable, and why?

Answer 7.  O2NCH2CH2O– is more stable.

Explanation:

Nitrogen being an electron-withdrawing group shows -I effect. It stabilises the molecule by lowering its negative electron charge. The methyl group, on the other hand,  is an electron donor group and exhibits a +I effect. This makes the molecule more negatively charged and makes it less stable.

Question 8. In which of the following compounds, the Carbon marked with asterisk is

expected to have the greatest positive charge?

(i) *CH3—CH2—Cl

(ii) *CH3—CH2—Mg+Cl–

(iii) *CH3—CH2—Br

(iv) *CH3—CH2—CH3

Answer 8. (i) *CH3—CH2—Cl

Explanation: According to the order of electronegativity: 

Cl > Br > C > Mg.

The more electronegative group attached to the Carbon will give a more positive charge. Therefore, the C-atom having -Cl group attached to it will have the greatest positive charge.

Question 9. Which of the following compounds contain all the carbon atoms in the same

hybridisation state?

(i) H—C ≡ C—C ≡ C—H

(ii) CH3—C ≡ C—CH3

(iii) CH2 = C = CH2

(iv) CH2 = CH—CH = CH2

Answer 9. (i) H—C ≡ C—C ≡ C—H   and (iv) CH2 = CH—CH = CH2

Explanation:

In these two compounds, all the carbon atoms have the same hybridisation state. In (i), all the carbon atoms are sp-hybridised, and in (iv), all the carbon atoms are sp2-hybridized. 

Question 10. In the organic compound CH2= CH – CH2– CH2– C ≡ CH, the pair of hybridised

orbitals involved in the formation of the C2– C3 bond is:

(a) sp – sp2

(b) sp – sp3

(c) sp2– sp3

(d) sp3– sp3

Answer 10. (b) sp– sp3

Explanation:

In the given compound, the carbon atoms can be numbered as:

6        5        4       3      2     1

CH2= CH – CH2– CH2– C ≡ CH

The hybridisation of carbon atoms 1, 2, 3, 4, 5, and 6 results in sp, sp, sp3, sp3, sp2, and sp2 atoms, respectively. Therefore, the pair of hybridised orbitals involved in forming the C2-C3 bond is sp-sp3.

Question 11. Explain why alkyl groups act as electron donors when attached to a π system.

Answer 11. When connected to a -system, an alkyl group acts as an electron-donor group due to hyperconjugation. Take propane as an illustration.

Due to hyperconjugation, the sigma electrons of the C-H bond become delocalised. An unsaturated system is immediately joined to the alkyl. Delocalisation occurs when an sp3-s sigma bond orbital partially overlaps with an empty p orbital of a nearby carbon atom’s n-bond.

This kind of overlap causes the electrons to be delocalised, also known as no-bond resonance, which increases the stability of the molecule.

Question 12. Explain, how is the electronegativity of carbon atoms related to their state of

hybridization in an organic compound?

Answer 12. As the s-character grows, electronegativity rises. This is due to the nucleus’s stronger attraction to s-electrons than p-electrons.

sp3 – 25% s-character, 75% p-character

sp2 – 33% s-character, 67% p-character 

sp – 50% s-character, 50% p-character

Thus, the order of electronegativity is sp3 < sp2 < sp

Question 13. What are electrophiles and nucleophiles? Explain with examples.

Answer 13. A reagent known as an electrophile, or an electron-loving pair, requires an electron pair—for instance, carbonyl groups, CH3CH2+, C=0 (due to the lone pair).

A reagent that contains an electron pair and is willing to contribute is known as a nucleophile. It is also referred to as a nucleus-loving reagent. These include NC–, OH–, and R3C– (carbanions).

Question 14. In DNA and RNA, nitrogen atoms are present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.

Answer 14. No, the heterocyclic rings in DNA and RNA include nitrogen that cannot be eliminated as ammonia.

As nitrogen contained in these structures cannot be transformed into ammonium sulfate, the Kjeldahl method cannot be used to estimate nitrogen present in rings, azo, or nitro groups.

Question 15. Explain the principle of paper chromatography.

Answer 15. Partition chromatography includes paper chromatography. Chromatography paper is a unique kind of paper that is used in paper chromatography. Water that has been trapped inside a chromatography paper serves as the stationary phase.

The solution of the mixture is spotted at the base of a strip of chromatography paper primarily suspended in a suitable solvent or mixture of solvents. The mobile phase is this solvent. Capillary action causes the solvent to ascend up the paper and run over the spot. According to how they were divided differently into the two phases, the paper preserves particular components only in some cases. The produced paper strip is referred to as a chromatogram.

On the chromatogram, the spots representing the separated coloured compounds are visible at various heights from the location of the initial spot. Using a suitable spray reagent, as described under thin layer chromatography, or by using UV light, one can see the spots of the separated colourless chemicals.

Question 16. Give a brief description of the principles of the following techniques taking an

example in each case.

(a) Crystallisation (b) Distillation (c) Chromatography

Answer 16.

Crystallisation: 

Crystallisation is a method used to purify solid organic molecules. The difference in the solubility of the substance and impurities in a particular solvent is the basis on which it operates. Since the impure product is only sparingly soluble at lower temperatures, it is made to dissolve in the solvent at a higher temperature. We keep doing this until the solution is practically saturated. We obtain its crystals by cooling and filtering the mixture. For instance, we can obtain pure aspirin by crystallising 2-4g of crude aspirin in 20mL of ethyl alcohol. If necessary, it is heated and then left alone until it crystallises. After being separated, the crystals are dried.

Distillation:

Non-volatile liquids are separated from volatile components using this technique. Additionally, it is employed when the boiling temperatures of the constituents differ significantly.

It operates under the premise that liquids with various boiling points vaporise at various temperatures. The generated liquids are then separated after they have been cooled.

Ex: In a flask with a circular bottom and a condenser, aniline (b.p. = 457 K) and chloroform (b.p = 334 K) are combined. Due to its high volatility, chloroform vaporises first when heated and is then forced to pass through a condenser where it cools. The flask with a circular bottom still holds the aniline.

Chromatography:

It is popularly used for the separation and purification of organic compounds.

It operates on the principle that each component of a mixture moves through the stationary phase under the influence of the mobile phase at a distinct rate.

Ex: A mixture of blue and red ink can be separated using chromatography. The component of the mixture less absorbed by the chromatogram is placed on the chromatogram. It causes the almost immobile component to travel more quickly up the paper than the other component.

Question 17. Give three points of differences between inductive effect and resonance effect.

Answer 17.

Question 18. Why does S03 act as an electrophile?

Answer 18. The sulphur atom is attached to three highly electronegative oxygen atoms, resulting in the Sulphur atom becoming electron-deficient. Additionally, Sulphur also takes up a positive charge due to resonance. Both these factors make S03 an electrophile.

Question 19. The empirical formula of a compound is CH2. Its one mole has a mass of 42 g. What is its molecular formula?

Answer 19.

Molecular formula = n × empirical formula,

 Therefore,  Molecular formula =4214 *  CH2 = C3H6 

Question 20. A mixture containing benzoic acid and nitrobenzene is given to you. Using an appropriate chemical reagent, how will you proceed to separate them?

Answer 20. When nitrobenzene reaches the organic layer, the mixture is agitated with a diluted NaHCO3 solution and extracted with ether or chloroform. The distillation of this will result in nitrobenzene. Dilute HCl is used to acidify the aqueous layer, and the solution is then cooled.

The final product is benzoic acid.

C6H5COOH + NaHCO3 → C6H5COONa + CO2 + H2O.

C6H5COONa + HCl (dilute) → C6H5COOH + NaCl.

Question 21. Discuss the chemistry of Lassaigne’s test.

Answer 21. Lassaigne’s test is used to identify the presence of phosphorus, halogens, nitrogen, and sulphur in an organic molecule. In an organic compound, these components are found in the covalent state. The chemical is fused with sodium metal to create the ionic form of these.

Boiling the fused mixture in distilled water releases the produced cyanide, sulfide, and sodium halide. The obtained extract is known as Lassaigne’s extract. Then, the extract from Lassaigne is examined for the presence of phosphorus, sulphur, nitrogen, and halogens.

(a)Test for nitrogen

A Lassaigne test is performed on sodium fusion extract by heating it with iron (II) sulphate and then acidifying it with sulfuric acid. The first reaction involves sodium cyanide interacting with iron sulfate (II) to form sodium hexacyanoferrate (II). Then, some iron (II) is heated with sulphuric acid to create iron (III) hexacyanoferrate (II), which is Prussian blue. Here are the chemical equations involved in the reaction:

(b) Test for sulphur

Acetic acid is used to make the sodium fusion extract acidic to perform Lassaigne’s test for sulphur in an organic compound. Next, lead acetate is added. Sulfur is present in the molecule because lead sulphide, which is black, precipitates out.

S2+     +Pb2+     →    PbS

                                 (Black)

Sodium nitroprusside is used to treat the sodium fusion extract. The compound’s appearance also indicates the presence of sulphur in the compound as violet.

S2+    +   [Fe(CN)5NO]2-   →  [Fe(CN)5NOS]4-

                                                     (Violet)

When both nitrogen and sulphur are present in an organic molecule, NaSCN is formed instead of NaCN.

Na + C + N + S    →   NaSCN

Fe3+    + SCN–    →    [Fe(SCN)]2+

                                     (Blood red)

The colour of this NaSCN (sodium thiocyanate) is blood red. The lack of free cyanide ions prevents the formation of Prussian colour.

NaSCN   +   2Na  →  NaCN  +  Na2S

(c)Test for halogens

To determine whether an organic molecule contains halogens, sodium fusion extracts are treated with silver nitrate after being acidified with nitric acid.

X–  +    Ag+   → AgX

Here, X represents a halogen Cl, Br or I.

As a result of heating Lassaigne’s extract, the nitrogen and sulphur are removed, which may interfere with the determination of halogens in organic compounds containing both elements.

Benefits of Solving Organic Chemistry Class 11 Important Questions

Organic Chemistry is essential because the majority of chemicals used in daily life are powered by organic chemistry. It also plays an essential role in producing common household chemicals, foods, polymers, and medications. Chemistry may be a tough subject and it becomes challenging for many students to comprehend a few advanced topics. To overcome this, students need to practice a lot of questions to self-assess their knowledge of the chapter. Extramarks organic chemistry class 11 extra questions along with our solutions is one of the best resources to practice exam-oriented questions. All key areas from the chapter are covered in our important questions and solutions. Our subject matter experts team have curated these questions with the CBSE exams in mind and many questions are most likely to be asked in exams. Students can gain a competitive advantage by regularly practising our Chapter 12 Class 11 Chemistry Important Questions.

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