NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

Question 1.
What are hybridisation states of each carbon atom in the following compounds?

Solution.

Question 2.
Indicate the σ and π bonds in the following molecules:

Solution.

Question 3.
Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Solution.

Question 4.
Give the IUPAC names following compounds:

Solution.

Question 5.
Which of the following represents the correct IUPAC name for the compounds concerned?

Solution.

Question 6.
Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H-COOH
(b) CH3COCH₃
(c) H-CH=CH₂
Solution.

Question 7.
Give condensed and bond line structure formulas and identify the functional groups present, if any, for:
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
Solution.

Question 8.
Identify the functional groups in the following compounds.

Solution.

Question 9.
Which of the two : O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?
Solution.

Question 10.
Explain why alkyl groups act as electron donors when attached to a π system.
Solution.

Question 11.
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C₆H₅OH
(b) C₆H₅NO₂
(c) CH₃CH=CHCHO
(d) C₆H₅-CHO
(e) C₆H₅-CH₂
(f) CH₃CH=CHH₂
Solution.

Question 12.
What are electrophiles and nucleophiles? Explain with examples.
Solution.

Question 13.
Identify the reagents shown in brackets in the following equations as nucleophiles or electrophiles:
(a) CH₃COOH + (HO^–) ➝ CH₃COO^– + H₂O
(b) CH₃COCH₃ + (CN^–) ➝ (CH₃)₂C(CN)(OH)
(c) C₆H₆ + (CH₃ O) ➝ C₆H5COCH₃
Solution.

Question 14.
Classify the following reactions in one of the reaction type studied in this unit.
(a) CH₃CH₂Br + HS^– ➝ CH₃CH₂SH + Br
(b) (CH₃)₂C = CH₂ + HCl ➝ (CH₃)₂CIC-CH₃
(c) CH₃CH₂Br + HO^– ➝ CH₂ = CH₂ + H₂O + Br^–
(d) (CH₃)₃C-CH₂OH + HBr ➝ (CH₃)₂CBrCH₂CH₃ + H₂O
Solution.

Question 15.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Solution.

Question 16.
For the following bond cleavages, use curved- arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

Solution.

Question 17.
Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

Solution.

Question 18.
Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography
Solution.
(a) Crystallisation : It is based on the difference in solubility of the compound and the impurities in a suitable solvent. While at room temperature, the compound is sparingly soluble and crystallizes out of solution but the impurities do not. As a result, they remain in solution and the compound is obtained as a crystal.
The impure compound is dissolved in a solvent and heated. At elevated temperature the compound dissolves as do the impurities. This solution is then gradually cooled. Being less soluble at room temperature it precipitates out in the form of crystals and pure compound is obtained.
(b) Distillation : This method is used to separate either :
(i) Volatile liquids from non-volatile impurities; and
(ii) Two liquids with different boiling points. The liquid mixture such as that of chloroform and aniline is taken in a round bottom flask fitted with a condenser.
Upon heating, the vapours of lower boiling liquid are formed first and collected through the condenser. The vapours of the higher boiling liquid are formed later. Thus, the two are separated.
(c) Chromatography: (i) It is applicable for the separation of virtually all inorganic and organic materials, except very insoluble polymers.
(ii) In this technique, the mixture of compounds which needs to be separated is applied onto a stationary phase, which may be a solid or a liquid. Another phase which may be a pure solvent, a mixture of solvents or a gas is allowed to more slowly over the stationary phase.
(iii) The components of the mixture which have different solubility in the moving phase, start moving. Since, they have different solubility, they move to different lengths on the stationary phase and become stable there,
(iv) Thus, the different components of the mixture are separated.

Question 19.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Solution.
Crystallization is the process that may be employed to seperate two compounds with different solubility in a given solvent at room temperature.
Upon heating such a solution to a sufficiently high temperature the solubility of the compound which is insoluble at room-temperature, increases and it dissolves. However, when this solution is cooled down to room temperature the lesser soluble or insoluble component precipitates out and is obtained as crystals while its soluble counterpart remains in solution. Thus, the separation is complete.

Question 20.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Solution.
Differences between distillation, distillation under reduced pressure and steam distillation may be summarized as :

Question 21.
Discuss the chemistry of Lassaigne’s test.
Solution.
The elements nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne’s test. The organic compound (N, S or halogens) is fused with sodium metal as to convert these elements into ionisable inorganic substances, i.e., nitrogen into sodium cyanide, sulphur into sodium sulphide and halogens into sodium halides.

Once the ions are formed, the inorganic tests can be applied to them and the compound can be analysed.
(i) Test for Nitrogen : The sodium fusion extract is boiled with iron(II) sulphate and then acidified with
acid. The formation of Prussian blue colour confirms the presence of nitrogen.

(ii) Test for Sulphur : The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. Ablack precipitate of lead sulphide indicates the presence of sulphur.

(iii) Test for Halogens : The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the. presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.

(iv) Test for Phosphorus : The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric add and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

Question 22.
Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Dumas method and
(ii) Kjeldahl’s method.
Solution.

Question 23.
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Solution.

Question 24.
Explain the principle of paper chromatography.
Solution.
The underlying principle of paper chromatography is that of partition chromatography which is based on continuous differential partitioning of components of mixture between stationary and mobile phases. In paper chromatography, the paper used has water trapped in it which acts as the stationary phase while a suitable solvent or a mixture of solvents is used as a mobile phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. As the mobile phase moves over the paper, it carries the mixture with it. Since the different components have different solubility, they travel to different extents on the paper and become stationary at different lengths on the paper and are thus, separated.

Question 25.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Solution.

Question 26.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Solution.
Lassaigne’s test is used for the detection of extra elements such as N, S and X by applying the inorganic tests of analysis to these. Since, in organic compounds, the elements are present in covalent form and .inorganic tests can be applied only to ions, therefore these extra elements are first converted into their inorganic (ionic) forms by fusing with sodium metal.

Question 27.
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Solution.
Camphor is sublimable compound while CaSO₄ being ionic is not. Therefore, the two can be separated by the method of sublimation. If a mixture of the two is heated in a China dish covered with a porous paper and an inverted funnel over it, we will find the crystals of camphor forming on the inside walls of the inverted funnel. Thus, pure CaSO4 will be left in the China dish.

Question 28.
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Solution.
We know that any liquid boils when its vapour pressure is equal to the atmospheric pressure. There are certain liquids such as aniline which need very high temperature in order to start boiling. It is quite likely that at such elevated temperatures the molecules may just disintegrate. Therefore, to prevent this, steam distillation is employed. Here, the mixture of organic liquids containing the high boiling liquid say, aniline is mixed with water and heated. On doing so, at a temperature close to but less than 100°C (b.p. of water) the vapour pressure of water equals the atmospheric pressure and it boils. Since, in the mixture, aniline is present in conjugation with water it vapourises and moves out of the mixture.
The mixture of water and aniline is separated using a separating funnel. Steam distillation is used extensively in perfumery to separate essential oils.

Question 29.
Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Solution.
AgNO₃ solution is ionic in nature. It contains Ag⁺ ions which when react with Ch ions produce a white ppt. of AgCl. In CCl₄ the Cl atoms are covalent in nature. They are not present as ions. Therefore, when AgNO₃ is added it, it does not produce a white ppt. of AgCl.

Question 30.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Solution.
KOH reacts with CO₂ to produce K₂CO3 which is a solid. The K₂CO₃ formed may be weighed and estimated to know the carbon content of the organic compound.

Question 31.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Solution.
Sulphuric acid cannot be used for acidification of sodium extract because it would oxidize the sulphur to sulphur dioxide which would not give the black ppt. of PbS, which is otherwise obtained upon reaction with lead acetate.

Question 32.
An organiccompound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Solution.

Question 33.
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Solution.

Question 34.
0 .3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Solution.

Question 35.
In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Solution.

Question 36.
In the organic compound


Solution.

Question 37.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of

Solution.

Question 38.
Which of the following carbocation is most stable?

Solution.

Question 39.
The best and latest technique for isolation, purification and separation of organic compounds is
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography
Solution.
(d) Chromatography

Question 40.
The reaction:
CH₃CH₂I + K0H_(aq) ➝ CH₃CH₂OH + KI is
classified as
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
Solution.

In the given reaction, the I^– from the alkyl iodide is replaced by the OH^– ion. Thus, it is substitution reaction.
The substitution is brought about by the OH^– ion which is a nucleophile.
∵ The reaction is a nucleophilic substitution reaction.

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NCERT Solutions for Class 11 Chemistry