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NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 12-SOME BASIC PRINCIPLES AND TECHNIQUES – Exercises

Chapter-12 Some Basics Principles and Techniques

Answer the following Questions.

1. Mention the hybridisation states of each carbon atom in the following compounds
CH2 = C = O, CH3 CH = CH2, (CH3)2CO, CH2 = CHCN, C6H6


Answer:
(i)chapter 12-Some Basic Principles and Techniques
C–1 is sp2 hybridised.
C–2 is sp hybridised.

(ii)chapter 12-Some Basic Principles and Techniques
C–1 is sp3 hybridised.
C–2 is sp2 hybridised.
C–3 is sp2 hybridised.

(iii)chapter 12-Some Basic Principles and Techniques
C–1 and C–3 are sp3 hybridised.
C–2 is sp2 hybridised.

(iv)chapter 12-Some Basic Principles and Techniques
C–1 is sphybridised.
C–2 is sphybridised.
C–3 is sp hybridised.

(v) C6H6
All the 6 carbon atoms in benzene are sp2 hybridised.

2. Indicate the σ and π bonds in the following molecules:
C6H6, C6H12, CH2Cl2, CH= C = CH2, CH3NO2, HCO NHCH3


Answer:
(i) C6H6
chapter 12-Some Basic Principles and Techniques
There are six C-C sigma bonds , six C-H bonds ,
and three C=C resonating bonds in the given compound.

(ii) C6H12
chapter 12-Some Basic Principles and Techniques

There are six C-C sigma bonds , twelve C-H bonds sigma in the given compound.

(iii) CH2Cl2,
chapter 12-Some Basic Principles and Techniques

There are two C-H sigma bonds , two C-Cl bonds sigma in the given compound.

(iv)CH= C = CH2,

chapter 12-Some Basic Principles and Techniques

There are two C-C sigma bonds , four C-H bonds and two
C=C π bonds in the given compound.

(v)CH3NO2,
chapter 12-Some Basic Principles and Techniques
There are three C-H sigma bonds , one C-N sigma bond and one N-O sigma bond and one
N=O π bond in the given compound.

(vi) HCONHCH3
chapter 12-Some Basic Principles and Techniques
There are two C-N sigma bonds , four C-H sigma bond , N-H sigma bond and one
C=O  π bond in the given compound.

3. What are the bond line formulas for the following compounds?
(a)2, 3–dimethyl butanal
(b) Heptan–4–one
(c) Isopropyl alcohol


Answer:
(a)2, 3–dimethyl butanal
chapter 12-Some Basic Principles and Techniques
 

(b) Heptan–4–one
chapter 12-Some Basic Principles and Techniques
 

(c) Isopropyl alcohol
chapter 12-Some Basic Principles and Techniques
 

4. Give the IUPAC names of the following compounds:
(a)

chapter 12-Some Basic Principles and Techniques

(b)

chapter 12-Some Basic Principles and Techniques

(c)

chapter 12-Some Basic Principles and Techniques

(d)
chapter 12-Some Basic Principles and Techniques

(e)

chapter 12-Some Basic Principles and Techniques

(f) Cl2 CH CH2 OH

Answer:
(a)

chapter 12-Some Basic Principles and Techniques
3–phenyl propane

(b)

chapter 12-Some Basic Principles and Techniques
2–methyl–1–cyanobutane

(c)

chapter 12-Some Basic Principles and Techniques
2, 5–dimethyl heptane

(d)

chapter 12-Some Basic Principles and Techniques
3–bromo–3–chloroheptane

(e)

chapter 12-Some Basic Principles and Techniques
3–chloropropanal

(f) Cl2 CH CH2OH
2,2- Dichloroethanol

5. Which of the following names of the respective compounds are the correct IUPAC-prescribed ones?
(a) 2,2-Dimethyl pentane or 2-Dimethyl pentane
(b) 2,4,7-Trimethyl octane or 2,5,7-Trimethyl octane
(c) 2-Chloro -4-methyl pentane or 4-Chloro-2-methyl pentane
(d) But-3-yn-1-ol or But-4-ol-1-yne


Answer:
(a)The prefix di shows that there are two methyl groups in the chain. Thus the correct IUPAC name would be 2,2-Dimethyl pentane.
(b)The locant number should start from the minimum. Here, 2,4,7 is lower than 2,5,7. Thus, the correct IUPAC name would be 2,4,7-Trimethyloctane.
(c)If the substituents in the chain are in equivalent positions, then the lower number is given to the substituent group in an alphabetical order.Thus, the correct IUPAC name would be 2-Chloro-4-methyl pentane.
(d)Out of the two functional groups present in the given compound, the alcoholic group is the principal functional group. Thus, the parent chain will have an –ol suffix. Since, the alkyne group is in C–3, the IUPAC name would be But–3–yn–1–ol.

6. What are the formulas for the first five members of each of the homologous series beginning with the below-given compounds?
(a) H– COOH
(b) CH3 COCH3
(c) H– CH = CH2


Answer:
The first five members of each homologous series beginning with the given compounds are
(a)
H – COOH :Methanoic acid
CH3 – COOH :Ethanoic acid
CH3– CH2– COOH :Propanoic acid
CH– CH2 – CH– COOH :Butanoic acid
CH3 – CH2 – CH2 – CH– COOH :Pentanoic acid

(b)
CHCOCH3 :Propanone
CH3 COCH2 CH3: Butanone
CHCOCH2 CH2 CH3 : Pentan-2-one
CH3 COCH2 CH2CH2 CH3 : Hexan-2-one
CH3 COCH2 CHCH2 CH2 CH3 : Heptan-2-one

(c)
H– CH = CH2 :Ethene
CH3– CH = CH2 : Propene
CH3 – CH2 – CH = CH2 : 1-Butene
CH3 – CH2 – CH2 – CH = CH2 : 1-Pentene
CH3 – CH2 – CH– CH– CH = CH2 : 1-Hexene

7. Write thebond line and condensed structural formulas and also find out the functional group for the following compounds.
(a)2,2,4-Trimethyl pentane
(b)2-Hydroxy-1,2,3-propanetri carboxylic acid
(c) Hexanedial


Answer:
(a) 2, 2, 4–trimethyl pentane
Condensed formula
(CH3)2 CHCH2C (CH3)3
Bond line formula :
chapter 12-Some Basic Principles and Techniques
(b) 2–hydroxy–1, 2, 3–propanetri carboxylic acid
Condensed Formula
(COOH) CH2C (OH) (COOH) CH2(COOH)
Bond line formula:

chapter 12-Some Basic Principles and Techniques
Functional groups:
Carboxylic acid(-COOH) and Alcoholic (-OH) groups

(c) Hexanedial
Condensed Formula
(CHO)(CH2)(CHO)
Bond line formula:

chapter 12-Some Basic Principles and Techniques
Functional groups:
Aldehyde(-CHO)

8. What are the functional groups present in the below given compounds?
(a)

chapter 12-Some Basic Principles and Techniques
(b)

chapter 12-Some Basic Principles and Techniques
(c)
chapter 12-Some Basic Principles and Techniques

Answer:
(a) Hydroxyl (–OH),Aldehyde (–CHO), Methoxy (–OMe),
C=C double bondchapter 12-Some Basic Principles and Techniques

(b)Ketone (C = O),Amino (–NH2),Diethylamine (N(CH5)2 )

(c) Nitro (–NO2),
C=C double bondchapter 12-Some Basic Principles and Techniques

9. Which of the two given compounds: O2 NCHCH2O– or CH3 CH2O– would you expect to be more stable and why?

Answer:
Since NO2 belongs to the electron-withdrawing group, it shows –I effect. NO2 tries to decrease the negative charge on the compound by withdrawing the electrons toward it. This stabilizes the compound whereas ethyl group belongs to the electron-releasing group and shows +I effect. This results in an increase in the negative charge on the compound thus destabilizing the compound. Hence, I would expect O2 NCHCH2Oto be more stable than CH3CH2O–.

10. Why do you think alkyl groups act as electron donors when they get attached to a π system?
Answer:
Due to hyperconjugation, an alkyl group behaves as an electron-donor group when attached to a π system.For example, look at propene.

chapter 12-Some Basic Principles and Techniques
The sigma electrons of the C-H bond get delocalized due to hyperconjugation. The alkyl is attached directly to an unsaturated system. The delocalization happens due to the partial overlap of a sp3-s sigma bond orbital with an empty p orbital of the n bond of an adjacent carbon atom.

This process can be shown as:
chapter 12-Some Basic Principles and Techniques

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the π electrons, making the molecule more stable.

chapter 12-Some Basic Principles and Techniques

11. Sketch the resonance structures of the below given compounds along with a curved arrow notation to show the electron shift.
(a) C6HOH
(b) C6HNO2
(c) CH3CH = CH – CHO
(d) C6H5 CHO
(e) C6H– CH2
(f)CHCH = CHCH2


Answer:
(a) The structure of C6H5 OH is:
chapter 12-Some Basic Principles and Techniques

Resonating structures:
chapter 12-Some Basic Principles and Techniques
 

(b) The structure of C6H5 NOis:
ncert solution

Resonating structures:
chapter 12-Some Basic Principles and Techniques
 

(c) CH3CH = CH – CHO

The resonating structures of the given compound are represented as:
chapter 12-Some Basic Principles and Techniques

(d) The structure of C6HCHO is:
chapter 12-Some Basic Principles and Techniques

Resonating structures:
chapter 12-Some Basic Principles and Techniques
 

(e) C6H5– CH2

Resonating structures:
chapter 12-Some Basic Principles and Techniques
(f) CH3CH=CHCH2
Resonating structures:
chapter 12-Some Basic Principles and Techniques
 

12. Define Nucleophiles and Electrophiles with the help of examples.
Answer:
A nucleophile is a reagent that has an electron pair and is willing to donate it. It is also known as a nucleus-loving reagent. Ex: NC, OH, R3C– (carbanions) etc.
An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair. Ex: Carbonyl groups, CH3CH2+(chapter 12-Some Basic Principles and TechniquesCarbocations), Neutral molecules( due to the presence of a lone pair).

13. Classify the following reagents as Electrophiles or Nucleophiles.
(a) CH3 COOH + HO → CH3COO + H2O
(b) CHCOCH3 +  C–N → (CH3)2 C(CN) + (OH)
(c) C6H5 + CH3C+ O → C6HCOCH3


Answer:
A nucleophile is a reagent that has an electron pair and is willing to donate it. It is also known as a nucleus-loving reagent.
An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair.

(a) CH3COOH + HO→ CH3COO + H2O
It is a nucleophile since HO- is electron rich in nature.

(b) CH3COCH3+ C–N → (CH3)2 C(CN) + (OH)
It is a nucleophile since C–N is electron rich in nature.

(c) C6H5 + CH3C + O → C6H5 COCH3
It is an electrophile since CH3C + O is electron-deficient in nature.

14. Find out the type of reaction happening in each of the following equations:
(a) CH3 CH2Br + HS– → CH3 CH2 SH + Br
(b) (CH3)2 C=CH2 + HCl→ (CH3)2 ClC – CH3
(c) CH3 CH2Br + HO → CH2 =CH2 + H2O + Br
(d) (CH3)3C–CH2 OH + HBr → (CH3)2 CBr CH2CH3 + H2O


Answer:
(a) Substitution reaction since bromine group gets substituted by –SH group.
(b) Addition reaction since two reactant molecules combines to form a single product.
(c) Elimination reaction since reaction hydrogen and bromine are removed to form ethene.
(d) Substitution reaction since rearrangement of atoms takes place.

15. Are the below given sets of compounds related to each other? If so, are they geometrical or structural isomers or resonance contributors?
(a)

chapter 12-Some Basic Principles and Techniques
(b)

chapter 12-Some Basic Principles and Techniques
(c)
chapter 12-Some Basic Principles and Techniques
Answer :
(a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).
chapter 12-Some Basic Principles and Techniques
In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.

(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.
chapter 12-Some Basic Principles and Techniques
In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.

(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.
chapter 12-Some Basic Principles and Techniques
 

16. For each of the below given bond cleavages, show the electron flow by using curved arrows and classify them as heterolysisor homolysis. Classify the intermediate produced as carbocation or free radical or carbanion.
(a)

chapter 12-Some Basic Principles and Techniques
(b)

chapter 12-Some Basic Principles and Techniques
(c)

chapter 12-Some Basic Principles and Techniques

(d)

chapter 12-Some Basic Principles and Techniques

Answer:
 (a) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
chapter 12-Some Basic Principles and Techniques
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.

(b)The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
chapter 12-Some Basic Principles and Techniques
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.

(c)The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
chapter 12-Some Basic Principles and Techniques
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.

(d)The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
chapter 12-Some Basic Principles and Techniques
It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.

17. Define the terms Electromeric and Inductive effects. Which effect of electron displacement do you think can explain the below given correct orders of acidity of the carboxylic acids?
(a) Cl3C COOH > Cl2CH COOH > ClCH2 COOH
(b) CH3CH2 COOH > (CH3)2 CH COOH > (CH3)3C. COOH


Answer:
Inductive effect
The permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect.
chapter 12-Some Basic Principles and Techniques
Inductive effect could be + I effect or – I effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess – I effect. For example,
chapter 12-Some Basic Principles and Techniques
When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess + I effect. For example,

Electrometric effect
It involves the complete transfer of the shared pair of π electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent. For example,
chapter 12-Some Basic Principles and Techniques
Electrometric effect could be + E effect or – E effect.

+ E effect:When the electrons are transferred towards the attacking reagent
– E effect:When the electrons are transferred away from the attacking reagent

(a)Cl3C COOH > Cl2CH COOH > ClCH2COOH
The order of acidity can be explained on the basis of Inductive effect (– I effect). As the number of chlorine atoms increases, the – I effect increases. With the increase in – I effect, the acid strength also increases accordingly.
chapter 12-Some Basic Principles and Techniques
 

(b)CH3 CH2 COOH > (CH3)2 CHCOOH > (CH3)3 C.COOH
The order of acidity can be explained on the basis of inductive effect (+ I effect). As the number of alkyl groups increases, the + I effect also increases. With the increase in + I effect, the acid strength also increases accordingly.
chapter 12-Some Basic Principles and Techniques
 

18. Explain the following techniques with the help of examples.
(a)Distillation
(b) Crystallisation
(c) Chromatography


Answer:
(a) Crystallisation
Crystallization is used to purify solid organic compounds.
Principle: The principle on which it works is the difference in the solubility of the compound and impurities in a given solvent. The impure compound is made to dissolve in the solvent at a higher temperature since it is sparingly soluble at lower temperatures. This is continued till we get an almost saturated solution. On cooling and filtering it, we get its’ crystals. Ex: By crystallizing 2-4g of crude aspirin in 20mL of ethyl alcohol, we get pure aspirin. It is heated if needed and left undisturbed until it crystallizes. The crystals are then separated and dried.

(b) Distillation
This method is used to separate non-volatile liquids from volatile impurities. It is also used when the components have a considerable difference in their boiling points.
Principle: The principle on which it works is that liquids having different boiling points vaporise at different temperatures. They are then cooled and the formed liquids are separated.
Ex: A mixture of aniline (b.p = 457 K) and chloroform (b.p = 334 K) is taken in a round bottom flask having a condenser. When they are heated, Chloroform, vaporizes first due to its high volatility and made to pass through a condenser where it cools down. The aniline is left behind in the round bottom flask.

(c) Chromatography
It is widely used for the separation and purification of organic compounds.
Principle: The principle on which it works is that individual components of a mixture move at different paces through the stationary phase under the influence of mobile phase.
Ex: Chromatography can be used to separate a mixture of blue and red ink. This mixture is placed on chromatogram where the component which is less absorbed by the chromatogram moves faster up the paper than the other component which is almost stationary.

19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Answer: Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent S. The process of fractional crystallisation is carried out in four steps.
(a) Preparation of the solution: The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. This saturated solution is then heated.
(b) Filtration of the solution: The hot saturated solution is then filtered through a filter paper in a China dish.
(c) Fractional crystallisation: The solution in the China dish is now allowed to cool. The less soluble compound crystallises first, while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently, the crystals of the more soluble compound are obtained.
(d) Isolation and drying: These crystals are separated from the mother liquor by filtration. Finally, the crystals are dried.

20. Discuss the chemistry of Lassaigne’s test.

Answer:
Lassaigne’s test
This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorous in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal.
chapter 12-Some Basic Principles and Techniques
The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called Lassaigne’s extract. This Lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens, and phosphorous.

(a)Test for nitrogen

chapter 12-Some Basic Principles and Techniques

Chemistry of the test
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as
chapter 12-Some Basic Principles and Techniques
(b)Test for sulphur
chapter 12-Some Basic Principles and Techniques
Chemistry of the test
In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is acidified with acetic acid and then lead acetate is added to it. The precipitation of lead sulphide, which is black in colour, indicates the presence of sulphur in the compound.
chapter 12-Some Basic Principles and Techniques

Chemistry of the test
The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour also indicates the presence of sulphur in the compound.
chapter 12-Some Basic Principles and Techniques
If in an organic compound, both nitrogen and sulphur are present, then instead of NaCN, formation of NaSCN takes place.

Na + C + N + S → NaSCN

This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed due to the absence of free cyanide ions.
chapter 12-Some Basic Principles and Techniques
(c)Test for halogens
chapter 12-Some Basic Principles and Techniques
Chemistry of the test
In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is acidified with nitric acid and then treated with silver nitrate.
chapter 12-Some Basic Principles and Techniques
If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test for halogens.

20. What is the difference between distillation, distillation under reduced pressure and steam distillation ?
Answer:
The differences among distillation, distillation under reduced pressure, and steam distillation are given in the following table.
 

 DistillationDistillation under reduced pressureSteam distillation
1.It is used for the purification of compounds that are associated with non-volatile impurities or those liquids, which do not decompose on boiling. In other words, distillation is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have sufficient difference in boiling points.This method is used to purify a liquid that tends to decompose on boiling. Under the conditions of reduced pressure, the liquid will boil at a low temperature than its boiling point and will, therefore, not decompose.It is used to purify an organic compound, which is steam volatile and immiscible in water. On passing steam, the compound gets heated up and the steam gets condensed to water. After some time, the mixture of water and liquid starts to boil and passes through the condenser. This condensed mixture of water and liquid is then separated by using a separating funnel.
2.Mixture of petrol and kerosene is separated by this method.Glycerol is purified by this method. It boils with decomposition at a temperature of 593 K. At a reduced pressure, it boils at 453 K without decomposition.A mixture of water and aniline is separated by steam distillation.

Conclusions for NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 12-SOME BASIC PRINCIPLES AND TECHNIQUES

An academic team of knowledgeable members of SWC has produced and published the NCERT Solutions for class 11’s Chemistry chapter for your use as a reference. You can get answers to all of the chapters of the NCERT Chemistry class 11 here at SWC. Please make use of the following NCERT answers that were created by SWC as a reference for this chapter. In addition to that, study the chapter’s theory before attempting to solve the NCERT problems.

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