SELINA Solutions for Class 10 Chemistry Chapter 12 - Organic Chemistry

Chapter 12 - Organic Chemistry Exercise Intext 1

Solution 1

(a) Organic chemistry may be defined as the chemistry of hydrocarbons and its derivatives.

(b) Vital Force Theory is a theory made by the Scientist Berzelius in 1809 which assumed that organic compounds are only formed in living cells and it is impossible to prepare them in laboratories.

It was discarded because Friedrich Wohler showed that it was possible to obtain an organic compound(urea) in the laboratory.

Solution 2

(a) Few sources of organic compounds are:

Plants

Animals

Coal

Petroleum

Wood

(b) The various applications of organic chemistry is:

It is used in the production of soaps, shampoos, powders and perfumes.

Various fuels like natural gas, petroleum are also organic compounds.

The fabrics that we use to make various dresses are also made from organic compounds.

Solution 3

Organic compounds are present everywhere. They are present in:

It is present in the production of soaps, shampoos, powders and perfumes.

It is present in the food we eat like carbohydrates, proteins, fats, vitamins etc.

Fuel like natural gas, petroleum are also organic compounds.

Medicines, explosives, dyes, insecticides are all organic compounds.

Thus we can say that organic compounds play a key role in all walks of life.

Solution 4

The unique properties shown by carbon are:

Tetravalency of carbon

Catenation

Isomerism

Solution 5

(a) Tetravalency: Carbon can neither lose nor gain electrons to attain octet. Thus it shares four electrons with other atoms. This characteristics of carbon by virtue of which it forms four covalent bonds, is called Tetravalency of carbon.

In structural form :

(b) Catenation: The property of self -linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation.

Carbon- carbon bond is strong so carbon can combine with other carbon atoms to form chains or rings and can involve single, double and triple bonds.

Solution 6

Four properties of organic compound that distinguish them from inorganic compounds are:

(i) Presence of carbon.

(ii) Solubility in the organic solvents.

(iii) Forming of covalent bonds.

(iv) Having low melting and boiling points.

Solution 7

Due to the unique nature of carbon atom, it gives rise to formation of large number of compounds. Thus this demands a separate branch of chemistry.

Solution 8

Hydrocarbons are compounds that are made up of only carbon and hydrogen.

Comparison of saturated and Unsaturated hydrocarbons:

Saturated Hydrocarbon

Unsaturated Hydrocarbon

1. Carbon atoms are joined only by single bonds.

Carbon atoms are joined by double or by triple bonds.

2. They are less reactive due to the non-availability of electrons in the single covalent bond.

They are more reactive due to presence of electrons in the double or the triple bond.

3. They undergo substitution reaction.

They undergo addition reaction.

Solution 9

Due to presence of unique properties of carbon like Tetravalency, catenation and Isomerism large number of organic compounds are formed.

Solution 10

(a) Single Bond compound: For example: In pentane

 

(b) Double bond compound: For example:- In pentene

 

(c) Triple bond compound: In case of Hexyne:

 

Solution 11

(a) Cyclic compound with single bond: cyclopentane

Structure:

(b) Cyclic compound with triple bond: cyclopentyne

Structure:

Solution 12

The member of each of the following is:

(a) Saturated Hydrocarbon: Hexane (C6H14)

(b) Unsaturated Hydrocarbon: Hexene (C6H12)

Solution 13

Substitution reaction: A reaction in which one atom of a molecule is replaced by another atom (or group of atoms) is called a substitution reaction.

Addition reaction: A reaction involving addition of atom(s) or molecules(s) to the double or the triple bond of an unsaturated compound so as to yield a saturated product is known as addition reaction.

Chapter 12 - Organic Chemistry Exercise Intext 2

Solution 1

A functional group is an atom or a group of atoms that defines the structure (or the properties of a particular family) of organic compounds.

 

The structural formula of

(a) Halides :-R-X

Example:

(b)Alcohols:- R-OH

Example:

(c) Aldehydes:-R-CH=O

Example:

Solution 2

a. Butyne; its formula is C4H6.

b. Butanol; its formula is C4H9OH.

Solution 3

(i) Physical properties: The alkyl group determines the physical properties.

(ii) Chemical properties: The functional group is responsible for the chemical properties.

Solution 4

The alkyl radical and the functional group are:


Sr.No

Formula

Name of alkyl radical

Name of Functional group

a

CH3OH

Methyl

Alcohol

b

C2H5OH

Ethyl

Alcohol

c

C3H7CHO

Propyl

Aldehyde

d

C4H9COOH

Butyl

Carboxyl

e

CH3COOH

CH3

COOH

f

HCHO

H

CHO

Solution 5

 

(a) An alkyl group is obtained by removing one atom of hydrogen from an alkane molecule. Alkyl group is named by replacing the suffix 'ane' of the alkane with the suffix -yl.

(b) The name of three alkyl radicals are:

Methyl

Ethyl

Propyl

 

They are formed by removing 1 hydrogen from an alkane.

CH4-CH3+H+

Methyl

CH3-CH3 CH3-CH2 -+ H+

Ethyl

CH3-CH2-CH3 CH3-CH2-CH2 -+ H+

Propyl

Solution 6

The names and the structural formula of first three members of the homologous series of alkane are:

(i)

CH4Methane

(ii)

C2H6Ethane

(iii)

C3H8Propane

Solution 7

(a) A homologous series is a group of organic compounds having a similar structure and similar chemical properties in which the successive compounds differ by a CH2 group.

(b) The difference in molecular formula of any two adjacent homologues is

(i) It differs by 14 a.m.u in terms of molecular mass.

(ii) It differs by three atoms. The kind of atoms it differs is one carbon and two hydrogen.

Chapter 12 - Organic Chemistry Exercise Ex. 12A

Solution 1(a)

2,2 dimethyl propane

Solution 1(b)

2-methyl butane

Solution 1(c)

Propene

Solution 1(d)

2,2-dimethyl pentane

Solution 1(e)

3-Pentyne

Solution 1(f)

3-methyl but-1-yne

Solution 1(g)

2,3-dimethyl pentane

Solution 1(h)

3-methyl heptane

Solution 1(i)

2-Butene

Solution 1(j)

Hept-2-yne

Solution 1(k)

5,5-dimethyl hexan-1-al

Solution 1(l)

Pentan-2-ol

Solution 1(m)

4-methyl pentan-1-oic acid

Solution 1(n)

2-bromo-2-methyl butane

Solution 1(o)

1-bromo-3-methyl butane

Solution 1(p)

Prop-1-yne

Solution 1(q)

Methanal

Solution 1(r)

1-propyne

Solution 1(s)

Propanol

Solution 1(t)

Ethanoic acid

Solution 1(v)

1,2-dichloroethane

Solution 2

The structure of the following compounds are:

(a) Prop-1-ene

CH3-CH=CH2

(b) 2,3-dimethylbutane

CH3-CH(CH3)-CH(CH3)-CH3

(c) 2-methylpropane

CH3-CH(CH3)-CH3

(d) 3-hexene

CH3-CH2-CH=CH-CH2-CH3

(e) Prop-1-yne

CH3-C?CH

(f) 2-methylprop-1-ene

CH3-C(CH3)=CH2

(g) Alcohol with molecular formula C4H10O

CH3-CH2-CH2-CH2-OH

Solution 3
  1. (iv) CnH 2n + 1 is the formula for an alkyl group. Hence it is C5H11.
  2. (i) A hydrocarbon of general CnH2n is C15H30
    As the formula of Alkene is CnH2n. Thus n + 2n = 72 3n = 72 n = 24. By filling value we get the molecular mass 72.
  3. (iv) The total number of carbon chains that four carbon atoms form an alkane is 2.
  4. (iv) Alcohol and ether are functional isomers as they have same molecular formula but different functional groups.
  5. (ii) The IUPAC name of this compound is: 3-methyl hexane.
Solution 4

(a) Propane and ethane are homologues.

(b) A saturated hydrocarbon does not participate in a/an addition reaction.

(c) Succeeding members of a homologous series differ by CH2.

(d) As the molecular masses of hydrocarbons increase, their boiling points Increase and melting point increase.

(e) C25H52 and C50H102 belong to the same homologous series.

(f) CO is an organic Compound.

(g) The chemical properties of an organic compound are largely decided by the functional group and the physical properties of an organic compound are largely decided by the number of carbon atoms. 

(h) CHO is the functional group of an aldehyde.

(i) The root in the IUPAC name of an organic compound depends upon the number of carbon atoms in Principal Chain.

(j) But-1-ene and but-2-ene are examples of position isomerism.

Solution 5

Chain isomerism

Chain isomerism arises due to the difference in arrangement of C atoms in the chain. For example, there are two isomers of butane, C4H10. In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.

Position isomerism

It is due to the difference in position of functional groups.

For example, there are two structural isomers with the molecular formula C3H7Br. In one of them, the bromine atom is on the end of the chain, whereas in the other it is attached in the middle.

Solution 6

(a)Isomerism: Compounds having the same molecular formula but different structural formula are known as isomers and the phenomenon as isomerism.

Two main causes of isomerism are:

Difference in mode of linking of atoms.

Difference in the arrangement of atoms or groups in space.

(b) 

(c)  

CH2=CHCH2CH3               H3C-CH=CHCH3

1-butene                              2-butene

Solution 7

(i) Ethane undergoes substitution reactions.

(ii) Ethene undergoes addition reactions.

Solution 8

a. 

i.  (Ethene)

 In the above structure, both carbons are bonded

 with double bonds.

ii.   (Ethyne)

 In the above structure, both carbons are bonded

 with triple bonds.

b. Addition reactions are common to both these compounds. Methane does not undergo this type of reaction because it is bounded with four hydrogen atoms, while in ethane, double bonds break and provide a site for addition.

(c) Methoxymethane

Solution 9

a. 

 

b. 

 

 

c. 

 

 

d. 

 

 

e. 

 

 

f. 

Solution 10

The alkanes form an (a) electrochemical homologous series with the general formula (b) CnH2n+2. The alkanes are (c) saturated (d) hydrocarbons which generally undergo (e) substitution reactions.

Solution 11

 a.

 

 

 b.

 

 c.

 

Solution 12

a. Ethanol

b. Ethanoic acid

c. Ethene

Solution 13

a. Propanal

b. Propanol

Solution 14

The homologous series of hydrocarbons are:

 

General Formula

CnH2n

CnH2n-2

CnH2n+2

IUPAC name of the homologous series

Alkenes

Alkynes

Alkanes

Characteristics bond type

Double bond

Triple Bond

Single Bond

IUPAC name of the first member of the series

Ethene

Ethyne

Methane

Type of reaction with chlorine

Addition

Addition

Substitution

 

Solution 15(a)

Alkenes are the (a) homologous series of (b) unsaturated hydrocarbons. They differ from alkanes due to presence of (c)single bonds. Alkenes mainly undergo (d) addition reactions.

Solution 15(b)

The organic compound which undergoes substitution reaction is C2H6.

Solution 15(c)

(c) Structural formulae of isomers ofButane are:

Butane2-methyl propane

Solution 16
  1. Ethane
  2. Propanol
  3. Ethyne 
Solution 1(u)

Ethanal 

Chapter 12 - Organic Chemistry Exercise Ex. 12B

Solution 1

Sources of alkane:

The principal sources of alkanes are Natural gas and petroleum.

Solution 2

Methane is a primary constituent of natural gas. It absorbs outgoing heat radiation from the earth, and thus contributes to the green house effect and so it is considered as a green house gas.

Solution 3

The general formula of alkane is :

CnH 2n+2

Solution 4

(a)The structures of isomers of butane are:

(i)

Common name:- n-Butane

IUPAC name:- Butane

(ii)

Common name:-iso butane

IUPAC name:- 2-methyl propane

(b) The structures of isomers of Pentane are:

(i)

Common name: n-pentane

IUPAC name:- Pentane

(ii)

Common name:- iso pentane

IUPAC name:- 2-methyl butane

(iii)

Common name- neo pentane

IUPAC name:- 2,2-dimethyl propane

Solution 5

For methane:

(a) Molecular formula is CH4

(b) Electron dot formula

(c) Structural formula

For ethane:

(a) Molecular formula is :- C2H6

(b) Electron dot formula:

(a) Structural Formula:

Solution 6

(a) Laboratory preparation of methane:

When the mixture of sodium ethanoate and soda lime is taken in a hard glass test tube and heated, the gas evolved is methane. It is collected by downward displacement of water.

CH3COONa+NaOHNa2CO3+CH4

(b) Laboratory preparation of ethane:

When the mixture of sodium propionate and soda lime is taken in the boiling tube and heated the ethane gas is evolved. It is also collected by downward displacement of water.

C2H5COONa+NaOHNa2CO3+C2H6

Solution 7

When methyl iodide is reduced by nascent hydrogen at ordinary room temperature then methane is formed.

CH3I+2[H] CH4+HI

When bromoethane is reduced by nascent hydrogen at ordinary room temperature then ethane is produced.

C2H5Br+2[H] C2H6+HBr

Solution 8

A reaction in which one atom of a molecule is replaced by another atom (or group of atoms)is called a substitution reaction.

When ethane reacts with chlorine

C2H6 +Cl2 C2H5Cl + HCl

Chloroethane

C2H5Cl + Cl2C2H4Cl2+HCl

Dichloroethane

C2H4Cl2 +Cl2 C2H3Cl3+HCl

Trichloroethane

C2H3Cl3 + Cl2 C2H2Cl4 + HCl

Tetrachloroethane

C2H2Cl4 +Cl2 C2HCl5 +HCl

Pentachloroethane

C2HCl5 +Cl2 C2Cl6 + HCl

Hexachloroethane

Solution 9

(a) Sufficient air: When methane burns in sufficient air, then carbon dioxide and water vapors are formed.

CH4 + 2O2 CO2+2H2O

(b) Insufficient air: When methane burns in insufficient air , then carbon monoxide and water is formed.

2CH4 + 3O2 2CO + 4H2O

Solution 10

(a)

(i) When methane reacts with chlorine in the presence of sunlight or UV light, it undergoes substitution reaction to form Tetrachloromethane.

(ii) When it reacts with bromine it forms Tetrabromomethane

CH4 + Br2CH3Br + HCl

CH3Br + Br2CH2Br2 + HCl

Dibromomethane

CH2Br2 + Br2CHBr3 + HCl

Tribromo methane

CHBr3 + Br2CBr4 + HCl

Tetrabromomethane

(b)

(i) When ethane reacts with chlorine it forms hexachoroethane.

C2H6 +Cl2 C2H5Cl + HCl

Chloroethane

C2H5Cl + Cl2 C2H4Cl2+HCl

Dichloroethane

C2H4Cl2 +Cl2 C2H3Cl3+HCl

Trichloroethane

C2H3Cl3 + Cl2 C2H2Cl4 + HCl

Tetrachloroethane

C2H2Cl4 +Cl2 C2HCl5 +HCl

Pentachloroethane

C2HCl5 +Cl2 C2Cl6 + HCl

Hexachloroethane

(ii) When ethane reacts with bromine it forms Hexabromoethane

C2H6 +Br2 C2H5Br + HBr

Bromoethane

C2H5Br + Br2 C2H4Br2+HBr

Dibromoethane

C2H4Br2 +Br2 C2H3Br3+HBr

Tribromoethane

C2H3Br3 + Br2 C2H2Br4 + HBr

Tetrabromoethane

C2H2Br4 +Br2 C2HBr5 +HBr

Pentabromoethane

C2HBr5 +Br2 C2Br6 + HBr

HexaBromoethane

Solution 11

(a) Ethane is prepared from sodium propionate.

C2H5COONa+NaOHNa2CO3+C2H6

(b) Methane is prepared from methyl iodide.

CH3I+2[H] CH4+HI

(c) Ethane is prepared from ethyl bromide.

C2H5Br+2[H] C2H6+HBr

Solution 12

(i) 

(ii)

Solution 13

(a) Methane into chloroform

CH4+Cl2CH3Cl+HCl

CH3Cl+Cl2 CH2Cl2+HCl

CH2Cl2+Cl2 CHCl3+HCl

 

(b) Sodium acetate into methane

 

CH3COONa+NaOHNa2CO3+CH4

(c) Methyl iodide into ethane

2CH3I +2NaCH3-CH3+2NaI

 

(d) Methane to methyl alcohol

 

Solution 14

(a) Methane: Three uses of methane are:

(i) Methane is a source of carbon monoxide and hydrogen

(ii) It is used in the preparation of ethyne, methanal, chloromethane, carbon tetrachloride.

(iii) It is employed as a domestic fuel.

(b) Ethane:

Three uses of ethane are:

(i) It is used in the preparation of ethene, ethanol, and ethanol.

(ii) It forms ethyl chloride, which is used to make tetraethyllead.

(iii) It is also a good fuel.

Solution 15

(a) When a mixture of ethane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

2C2H6 +O22C2H5OH

(b) When mixture of ethane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Acetaldehyde.

C2H6 +O2 CH3CHO+H2O

(c) Ethanol formed from ethane gets oxidized to acetic acid.

2C2H6 +O22C2H5OH

C2H5OH + O2 CH3COOH+H2O

Solution 16

Ethane can be oxidized as follows:

When a mixture of ethane and oxygen in the ratio 9:1 by volume is compressed to about 120 atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.

2C2H6 + O2 2C2H5OH

When a mixture of ethane and oxygen is passed through heated MoO, the mixture is oxidized to ethanal.

C2H6+ O2 CH3CHO + H2O

When a manganese based catalyst is used 100oC, ethane can be oxidized to ethanoic acid.

2C2H6 + 3O2 2CH3COOH + 2H2O

Chapter 12 - Organic Chemistry Exercise Ex. 12C

Solution 1

(a) The molecular formula of ethene is C2H4

(b) Electron dot formula of ethene is:

(c) Structural formula of ethene:

Solution 2

(a) n signifies the number of carbon atoms and 2n signifies the number of hydrogen atoms.

(b) The name of alkene when n=4 is Butene.

(c) The molecular formula of alkene when n=4 is C4H8.

(d) The molecular formula of alkene when there are 10 H atom in it C5H10.

(e) The structural formula of the third member of alkene is

(f) Lower homologus of alkene which contain four carbons is C3H6.

Higher homologus of alkene which contain four carbons is C5H10.

Solution 3(a)

Ethane

Ethene

It has carbon -carbon single bond.

It has carbon-carbon double bond

It is saturated.

It is unsaturated

Alkanes undergo substitution reaction.

Alkenes undergo addition reaction.

Solution 3(b)

There are two isomers are possible for butane:

Solution 4

Balanced Equation of ethylene:

CH3-CH2OH + H2SO4 CH3-CH2HSO4+H2O

CH3-CH2HSO4CH2=CH2

The gas is collected by downward displacement of water.

Solution 5

(a) Dehydrohalogenation reaction:

C2H5Cl + KOH(alc.and hot) C2H4 + KCl + H2O

Ethene

(b) Dehydration reaction:

C2H5OH C2H4+H2O

Ethene

Solution 6(a)

Chlorine and bromine are added to the double bond of ethene to form saturated ethylene chloride and ethylene bromide respectively.

CH2 = CH2 + Cl2 CH2(Cl)-CH2(Cl)

1,2-dichloro ethane

 

CH2 = CH2 + Br2 CH2(Br)-CH2(Br)

1,2-dibromo ethane

 

Solution 6(b)

When ethene and hydrogen are passed over finely divided catalyst such as platinum or palladium at ordinary temperature or nickel at 200o C, the two atom of hydrogen molecule are added to the unsaturated molecule, which thus becomes a saturated one.

 

C2H4 +H2 C2H6

Solution 7

Conversion of ethanol to ethene by using

(a) Solid dehydrating agent:

 

(b) Hot conc. H2SO4:

Solution 8

(a) Physical state: Ethene is a colourless and inflammable gas.

(b) Odour: It has faint sweetish odour.

(c) Density as compared to air: It has density less than one hence it is lighter than air.

(d) Solubility: It is sparingly soluble in water but highly soluble in organic solvents like alcohol, ether and chloroform.

Solution 9

(a) Ethene into 1, 2 -dibromoethane: Ethene reacts with bromine at room temperature to form saturated ethylene chloride.

CH2=CH2 + Br2 CH2(Br)-CH2(Br)

1,2-dibromo ethane

(b) Ethene into ethyl bromide: When ethene is treated with HBr bromoethane is formed.

CH2=CH2 + HBr CH3-CH2Br

Ethyl bromide

Solution 10

(a) C2H4+3O2 2CO2 +2H2O + heat

(b) CH2=CH2+Cl2 CH2(Cl)-CH2(Cl)

(c) CH2=CH2 + HCl CH3-CH2-Cl

(d) C2H4 +H2 C2H6

Solution 11

(a) CH4CH3ClCH2Cl2CHCl3CCl4

A= monochloromethane

B= dichloromethane

C=Trichloromethane

D=Tetrachloromethane

(b) C2H2C2H4C2H6C2H5BrC2H4Br

A= Ethene

B=ethane

C=bromoethane

D=dibromoethane

(c) C2H4 +H2C2H6

B= hydrogen

Solution 12

(a) C2H4 +Cl2 CH2(Cl)-CH2(Cl)

1,2- dichloro ethane

(b) C2H5Br +KOH (alc.) C2H4 +KBr +H2O

Ethane

(c) CH2=CH2 CH2(OH)-CH2(OH)

1,2- Ethanediol

(d) CH2=CH2+HBr CH3-CH2Cl

chloroethane

Solution 13

When ethylene is passed through alkaline KMnO4 solution 1, 2-Ethanediol is formed. The Purple color of KMnO4 decolorizes.

 

CH2=CH2+H-O-H +[O] CH2(OH)-CH2(OH)

Cold alkaline

KMnO4 solution

Solution 14

Three compounds formed by ethylene are:

 

Polythene

Ethanol

Epoxyethane

 

Uses of above compounds:

 

Polythene is used as carry bags.

Ethanol is used as a starting material for other products, mainly cosmetics and toiletry preparation.

Epoxyethane is used in the manufacture of detergents.

Chapter 12 - Organic Chemistry Exercise Ex. 12D

Solution 1

Natural gas and Petroleum are sources for alkynes.

The general formula of alkynes are:

CnH2n-2

Solution 2

Butyne is an example, its isomers are:

IUPAC name: But-2-yne But-1-yne

Solution 3

(a) Diagram of acetylene preparation:

(b) CaC2 +2H2O Ca(OH)2 +C2H2

(c) The pure dry gas is collected by downward displacement of water, since it is insoluble in water.

Solution 4

When 1,2 -dibromoethane is boiled with alcoholic potassium hydroxide ,ethyne is formed.

Solution 5

(a) The hydrocarbon which is tetrahedral is Methane.

(b) The hydrocarbon which is planar molecule is ethene.

(c) The hydrocarbon which is a linear molecule is Ethyne.

(d) The hydrocarbon which forms a red precipitate with ammoniacal solution of copper chloride is acetylene.

(e) Alkanes are also called as paraffin.

(f) Alkenes are also called olefin.

(g) Calcium carbide

Solution 6

The following compounds can be classified as:

C3H4:- Alkynes

C3H8:- Alkanes

C5H8:- Alkynes

C3H6:- Alkenes

Solution 7

Chemical test to distinguish :

(b) Ethane and ethene:

 

S.No.

Test

Ethane

Ethene

1.

On adding a few drops of bromine solution in carbon tetrachloride to the hydrocarbon

No change is observed

The reddish brown colour gets decolorized

2.

On adding a few drops of alkaline potassium permanganate (purple colour) to the hydrocarbon

No change is observed

The purple colour fades.

 

 

(c) Ethene and ethyne:

S.No.

Test

Ethene

Ethyne

1.

On adding a few drops of ammonical cuprous chloride to the hydrocarbon

No change is observed

Red precipitate of copper acetylide is formed

2.

On adding ammonical silver nitrate

No observation

White precipitate of silver acetylide is formed.

 

Solution 8

(a) HC≡CH

(b) Brown colour of CCl4 disappeared due to formation of addition product, i.e. 1, 2-dibromo ethane.

Solution 9
  1. CH2 =CH2  + H2   CH3 -CH3
  2. CH2 =CH2  + H2O  CH3 -CH2-OH
  3. CH ≡CH + H2   CH2 =CH2  
Solution 10

(a) Ethyne in an inert solvent of carbon tetrachloride adds chlorine to change into 1,2-dichloro ethene with carbon-carbon double bond, and then to an 1,1,2,2-tetrachloro ethane with carbon-carbon single bond.

C2H2C2H2Cl2C2H2Cl4

1,2-dichloro ethene1,1,2,2 -tetrachloro ethane

 (b) Ethyne in an inert solvent of carbon tetrachloride adds bromine to change into 1,2-dibromo ethene and then to 1,1,2,2 -tetrabromo ethane .

 C2H2C2H2Br2C2H2Br4

(c) Iodine reacts slowly in the presence of alcohol to form di-iodo ethene

CHCH +I2 ICH=CHI

1,2-di-iodoethene

(d) In the presence of nickel, platinum or palladium ethyne change to ethene and then to ethane.

CHCH CH2=CH2CH3-CH3

(e) begin mathsize 12px style table attributes columnalign left end attributes row cell straight C subscript 2 straight H subscript 2 plus HCl rightwards arrow with blank on top CH subscript 2 CHCl rightwards arrow with plus HCl on top CH subscript 3 CHCl subscript 2 end cell row cell ethyne space space Chloro space space ethane space space space 1 comma 1 minus dichloro space ethane end cell end table end style

Solution 11

Substitution reactions are characteristic reactions of alkanes.

Solution 12

  (a)  i. begin mathsize 12px style table attributes columnalign left end attributes row cell CaC subscript 2 plus 2 straight H subscript 2 straight O rightwards arrow with blank on top Ca left parenthesis OH right parenthesis subscript 2 plus straight C subscript 2 straight H subscript 2 straight H upwards arrow end cell row cell Calcium space space space space space space space space space space space space space space space space space space Calcium space acetylene end cell row cell carbide space space space space space space space space space space space space space space space space space space space hydroxide end cell end table end style

ii. begin mathsize 12px style table attributes columnalign left end attributes row cell straight C subscript 2 straight H subscript 5 Br plus NaOH rightwards arrow with boil on top straight C subscript 2 straight H subscript 5 OH plus NaBr end cell row cell Ethyl space left parenthesis aq. right parenthesis space space space space space space space space space space space space space space space space space ethyl end cell row cell bromide space space space space space space space space space space space space space space space space space space space space space alcohol end cell end table end style

(b) When bromine in carbon tetrachloride is added to ethyne, the orange colour of the bromine disappears due to the formation of the colourless ethylene bromide.

(c) Water reacts with ethene to form ethanol.

CH2=CH2 +H2OC2H5OH


 
 
  

Solution 13

(a) Ethyne is a highly reactive compound than ethene because of the presence of a triple bond between its two carbon atoms.

(b) Ethene is a highly reactive compound than ethane because of the presence of a double bond between its two carbon atoms.

(c) Hydrocarbons such as alkanes undergo combustion reactions with oxygen to produce carbon dioxide and water vapour. Alkanes are flammable which makes them excellent fuels.

Methane for example is the principal component of natural gas.

CH4 + 2O2 → CO2 + 2H2O

Solution 14

a. 

i. 

ii. 

b. 

i.   

ii. 

  

iii. 

 

iv. 

 

Solution 15

(a)Preparation of carbon tetrachloride from methane:

 

CH4+Cl2 CH3Cl +HCl

CH3Cl + Cl2 CH2Cl2 +HCl

CH2Cl2 +Cl2 CHCl3 +HCl

CHCl3 + Cl2 CCl4 +HCl

 

(b)Structural formula of ethyne:

 

 

(c)Alkynes contain triple bond where as alkenes contain double bond.

Chapter 12 - Organic Chemistry Exercise Ex. 12E

Solution 1

a. ii. They can undergo addition as well as   substitution reactions.

b. ii. Ethanol.


c. C5H10 is an alkene, while the rest are alkanes.

Solution 2

i. 

ii. 

iii. 

iv. 

Solution 3

i. By treating ethyl chloride with aqueous KOH 

ii. By heating ethyl chloride with alcoholic KOH 

iii. By passing ethene into conc. H2SO4 at 80°C and high pressure

iv. By heating ethyl alcohol with conc. H2SO4 at 170°C

Solution 4

i. Isomerism: Compounds having the same molecular formula but different structural formulae are known as isomers, and the phenomenon is known as isomerism.

ii. The IUPAC name of the branched isomer of butane is 2-methyl propane

Solution 5

(a)The melting and boiling point of the successive members of the homologous series of alcohols increase with the increase in molecular mass.

(b)When ethanol reacts with acetic acid ethyl acetate is formed.

 C2H5OH + CH3COOH CH3COOC2H5 + H2O

(c) This reaction is known as esterification reaction.

Solution 6

(a) CHCH +H2 CH2=CH2 +H2 CH3-CH3

(b) C2H4 + Cl2 CH2(Cl)-CH2(Cl)

(c) C2H4 + HCl CH3-CH2Cl

(d) CaC2 + 2H2O C2H2+Ca(OH)2

(e) C2H2 + Br2 H(Br)C=C(Br)H

(f) C2H5OH CH3CHO

Solution 7

Ethanol affects that part of the brain which controls our muscular movements and then gives temporary relief from tiredness. But it damages the liver and kidney too.

Solution 8

(a) Absolute alcohol: Absolute alcohol may be obtained by distilling moist alcohol with benzene. The mixture of water and benzene distills off and anhydrous alcohol is left behind.

(b) Spurious alcohol: It is made by improper distillation. It contains large portions of methanol in a mixture of alcohols.

(c) Methylated spirit: Methylated spirit or denatured alcohol is ethyl alcohol with 5%methyl alcohol, a coloured dye and some pyridine.

Solution 9

(a) Sodium reacting with ethyl alcohol:

2C2H5OH + 2Na 2C2H5ONa + H2

When sodium reacts with ethyl alcohol hydrogen is evolved with formation of sodium ethoxide.

 

(b) Ethanol oxidized by K2Cr2O7:

C2H5OHCH3CHO+H2OCH3COOH

Alcohols gets oxidized and get converted into ethanal and then into acetic acid.

Solution 10

S No

Formula

Common Name

IUPAC

1

C3H6

Propylene

Propene

2

C2H4

Ethylene

Ethene

3

C2H2

Acetylene

Ethyne

4

CH3OH

Methyl alcohol

Methanol

5

C2H5OH

Ethyl alcohol

Ethanol

Solution 11

C2H5OHCH3CHO+H2OCH3COOH

The oxidizing agents that can be used are potassium dichromate and potassium permanganate.

Solution 12
  1. Used for illuminating country houses : Ethyne
  2. Used for making a household plastic material: ethyne
  3. Called 'wood spirit' : Methanol
  4. Poisonous: Methanol
  5. Consumed as a drink: Ethanol
  6. Made from water gas: Methanol
  7. Ethanol
  8. Ethyl alcohol
Solution 13

(a) The conversion of ethanol into ethene is an example ofDehydration.

(b) Converting ethanol into ethene requires the use of Conc. H2SO4.

(c) The conversion of ethene into ethane is an example of hydrogenation.

(d) The catalyst used in the conversion of ethene into ethane is commonly nickel.

Solution 14

(a)Ethane from sodium propionate

C2H5COONa + NaOH Na2CO3 + C2H6

 

(b)Ethene from iodoethane

C2H5 I +KOH(alcoholic) C2H4 +KI + H2O

 

(c)Ethyne from calcium carbide

CaC2 +2H2O Ca(OH)2 + C2H2

 

(d)Methanolfrom iodoethane

CH3I + NaOH CH3OH + NaI

Solution 15

(i) C2H5COONa +NaOHNa2CO3 + C2H6

(ii) CH3I +2[H] CH4 +HI

(iii) C2H5Br + KOH C2H4 +KBr +H2O

(iv) CO + 2H2 CH3OH

(v) CaC2 +2H2O Ca(OH)2 +C2H2

Solution 16

(a) Calcium carbide and water:

CaC2 + 2H2O Ca(OH)2 + C2H2

(b) Ethene and water:

CH2=CH2 +H2O C2H5OH

(c) Bromoethane and aqueous solution of sodium hydroxide

C2H5Br +NaOH C2H5OH +NaBr

Chapter 12 - Organic Chemistry Exercise Ex. 12F

Solution 1

An organic compound containing the carboxyl group(COOH) is known as carboxylic acid.

The general formula: CnH2n+1COOH

Solution 2

(a) First three members of carboxylic acids are:

Methanoic acid

Ethanoic acid

Propanoic acid

(b) Three compounds that can be oxidized directly or in stages to produce acetic acid are:

Ethanol

Acetylene

Ethanal

Solution 3

Structural formula of acetic acid:

IUPAC name of acetic acid is:

Ethanoic acid

Glacial acetic acid is the pure form of acetic acid. It does not contain water.

Solution 4

Vinegar commonly called Sirka is a dilute solution of acetic acid. The presence of colouring matter gives it a greyish colour while the presence of some other organic acids and organic compounds impart it the usual taste and flavour.

Solution 5

(a) Ethanol

(b) Acetic acid

(c) Propanoic acid

Solution 6

(a)It is prepared in the lab by the oxidation of ethanol with acidified potassium dichromate.

C2H5OHCH3CHO CH3COOH

(b)Acetylene is first converted to acetaldehyde by passing through 40% H2SO4 at 60°C in the presence of 1% HgSO4.

The acetaldehyde is then oxidised to acetic acid in the presence of catalyst manganous acetate at 70°C.

C2H2  + H2begin mathsize 14px style rightwards arrow from HgSO subscript 4 to space space space space straight H 2 SO 4 left parenthesis dil right parenthesis space space space space of end style CH3CHO

CH3CHO + O2 begin mathsize 14px style rightwards arrow from space space space space Catalyst space space space space space to space space space increment space space space space of end style 2CH3COOH

 

Solution 7

(a) When acetic acid reacts with litmus it turns blue litmus red.

(b) When acetic acid reacts with metals hydrogen is evolved.

2CH3COOH + Zn (CH3COO)2Zn + H2

(c) When acetic acid reacts with alkalies it forms salt

CH3COOH + NaOH CH3COONa + H2O

(d) Acetic acid reacts with alcohols forming esters

CH3COOH + C2H5OH CH3COOC2H5 + H2O

Solution 8

(a) 2CH3COOH + Zn (CH3COO)2Zn + H2

(b) CH3COOH + NaOH CH3COONa + H2O

(c) 2CH3COOH + Na2CO3 2CH3COONa +H2O+ CO2

(d) CH3COOH + NaHCO3 CH3COONa + H2O +CO2

Solution 9

(a) When acetic acid is added to sodium bicarbonate, carbondioxide is liberated.

CH3COOH + NaHCO3 CH3COONa + H2O +CO2

(b) When acetic acid is added to ethyl alcohol in presence of sulphuric acid ester (ethyl acetate) is formed.

CH3COOH + C2H5OH CH3COOC2H5 + H2O

(c) When acetic acid is added to neutral FeCl3, wine red color is produced.

Solution 10

(a) When acetic acid and ethanol react it results in the formation of ethyl acetate.

(b) Lithum aluminium hydride(LiAlH4) is used to convert acetic acid to ethanol.

(c) Phosphorous pentoxide(P2O5) is heated along with acetic acid to form acetic anhydride.

Chapter 12 - Organic Chemistry Exercise Misc. Ex.

Solution 5

(a) X = Ethanol, C2H5OH

 Y = Acetic acid, CH3COOH

 Z = Conc. H2SO4

(b)  

(c) Compound P is an ester named ethyl acetate.

(d) Esterification

(e) 

Solution 2010

a. iii. Ethyne

b. i. Formic acid

c. i. Methanol

d. i.   

ii    

e.  

i.   

ii. Addition reaction

iii. Bromine solution gets decolourised

iv. Ethanol 

v. By heating it (ethanol) with concentrated sulphuric acid at 170°C

Solution 2011

a. iv. Carboxyl - COOH

b. iii. An addition reaction

c. iii Six

d. 

i. Nickel

 ii.  Acetic acid

 iii. Esterification

 iv.   

  

 v.  Ethanol

 

e.  

i. 

ii.   

iii.   

iv.  CaC2 + 2H2O Ca(OH)2 + C2H2 

v. 2C2H5OH + 2Na 2C2H5ONa + H2 

Solution 2012

a. i.

  

ii.

 

 

 iii.   

iv. 

 

 

v.

 

 

b.  i.  Ethyne

 ii.  Acetic acid

 iii. Ethene

 iv Ethanol

 

c. i. Pure acetic acid is called glacial acid because it forms an ice-like solid when cooled.

ii.   

Solution 2013

a. 

i. Ethene gas decolourises the purple colour of KMnO4, whereas ethane does not decolourise KMnO4 solution.

 ii. They can undergo both substitution as well as addition reactions.

b. 

 i. Isomer of n-butane: Isobutane

  

 ii. 2-propanol

 

Solution 2014

a. (iv) ethyne

b. Ketones

c.  

  

d. Ethene gas decolourises the purple colour of KMnO4, whereas ethane does not decolourise KMnO4 solution.

 

e. 

 i. Ethanol:

   

 ii. 1-propanal:

  

 iii. Ethanoic acid:

 

 iv. 1, 2-dichloroethane:

   

f.  

 The balanced chemical equation for the preparation

 of ethanol from monochloroethane and aqueous

 sodium hydroxide:

 C2H5-Cl + NaOH (aq.)   C2H5OH + NaCl

Solution 2015

(a)  

(i) Ethanoic acid to ethyl ethanoate

 CH3COOH + C2H5OH    CH3COOC2H5 + H2O

(ii) Calcium carbide to ethylene

 CaC2 + 2 H2O  Ca(OH)2 + CH º CH

(iii) Sodium ethanoate to methane

CH3COONa + NaOH   Na2CO3 + CH4

(b)  

(i) Dimethyl ether

   

(ii) Propanone 

 

(c) Name the following:

(i) Hydrogenation

(ii) Methane

(iii) Esterification 

(iv) Catenation

(v) Dehydrohalogenation