9 Carbon Compounds Exercise, Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds Exercise, 9 Carbon Compounds Class 10 question and answers, 9 Carbon Compounds class 10 important questions and answers PDF

9 Carbon Compounds Exercise Solutions Science 1 – Maharashtra board

Question No.1]

Match the pairs.

Answer:

9 Carbon Compounds Exercise

Question No.2] 

Draw an electron dot structure of the following molecules. (Without showing the circles)

a. Methane.
Answer:
Molecular formula: CH₄

b. Ethene.
Answer:
Molecular formula: H₂C = CH₂

c. Methanol.
Answer:
Molecular formula: H₃C – OH

d. Water.
Answer:
Molecular formula: H₂O

9 Carbon Compounds Exercise

Question No.3]

Draw all possible structural formulae of compounds from their molecular formula given below.

a. C₃H₈
b. C₄H₁₀
c. C₃H₄
Answer:
a. C₃H₈ Propane:

b. C₄H₁₀ Butane:

c. C₃H₄ Propyne:

9 Carbon Compounds Exercise

Question No.4]

Explain the following terms with example.

a. Structural isomerism.
Answer:
The phenomenon in which compounds having different structural formulae have the same molecular formula is called structural isomerism. Butane is represented by two different compounds as their structural formulae are different. The first compound is a straight chain compound and the second compound is a branched chain compound. These two different structural formulae have the same molecular formula i.e. C₄H₁₀.

b. Covalent bond.
Answer:
The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
Example:
1. Hydrogen molecule formation: The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

2. Formation of oxygen molecule:
(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

c. Hetero atom in a carbon compound.
Answer:
Carbon compounds are formed by formation of bonds of carbon with other elements such as halogens, oxygen, nitrogen, sulfur. The atoms of these elements substitute one or more hydrogen atoms in the hydrocarbon chain and thereby the tetravalency of carbon is satisfied. The atom of the element which is substitute for hydrogen is referred to as a heteroatom. Sometimes hetero atoms are not alone but exist in the form of certain groups of atoms.

d. Functional group.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.

e. Alkane.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: In methane, four hydrogen atoms are bonded to carbon atom by four single covalent bonds.

f. Unsaturated hydrocarbons.
Answer:
The carbon compounds having a double bond or triple bond between two carbon atoms are called unsaturated hydrocarbons. The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example: Ethene (CH₂ = CH₂), Propene (CH₃ – CH = CH₂).
The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH ≡ CH).

g. Homopolymer.
Answer:
The polymers formed by repetition of single monomer are called homopolymer.

Example: polyethylene (CH₂ – CH₂)n.

h. Monomer.
Answer:
The small unit that repeats regularly to form a polymer is called monomer.
Example: Ethylene.

i. Reduction.
Answer:
In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.

j. Oxidant.
Answer:
An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction. An oxidant may also be called an oxidizer or oxidizing agent. When the oxidant includes oxygen, it may be called an oxygenation reagent or oxygen-atom transfer (OT) agent.
Examples of oxidants include:

1. Hydrogen peroxide
2. Ozone
3. Nitric acid
4. Sulfuric acid

9 Carbon Compounds Exercise

Question No.5] 

Write the IUPAC names of the following structural formulae.

a. CH₃ – CH₂ – CH₂ – CH₃
Answer:
The number of carbon atopic in the longest chain: 4
Parent alkane: Butane IUPAC name: n-Butane

b. CH₃ – CHOH – CH₃ (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3

Parent alkane: Propane
Functional group: -OH (ol)
Assign the number: 2
The carbon atom to which the -OH group is attached is numbered as C₂. If the carbon chain of the compound contains a -OH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘ol’. (ol stands for alcohol)
Parent suffix: Propan-2-ol
IUPAC name: Propan-2-ol

c. CH₃ – CH₂ – COOH (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane: Propane
Functional group: -COOH (-oic acid)
If the carbon chain of the compound contains a -COOH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘oic acid’.
Parent suffix: Propanoic acid
IUPAC name: Propanoic acid

d. CH₃ – CH₂ – NH₂
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -NH₂ (amine)
If the carbon chain of the compound contains a -NH₂ group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘amine’.
Parent suffix: Ethanamine
IUPAC name: Ethanamine.

e. CH₃ – CHO
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -CHO (al)
If the carbon chain of the compound contains a -CHO group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘al’.
Parent suffix: Ethanal
IUPAC name: Ethanal

f. CH₃ – CO – CH₂ – CH₃
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CO- (one)
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the function group.
If the carbon chain of the compound contains a (-CO-) group, then change the ending of the parent name, i.e., ‘e’ of butane is replaced by ‘one’.
Parent suffix: Butan-2-one
IUPAC name: Butan-2-one

9 Carbon Compounds Exercise

Question No.6]

Identify the type of the following reaction of carbon compounds.

1. CH₃ – CH₂ – CH₂ – OH + (O) → CH₃ – CH₂ – COOH
2. CH₃ – CH₂ – CH₃ + O₂ → 3CO₂ + 4H₂O
3. CH₃ – CH = CH – CH₃ + Br₂ → CH₃ – CHBr – CHBr – CH₃
4. CH₃ – CH₃ + Cl₂ → CH₃ – CH₂ – Cl + HCl
5. CH₃ – CH₂ – CH₂ – CH₂ – OH → CH₃ – CH₂ – CH = CH₂ + H₂O
6. CH₃ – CH₂ – COOH + NaOH → CH₃ – CH₂ – COO¯Na⁺ + H₂O
7. CH₃ – COOH + CH₃ – OH → CH₃ – COO – CH₃ + H₂O
Answer:

9 Carbon Compounds Exercise

Question No.7] 

Write the structural formulae for the following IUPAC names:

a. Pent-2-one
Answer:
Pent-2-one.
(1) Pent stands for 5 carbon atoms in a chain.
Number the carbon atoms in a chain as 1, 2, 3,…..

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

b. 2-Chlorobutane
Answer:
(1) In 2-chlorobutane, butane is parent alkane stands for 4 carbon atoms and number the carbon atoms in a chain as 1, 2, 3,….

(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C₂

(3) Now satisfy the valencies of each carbon atom

c. Propan-2-ol
Answer:
(1) Propan stands for 3 carbon atoms in a chain. Number the carbon atom in a chain as 1, 2, 3.

(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C₂.

(3) Now satisfy the valencies of each carbon atom

d. Methanal
Answer:
(1) Meth – stands for one carbon atom and assigned the number ‘1’ to carbon in the functional group -CHO.
(2) ‘-al’ stands for functional group (-CHO) aldehyde.
(3) Now satisfy the valencies of carbon in -CHO.

e. Butanoic acid
Answer:
(1) But stands for 4 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,….

‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.

Now satisfy the valencies of each carbon atom

f. 1-Bromopropane.
Answer:
(1) In 1-bromopropane, propane is parent alkane stands for 3 carbon atoms and number the carbon atoms in a chain as 1, 2, 3…….

(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C₁.

(3) Now satisfy the valencies of each carbon atom

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
– C – C –
(2) ‘amine’ stands for (- NH₂) amino group. Show the amino (-NH₂) at any carbon atom.
– C – C – NH₂
(3) Now satisfy the valencies of each carbon atom

h. Butanone.
Answer:
(1) But stands for 4 carbon atoms in a chain and the parent alkane is butane. Number the carbon atoms in a chain 1, 2, 3,….

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

9 Carbon Compounds Exercise

Question No.8]

Write answers as directed.

a. What causes the existance of very large number of carbon compound?
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(2) One, two or three covalent bonds can bond together two carbon atoms. These bonds are called single covalent bond, double covalent bond and triple covalent bond respectively. Due to the ability of carbon atoms to form multiple bonds as well as single bonds, the number of carbon compounds increases. For example, there are three compounds, namely, ethane (CH₃ – CH₃), ethene (CH₂ = CH₂) and ethyne (CH = CH) which contain two carbon atoms.

(3) Carbon being tetravalent, one carbon atom can form bonds with four other atoms (carbon or any other). This results in formation of many compounds. These compounds possess different properties as per the atoms to which carbon is bonded. For example, five different compounds are formed using one carbon atom and two monovalent elements hydrogen and chlorine: CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄. Similarly carbon atoms form covalent bonds with atoms of elements like O, N, S, halogen and P to form different types of carbon compounds in large number.

(4) Isomerism is one more characteristic of carbon compound which is responsible for large number of carbon compounds.

b. Saturated hydrocarbons are classified into three types. Write these names giving one example each.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called saturated hydrocarbons. Methane molecule contains only one carbon atom. In methane, four hydrogen atoms are bonded to carbon atom by four covalent bonds.

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

e. Give names of three natural polymers. write the place of their occurance and names of monomers from which they are formed.
Answer:

1. Poly saccharide is a natural polymer. It occurs in starch/carbohydrates. It is formed from monomer glucose.
2. Protein is a natural polymer. It occurs in muscles, hair, enzymes, skin, egg. It is formed from alpha amino acids.
3. Rubber is a natural polymer. It occurs in latex of rubber tree. It is formed from monomer isoprene.

f. What is meant by vinegar and gasohol? What are their uses?
Answer:
(1) Vinegar is a 5 – 8% aqueous solution of acetic acid. It is used as preservative in pickles. It is used to cook meat. 1t is used as a salad dressing.
(2) To increase the efficiency of petrol, it is mixed with 10% anhydrous ethanol, such a fuel is called gasohol. It is used as a fuel in cars and other vehicles.

g. what is a catalyst ? write any one reaction which is brought about by use of catalyst?
Answer:
Catalyst is a substance, which changes the rate of reaction, without causing any disturbance to it. Vegetable oil (unsaturated compound) undergoes addition reaction with hydrogen in the presence of nickel catalyst to form vanaspati ghee (saturated compound).

10 Space Missions Exercise Solutions Science 1 – Maharashtra board