Class 11 - Chemistry - Organic Chemistry

Question 12.1

What are hybridization states of each carbon atom in the following compounds?

 

 Class_11_Chemistry_Organic_Chemistry_Equation_22

Question 12.2

Indicate the σ and π bonds in the following molecules:

Class_11_Chemistry_Organic_Chemistry_Equation_23

 Class_11_Chemistry_Organic_Chemistry_Equation_24

Class_11_Chemistry_Organic_Chemistry_Equation_25

Question 12.3

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

Answer.

 Class_11_Chemistry_Organic_Chemistry_Equation_21

Question 12.4

Give the IUPAC names of the following compounds:

Class_11_Chemistry_Organic_Chemistry_Equation_20

Answer.

  • Propylbenzene
  • 3-Methylpentanenitrile
  • 2,5-Dimethylheptane
  • 3-Bromo-3-chloroheptane
  • 3-Chloropropanal
  • 2,2-Dichloroethanol

 

 

Question 12.5

Which of the following represent the correct IUPAC name of the compounds concerned?

  • 2,2 – Dimethylpentane or 2 - Dimethylpentane
  • 2,4,7 – Trimethyloctane or 2,5,7 – Trimethyloctane
  • 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
  • But-3-yn-1-ol or But-4-ol-1-yne

Answer.

  • 2,2 – Dimethylpentane
  • 2,4,7 – Trimethyloctane. For 2 alkyl groups on the same carbon its locant is repeated twice.2,4,7-locant set is lower than 2,5,7.
  • 2-Chloro-4-methylpentane
  • But-3-yn-1-ol. Lower locant for the principal functional group, i.e. alcohol.

 

 

Question 12.6

Draw formulas for the first five members of each homologous seris beginning with the following compounds.

(a) H-COOH (b) CH3COCH3 (c) HCH=CH2

Answer.

 

Class_11_Chemistry_Organic_Chemistry_Equation_19 

Question 12.7

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

  • 2,2,4-Trimethylpentane
  • 2-Hydroxyl-1,2,3-propanetricarboxylic acid
  • Hexanedial

Answer.

 Class_11_Chemistry_Organic_Chemistry_Equation_18

 

Question 12.8

Identify the functional groups in the functional compounds.

Answer.

 

 Class_11_Chemistry_Organic_Chemistry_Equation_17

Question 12.9

Which of the two: O2NCH2CH2O- or CH3CH2O- is expected to be more stable and why?

Answer.

O2NCH2CH2O- is more stable than CH3CH2O- because NO2 has I-effect and hence, it tends to disperse the negative charge on oxygen atom.

CH3CH2 exerts a +Ieffect and it tends to intensify the negative charge and destabilize it.

 

  

Question 12.10

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

Answer.

Due to hyperconjugation, alkyl groups act as electron donor which when attached to pi-system.

Ncert_Solutions_Class_11_Chemistry_Organic_Chemistry_Diagram1

The hydrogen which is shown by dotted line is firmly attached.  

Question 12.11

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

Class_11_Chemistry_Organic_Chemistry_Equation_14

Answer.

  • Phenol

Class_11_Chemistry_Organic_Chemistry_Equation_15

 

Question 12.12

What are electrophiles and nucleophiles? Explain with 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+(Carbocations), Neutral molecules( due to the presence of a lone pair).

 

Question 12.13

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

Class_11_Chemistry_Organic_Chemistry_Equation_13

Answer.

Nucleophiles (OH) and  (CN) are (a) and (b)

Electrophiles(CH3C+O) are (c)

 

 

Question 12.14

Classify the following reactions in one of the reaction type studied in this unit.

Class_11_Chemistry_Organic_Chemistry_Equation_12

Answer.

  • Nucleophile substitution since bromine group gets substituted by –SH group.
  • Electrophilic addition reaction because two reactant molecules combines to form a single product.
  • Bimolecular elimination reaction because hydrogen and bromine are removed to form ethene.
  • Nucleophilic substitution because rearrangement of atoms takes place.

 

 

Question 12.15

What is the relationship between the members of following pairs of structures?

Are they structural or geometrical isomers or resonance contributions?

Class_11_Chemistry_Organic_Chemistry_Equation_11

Answer.

Structural isomers. They have the same molecular formula but have different structures.

  • These compounds differ in the position of the ketone group. For the first structure, it is in C-3 whereas, for the 2nd one, it is in C-2.
  • Geometrical isomers they have the same molecular formula, sequence of covalent bonds and the
  • same constitution but differ in the relative positioning of the atoms in space.
  • These compounds differ in the positioning of the Deutrium and Hydrogen.
  • Resonance contributors because they differ in the position of electrons but not atoms.

 

 

Question 12.16

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.

Class_11_Chemistry_Organic_Chemistry_Equation_10

Answer.

 

 Class_11_Chemistry_Organic_Chemistry_Equation_9

Question 12.17

Explain the terms Inductive and Electromeric effects.

Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

Class_11_Chemistry_Organic_Chemistry_Equation_8

Answer.

Inductive effect is polarization of sigma bond due to electron donating or withdrawing nature of group attached.

For example: Attachment of chlorine (which is electronegative atom) to ethane molecule causes carbons to carry a partial positive charge.

This is induced polarity due to presence of Cl atoms.

Class_11_Chemistry_Organic_Chemistry_Equation_7

Electromeric effect is the polarity produced in a multiple bonded compound as it is approached by a reagent.

Class_11_Chemistry_Organic_Chemistry_Equation_6

It is temporary effect and original condition is restored if the reagent is removed.

  • –I – effect

As the number of halogen atoms decreases, overall negative inductive effective decreases. Hence, the acid strength decreases.

  • + I – effect

As the number of alkyl groups increases, positive inductive effect increases and acid strength decreases.

 

 

Question 12.18

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

  • Crystallization (b) Distillation (c) Chromatography

Answer.

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 (boiling point = 457 K) and chloroform (boiling point = 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.

 

Question 12.19

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

Answer.

Fractional crystallization is used to separate two compounds with different solubilities in solvent.

A hot saturated solution of two compounds is allowed to cool.

Less soluble crystallizes out while the more soluble remains in the solution.

Crystals are separated from mother liquor and mother liquor is again concentrated and hot solution again

allowed to cool down when crystals of second compound are obtained. These are filtered and dried.

 

 

Question 12.20

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

Answer.

Distillation is used for those liquids which are sufficiently stable at their boiling points and contain non-volatile impurities.

Distillation under reduced pressure is used to purify liquid which have very

high boiling point and decompose at or below their boiling points.

Steam distillation is used to purify volatile liquids associated with water immiscible impurities.

 

 

Question 12.21

Discuss the chemistry of Lassaigne’s test.

Answer.

Lassaigne’s test tests for nitrogen, sulphur, halogens and phosphorous present in an organic compound.

Initially, all the compounds are converted to ionic form by fusing compound with sodium metal.

Class_11_Chemistry_Organic_Chemistry_Equation_5

Here, X = Cl, Br, I

Cyanide, sulphide or halide of sodium are extracted from fused mass by boiling with distilled water.

This is known as sodium fusion extract.

 

 

Question 12.22

Difference between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method

Answer.

  • Dumas method: In this method, organic compound is heated strongly with
  • excess of cupric oxide (CuO) in an atmospheric carbon dioxide when free nitrogen, CO2 and H2O are obtained.
  • Kjeldahl’s method: In this method a known mass of organic compound is heated strongly with concentrated sulphuric acid,
  • a little potassium sulphate and mercury. Mercury is a catalyst.
  • As a result, the nitrogen present in organic compound is converted to ammonium sulphate.

 

 

Question 12.23

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

Answer.

Estimation of halogens:

On heating an organic compound with fuming HNO3 in the presence of AgNO3,

the halogen present in the compound forms silver halide which is collected, filtered, washed, dried and weighed.

If, weight of organic compound = W grams

Weight of silver halide = x gram

Class_11_Chemistry_Organic_Chemistry_Equation_2

Estimation of sulphur:

The organic compound is first heated with fuming nitric acid.

It oxidizes sulphur to sulphuric acid which is precipitated as BaSO4 upon reaction with Ba(OH)2.

Class_11_Chemistry_Organic_Chemistry_Equation_3

Estimation of Phosphorous:

A known mass of organic compound is heated with fuming nitric acid.

Phosphorous present in the compound is oxidised to phosphoric acid.

Phosphoric acid is precipitated as ammonium phosphomolybdate, (NH4)3PO4.12MOO3, by the addition of ammonia and

ammonium molybdate solution which is then separated, dried and weighed.

 Class_11_Chemistry_Organic_Chemistry_Equation_4

 

Question 12.24

Explain the principle of paper chromatography.

Answer.

Paper chromatography is the simplest form of chromatography. A strip of paper acts as an adsorbent.

It is based on the principle of partial adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them.

This acts as stationary phase. Mobile phase is the mixture of the components to be identified prepared in s suitable solvent.

 

 

Question 12.25

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

Answer.

Nitric acid is added to sodium extract.

NaCN + HNO3 ---> NaNO3 + HCN

Na2S + 2HNO3 ---> 2NaNO3 + H2S

 

 

Question 12.26

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

Answer.

Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na2S, Na3PO4.

These ionic compounds become more reactive and can be easily tested by suitable reagents.

 

 

Question 12.27

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

Answer.

This is the technique of sublimation because camphor can sublime whereas CaSO4 does not.

 

 

Question 12.28

Explain, why an organic liquid vaporizes at a temperature below its boiling point in its steam distillation?

Answer.

Organic liquid vaporizes at a temperature below its boiling point in its steam distillation

because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

 

 

Question 12.29

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

Answer.

CCl4 is completely non-polar covalent compounds whereas AgNO3 is ionic in nature.

They do not react and thus white precipitates of silver chloride will not be formed. As CCl4 is a covalent 

compound, therefore, it does not ionize to give Clions needed for the formation of ppt. of AgCl.

 

 

Question 12.30

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during

the estimation of carbon present in an organic compound?

Answer.

Carbon dioxide is acidic in nature and it reacts with the strong base KOH to form K2CO3.

2KOH + CO2 --->K2CO3 + H2O

The increase in the mass of U-tube containing KOH then gives the mass of CO2 produced and from its mass, the percentage 

of carbon in the organic compound can be calculated as:-

%C = (12/44) x (Mass of COformed )/(Mass of substance taken) x 100

 

Question 12.31

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.

For testing sulphur, sodium extract is acidified with acetic acid.

This is because lead acetate is soluble and does not interfere with the test.

 

Pb(OCOCH3)2 + H2SO4 ---> PbSO4(ppt) + 2CH3COOH

If  H2SOwere used, lead acetate itself react with H2SO4 to form white ppt. of lead sulphate which will interfere 

with the test. 

 

Question 12.32

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.

Percentage of C = 69

Percentage of H = 4.8

Percentage of O = 100 – (69 + 4.8) = 26.2

Mass of compound = 0.2 g

100 g of compound contains 69 g of C.

0.2 g of compound contains

=(69 x 0.2)/(100)

= 0.138 g of C

12g of C produces 44 g of CO2.

0.138 g of C will give

=(44 x 0.138)/(12)

= 0.506 g of CO2

100 g of compound contains 4.8 g of H

Therefore, 0.2 g of compounds contains =(48 x 0.2)/(100)

= 0.0096 g of H

2 g of H produces 18 g of H2O

0.0096 g of H produces

=(4.8 x 0.2)/(100)

= 0.0864 g of H2O

Mass of CO2 produced = 0.506 g

Mass of H2O produced = 0.0864 g

 

 

Question 12.33

A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.

The ammonia evolved gas absorbed in 50 mL of 0.5 M H2SO4.

The residual acid required 60 Ml of 0.5 M solution of NaOH for neutralization.

Find the percentage composition of nitrogen in the compound.

Answer.

Mass of compound taken = 0.50 g

Volume of H2SO4 = 50 mL

Molarity of H2SO4 = 0.5 M

Volume of NaOH required = 60 mL

Required, Molarity of NaOH = 0.5 M

Using Kjeldahl’s method

% of N = [1.4 x Mx 2[V-(V1 /2)]/(m)

Put, M = 0.5 M

V = 50 mL

V1 = 60 mL

m = 0.5 g

% of N = [1.4 x 0.5 x 2(50-(60/2)]/(0.5)

= 56

Percentage of N in given compound = 56.

 

 

Question 12.34

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.

Mass of organic compound = 0.3780 g

Mass of silver chloride = 0.5740 g

1 mol of AgCl = 143.5 g

143.5 g of AgCl have 35.5 g of Cl

0.5740 g of AgCl have

= (35.5 x 0.574)/(143.5)= 0.142 g

0.142 g of Cl present in 0.3780 g of compound =[(0.142)/(0.378)] x 100 = 37.56%

 

 

Question 12.35

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 per centage of sulphur in the given compound.

Answer.

Mass of organic sulphur compound = 0.468 g

Mass of BaSO4 = 0.668 g

1 mole of BaSO4 = 233 g

233 g of BaSO4 conatisn 32 g of S.

0.668 g of BaSO4 contains

 =(0.668 x 32)/(233)

= 0.09174 g

Percentage of S in compound

=(0.09174)/(0.468)

= 19.6%

 

Question 12.36

In the organic compound CH2=CH-CH2-CH2-C=CH, the pair of hybridized orbitals involved in the formation of C2-C3 bond is:

(a) sp-sp2 (b) sp-sp3 (c) sp2-sp3 (d) sp3-sp3

Answer.

Correct option is (c)

 

Question 12.37

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

  • Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4

Answer.

Correct option is (b)

Prussian blue color in the Lassaigne’s test is due to the formation of Fe4 [Fe(CN)6]3

 

Question 12.38

Which of the following carbocation is most stable?

Class_11_Chemistry_Organic_Chemistry_Equation_1

Answer.

Correct option is (b)

Most stable is tertiary carbocation.

 

 

Question 12.39

The best and latest techniques for isolation, purification and separation of organic compounds is

  • Crystallization (b) Distillation (c) Sublimation (d) Chromatography

Answer.

Correct option is (d)

  

Question 12.40

The reaction

CH3CH2I + KOH (aq) --> CH3CH2OH + KI

Is classified as

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

Answer.

Correct option is (b). As the nucleophile I is replaced by the nucleophile OH- ion.

KOH provides OH- ion for nucleophile attack.

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