Organic Chemistry Class-10 Goyal Brothers ICSE Solutions Ch-11

Organic Chemistry Class-10 Goyal Brothers ICSE Solutions Ch-11. Step by Step Solutions of Exercise and Objective Type Questions of Goyal Brothers Prakashan Chapter-11 Organic Chemistry for ICSE Class 10 .

Organic Chemistry –  Test , use, observation seen , preparation , properties (physical and chemical) of Organic Compound explain with suitable chemical reaction. Visit official Website CISCE  for detail information about ICSE Board Class-10 Chemistry .

Organic Chemistry Class-10 Goyal Brothers ICSE Solutions Ch-11

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Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Objective Type Questions

Exercise 1 Page –191

Organic Chemistry Class-10 Goyal Brothers ICSE Solutions Ch-11

Question 1. What is vital force theory? Why was it discarded?

Answer :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.


Question 2. Name the scientist who disproved the vital force theory.
Answer :  vital force theory was discarded because Friedrich Wohler showed that it was possible to obtain an organic compound (urea) in the laboratory

Question 3. What do you understand by the following terms?

(i) Organic chemistry

(ii) Organic compounds

(iii) Catenation

Answer :

(i) Organic chemistryOrganic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which include not only hydrocarbons but also compounds with any number of other elements, including hydrogen (most compounds contain at least one carbon–hydrogen bond), nitrogen, oxygen

(ii) Organic compounds– Organic compound, any of a large class of chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides

(iii) Catenation- Catenation is the binding of an element to itself through covalent bonds to form chain or ring molecules. Examples: Carbon is the most common element that exhibits catenation. It can form long hydrocarbon chains and rings like

Question 4. What are the unique properties of carbon?

Answer :

Unique properties of carbon

1) Catenation: It is the ability to form bonds with other carbon atoms. 2) Tetravalency: Due to 4 valency of carbon, it is capable of bonding with four other atoms.

3) A carbon atom can bond with another carbon atom two or three times to make double and triple covalent bonds between two carbon atoms

Question 5.  Why are there very large number of organic compounds?


Answer :Carbon forms large number of compound due to following reasons: 1. Catenation: The unique property of the ‘C’ element to be able to form continuous links with other ‘C’ atoms through covalency called catenation, is one reason for the existence of a large number of organic compounds

Question 6. Distinguish between organic and inorganic compounds on the basis of solubility.
Answer : Generally, organic compounds are less soluble in water than inorganic compoundsOrganic compounds are more inflammable (more volatile) but are poorer conductors of heat and electricity than inorganic compounds

Question 7. With reference to butane, explain what do you understand by the following terms:

(i) Molecular formula

(ii) Condensed formula

(iii) Structural formula
Answer :

(i) Molecular formula–C4H10

(ii) Condensed formula –  CH3-CH2-CH2-CH3.

(iii) Structural formula –

structure formula of butane

Question 8. Name three major classes of aliphatic hydrocarbon.

Answer :Aliphatic hydrocarbons are divided into three main groups according to the types of bonds they contain:

1. alkanes,


3. alkynes

Question 9.  How do paraffins differ from unsaturated hydrocarbons?

Answer :Alkanes – Are saturated hydrocarbons that therefore contain only hydrogen and carbon atoms bonded to each other, and typically follow the chemical formula CnH2n+2. A common example is paraffin. 2. Alkenes – These unsaturated hydrocarbons are molecules that contain at least one carbon-to-carbon double bond

Question 10.  What do you understand by alkyl group? How are they formed?

Answer :Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl

Question 11.  What is functional group? Give definition with example.

Answer :Functional groups are collections of atoms that attach the carbon skeleton of an organic molecule and confer specific properties. .and  Functional groups include: hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl.

Question 12.  identify and name the functional groups of the following organic compounds
(i) C2H5OH

(ii) CH3Cl

(iii) CH3CHO
(v) CH3COCH3
(vi) HCOOH

Answer :

(i) C2H5OH– Alcohol

(ii) CH3Cl– Alkyl Halide

(iii) CH3CHO– Aldehyde
(iv) CH3CH2COOH– Carboxylic Acid
(v) CH3COCH3–  Ketone
(vi) HCOOH– Carboxylic Acid

Exercise 2 Page –197

Organic Chemistry Class-10 Goyal Brothers ICSE Solutions Ch-11

Question 1. Write the IUPAC name of the following compounds:

Write the IUPAC Name.

Answer :

(i) 2 ,2 Di Methyl Butane

(ii) 2-methyl Propane

(iii) Prop-1-ene

(iv) Butan-2-yne
(v) 2, Chloro-butane

(vi) 3-methyl pentane

(vii) propanal

(viii) propanol

(ix) butanoic acid

(x) propanone

(xi) ethenal

(xii) methenoic acid

Question 2. Write the structural formulae for the following compounds:
(i) Hexane

(ii) 2-methyl butane

(iii) Pent-2-yne

(iv) Butan-2-ol
(v) Prop-1-ene

(vi) Formic acid

(vii) Acetic acid

(viii) Acetone

Answer :

structural formulae

(i) Hexane– CH3- CH2- CH2- CH2- CH2- CH3

(ii) 2-methyl butane

structural formula of 2-methyl butane

(iii) Pent-2-yne

structural formula of Pent-2-yne

(iv) Butan-2-ol

structural formula of Butan-2-ol
(v) Prop-1-ene

structural formula of Prop-1-ene

(vi) Formic acid

structural formula of Formic acid

(vii) Acetic acid

structural formula of Acetic acid

(viii) Acetone

structural formula of Acetone

Question 3. Give the IUPAC name, molecular formula, condensed formula, general formula and structural formula of the following hydrocarbons:

(i) Propane
(ii) n-Butene
(iii) Ethene
Answer :

(i) Propane—

IUPAC name-Alkane

molecular formula- C5H12

condensed formula- CH3-CH2-CH2-CH2-CH2-CH3

general formula -Cn2n+2

structural formula-.

structural formula of propane
(ii) n-Butene—

IUPAC name, –Alkine

molecular formula,= C4H8

condensed formula,–CH2=CH2-CH2-CH2-CH3

general formula –Cn2n

structural formula

structural formula of Butine
(iii) Ethene—

IUPAC name,–Ethene

molecular formula,-C2H4

condensed formula,-CH2=CH2

general formula –Cn2n

structural formula-

structural formula of Ethene

Question 4. What do you understand by the term homologous series? Give four characteristics of homologous series.

Answer :

Characteristics are : The members of the homologous series have same functional group. Members have the same general formula. Members have the almost same chemical properties due to same functional group. Members have common general method of preparation and deffer in molecular mass by CH2 (14amu)

Question 5.  Give names and molecular formulae of first four homologous of alkynes and alcohols.

Answer :

alkynes — ethyne (C2H2), propyne (C3H4), butyne (C4H6)

alcohols.– methenol  (CH3OH)  , ethenol (C2H5OH), propenol (C3H7OH)

Question 6.  What do you understand by isomerism?

Answer : Isomerism, the existence of molecules that have the same numbers of the same kinds of atoms (and hence the same formula) but differ in chemical and physical properties

Question 7. Draw chain isomers of the following-
(i) C5H12

(ii) C6H14

Hexane has five isomers:

  • Hexane, CH3CH2CH2CH2CH2CH3, a straight chain of six carbon atoms.
  • 2-Methylpentane (Isohexane), CH3CH(CH3)CH2CH2CH3, a five-carbon chain with one methyl branch on the second.
  • 3-Methylpentane, CH3CH2CH(CH3)CH2CH3, a five-carbon chain with one methyl branch on the third.
  • 2,3-Dimethylbutane, CH3CH(CH3)CH(CH3)CH3, a four-carbon chain with one methyl branch on the second and third.
  • 2,2-Dimethylbutane (neohexane), CH3C(CH3)2CH2CH3, a four-carbon chain with two methyl branches on the second.

Question 8. Draw position isomers of the following:

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