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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

Edited By Shivani Poonia | Updated on Mar 22, 2025 11:03 PM IST

Some Basic Principles and Techniques is one of the most important chapters in Class 11 Chemistry as it forms the foundation of organic Chemistry which is going to help students throughout their career in Chemistry. This chapter helps students to understand topics like the shapes of compounds, their representation, classification of compounds, nomenclature of compounds and purification of organic compounds. Some Basic Principles And Techniques is an important chapter for final exams as well as competitive exams like JEE, NEET, and BITSAT. Chemistry chapter 8 is a blend of theoretical and practical concepts. The concepts provided in this chapter help us to understand some phenomena from our day-to-day lives, like the purification of organic compounds by distillation.

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This Story also Contains
  1. NCERT Solutions For Class 11 Chemistry
  2. More About Class 11 Chemistry NCERT Chapter 8
  3. Topics of Organic Chemistry- Some Basic Principles and Techniques
  4. What is a Functional Group?
  5. Some important points of Chapter 8 of Class 11 Chemistry-
  6. NCERT Solutions for Class 11- Subject-wise
  7. Also Check NCERT Books and NCERT Syllabus here:

Chapter 8 solutions of Class 11 Chemistry are prepared by subject experts in a very comprehensive and systematic way, which helps students develop a clear understanding of the topics used to solve particular problems. Solutions provided help students in their CBSE class 11 as well as competitive exams like JEE, NEET, and BITSAT. By referring to the NCERT solutions for class 11, students can understand all the important concepts well enough before their examination.

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To perform well in upcoming exams like CBSE exams, state board exams, and competitive exams, students need to solve every question in NCERT class 11 chapter 8. The solutions provided below cover key concepts in a very concise and systematic way which helps students to gain confidence in their preparation.

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NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry Some Basic Principles and Techniques


Q8.1(i) What are the hybridization states of each carbon atom in the following compounds? CH2=C=O

Answer:

(i)CH2=C=O
c1 c2

Here C1 carbon is sp2 hybridised and the C2 carbon is sp hybridised.

Q 8.1(ii) What are the hybridisation states of each carbon atom in the following compounds?

(ii)CH3CH=CH2

Answer:

(ii)CH3CH=CH2
c1 c2 c3

Here C1 caron is sp3 hybridised, C2 carbon is sp2 and the third carbon C3 atom is sp2 hybridise

Q 8.1(iii) What are the hybridisation states of each carbon atom in the following compounds?

(iii)(CH3)2CO

Answer:
Here is the structure

1650540855701 Here C1 and C3 carbon atom is in sp3 hybridistaion and the C2 carbon is sp2

Q 8.1(iv) What are the hybridisation states of each carbon atom in the following compounds ?

(v)C6H6

Answer:
In benzene, all the six carbon atoms are sp2 hybridised.


Q 8.1(v) What are the hybridisation states of each carbon atom in the following compounds ?

(vi)CH2=CHCN

Answer:

CH2=CHCN
1 2 3

Here, in the above compound the C1 carbon atom is sp2 hybridised, C2 carbon atom is sp2 hybridised and the C3 carbon is sp hybridiesd

Q 8.2(i) Indicate the σ and π bonds in the following molecule :

(i)C6H6

Answer:

yrtdgerfsd
There are six C-C sigma bonds, six C-H sigma bonds and three π -bonds between the carbon atom in the benzene.

Q 8.2(ii) Indicate the σ and π bonds in the following molecule:

(ii)C6H12

Answer

trgearfds
In C6H12, there are six sigma bonds between C-Cand and twelve sigma bonds between C-H bonds in the given compound.

Q 8.2(iii) Indicate the σ and π bonds in the following molecule :

(iii)CH2Cl2

Answer:

In CH2Cl2 ,
1650540886373 There are two sigma C-H bonds and two C-Cl sigma bonds.


Q 8.2 (iv) Indicate the σ and π bonds in the following molecule :

(iv)CH2=C=CH2


Answer:

1650540914961 In this compound, there are two C-C sigma bonds, four C-H sigma bonds and two π bonds between carbon atoms.

Q 8.2(v) Indicate the σ and π bonds in the following molecule :

(v)CH3NO2

sdffhf
There are three C-H sigma bonds, one C-N sigma bond, one N-O sigma bond and one N-O π -bond.

Q 8.2(vi) Indicate the σ and π bonds in the following molecule :

(vi)HCONHCH3


Answer:

tgrfsed
Two C-N sigma bonds, four C-H sigma bonds, one N-H sigma bond and one C=O π -bond in the above compound.

Q 8.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4- one.

Answer:

Bond line formula of the following compounds-

Isopropyl alcohol
sdfgf
2, 3-Dimethyl butanal
jdthsdgrfs
Heptan-4-one
hrtgsfds


Q 8.4(a) Give the IUPAC names of the following compounds:

eards


Answer:

gsdfxbd
The IUPAC name of the above compound is 1-phenylpropane

Q 8.4(c) Give the IUPAC names of the below compound :

ggr

Answer:

1650540952156 The IUPAC name of the above structure is 2, 5-dimethyl heptane.

Q 8.4(f) Give the IUPAC names of the following compounds :

(e)Cl2CHCH2OH

Answer: (e)Cl2CHCH2OH
The IUPAC name of the compound is 1, 1-dichloro-2-ethanol

Q 8.5 Which of the following represents the correct IUPAC name for the compounds concerned?

(a) 2,2-Dimethylpentane or 2-Dimethylpentane

(b) 2,4,7- Trimethyloctane or 2,5,7-Trimethyloctane

(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane

(d) But-3-yn-1-ol or But-4-ol-1-yne .

Answer:

(a) The prefix di in the IUPAC nomenclature indicates two identical substituent groups in the parent chain. Since two methyl groups are present at C2 of the parent chain. So, the correct IUPAC name is 2, 2-dimethylpentane.

(b) In IUPAC nomenclature the sum of locant number should be minimum. Here the sum of 2,4,7 (=13)is less than the sum of 2, 5, 7 (=14). Thus the correct IUPAC name is 2,4,7- Trimethyloctane

(c) In IUPAC nomenclature, if the substituent group acquires the equivalent position of the parent chain then the lower number is given to one that comes first in alphabetical order. Hence 2-chloro-4-methyl pentane is the correct IUPAC name of the compound.

(d) If the two functional groups are present in the parent chain then the suffix of the IUPAC name depends on the principal functional group. Here alcohol is the principal function group so the suffix should be ol and the alkyne group is considered as a substituent group. Therefore the correct IUPAC name of the compound is But-3-yn-1-ol.

Q 8.6(a) Draw formulas for the first five members of each homologous series beginning with the following compound.

(a)HCOOH

Answer:

the first five members of each homologous series beginning with the following compound are shown as-

(a)HCOOH ---methanoic acid

  • ethanoic acid
    CH3COOH
  • propanoic acid
    CH3CH2COOH
  • butanoic acid
    CH3CH2CH2COOH
  • pentanoic acid
    CH3CH2CH2CH2COOH

Q 8.6 (b) Draw formulas for the first five members of each homologous series beginning with the following compound.
(b)CH3COCH3

Answer: the first five members of each homologous series beginning with the following compound are shown as-
(b)CH3COCH3
(propanone)

  • Butanone
    CH3COCH2CH3
  • Pentan-2-one
    CH3COCH2CH2CH3
  • hexan-2-one
    CH3COCH2CH2CH2CH3
  • Hept-2-one
    CH3COCH2CH2CH2CH2CH3

Q 8.6 (c) Draw formulas for the first five members of each homologous series beginning with the following compound.

(c)HCH=CH2

Answer: The first five members of each homologous series beginning with the following compound are shown as-
(c)HCH=CH2 ..........(Ethene)

  • Propene
    CH3CH=CH2
  • 1-Butene
    CH3CH2CH=CH2
  • 1-Pentene
    CH3CH2CH2CH=CH2
  • 1-Hexene
    CH3CH2CH2CH2CH=CH2

Q 8.7(a) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
2,2,4-Trimethylpentane

Answer: 2,2,4-Trimethylpentane

Condensed formula- (CH3)2CHCH2C(CH3)3

Bond line formula-
1650540999238

Q 8.7(b) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
2-Hydroxy-1,2,3-propanetricarboxylic acid

Answer:

The condensed formula of 2-Hydroxy-1,2,3-propanetricarboxylic acid
(COOH)CH2C(OH)(COOH)CH2(COOH)

The bond line structure -

sferdgf

Q 8.7(c) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
Hexanediol

Answer:

the condensed formula of hexanediol

(CHO)(CH2)4(CHO)

The bond line structure is-
1650541037679

Q 8.8 Identify the functional groups in the following compounds

aedsfa gerwfdes uuyutdyrd


Answer:

sdfedaxzs
The functional groups in the above structure are-

  1. Aldehyde ( CHO )
  2. HYdroxyl ( OH )
  3. Methoxy ( OMe )
  4. C=C double bond

(b)
1650541078949 The following functional groups are present-

  1. Amino, primary amine ( NH2 )
  2. Ester ( OCO )
  3. Tertiary amine ( NR2 ) R = ethyl group

sdfxcv
Here,

  1. Nitro group ( NO2 )
  2. C=C double bond

Q 8.9 Which of the two: O2NCH2CH2O or CH2CH2O is expected to be more stable and why ?

Answer: Since the Nitro group is an electron-withdrawing group. So, it shows I effect, By reducing the electron negative charge of the compound, it stabilises the compound. On the other hand, the methyl group is an electron donor group so it shows +I effect. This increases the negative charge on the compound and destabilises the compound.

Hence O2NCH2CH2O is more stable than the CH2CH2O .

Q 8.10 Explain why alkyl groups act as electron donors when attached to a π system.

Answer: When an alkyl group is attached to the π system, it acts as an electron donor group by the property of hyperconjugation.
for example in propene,
awfdsfa
In the figure, you can see that the sigma electrons of C-H bonds of the alkyl group are delocalised because of the partially overlapping of the sp3s sigma bond orbital with the empty p orbital of the π bond of the adjacent carbon atom. It is also known as no-bond resonance.
sdd

Q 8.11(a) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(a)C6H5OH

Answer: The resonating structure of the phenol--
The lone pair of electrons start shifting to an oxygen-carbon bond and form a double bond character and π electron of the C-C bond shift to the next C-C single bond.

adws


Q 8.11 (b) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(b)C6H5NO2

Answer: The resonating structure of the nitrobenzene-

Here the electrons of N-O bonds shift to the Oxygen atom (more electronegative) and then the π electron of carbon-carbon double bonds starts delocalising towards the N atom.

rwfdasd

Q 8.11(c) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(c)CH3CH=CHCHO

Answer: The resonating structure of the But-2-ene-1-al-
Here the π electrons of C-O bonds shift to the Oxygen atom (more electronegative) and then the π electron of carbon-carbon double bonds starts shifting towards the next C-Cbond, introducing a partial double bond character.
1650541150092

Q 8.11(d) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(d)C6H5CHO

Answer: The resonating structure of the benzaldehyde--
Here the π electron of the carbon-oxygen double bond starts shifting towards the electronegative oxygen atom and the π -electrons of the C-C double bonds shift towards the carbonyl group (as shown in figures).
1650541199113

Q 8.11(e) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(e)C6H5CH2

Answer: The resonating structure of Benzyl carbocation-
The π -electrons of the C-C double bond shift towards the CC+ bond to minimise the deficiency of electron density and the rest of the π -electrons follow the same process.
1650541233612

Q 8.11(f) Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(f)CH3CH=CHCH2

Answer: The resonating structure of But-2-ene-1-yl carbocation-
The π -electrons of the C-C double bond shift towards the CC+ bond to minimise the deficiency of electron density at CH2+. So, this way only one resonant is possible.
1650541262927

Q 8.12 What are electrophiles and nucleophiles? Explain with examples.

Answer:

Electrophile- It is an electron-deficient species, which seeks an electron pair. This reagent takes away an electron pair. It is denoted as E+. For example, carbocations CH3CH2+ neutral molecules having functional groups such as the carbonyl group are an example of the electrophile.

A nucleophile is a reagent that brings an electron pair. In other words, it is nucleus- seeking reagent called nucleophile.
For examples- OH,NC and carbanions ( R3C ) etc. Ammonia and water also act as a nucleophile due to the presence of lone pair of electrons.

Q 8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

(a)CH3COOH+HOCH3COO+H2O

(b)CH3COCH3+CN(CH3)2C(CN)(OH)

(c)C6H6+CH3COC6H5COCH3

Answer:

Electrophiles are electron deficient species and can be seeking for an electron pair. On the other hand, the nucleophile is electron rich reagents and they can electron donor(nucleus seeking).

Therefore,
(a) OH have one lone pair of an electron, electron-rich species. So, it is a nucleophile

(b) CN acts as a nucleus-seeking reagent and acts as a nucleophile

(c) CH3C+O, it is an electron-deficient species, So, it is an electrophile.


Q 8.14. (a) Classify the following reactions in one of the reaction types studied in this unit.

(a)CH3CH2Br+HSCH3CH2SH+Br


Answer:

(a)CH3CH2Br+HSCH3CH2SH+Br
It is a substitution reaction as in this above equation the bromine group is replaced by the SH group.


Q 8.14.(b) Classify the following reactions in one of the reaction types studied in this unit.

(b)(CH3)2C=CH2+HCl(CH3)2ClCCH3

Answer: (b)(CH3)2C=CH2+HCl(CH3)2ClCCH3

The given reaction is an example of an addition reaction because in this reaction the two reactant molecules combined to form a single product. Also, we can say that the Hydrogen and chlorine of HCl is added in the two different carbon atom of the same compound.

Q 8.14(c) Classify the following reactions in one of the reaction types studied in this unit.

(c)CH3CH2Br+HOCH2=CH2+H2O+Br

Answer: (c)CH3CH2Br+HOCH2=CH2+H2O+Br

In this reaction, the hydrogen and bromine are removed from the original compound and formed ethene. So, this is an elimination reaction.

Q 8.14(d) Classify the following reactions in one of the reaction type studied in this unit.

(d)(CH3)3CCH2OH+HBr(CH3)2CBrCH2CH2CH3+H2O

Answer: (d)(CH3)3CCH2OH+HBr(CH3)2CBrCH2CH2CH3+H2O

In the above reaction, the substitution takes place between the two reactants, followed by the rearrangement of atoms and the groups of atoms.

Q 8.15 What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

erfeads

Answer:

eagrfdsa
These pairs are structural isomers.
Compounds having the same molecular formula but different structures are called structural isomers. The above compounds have the same molecular formula but the structures are different due to the difference in the position of the carbonyl group.
In structure one -CO group is present at C3 position and in second the -CO group is present at C2 position.

Q 8.16 (a) 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.

(a) ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Answer:

Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as
1650541439122 It is an example of hemolysis because the two electrons are equally divided into the products. (see in fig)

Q 8.16(b) 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.

1650541480519

Answer:

Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as;
1650541503670 It is an example of heterolytic cleavage as the bond breaks in such a manner that the electron pair will remain with the carbon of propanone. The reaction intermediate is carbanion.

Q 8.16(c) 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.

1650541539381

Answer: Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as;

1650541562384 It is heterolysis as shared pair of an electron is distributed to only bromine ion. Here the reaction intermediate is carbocation.

Q 8.16(d) 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.

1650541607370

Answer: Bond cleavage by using curved arrows to show the electron flow of the given reaction can be represented as

1650541626097 The reaction intermediate is carbocation. It is a heterolytic cleavage as the bonds break in such a manner that the shared; electron pair will remain with the one species.

Q 8.17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

(a)Cl3CCOOH>Cl2CHCOOH>ClCH2COOH

(b)CH3CH2COOH>(CH3)2CHCOOH>(CH3)3C.COOH

Answer: Inductive effect- The permanent displacement of sigma ( σ ) electrons along the saturated chain, whenever an electron donating or withdrawing group is present, is known as the inductive effect. It could be +I/I effect.

Electromeric effect- The complete transfer of the shared pair of π -electrons to one of the atoms linked by multiple bonds on the demand of attacking reagent. It is a temporary effect. It may be +E(electrons transferred towards the attacking reagent) or -E effect (electrons transferring away from the attacking reagent).

(a) The order of acidity can be explained by the negative inductive effect ( I ). As the no. of chlorine atoms increases, the I effect also increases and so, the acidic strength also increases.
Cl3CCOOH>Cl2CHCOOH>ClCH2COOH

(b) This can be explained by the +I effect of the alkyl group. As the number of electron donor groups increases, the +I effect will also increase. With the increases in the +I effect, the acidic character decreases accordingly.
CH3CH2COOH>(CH3)2CHCOOH>(CH3)3C.COOH

Q 8.18 (a) Give a brief description of the principles of the following techniques taking an example in each case.

Crystallisation

Answer: Crystallisation- It is one of the most commonly used techniques for the purification of solid organic compounds. Its principle is based on the difference in the solubilities of the compounds and the solvent's impurities. The impure compounds are dissolved in solvent but they are sparingly soluble at room temperature but soluble at the higher temperature. On cooling the compound, the pure compounds crystallise and are removed by filtration.
For example - pure aspirin is obtained by recrystallising crude aspirin. Around 2 - 4 g of crude aspirin is dissolved in 20 mL of ethyl alcohol and the solution is heated for complete dissolution. Then after crystal formation, they can filter out and dried.

Q 8.18 (b) Give a brief description of the principles of the following techniques taking an example in each case.

Distillation

Answer: Distillation- This method is used for the purification of liquids from non-volatile impurities. It is based on the fact that fluids of having different boiling points vaporise at different temperatures. The impure liquid is boiled in a flask, and initially, the vapours of lower boiling points component are formed. The vapours are condensed by using a condenser, and the liquid is collected in a receiver.
For example- Organic liquids such as benzene, toluene, xylene etc can be purified by this method.

Q 8.18 (c) Give a brief description of the principles of the following techniques taking an example in each case.

Chromatography

Answer: Chromatography- It is one of the extensive methods for the separation and purification of organic compounds. It is based on the difference in the movements of components of mixtures through the stationary phase under the influence of the mobile phase. The stationary phase can be solid or liquid. While the mobile phase is only liquid or gas.
For example- This technique can separate a mixture of blue and red ink.

Q 8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Answer: Fractional distillation can be used to separate the two compounds with different solubilities in a solvent S. The following steps are carried out in this process-

  1. The powdered mixture is taken in a flask, and the solvent is added into it and stirred simultaneously. We have to prepare a saturated solution and then heat it.
  2. After heating, this hot saturated solution can be filtered with filter paper in a china dish.
  3. Now the solution is allowed to cool. The less soluble compounds crystalise first, and more soluble compounds remain in the solution. After removing these crystals, the latter is concentrated once again. The hot solution is allowed to cool, and then the crystals of the more soluble compound are obtained.
  4. The last step is the isolation and drying of crystals from the mother liquor.

Q 8.20 What is the difference between distillation, distillation under reduced pressure and steam distillation ?

Answer: The differences between distillation, distillation under reduced pressure and steam distillation are-

Distillation
Distillation under reduced pressure
Steam Distillation
1. Used for the purification of compounds that are non-volatile impurities or liquids, which don't decompose on boiling.

1. Used to purify liquids, which tend to decompose on boiling. Under the condition of reduced pressure.

1. Used to purify the organic compounds, which are steam volatile and immiscible with water.

Simply, to separate the volatile liquids from non-volatile impurities. Or a mixture of liquids having sufficient Boiling point difference.
The liquid will boil at a lower temperature than its boiling points, therefore, it does not decompose.
The mixture of water and aniline can be separated by this method.

The mixture of petrol and kerosene is separated by this method.
Glycerol is purified by this method.

Q 8.21. Discuss the chemistry of Lassaigne’s test.

Answer: Lassaigne’s test.-
This test is used to detect the presence of nitrogen, sulphur, halogens and sulphur in organic compounds. These elements are present in the covalent form in an organic compound. So, they are converted into the ionic form by fusing the compound with the sodium metal.
Na+C+NNaCN2Na+SNa2SNa+XNaX (X = Cl, Br, I)

Cyanide, sulphide or halide of sodium are extracted from the fused mass by boiling it with distilled water. This Is known as sodium extract method or Lassaigne's extract.

Q 8.22(i) Differentiate between the principle of estimation of nitrogen in an organic compound by

Dumas method

Answer: In the Dumas method, Nitrogen-containing organic compound is heated with copper oxide in a carbon dioxide atmosphere, yielding free nitrogen in addition to CO2 and water.

CxHyNz+(2x+y2)CuOxCO2+y2H2O+z2N2+(2x+y2)Cu
Traces of nitrogen oxides can also be formed in the reaction, which can be reduced to nitrogen by passing the gaseous mixture over the heated copper gauge. The produced mixture of gases is collected over by an aqueous solution of KOH , it absorbs CO2. Nitrogen is collected in the upper part of the graduated tube.

7tds


Q 8.22(ii) Differentiate between the principle of estimation of nitrogen in an organic compound by

Kjeldahl’s method.

Answer:

In Kjeldahl’s method, the nitrogen-containing organic compound is heated with concentrated sulphuric acid. Nitrogen is converted into ammonium sulphate. It is then passed into the known volume of sulphuric acid. The amount of ammonia produced can be estimated by the amount of H2SO4 consumed in the reaction.
O.C+H2SO4(NH4)2SO4(NH4)2SO4+2NaOHNa2SO4+2NH3+2H2O2NH3+H2SO4(NH4)2SO4
It is done by estimating the amount of unreacted H2SO4 left after the absorption of ammonia by titrating it with a standard alkali solution. This method is not applicable for the compound containing nitrogen in -nitro form or nitrogen present in the ring structure.
tdyrtsrfsdaa

Q 8.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Answer:

Estimation of halogen is done by the Carius method. In this method, a known quantity of organic compound is heated with fuming HNO3 (nitric acid) with the presence of silver nitrate, contained in a hard glass tube, known as a carius tube. C and H present in the compound are oxidised to carbon dioxide and water. Halogens are into to form AgX and then it is filtered, dried, and weighed.

Let the mass of the organic compound be m gram.
Mass of AgX formed = m1 gram
1 mol of AgX contains 1 mol of X.
Therefore, Mass of halogen inm1 g of AgX = (Atomic mass of Xm1 ) / (Molecular mass of AgX )

Thus, % of halogen will be = (Atomic mass of X(m1×100)/ mo l. wt of AgX(m)

Estimation of sulphur- In this method, Organic compound is heated either fuming nitric acid or sodium peroxide in a hard glass tube called carius tube. Sulphur present in the compound is oxidised to form sulphuric acid. It is precipitated by as barium sulphate by adding barium chloride solution in water. Then ppt is filtered, washed and weighed.

Let the mass of organic compound taken = m g
and the mass of barium sulphate formed = m1 g

1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur

m1 g BaSO4 contains = 32×m1233 g sulphur
Percentage (%)of sulphur =
32×m1×100233×m

Estimation of phosphorus- In this process, a known mass of an organic compound is heated with fuming nitric acid and the phosphorus gets oxidised to phosphoric acid. By adding ammonia and ammonium molybdate, phosphorus can be precipitated as ammonium phosphomolybdate, (NH4)3PO4.12MoO3. It can be also estimated as by precipitating it as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7.

Let the mass of organic compound taken = m g and the mass of ammonium phosphomolybdate = m1 g

Molar mass of (NH4)3PO4.12MoO3=1877 g
Percentage(%) of phosphorus = 31×m1×1001877×m

If phosphorus is estimated as Mg2P2O7 ,
Percentage(%) of phosphorus = 62×m1×100222×m %

Q 8.24 Explain the principle of paper chromatography.

Answer: In paper chromatography, chromatography paper is used. It contains water trapped in it, which acts as the stationary phase. The solution of the mixture is spotted on this base of chromatography paper. The strip of paper is suspended in a suitable solvent, which is the mobile phase. Due to capillary action, the solvent rises up in the paper and it flows over the spot. The spots of the different component travel with the mobile phase to a different level of heights. The obtained paper is called chromatogram.
gsfddzdv

Q 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Answer: Nitric acid was added to sodium extract before adding silver nitrate for testing halogens to decompose the NaCN to HCN and Na2S to H2S and to expel these gases. If any amount of nitrogen and sulphur are present in the form of NaCN and Na2S, then they are removed.

NaCN+HNO3NaNO3+HCNNa2S+HANO32NaNO3+H2S

Q 8.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Answer:

In Organic compounds, nitrogen, sulphur and halogens are covalently bonded. To detect them, they have to convert into ionic form. This is done by the fusing of an organic compound by the sodium metal. This is known as Lassaigne's test.
The following chemical reactions are-
Na+C+NNaCNNa+S+C+NNaSCN2Na+SNa2SNa+XNaX
Here X = halogen atoms

Q 8.27. Name a suitable technique for the separation of the components from a mixture of calcium sulphate and camphor.

Answer:

To separate a mixture of calcium sulphate and camphor, we use the sublimation method. In this method, the sublimable compound is converted to vapour state from the solid state without achieving its liquid states. Here camphor is sublimable compound and calcium sulphate is not.

Q 8.28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ?

Answer:

In steam distillation, Organic liquid starts boiling when the total sum of the vapour pressure of an organic liquid (P') and the of water (P'') becomes equal to the atmospheric pressure (P),
it means P = P' + P''. since P'< P'', the organic liquid will vaporise at a lower temperature than its boiling point.

Q 8.29. Will CCl4 givea white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Answer:

No, CCl4 does not give white ppt of AgCl on heating with the silver nitrate. This is because the chlorine atoms are covalently bonded with the carbon atom. To get ppt, it should be present in ionic form.

Q 8.30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolves during the estimation of carbon present in an organic compound?

Answer:

Carbon dioxide is acidic in nature and KOH is a strong base. Thus, carbon dioxide reacts with the potassium hydroxide by acid-base reaction and form a salt known as potassium carbonate and water as a by-product.

KOH+CO2K2CO3+H2O

Thus, due to the increase in KOH the mass of U-tube is also increases. This increased mass of the tube gives the mass of carbon dioxide produced. From its mass, the % of carbon in the Organic compound can be estimated.

Q 8.31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Answer:

It is necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test because in case of sulphuric acid the complete ppt formation of lead sulphate does not take place. While the addition of acetic acid will ensure a complete ppt. formation of sulphur in the form of lead sulphate due to the common ion effect.

Q 8.32. An organic compound 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.

Answer:

Given,
Percentage of carbon = 69%
% of Hydrogen = 4.8%, and
% of Oxygen = 26.2 %

So, 0.2 g of OC have (0.2×0.69) g of Carbon = 0.138g
we know that,
molecular weight of CO2 = 44 g
12 g of C is present in 44 g of CO2

Therefore, 0.138 g of C present in 0.138×4412 = 0.506 g of CO2

Hence, from 0.2 g of OC 0.506 g of CO2 will be produced.

Similarly,
100 g of OC contains 4.8 g of H
So, 0.2 g of OC contains (0.2×0.048) = 0.0096 g of H
We know the mol. wt. of water = 18 g

Therefore 0.0096 g of H will present in 18×0.00962 = 0.0864 g of water

Hence 0.2 gram of OC will produce 0.0864 g of water on complete combustion of


Q 8.33. 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 MH2SO4 . The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer:

Given that,the
total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH is required for neutralisation by residual acid.

60 mL of 0.5 M NaOH solution = 602 mL of 0.5M H2SO4 = 30 mL of 0.5 M H2SO4

Therefore,
Acid consumed in the absorption of evolved ammonia is (50-30) mL = 20 mL

Again, 20 mL of 0.5 M H2SO4 = 40 mL of 0.5 M NH3

Also, since 1000 mL of 1 M NH3 contains 14 g of nitrogen,
So, 40 mL of 0.5 M NH3 will contain = 14×40×0.51000=0.28 g of Nitrogen

Therefore, percentage(%) of nitrogen in 0.50 g of organic compound = (0.28/0.5)×100 = 56 %

Q 8.34. 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.

Answer:

Given that,
Mass of organic compound = 0.3780 g.
Mass of AgCl formed = 0.5740 g

It is known that,
1 mol of AgCl contains 1 mol of Cl.

Thus, the mass of chlorine in 0.5740 g of AgCl

=35×0.5740143.32

= 0.1421 g

∴ Percentage(%) of chlorine = =0.14210.3780×100 = 37.59%

Hence, the percentage of chlorine present in the given organic compound is 37.59%.

Q 8.35. In the estimation of sulphur by the 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.

Answer:

Given data,

The total mass of the OC = 0.468 g
Mass of barium sulphate formed = 0.668 g

We know that,

1 mol of BaSO 4 = 233 g of BaSO4 = 32 g of sulphur
Thus, 0.668 g of BaSO4 contains = 32×0.668233 g of sulphur = 0.0917 g of sulphur

Therefore, the percentage(%) of sulphur
=0.01970.468×100
= 19.59 %

Hence, the percentage of sulphur in the given compound is 19.59 %.

Q 8.36. In the organic compound CH2=CHCH2CH2CCH , the pair of hydridised orbitals involved in the formation of: C2C3 bond is:

&nbnbsp; (a)spsp2 (b)spsp3 (c)sp2sp3 (d)sp3sp3

Answer:

In the organic compounds-
CH2=CHCH2CH2CCH
1 2 3 4 5

In C2 and C3 carbon atom the hybridisation is sp2sp3
So, the correct option is (c)

Q 8.37. 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:

The Prussian blue colour is due to the formation of Fe4[Fe(CN)6]3.xH2O in Lassaigne's test.

6CN+Fe2+[Fe(CN)6]4

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

So, the correct option is (b)


Q 8.38. Which of the following carbocation is most stable?

(a)(CH3)3C.CH2 (b)(CH3)3C (c)CH3CH2CH2 (d)CH3CHCH2CH3
Answer:

We know the stability of carbocation order is-
3o>2o>1o>CH3+ (It is due to the tendency of the methyl group to electron release and stabilise the carbocation)

So, the correct option is (b)

Q 8.39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

Answer: Chromatography is the latest technique of separation and purification of organic compounds.
So, the correct option is (d)

Q 8.40. The reaction:
CH3CH2I+KOH(aq)CH3CH2OH+KI is classified as :

(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition

Answer:

CH3CH2I+KOH(aq)CH3CH2OH+KI
Here, OH attacks ethyl iodide and substitutes the iodide ion ion. OH has a lone pair of an electron, so it acts as a nucleophile. Hence it is a nucleophilic substitution reaction.
So, the correct option is (b)


NCERT Solutions For Class 11 Chemistry


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More About Class 11 Chemistry NCERT Chapter 8

Chapter 8 Some Basic Principles and Techniques is an important chapter for a chemistry student from the perspective of getting hold of basic concepts of organic chemistry which will eventually help in subsequent chapters. Since this is a theoretical chapter and many concepts and facts will be learned from exercise questions directly, hence It is must to go through NCERT solutions for Class 11 Chemistry Chapter 8 . In this chapter, there are a total of 40 questions in the exercise.
This chapter along with NCERT solutions for Class 11 Chemistry Chapter 8 Organic Chemistry Some Basic Principles and Techniques discusses some basic concepts in the reactivity and structure of organic compounds which are formed due to the covalent bonding, IUPAC nomenclature and classification of organic compounds, electromeric effect, inductive effect, hyperconjugation, and resonance. All related information about the heterolytic fission and homolytic fission of a covalent bond- carbocations, carbanions, free radicals, nucleophiles, and electrophiles is discussed in the later part of the NCERT syllabus chapter.

Organic compounds are essential for sustaining life's on the earth and include complex molecules like protein and DNA that constitutes important compounds of our blood, skin, and muscles. Some important areas of application of organic compounds are medicines, dyes, fuels, polymers, and clothing. These compounds can be classified on the basis of their functional group or the structure.

Topics of Organic Chemistry- Some Basic Principles and Techniques

8.1 General Introduction

8.2 Tetravalence of Carbon: Shapes of Organic Compounds

8.3 Structural Representations of Organic Compounds

8.4 Classification of Organic Compounds

8.5 Nomenclature of Organic Compounds

8.6 Isomerism

8.7 Fundamental Concepts in Organic Reaction Mechanism

8.8 Methods of Purification of Organic Compounds

8.9 Qualitative Analysis of Organic Compounds

8.10 Quantitative Analysis

What is a Functional Group?

A functional group is defined as an atom or group of atoms bonded together in a specific manner, which gives the chemical and physical properties of the organic compounds and they are the centres of the chemical reactivity.

Some important points of Chapter 8 of Class 11 Chemistry-

1. Organic compounds are formed due to covalent bonding.

2. According to orbital hybridization concept carbon can have sp3,sp2, and sp hybridized orbitals.

3. The 3D representation of organic compounds on paper can be drawn by dash and wedge formula.

4. Compounds having the same molecular formula but differ in their physical and chemical properties are known as isomers and the phenomenon is called isomerism.

5. Addition, substitution, elimination and rearrangement reactions are the types of organic reactions.

NCERT Solutions for Class 11- Subject-wise

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What are the important topics NCERT class 11 chemistry chapter 9 ?

- classification and IUPAC nomenclature of organic compounds

- Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyperconjugation

- Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles

- Types of organic reactions.

2. Where can I find complete solutions of NCERT class 11 Chemistry?

For complete solutions of NCERT, students can refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry 

3. What are homolytic and heterolytic fission in organic chemistry?

 In  Homolytic, the bond between two atoms breaks evenly. While in heterolytic fission bonds between atoms break unevenly.

4. What are nucleophiles in organic chemistry?

Nucleophiles are electron-rich species and they tend to donate electrons to electron-deficient molecules.

5. Why is the inductive effect important in organic reactions?

The inductive effect is important because it helps to explain how electrons move in a molecule. Due to this movement, some molecules have a positive charge while some have a negative charge

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Upcoming School Exams

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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