CG Board Class 9 Science Chapter 15 Solutions: Hydrocarbons (हाइड्रोकार्बन) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Science Chapter 15 Solutions: Hydrocarbons (हाइड्रोकार्बन) in Bilingual Format (English & Hindi Medium).

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Complete step-by-step solutions for all in-text questions, activity prompts, table completions, and end-of-chapter exercises from CG SCERT Class 9 Science Chapter 15: ‘Hydrocarbons’ (2026-27 Official Edition).

Part 1

In-Text Questions, Tables & Discussion Prompts

Section 15.4

Naming of Straight Chain Alkanes

Question 1

Draw the structural formula of an alkane where…


📝 Question:

**: Draw the structural formula of an alkane where n = 3. Name it.

💡 Solution:

Draw the structural formula of an alkane where n = 3. Name it.

  • For n = 3, the general formula of alkanes (Cn H2n+2) gives C3 H2(3)+2 = C3 H8 (Propane).
  • Structural Formula

H H H

H – C – C – C – H

H H H

  • Name: Root (Prop) + Suffix (-ane) = Propane.

Question 2

Write the condensed structural formula of CHsub3sub CHsub2sub…


📝 Question:

**: Write the condensed structural formula of CH3 CH2 CH2 CH2 CH3.

💡 Solution:

Write the condensed structural formula of CH3 CH2 CH2 CH2 CH3.

  • The repeated group inside the carbon chain is -CH2– (repeating 3 times).
  • Condensed Structural Formula: CH3 (CH2)3 CH3 (Pentane).

Question 3

Csub7sub Hsub16sub is called heptane Identify the root…


📝 Question:

**: C7 H16 is called heptane. Identify the root part and suffix in the name.

💡 Solution:

C7 H16 is called heptane. Identify the root part and suffix in the name.

  • Root part: Hept (denotes 7 carbon atoms).
  • Suffix: -ane (denotes saturated hydrocarbon with single bonds).

  • English: Calculate the molecular weights of methane and ethane and find the difference between them. Also find the difference in molecular weight between ethane and propane, propane and butane, and butane and pentane. Can you see any relation between the molecular weights of members of a homologous series of alkanes?

Step-by-step Solution /

  • English:
  • Atomic masses: Carbon (C) = 12 u, Hydrogen (H) = 1 u.
  1. Methane (CH4): 12 + (4 × 1) = 16 u.
  2. Ethane (C2 H6): (2 × 12) + (6 × 1) = 30 u.
  • Difference (Ethane – Methane) = 30 – 16 = 14 u.
  1. Propane (C3 H8): (3 × 12) + (8 × 1) = 44 u.
  • Difference (Propane – Ethane) = 44 – 30 = 14 u.
  1. Butane (C4 H10): (4 × 12) + (10 × 1) = 58 u.
  • Difference (Butane – Propane) = 58 – 44 = 14 u.
  1. Pentane (C5 H12): (5 × 12) + (12 × 1) = 72 u.
  • Difference (Pentane – Butane) = 72 – 58 = 14 u.
  • Conclusion / Relation: Two consecutive members of a homologous series differ by a -CH2– (methylene) group, which corresponds to a constant molecular weight difference of 14 u (12 + 2 = 14 u).
  • Hindi:
  1. (CH4): 12 + (4 × 1) = 16 u
  2. (C2 H6): (2 × 12) + (6 × 1) = 30 u
  3. (C3 H8): (3 × 12) + (8 × 1) = 44 u
  4. (C4 H10): (4 × 12) + (10 × 1) = 58 u
  5. (C5 H12): (5 × 12) + (12 × 1) = 72 u
Section 15.5

Trends in Physical Properties

Question 1

What is a homologous series Explain through examples


📝 Question:

**: What is a homologous series? Explain through examples.

💡 Solution:

What is a homologous series? Explain through examples.

  • Definition: A series or family of organic compounds having similar chemical properties, represented by the same general formula, where consecutive members differ from each other by a -CH2– (methylene) unit and 14 u in molecular weight.
  • Example: Alkanes form a homologous series with the general formula Cn H2n+2
  • Methane (CH4)
  • Ethane (C2 H6)
  • Propane (C3 H8)

Question 2

Which among butane propane and pentane will have…


📝 Question:

**: Which among butane, propane, and pentane will have the highest boiling point? Explain.

💡 Solution:

Which among butane, propane, and pentane will have the highest boiling point? Explain.

  • Highest Boiling Point: Pentane (36°C).
  • Reason: In a straight-chain alkane series, as the length of the carbon chain and molecular weight increases, the surface area increases. This leads to stronger intermolecular van der Waals forces of attraction, requiring more heat energy to vaporize the substance.
  • Boiling points: Propane (-42°C) < Butane (-0.5°C) < Pentane (36°C).

Section 15.6

Branching and Structural Isomerism

Question 1

Draw the isomers of Csub5sub Hsub12sub Hint three…


📝 Question:

**: Draw the isomers of C5 H12. (Hint: three isomers are possible)

💡 Solution:

Draw the isomers of C5 H12. (Hint: three isomers are possible)

The molecular formula C5 H12 has 3 structural isomers

  1. Pentane (n-pentane)
  • Condensed Formula: CH3 – CH2 – CH2 – CH2 – CH3
  1. 2-Methylbutane (Isopentane)
  • Condensed Formula

CH3 – CH – CH2 – CH3

CH3

  1. 2,2-Dimethylpropane (Neopentane)
  • Condensed Formula

CH3

CH3 – C – CH3

CH3

Section 15.7

Alkene and Alkyne Completion

  • Six Carbon Alkene Name: Hexene (C6 H12).
  • Table 3 Completion (n = 6 Alkene):
  • Condensed Structure: H_2C = CH (CH2)3 CH3
  • Molecular Formula: C6 H12
  • Name: Hex + ene = Hexene
  • Table 4 Completion (Alkynes for n = 5 and n = 6):
  • n = 5:
  • Condensed Structure: CH3 CH2 CH2 C ≡ CH (or CH3 (CH2)2 C ≡ CH)
  • Molecular Formula: C5 H8
  • Name: Pent + yne = Pentyne
  • n = 6:
  • Condensed Structure: CH3 (CH2)3 C ≡ CH
  • Molecular Formula: C6 H10
  • Name: Hex + yne = Hexyne
Section 15.9

Types of Hydrocarbons on Basis of Bonding

Question 1

How many straight chain isomers are possible for…


📝 Question:

**: How many straight chain isomers are possible for the molecular formula C5 H8? Draw their structures.

💡 Solution:

How many straight chain isomers are possible for the molecular formula C5 H8? Draw their structures.

For C5 H8, two straight-chain alkyne position isomers are possible

  1. Pent-1-yne: HC ≡ C – CH2 – CH2 – CH3
  2. Pent-2-yne: CH3 – C ≡ C – CH2 – CH3

Question 2

Group the following into saturated and unsaturated hydrocarbons


📝 Question:

**: Group the following into saturated and unsaturated hydrocarbons

💡 Solution:

Group the following into saturated and unsaturated hydrocarbons

(Methane, Ethene, Ethyne, Propane, Cyclopropane, Butene, Propyne)

  • Saturated Hydrocarbons (Only C-C single bonds): Methane, Propane, Cyclopropane.
  • Unsaturated Hydrocarbons (Contains C=C double or C ≡ C triple bonds): Ethene, Ethyne, Butene, Propyne.

Part 2

End-of-Chapter Exercises

Exercise Question 1

Choose the correct option

Choose the term from the box that fits each description:

(Position isomerism, homologous series, alkanes, isomers, chain isomerism, catenation)

(i) This hydrocarbon only has single bonds.

–

(ii) Carbon has the ability to form strong bonds with atoms of other elements. But what is more special is its ability to bond strongly with other carbon atoms.

–

(iii) Molecules having same molecular formula but different structures.

–

(iv) Butane and 2-methylpropane exhibit this type of isomerism.

–

(v) Isomerism that can be exhibited by alkenes and alkynes but not alkanes.

–

(vi) Members of this series show a regular gradation in their physical properties.

–

Exercise Question 2

Fill in the blanks

(i) Hex-1-ene and hex-2-ene are ……………….. isomers.

–

(ii) The boiling point of butane is …………. than propane.

–

Answer / : More (Higher) / .

(iii) The number of carbon atoms in the parent chain of 2-methylpropane is ……..

–

(iv) The number of hydrogen atoms in cyclobutane is ………….

–

Answer / : Eight / .

  • Verification: Cyclobutane (C4 H8) has 4 × 2 = 8 Hydrogen atoms.

(v) C2 H6, C3 H8, C4 H10 are members of the ……………. homologous series.

–

SECTION / EXERCISE

Exercise Question 3

Draw the structures of the following: 2-methylbutane, prop-1-yne, pent-2-ene.

  • English:
  1. 2-Methylbutane:

`

CH3 – CH – CH2 – CH3

|

CH3

`

  1. Prop-1-yne:

`

HC ≡ C – CH3

`

  1. Pent-2-ene:

`

CH3 – CH = CH – CH2 – CH3

`

  • Hindi:

`

CH3 – CH – CH2 – CH3

|

CH3

`

`

HC ≡ C – CH3

`

`

CH3 – CH = CH – CH2 – CH3

`

SECTION / EXERCISE

Exercise Question 4

Write any three characteristics of the alkane family.

  • English:
  1. Saturated Nature: All carbon atoms are linked to each other and to hydrogen atoms by single covalent bonds only.
  2. General Formula: All members follow the general molecular formula Cn H2n+2.
  3. Gradation in Physical Properties: As molecular weight increases, physical properties like boiling point and melting point show a gradual increase.
  • Hindi:
SECTION / EXERCISE

Exercise Question 5

Differentiate between position and chain isomerism with examples.

  • English:
  • Chain Isomerism: Arises due to a difference in the arrangement of the main carbon chain/skeleton.
  • Example: Butane (CH3-CH2-CH2-CH3) and 2-Methylpropane (CH3-CH(CH3)-CH3) both have the formula C4 H10.
  • Position Isomerism: Arises due to a difference in the position of a double bond, triple bond, or functional group on the same carbon skeleton.
  • Example: But-1-ene (CH2=CH-CH2-CH3) and But-2-ene (CH3-CH=CH-CH3) both have the formula C4 H8.
  • Hindi:
  • **: (CH3-CH2-CH2-CH3) 2- (CH3-CH(CH3)-CH3) C4 H10
  • **: -1- (CH2=CH-CH2-CH3) -2- (CH3-CH=CH-CH3) C4 H8
SECTION / EXERCISE

Exercise Question 6

What is the relation between the boiling points and number of carbon atoms present in straight chain alkane? Explain.

  • English:
  • Relation: In straight-chain alkanes, the boiling point is directly proportional to the number of carbon atoms. As the number of carbon atoms increases, the boiling point increases.
  • Explanation: Increasing the number of carbon atoms increases molecular weight and surface area. Larger surface area enhances van der Waals attractive forces between molecules, requiring more energy to break them apart into vapor.
  • Hindi:
SECTION / EXERCISE

Exercise Question 7

Alkanes, alkynes and alkenes with three or less carbon atoms do not show structural isomerism. Explain.

  • English:
  1. No Chain Isomerism: For n = 1, 2, 3, carbon atoms can only be connected in a single continuous linear sequence. Branching requires a central carbon atom to be bonded to at least three other carbon atoms, which needs a minimum of 4 carbon atoms (e.g., C4 H10).
  2. No Position Isomerism: In alkenes (C2 H4, C3 H6) or alkynes (C2 H2, C3 H4), numbering the double/triple bond from either terminal carbon always places the multiple bond at position 1 (e.g., Propene is always Prop-1-ene). Position shift requires at least 4 carbon atoms (e.g., C4 H8).
  • Hindi:
SECTION / EXERCISE

Exercise Question 8

How many structures/isomers are possible for the molecular formula C4 H8? Draw their structures. (Hint – three are possible).

  • English:

The molecular formula C4 H8 represents 3 acyclic (alkene) structural isomers:

  1. But-1-ene:

`

CH2 = CH – CH2 – CH3

`

  1. But-2-ene:

`

CH3 – CH = CH – CH3

`

  1. 2-Methylpropene (Isobutene):

`

CH2 = C – CH3

|

CH3

`

  • Hindi:

`

CH2 = CH – CH2 – CH3

`

`

CH3 – CH = CH – CH3

`

`

CH2 = C – CH3

|

CH3

`

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