Maharashtra Board Class 10 Science Chapter 9 Carbon Compounds Solutions

Question 1.Match the pairs.

Group A	Group B
a. C2H6	1. Unsaturated hydrocarbon
b. C2H2	2. Molecular formula of an alcohol
c. CH4O	3. Saturated hydrocarbon
d. C3H6	4. Triple bond

Answer:
(a) C2H6 - 3. Saturated hydrocarbon
(b) C2H2 - 4. Triple bond
(c) CH4O - 2. Molecular formula of an alcohol
(d) C3H6 - 1. Unsaturated hydrocarbon
In simple words: This question tests your knowledge of different types of hydrocarbons and their basic properties or classifications, matching molecular formulas with descriptive characteristics.

🎯 Exam Tip: For matching pairs, ensure you know the definition and examples of each type of compound (saturated/unsaturated hydrocarbons, alcohols, triple bonds) to correctly associate them.

 

Question 2.Draw an electron dot structure of the following molecules. (Without showing the circles)
a. Methane.
Answer:
Molecular formula: CH4
ℹ️ चित्र व्याख्या (Diagram Explanation): यह मीथेन अणु की इलेक्ट्रॉन डॉट संरचना को दर्शाता है। केंद्रीय कार्बन परमाणु चार हाइड्रोजन परमाणुओं से घिरा है, जहाँ प्रत्येक हाइड्रोजन परमाणु एक इलेक्ट्रॉन साझा करके कार्बन के साथ एकल सहसंयोजक बंधन बनाता है।
In simple words: Methane (CH4) has a central carbon atom forming single covalent bonds with four hydrogen atoms, sharing one electron pair with each.

🎯 Exam Tip: Practice drawing electron dot structures for simple molecules as it's a common question type that assesses understanding of covalent bonding and valency.

 

b. Ethene.
Answer:
Molecular formula: H2C = CH2
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथीन अणु की इलेक्ट्रॉन डॉट संरचना और रेखीय संरचना को दर्शाता है। इसमें दो कार्बन परमाणु एक-दूसरे से दो इलेक्ट्रॉन जोड़े साझा करके एक दोहरा बंधन बनाते हैं, और प्रत्येक कार्बन परमाणु दो हाइड्रोजन परमाणुओं से एकल बंधन द्वारा जुड़ा होता है।
In simple words: Ethene (H2C=CH2) features a double bond between two carbon atoms, with each carbon also bonded to two hydrogen atoms.

🎯 Exam Tip: When drawing electron dot structures, especially for unsaturated compounds like ethene, ensure correct representation of double or triple bonds by showing the shared electron pairs accurately.

 

c. Methanol.
Answer:
Molecular formula: H3C-OH
ℹ️ चित्र व्याख्या (Diagram Explanation): यह मेथनॉल (\(CH_3OH\)) की संरचना को दर्शाता है। इसमें एक कार्बन परमाणु तीन हाइड्रोजन परमाणुओं और एक हाइड्रॉक्सिल समूह (-OH) से एकल सहसंयोजक बंधन द्वारा जुड़ा होता है।
In simple words: Methanol (\(CH_3OH\)) consists of a carbon atom bonded to three hydrogen atoms and one hydroxyl (-OH) group via single bonds.

🎯 Exam Tip: Pay close attention to the functional group in alcohols (-OH) and ensure all atoms satisfy their valencies when drawing structural formulas.

 

d. Water.
Answer:
Molecular formula: H2O
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पानी (\(H_2O\)) अणु की संरचना को दर्शाता है। इसमें एक ऑक्सीजन परमाणु दो हाइड्रोजन परमाणुओं से एकल सहसंयोजक बंधन द्वारा जुड़ा होता है, जिसमें ऑक्सीजन पर दो अकेले इलेक्ट्रॉन जोड़े भी होते हैं।
In simple words: A water molecule (\(H_2O\)) has an oxygen atom forming single bonds with two hydrogen atoms and having two lone pairs of electrons.

🎯 Exam Tip: Remember that oxygen typically forms two bonds and has two lone pairs of electrons in a stable molecule like water, which is important for understanding its shape.

 

Question 3.Draw all possible structural formulae of compounds from their molecular formula given below.
a. C3H8
Answer:
a. C3H8 Propane:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेन (\(C_3H_8\)) की सीधी श्रृंखला संरचना को दर्शाता है। इसमें तीन कार्बन परमाणु एक सीधी रेखा में एक-दूसरे से एकल बंधनों द्वारा जुड़े होते हैं, और शेष वैलेंसी हाइड्रोजन परमाणुओं द्वारा संतुष्ट होती हैं।
In simple words: Propane (\(C_3H_8\)) has a straight chain of three carbon atoms, with each carbon bonded to enough hydrogen atoms to satisfy its valency.

🎯 Exam Tip: For alkanes, ensure that each carbon atom forms four single bonds, either with other carbon atoms or with hydrogen atoms.

 

b. C4H10
Answer:
b. C4H10 Butane:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ब्यूटेन (\(C_4H_{10}\)) की सीधी श्रृंखला संरचना को दर्शाता है। इसमें चार कार्बन परमाणु एक सीधी रेखा में एक-दूसरे से एकल बंधनों द्वारा जुड़े होते हैं, और सभी शेष वैलेंसी हाइड्रोजन परमाणुओं द्वारा संतुष्ट होती हैं।
In simple words: Butane (\(C_4H_{10}\)) can exist as a straight chain of four carbon atoms, with all available bonds occupied by hydrogen atoms.

🎯 Exam Tip: When drawing structural isomers, remember to explore different carbon chain arrangements (straight, branched) while keeping the molecular formula constant.

 

c. C3H4
Answer:
c. C3H4 Propyne:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपाइन (\(C_3H_4\)) की संरचना को दर्शाता है। इसमें तीन कार्बन परमाणु होते हैं, जिनमें से दो के बीच एक तिहरा बंधन होता है, और शेष वैलेंसी हाइड्रोजन परमाणुओं द्वारा पूरी की जाती है।
In simple words: Propyne (\(C_3H_4\)) is an alkyne with a three-carbon chain containing a triple bond between two of the carbon atoms.

🎯 Exam Tip: For alkynes, ensure one triple bond is present between carbon atoms and that each carbon satisfies its tetravalency with hydrogen atoms.

 

Question 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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ब्यूटेन (\(C_4H_{10}\)) के दो संरचनात्मक समावयवों को दर्शाता है: एक सीधी श्रृंखला वाला नॉर्मल ब्यूटेन और दूसरा एक शाखित श्रृंखला वाला आइसोब्यूटेन। दोनों में कार्बन और हाइड्रोजन परमाणुओं की संख्या समान है लेकिन उनके परमाणुओं की व्यवस्था भिन्न है।
In simple words: Structural isomerism occurs when compounds have the same molecular formula but different arrangements of atoms, leading to different structural formulas. Butane is a common example, existing as a straight chain or a branched chain.

🎯 Exam Tip: When explaining structural isomerism, always provide an example with the molecular formula and different structural forms to clearly illustrate the concept.

 

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो हाइड्रोजन परमाणुओं से एक हाइड्रोजन अणु के निर्माण को दर्शाता है। प्रत्येक हाइड्रोजन परमाणु एक इलेक्ट्रॉन साझा करता है, जिससे उनके बीच एक एकल सहसंयोजक बंधन बनता है और प्रत्येक परमाणु हीलियम जैसी स्थिर इलेक्ट्रॉनिक विन्यास प्राप्त करता है।
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 (=).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो ऑक्सीजन परमाणुओं से एक ऑक्सीजन अणु (\(O_2\)) के निर्माण को दर्शाता है। प्रत्येक ऑक्सीजन परमाणु दो इलेक्ट्रॉनों को साझा करता है, जिससे उनके बीच एक दोहरा सहसंयोजक बंधन बनता है और प्रत्येक ऑक्सीजन परमाणु नियॉन जैसा स्थिर इलेक्ट्रॉनिक विन्यास प्राप्त करता है।
In simple words: A covalent bond is formed when two atoms share valence electrons, like in hydrogen (\(H_2\)) where a single bond forms, or in oxygen (\(O_2\)) where a double bond forms, allowing both atoms to achieve a stable electron configuration.

🎯 Exam Tip: When defining covalent bonds, always include examples like \(H_2\) and \(O_2\) and explain how electron sharing leads to stable configurations, highlighting single and double bonds.

 

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.

Structural formula	Hetero atom
-X (-Cl, Br, -I)	Halogen
-O-H	Oxygen
-O-	Oxygen
-NH2	Nitrogen
-O-	Oxygen
-C-	Oxygen

In simple words: A heteroatom is any atom other than carbon or hydrogen that is present in a carbon compound, often substituting for hydrogen, and influencing the compound's properties.

🎯 Exam Tip: Understand that heteroatoms (like O, N, S, halogens) are crucial for determining the functional group and chemical properties of organic compounds.

 

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.
In simple words: A functional group is a specific arrangement of atoms within a molecule that is responsible for its characteristic chemical reactions and properties, regardless of the carbon chain length.

🎯 Exam Tip: Memorize common functional groups (e.g., -OH for alcohol, -COOH for carboxylic acid) and how they impact the chemical behavior of organic compounds.

 

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो सरल एल्केन अणुओं, मीथेन (\(CH_4\)) और इथेन (\(C_2H_6\)) की संरचनाओं को दर्शाता है। मीथेन में एक कार्बन परमाणु चार हाइड्रोजन से जुड़ा होता है, जबकि इथेन में दो कार्बन परमाणु एक-दूसरे से और शेष हाइड्रोजन से एकल बंधों द्वारा जुड़े होते हैं।
In simple words: Alkanes are saturated hydrocarbons where all carbon atoms are linked by single bonds, satisfying each carbon's four valencies. Methane and ethane are basic examples.

🎯 Exam Tip: For alkanes, the key is the presence of only single bonds between carbon atoms, making them saturated hydrocarbons with the general formula \(C_nH_{2n+2}\).

 

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. e.g. Ethene (CH2 = CH2), Propene (CH3-CH = CH2). The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH = CH).
In simple words: Unsaturated hydrocarbons contain at least one carbon-carbon double or triple bond, unlike saturated hydrocarbons which only have single bonds.

🎯 Exam Tip: Differentiate between alkenes (double bond) and alkynes (triple bond) as both are types of unsaturated hydrocarbons, and their presence indicates higher reactivity.

 

g. Homopolymer.
Answer:The polymers formed by repetition of single monomer are called homopolymer. e.g. polyethylene (CH2-CH2)n.
In simple words: A homopolymer is a polymer made from the repeated linking of identical monomer units.

🎯 Exam Tip: Remember that homopolymers are distinct from copolymers, which are formed from two or more different types of monomers.

 

h. Monomer.
Answer:The small unit that repeats regularly to form a polymer is called monomer. Example: Ethylene.
In simple words: A monomer is a single, small molecule that can chemically bond with other identical or similar monomers to form a larger polymer chain.

🎯 Exam Tip: Understanding monomers is fundamental to grasping how large polymeric materials are constructed through polymerization reactions.

 

i. Reduction.
Answer:In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.
In simple words: Reduction is a chemical process involving either the gain of hydrogen atoms or the loss of oxygen atoms by a molecule.

🎯 Exam Tip: Remember reduction as the opposite of oxidation; it can also be defined in terms of electron gain, which is useful in redox reactions.

 

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
In simple words: An oxidant, or oxidizing agent, is a substance that causes another substance to lose electrons (get oxidized) in a chemical reaction.

🎯 Exam Tip: Oxidants are key components in redox reactions; they themselves get reduced while oxidizing another species.

 

Question 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
In simple words: This compound is an alkane with four carbon atoms in a straight chain, named n-Butane according to IUPAC rules.

🎯 Exam Tip: For straight-chain alkanes, simply count the carbons and use the appropriate prefix (meth-, eth-, prop-, but-, pent-, etc.) followed by "-ane".

 

b. CH3-CHOH-CH3 (Practice Activity Sheet-3)
Answer:The number of carbon atoms in the longest chain: 3
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेन-2-ol की संरचना को दर्शाता है। इसमें तीन कार्बन परमाणुओं की एक श्रृंखला होती है, जिसमें दूसरे कार्बन परमाणु पर एक हाइड्रॉक्सिल (-OH) समूह जुड़ा होता है, जो इसे एक अल्कोहल बनाता है।
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
In simple words: This alcohol has a three-carbon chain with a hydroxyl group on the second carbon, giving it the IUPAC name Propan-2-ol.

🎯 Exam Tip: When naming alcohols, identify the longest carbon chain containing the -OH group, number it to give the -OH group the lowest possible number, and replace "-e" of the alkane with "-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
In simple words: This compound is a carboxylic acid with a three-carbon chain and a -COOH functional group, so its IUPAC name is Propanoic acid.

🎯 Exam Tip: For carboxylic acids, the carbon in the -COOH group is always carbon #1. Replace "-e" of the corresponding alkane with "-oic acid" for the IUPAC name.

 

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.
In simple words: This compound is an amine with a two-carbon chain, where the amino group is attached, giving it the IUPAC name Ethanamine.

🎯 Exam Tip: Amines are named by replacing the "-e" of the parent alkane with "-amine". Ensure correct numbering if the amino group can be in different positions.

 

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
In simple words: This compound is an aldehyde with a two-carbon chain, featuring the -CHO functional group, and is named Ethanal.

🎯 Exam Tip: For aldehydes, the carbon in the -CHO group is always carbon #1. Replace "-e" of the corresponding alkane with "-al" for the IUPAC name.

 

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: 1 2 3 4 CH3-CO-CH2-CH3 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
In simple words: This compound is a ketone with a four-carbon chain, and the carbonyl group is located on the second carbon, hence it is called Butan-2-one.

🎯 Exam Tip: For ketones, number the carbon chain to give the carbonyl (-CO-) group the lowest possible number. Replace "-e" of the alkane with "-one".

 

Question 6.Identify the type of the following reaction of carbon compounds.
1. CH3-CH2-CH2-OH + (O) \( \implies \) CH3-CH2-COOH
2. CH3-CH2-CH3 + O2 \( \implies \) 3CO2 + 4H2O
3. CH3-CH=CH-CH3 + Br2 \( \implies \) CH3-CHBr-CHBr-CH3
4. CH3-CH3 + Cl2 \( \implies \) CH3-CH2-Cl + HCl
5. CH3-CH2-CH2-CH2-OH \( \implies \) CH3-CH2-CH=CH2 + H2O
6. CH3-CH2-COOH + NaOH \( \implies \) CH3-CH2-COONa+ + H2O
7. CH3-COOH + CH3-OH \( \implies \) CH3-COO-CH3 + H2O
Answer:
(1) CH3-CH2-CH2-OH \( \xrightarrow[]{(O) \atop alk.KMnO_4} \) CH3CH2-COOH
Chemical reaction: Oxidation reaction
(2) CH3-CH2-CH3 + O2 \( \xrightarrow[]{heat} \) 3CO2 + 4H2O
Chemical reaction: Combustion
(3) CH3-CH=CH-CH3 + Br2 \( \implies \) CH3-CH-CH-CH3
|
Br Br
Chemical reaction: Addition reaction
(4) CH3-CH3 + Cl2 \( \xrightarrow[]{Sunlight} \) CH3-CH2-Cl + HCl
Chemical reaction: Substitution reaction
(5) CH3-CH2-CH2-CH2-OH \( \xrightarrow[]{Con. H_2SO_4 \atop \Delta} \) CH3-CH2-CH=CH2 + H2O
Chemical reaction: Dehydration reaction
(6) CH3-CH2-COOH + NaOH \( \implies \) CH3-CH2-COO-Na+ + H2O
Chemical reaction: Neutralization
(7) CH3-COOH + CH3-OH \( \xrightarrow[]{acid \atop catalyst} \) CH3-COO-CH3 + H2O
Chemical reaction: Esterification
In simple words: Each reaction demonstrates a distinct chemical transformation, such as oxidation of alcohol to carboxylic acid, combustion of hydrocarbons, addition to unsaturated compounds, substitution of alkanes, dehydration to form alkenes, neutralization of acid with base, and esterification forming an ester.

🎯 Exam Tip: For reaction identification, focus on recognizing changes in functional groups, presence of reagents (e.g., \(O_2\), \(Br_2\), \(NaOH\)), and conditions (heat, sunlight, catalyst) to determine the reaction type.

 

Question 7.Write the structural formulae for the following IUPAC names:
a. Pentan-2-one
Answer:Pentan-2-one.
(1) Pent stands for 5 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,.....
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पेंटेन-2-ओन की संरचना को चरण-दर-चरण दर्शाता है। पहले 5-कार्बन श्रृंखला को दर्शाया गया है, फिर दूसरे कार्बन पर कीटोन समूह (\(C=O\)) को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) 'one' stands for functional group
\[\begin{array}{l} \mathrm{O} \\ \mathrm{||} \\ (-\mathrm{C}-) \end{array}\] ketone. The number assigned for the ketone group is 2. Show the ketone group at C2.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 5-कार्बन श्रृंखला में दूसरे कार्बन पर कीटोन समूह को स्पष्ट रूप से दर्शाता है। यह पेंटेन-2-ओन के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पेंटेन-2-ओन की पूर्ण संरचना को दर्शाता है, जिसमें 5 कार्बन परमाणुओं की श्रृंखला है, दूसरे कार्बन पर एक कीटोन समूह (\(C=O\)) है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: Pentan-2-one is a ketone with a five-carbon chain, where the carbonyl group is located on the second carbon atom.

🎯 Exam Tip: When drawing structures from IUPAC names, first identify the parent chain and functional group, then position the functional group correctly, and finally add hydrogen atoms to satisfy all valencies.

 

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,....
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-क्लोरोब्यूटेन की संरचना को चरण-दर-चरण दर्शाता है। पहले 4-कार्बन श्रृंखला को दर्शाया गया है, फिर दूसरे कार्बन पर क्लोरीन परमाणु को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C2
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 4-कार्बन श्रृंखला में दूसरे कार्बन पर क्लोरीन परमाणु को स्पष्ट रूप से दर्शाता है। यह 2-क्लोरोब्यूटेन के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-क्लोरोब्यूटेन की पूर्ण संरचना को दर्शाता है, जिसमें 4 कार्बन परमाणुओं की श्रृंखला है, दूसरे कार्बन पर एक क्लोरीन परमाणु है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: 2-Chlorobutane is a butane derivative where a chlorine atom is attached to the second carbon atom in the four-carbon chain.

🎯 Exam Tip: For haloalkanes, ensure the halogen atom is placed at the correct numbered carbon, and remember that halogens act as prefixes in IUPAC naming.

 

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेन-2-ol की संरचना को चरण-दर-चरण दर्शाता है। पहले 3-कार्बन श्रृंखला को दर्शाया गया है, फिर दूसरे कार्बन पर हाइड्रॉक्सिल (-OH) समूह को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) '-ol' stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C2.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 3-कार्बन श्रृंखला में दूसरे कार्बन पर हाइड्रॉक्सिल (-OH) समूह को स्पष्ट रूप से दर्शाता है। यह प्रोपेन-2-ol के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेन-2-ol की पूर्ण संरचना को दर्शाता है, जिसमें 3 कार्बन परमाणुओं की श्रृंखला है, दूसरे कार्बन पर एक हाइड्रॉक्सिल (-OH) समूह है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: Propan-2-ol is an alcohol with a three-carbon chain where the hydroxyl group (-OH) is attached to the second carbon atom.

🎯 Exam Tip: Alcohols are named by replacing the "-e" of the parent alkane with "-ol," with a number indicating the position of the hydroxyl group on the main chain.

 

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह मेथनल (\(HCHO\)) की संरचना को दर्शाता है। इसमें एक कार्बन परमाणु ऑक्सीजन के साथ दोहरा बंधन और दो हाइड्रोजन परमाणुओं के साथ एकल बंधन बनाता है।
In simple words: Methanal, also known as formaldehyde, is the simplest aldehyde, consisting of one carbon atom double-bonded to an oxygen and single-bonded to two hydrogens.

🎯 Exam Tip: Aldehydes always have the -CHO group at the end of a carbon chain, and the carbon in this group is always numbered 1.

 

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.
In simple words: Butanoic acid is a carboxylic acid with a four-carbon chain, where the -COOH functional group is located at the end.

🎯 Exam Tip: When naming carboxylic acids, the carbon atom of the -COOH group is always designated as carbon 1 of the parent chain.

 

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.......
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1-ब्रोमोप्रोपेन की संरचना को चरण-दर-चरण दर्शाता है। पहले 3-कार्बन श्रृंखला को दर्शाया गया है, फिर पहले कार्बन पर ब्रोमीन परमाणु को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C1.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 3-कार्बन श्रृंखला में पहले कार्बन पर ब्रोमीन परमाणु को स्पष्ट रूप से दर्शाता है। यह 1-ब्रोमोप्रोपेन के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1-ब्रोमोप्रोपेन की पूर्ण संरचना को दर्शाता है, जिसमें 3 कार्बन परमाणुओं की श्रृंखला है, पहले कार्बन पर एक ब्रोमीन परमाणु है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: 1-Bromopropane is a three-carbon alkane with a bromine atom attached to the first carbon atom.

🎯 Exam Tip: For haloalkanes, the position of the halogen is indicated by a number, and the halogen is named as a prefix (e.g., bromo-, chloro-).

 

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथानामाइन की संरचना को चरण-दर-चरण दर्शाता है। पहले 2-कार्बन श्रृंखला को दर्शाया गया है, फिर एक कार्बन पर एमिनो (-NH2) समूह को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) 'amine' stands for (-NH2) amino group. Show the amino (-NH2) at any carbon atom.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-कार्बन श्रृंखला में एमिनो (-NH2) समूह को स्पष्ट रूप से दर्शाता है। यह एथानामाइन के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथानामाइन की पूर्ण संरचना को दर्शाता है, जिसमें 2 कार्बन परमाणुओं की श्रृंखला है, एक कार्बन पर एक एमिनो (-NH2) समूह है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: Ethanamine is a two-carbon compound where an amino group (-NH2) is attached, making it an amine.

🎯 Exam Tip: Amines are named by replacing the "-e" of the corresponding alkane with "-amine". For simple straight-chain amines, the position of the -NH2 group might not need a number if it can only be on the first carbon.

 

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,....
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ब्यूटेनोन की संरचना को चरण-दर-चरण दर्शाता है। पहले 4-कार्बन श्रृंखला को दर्शाया गया है, फिर दूसरे कार्बन पर कीटोन समूह (\(C=O\)) को जोड़ा गया है, और अंत में हाइड्रोजन परमाणुओं के साथ सभी वैलेंसी को पूरा किया गया है।
(2) 'one' stands for functional group
\[\begin{array}{l} \mathrm{O} \\ \mathrm{||} \\ (-\mathrm{C}-) \end{array}\] ketone. The number assigned for the ketone group is 2. Show the ketone group at C2.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 4-कार्बन श्रृंखला में दूसरे कार्बन पर कीटोन समूह (\(C=O\)) को स्पष्ट रूप से दर्शाता है। यह ब्यूटेनोन के संरचनात्मक सूत्र का एक मध्यवर्ती चरण है।
(3) Now satisfy the valencies of each carbon atom
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ब्यूटेनोन की पूर्ण संरचना को दर्शाता है, जिसमें 4 कार्बन परमाणुओं की श्रृंखला है, दूसरे कार्बन पर एक कीटोन समूह (\(C=O\)) है, और सभी कार्बन परमाणुओं की वैलेंसी को संतुष्ट करने के लिए पर्याप्त हाइड्रोजन परमाणु जुड़े हुए हैं।
In simple words: Butanone is a ketone with a four-carbon chain, where the carbonyl group is located on the second carbon atom.

🎯 Exam Tip: For ketones, ensure the carbonyl carbon is not at the end of the chain. Number the chain to give the carbonyl group the lowest possible position number.

 

Question 8.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.
In simple words: The vast number of carbon compounds is due to carbon's unique properties: catenation (forming long chains and rings), tetravalency (forming four bonds), and ability to form multiple bonds (single, double, triple) and isomers.

🎯 Exam Tip: Focus on the four key properties of carbon-catenation, tetravalency, multiple bond formation, and isomerism-as they are fundamental to explaining the diversity of organic 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 single covalent bonds.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो सरल एल्केन अणुओं, मीथेन (\(CH_4\)) और इथेन (\(C_2H_6\)) की संरचनाओं को दर्शाता है। मीथेन में एक कार्बन परमाणु चार हाइड्रोजन से जुड़ा होता है, जबकि इथेन में दो कार्बन परमाणु एक-दूसरे से और शेष हाइड्रोजन से एकल बंधों द्वारा जुड़े होते हैं।
In simple words: Saturated hydrocarbons are compounds where carbon atoms are linked only by single bonds. They are classified into straight chain alkanes (e.g., Methane, Ethane), branched chain alkanes, and cyclic alkanes.

🎯 Exam Tip: When asked to classify saturated hydrocarbons, remember to mention straight-chain, branched-chain, and cyclic forms, providing a clear example for each.

Question 8. 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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मीथेन और इथेन के संरचनात्मक सूत्रों को दर्शाता है। मीथेन (H-C-H) में एक कार्बन परमाणु चार हाइड्रोजन परमाणुओं से जुड़ा होता है, जबकि इथेन (H-C-C-H) में दो कार्बन परमाणु एक दूसरे से और शेष हाइड्रोजन परमाणुओं से एकल सहसंयोजक बंधों द्वारा जुड़े होते हैं। ये संरचनाएं संतृप्त हाइड्रोकार्बन को दर्शाती हैं।
In simple words: Saturated hydrocarbons have only single bonds between carbon atoms, meaning each carbon is bonded to the maximum number of other atoms. Examples include methane and ethane, which are very stable.

🎯 Exam Tip: When asked to classify and provide examples, ensure the examples clearly illustrate the definition. Structural formulas help in understanding the bonding.

Question 8. c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

Hetero Atom	Name	Structural Formula	Condensed Structural Formula	Example	Name
Oxygen	1. Alcohol	H | H-C-OH | H	-OH	H | H-C-OH | H	Methanol
	2. Aldehyde	O || -C-H	-CHO	H | H-C = O | H	Acetaldehyde
	3. Ketone	O || -C-	-CO-	H O H | H-C-C-C-H | H H	Acetone
	4. Carboxylic	O || -C-O-H	-COOH	H O | H-C-C-OH | H	Acetic acid
	5. Ether	-O-	-O-	H H | H-C-O-C-H | H H	Dimethyl ether

In simple words: Functional groups with oxygen like alcohol, aldehyde, ketone, carboxylic acid, and ether determine the specific chemical properties of organic compounds. Each has a distinct structural arrangement involving oxygen.

🎯 Exam Tip: Memorize the common oxygen-containing functional groups, their structural formulas, and one example for each. This is fundamental for organic chemistry nomenclature and reactions.

Question 8. d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

Hetero Atom	Name	Structural Formula	Condensed Structural Formula	Example
(1) Halogen, [Chlorine Bromine, Iodine]	Halo	-X (-Cl, -Br, -I)	-X(-Cl, -Br, -I)	CH3Cl Chloromethane CH3-CH2-Br Bromoethane
(2) Oxygen	Alcohol	-O-H	-OH	CH3OH Methanol CH3-CH2-OH Ethanol
(3) Nitrogen	Amine	-N-H | H	-NH2	CH3-NH2 Methyl amine CH3-CH2-NH2 Ethylamine

In simple words: Functional groups with different heteroatoms like halogens (e.g., chlorine in chloromethane), oxygen (e.g., -OH in methanol), and nitrogen (e.g., -NH2 in methylamine) impart unique chemical characteristics to organic compounds.

🎯 Exam Tip: Pay attention to the distinction between structural and condensed structural formulas. Be ready to provide specific examples for each functional group as requested.

Question 8. 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.
In simple words: Natural polymers are large molecules found in nature, like polysaccharides (from glucose in starch), proteins (from amino acids in muscles), and natural rubber (from isoprene in trees).

🎯 Exam Tip: For natural polymers, remember both their common sources and the specific monomer units that build them up. This demonstrates a complete understanding.

Question 8. 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. It 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.
In simple words: Vinegar is a dilute solution of acetic acid used for food preservation and cooking. Gasohol is a blend of petrol and ethanol, used as an alternative fuel to enhance engine efficiency.

🎯 Exam Tip: When defining everyday chemical terms, include their chemical composition (e.g., % solution) and practical applications/uses to show comprehensive knowledge.

Question 8. 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).
In simple words: A catalyst speeds up a chemical reaction without being consumed itself. For instance, nickel acts as a catalyst in the hydrogenation of vegetable oil to produce vanaspati ghee.

🎯 Exam Tip: Understand that catalysts influence reaction *rates* but not the equilibrium position or the products themselves. Providing a relevant chemical reaction illustrates the concept effectively.

Project: Prepare A Chart Giving Detailed Information Of Carbon Compounds In Everyday Use. Display It In The Class And Discuss.

Can You Recall? (Text Book Page No. 110)

Question 1. What are the types of compounds?
Answer: Organic and inorganic compounds are the two important types of compounds.
In simple words: Compounds are broadly classified into organic, which contain carbon, and inorganic, which generally do not.

🎯 Exam Tip: This is a basic classification; ensure you can quickly identify common examples for each type.

Question 2. Objects in everyday use such as foodstuff, fibres, paper, medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Answer: The constituent elements common in these compounds are carbon (C), hydrogen (H) and oxygen (O).
In simple words: Many everyday materials like food, paper, and wood primarily consist of carbon, hydrogen, and oxygen, forming various organic compounds.

🎯 Exam Tip: Recognize that carbon, hydrogen, and oxygen are the backbone elements of most organic substances we encounter daily.

Question 3. To which group in the periodic table does the element carbon belongs? Write down the electronic configuration of carbon and deduce the valency of carbon.
Answer: The element carbon belongs to group 14 and its electronic configuration is 2, 4. The valency of carbon is 4.
In simple words: Carbon is in Group 14, has an electronic configuration of 2, 4, and a valency of 4, meaning it can form four bonds.

🎯 Exam Tip: Knowing carbon's position, electronic configuration, and valency (tetravalency) is crucial for understanding its bonding behavior in organic compounds.

Use Your Brain Power! (Text Book Page No. 115)

Question 1. The molecular formula of ethyne is C2H2. From this draw its structural formula and electron-dot structure.
Answer: Ethyne: Molecular formula: C2H2
H-C≡C-H H:C:::C:H
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र इथाइन अणु की संरचना को दर्शाता है। इसमें दो कार्बन परमाणु एक ट्रिपल बॉन्ड (त्रिबंध) से जुड़े होते हैं, और प्रत्येक कार्बन परमाणु एक-एक हाइड्रोजन परमाणु से एकल बॉन्ड द्वारा जुड़ा होता है। इलेक्ट्रॉन-डॉट संरचना कार्बन और हाइड्रोजन के बीच और कार्बन-कार्बन के बीच साझा किए गए वैलेंस इलेक्ट्रॉनों को दिखाती है, जिसमें कार्बन-कार्बन त्रिबंध में छह इलेक्ट्रॉन साझा होते हैं।
In simple words: Ethyne (C2H2) has a linear structure with a triple bond between the two carbon atoms, and each carbon is singly bonded to a hydrogen atom.

🎯 Exam Tip: For electron-dot structures, always ensure each atom satisfies its octet (or duet for hydrogen) by correctly showing shared and unshared electron pairs. Triple bonds involve sharing six electrons.

Question 2. How many bonds have to be there in between the carbon atoms in ethyne so as to satisfy their tetra valency?
Answer: To satisfy their tetravalency, three double bonds have to be there in between two carbon atoms in ethyne.
In simple words: Ethyne requires a triple bond between its carbon atoms to satisfy carbon's tetravalency, allowing each carbon to form a total of four bonds.

🎯 Exam Tip: Carbon's tetravalency (forming four bonds) is a fundamental principle. Ensure that structural formulas always reflect this, whether through single, double, or triple bonds.

Use Your Brain Power! (Text Book Page No. 116)

Question 1. Draw the electron-dot structure of cyclohexane.
Answer: Cyclohexane: Molecular formula: C6H12
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र साइक्लोहेक्सेन के संरचनात्मक सूत्र और इलेक्ट्रॉन-डॉट संरचना को दर्शाता है। साइक्लोहेक्सेन में छह कार्बन परमाणु एक रिंग में जुड़े होते हैं, और प्रत्येक कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़ा होता है। इलेक्ट्रॉन-डॉट संरचना में, प्रत्येक कार्बन-कार्बन और कार्बन-हाइड्रोजन एकल सहसंयोजक बंध को साझा इलेक्ट्रॉनों के जोड़े के रूप में दिखाया गया है, जिससे प्रत्येक कार्बन की चतुर्योग्यता पूरी होती है।
In simple words: Cyclohexane (C6H12) forms a ring of six carbon atoms, with each carbon atom bonded to two hydrogen atoms, satisfying its tetravalency.

🎯 Exam Tip: For cyclic compounds, clearly show the ring structure and ensure all carbon atoms maintain their tetravalency by bonding to the correct number of hydrogen atoms.

Use Your Brain Power! (Text Book Page No. 112)

Question 1. Atomic number of chlorine is 17. What is the number of electrons in the valence shell of the chlorine?
Answer: There are seven electrons in the valence shell of the chlorine.
In simple words: Chlorine, with an atomic number of 17, has seven electrons in its outermost shell.

🎯 Exam Tip: Remember that valence electrons, the electrons in the outermost shell, determine an element's chemical reactivity and bonding behavior.

Question 2. Molecular formula of chlorine is Cl2. Draw electron-dot and line structure of a chlorine molecule.
Answer:
:Cl̈:Cl̈: \( \implies \) Cl-Cl
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्लोरीन अणु (Cl2) के इलेक्ट्रॉन-डॉट और लाइन संरचना को दर्शाता है। इसमें दो क्लोरीन परमाणु एक एकल सहसंयोजक बंध द्वारा जुड़े होते हैं, और प्रत्येक क्लोरीन परमाणु पर तीन अकेले इलेक्ट्रॉन जोड़े (lone pairs) होते हैं। इलेक्ट्रॉन-डॉट संरचना में साझा इलेक्ट्रॉन और अकेले जोड़े स्पष्ट रूप से दिखाए गए हैं, जबकि लाइन संरचना में साझा जोड़े को एक डैश से दर्शाया गया है।
In simple words: A chlorine molecule (Cl2) consists of two chlorine atoms joined by a single covalent bond, with each chlorine atom having three lone pairs of electrons.

🎯 Exam Tip: When drawing electron-dot structures for diatomic molecules, remember to show both shared and unshared (lone pair) electrons to satisfy the octet rule for each atom.

Question 3. The molecular formula of water is H2O. Draw electron-dot and line structures for triatomic molecule. (Use dots for electron of oxygen atom and crosses for electrons of hydrogen atoms.)
Answer:
H × Ö × H
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पानी के अणु (H2O) के इलेक्ट्रॉन-डॉट संरचना को दर्शाता है। इसमें एक ऑक्सीजन परमाणु दो हाइड्रोजन परमाणुओं से एकल सहसंयोजक बंधों द्वारा जुड़ा होता है। ऑक्सीजन के इलेक्ट्रॉनों को बिंदुओं से और हाइड्रोजन के इलेक्ट्रॉनों को क्रॉस से दर्शाया गया है, जिसमें ऑक्सीजन पर दो अकेले इलेक्ट्रॉन जोड़े भी हैं, जिससे सभी परमाणुओं की वैलेंसी संतुष्ट होती है।
In simple words: A water molecule (H2O) has a central oxygen atom bonded to two hydrogen atoms via single covalent bonds, with oxygen also having two lone pairs.

🎯 Exam Tip: When using different symbols (dots and crosses) for electrons from different atoms, ensure consistency. Always check that the central atom and terminal atoms satisfy their valencies and octets/duets.

Question 4. The molecular formula of ammonia is NH3. Draw electron-dot and line structures for ammonia molecule.
Answer:
H:N: H \( \implies \) H-N-H
H H
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अमोनिया अणु (NH3) के इलेक्ट्रॉन-डॉट और लाइन संरचना को दर्शाता है। इसमें एक नाइट्रोजन परमाणु तीन हाइड्रोजन परमाणुओं से एकल सहसंयोजक बंधों द्वारा जुड़ा होता है, और नाइट्रोजन पर एक अकेला इलेक्ट्रॉन जोड़ा (lone pair) भी होता है। इलेक्ट्रॉन-डॉट संरचना में साझा इलेक्ट्रॉन और नाइट्रोजन पर अकेले जोड़े स्पष्ट रूप से दिखाए गए हैं।
In simple words: An ammonia molecule (NH3) features a central nitrogen atom bonded to three hydrogen atoms with a lone pair of electrons on the nitrogen.

🎯 Exam Tip: For electron-dot structures, always remember to include lone pairs on the central atom if present, as they influence molecular geometry and reactivity.

Question 5. The molecular formula of carbon dioxide is CO2. Draw the electron-dot structure (without showing circle) and line structure for CO2.
Answer:
:Ö::C::Ö: \( \implies \) O=C=O
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कार्बन डाइऑक्साइड अणु (CO2) के इलेक्ट्रॉन-डॉट और लाइन संरचना को दर्शाता है। इसमें एक कार्बन परमाणु दो ऑक्सीजन परमाणुओं से डबल बॉन्ड (द्विबंध) द्वारा जुड़ा होता है। प्रत्येक ऑक्सीजन परमाणु पर दो अकेले इलेक्ट्रॉन जोड़े भी होते हैं, जिससे कार्बन और ऑक्सीजन दोनों की वैलेंसी संतुष्ट होती है।
In simple words: Carbon dioxide (CO2) has a central carbon atom double-bonded to two oxygen atoms, forming a linear molecule.

🎯 Exam Tip: For CO2, recognizing the double bonds is crucial to satisfy the octet rule for carbon and oxygen. Lone pairs on oxygen are also important.

Question 6. With which bond C atom in CO2 is bonded to each of the O atoms?
Answer: In CO2, carbon atom is bonded to each of the O atoms by double bond.
In simple words: The carbon atom in carbon dioxide forms double bonds with each of the two oxygen atoms.

🎯 Exam Tip: Understanding the type of bond (single, double, triple) between atoms is essential for predicting molecular geometry and chemical reactivity.

Question 7. The molecular formula of sulfur is S8 in which eight sulphur atoms are bonded to each other to form one ring. Draw electron-dot structure for S8 without showing the circles.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सल्फर अणु (S8) के इलेक्ट्रॉन-डॉट संरचना को दर्शाता है, जिसमें आठ सल्फर परमाणु एक क्राउन-आकार की रिंग संरचना में जुड़े होते हैं। प्रत्येक सल्फर परमाणु दो अन्य सल्फर परमाणुओं से एकल सहसंयोजक बंधों द्वारा जुड़ा होता है, और प्रत्येक सल्फर परमाणु पर दो अकेले इलेक्ट्रॉन जोड़े (lone pairs) होते हैं, जिससे उसकी वैलेंसी संतुष्ट होती है।
The above S8 molecule of sulphur has crown shaped structure. One molecule of sulpur is made up of eight atoms of sulphur.
In simple words: An S8 sulfur molecule forms a crown-shaped ring where eight sulfur atoms are linked together by single bonds, each sulfur having two lone pairs.

🎯 Exam Tip: For elemental structures like S8, visualize the ring shape. Ensure each sulfur atom forms two bonds and has two lone pairs to achieve stability, fulfilling its tendency for two bonds.

Use Your Brain Power! (Text Book Page No. 113)

Question 1. Hydrogen peroxide decomposes of its own by the following reaction:
2H - O - O - H \( \implies \) 2H - O - H + O2
From this, what will be your inference about the strength O - O covalent bond? Tell from the above example whether oxygen has catenation power or not.
Answer: In hydrogen peroxide (H2O2), the O - O covalent bond is not strong as oxygen has no catenation power.
In simple words: The decomposition of hydrogen peroxide shows that the oxygen-oxygen bond is weak, indicating that oxygen does not have strong catenation power.

🎯 Exam Tip: Understand that catenation refers to an atom's ability to form long chains with itself. For oxygen, unlike carbon, this ability is very limited, leading to weak O-O bonds.

Question 1. The above table shows the homologous series of alkenes. Inspect the molecular formulae of the members of this series. Do you find any relationship, in the number of carbon atoms and the number of hydrogen atoms in the molecular formulae?
Answer: In the above homologous series, if we observe the molecular formulae of alkenes then the number of carbon atoms are half the number of hydrogen atoms.
In simple words: In the alkene homologous series, the number of carbon atoms is always half the number of hydrogen atoms, following the CnH2n formula.

🎯 Exam Tip: For homologous series, always look for patterns in the number of carbon and hydrogen atoms to deduce the general formula. This helps in understanding the series' characteristics.

Question 2. If the number of carbon atoms in the molecular formulae of alkenes is denoted by 'n' what will be the number of hydrogen atoms?
Answer: If the number of carbon atoms in the molecular formulae of alkenes is denoted by 'n' then the number of hydrogen atoms would be 2n.
In simple words: For alkenes, if 'n' is the number of carbon atoms, then there are '2n' hydrogen atoms.

🎯 Exam Tip: The general formula (CnH2n) is a key characteristic of homologous series like alkenes and allows you to predict molecular formulas for any member.

Question 3. What would be the general formula for the molecular formulae of the members of the homologous series of alkanes? What would be the value of 'n' for the first member of this series?
Answer: The general formula for the homologous series of alkane is CnH2n + 2. The value of 'n' for the first member of homologous series is 1.
CnH2n+2 = C1H2 x 1 + 2 = CH4
In simple words: The general formula for alkanes is CnH2n+2, and for the first member, methane, 'n' is 1.

🎯 Exam Tip: Know the general formulas for alkanes, alkenes, and alkynes, as they are frequently asked. Remember that 'n' starts from 1 for alkanes and 2 for alkenes/alkynes.

Question 4. The general molecular formula for the homologous series of alkynes is CnH2n - 2. Write down the individual molecular formulae of the first, second and third members by substituting the values 2, 3 and 4 respectively for 'n' in this formula.
Answer: The general molecular formula for the homologous series of alkynes is CnH2n - 2
n = 2 C2H2 x 2 - 2 = C2H2 Ethyne
n = 3 C3H2 x 3 - 2 = C3H4 Propyne
n = 4 C4H2 x 4 - 2 = C4H6 Butyne
In simple words: For alkynes, using the general formula CnH2n-2, the first three members are Ethyne (C2H2), Propyne (C3H4), and Butyne (C4H6) for n=2, 3, and 4 respectively.

🎯 Exam Tip: Practice substituting 'n' values into general formulas for different homologous series to quickly derive molecular formulas of specific members. Ensure correct calculations.

Question 5. Write down structural formulae of the first four members of the various homologous series formed by making use of the functional groups.
Answer:

Functional group	Functional group	Functional group	Functional group
Halo – X (Cl, Br, -I)	Aldehyde – CHO	Carboxylic acid – COOH	Amine -NH2
CH3Cl Chloromethane	HCHO Methanal	HCOOH Methanoic acid	CH3NH2 Methenamine
CH3-CH2 - Cl Chloroethane	CH3CHO Ethanal	CH3COOH Ethanoic acid	CH3CH2NH2 Ethanamine
CH3 – CH2 – CH2 – Cl 1-Chloropropane	CH3CH2CHO Propanal	CH3CH2COOH Propanoic acid	CH3CH2CH2NH2 Propanamine
CH3-CH2-CH2-CH2 - Cl 1-Chlorobutane	CH3CH2CH2 CHO Butanal	CH3CH2CH2COOH Butanoic acid	CH3CH2CH2CH2NH2 Butanamine

In simple words: Different functional groups like halo, aldehyde, carboxylic acid, and amine can be attached to carbon chains to form homologous series, each with distinct structural and molecular formulas.

🎯 Exam Tip: Practice drawing structural formulas for the first few members of each homologous series. Pay close attention to the positioning of the functional group and ensuring carbon's tetravalency.

Question 6. General formula of the homologous series of alkanes is CnH2n + 2. Write down the molecular formula of the 8th and 12th member using this.
Answer: General formula of alkanes is CnH2n + 2
n = 8 C8H2 x 8 + 2 = C8H18 Octane
n = 12 C12H2 × 12 + 2 = C12H26 Dodecane
In simple words: Using the alkane general formula CnH2n+2, the 8th member is C8H18 (Octane) and the 12th member is C12H26 (Dodecane).

🎯 Exam Tip: Accurately apply the general formula by substituting the given 'n' values. Remember that for alkanes, the number of hydrogen atoms is always two more than twice the number of carbon atoms.

Use Your Brain Power! (Text Book Page No. 121)

Question 1. Draw three structural formulae having molecular formula C5H12. Give the names n-pentane, i-pentane and neo-pentane to the above structural formulae.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र C5H12 के तीन संरचनात्मक आइसोमर्स - n-पेंटेन, i-पेंटेन (आइसोपेंटेन), और नियो-पेंटेन को दर्शाता है। n-पेंटेन एक सीधी श्रृंखला वाला हाइड्रोकार्बन है। i-पेंटेन में एक शाखित श्रृंखला होती है जहाँ दूसरा कार्बन परमाणु एक मिथाइल समूह से जुड़ा होता है। नियो-पेंटेन में केंद्रीय कार्बन परमाणु से चार मिथाइल समूह जुड़े होते हैं, जो इसे अत्यधिक शाखित बनाता है। ये सभी संरचनाएं कार्बन परमाणुओं की चतुर्योग्यता को बनाए रखती हैं।
In simple words: For C5H12, three structural isomers exist: n-pentane (a straight chain), i-pentane (a branched chain with a methyl group on the second carbon), and neo-pentane (a highly branched structure with a central carbon bonded to four methyl groups).

🎯 Exam Tip: When drawing isomers, ensure all atoms maintain their correct valency (carbon-4, hydrogen-1). Systematically explore different chain and branching arrangements to find all possible isomers without repetition.

Question 2. Draw all possible structural formulae having molecular formula C6H14. Give names to all the isomers.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र C6H14 के तीन संरचनात्मक आइसोमर्स - n-हेक्सेन, i-हेक्सेन (2-मिथाइलपेंटेन), और नियो-हेक्सेन (2,2-डाइमिथाइलब्यूटेन) को दर्शाता है। n-हेक्सेन एक सीधी श्रृंखला वाला हाइड्रोकार्बन है। i-हेक्सेन में एक शाखा होती है। नियो-हेक्सेन में अधिक शाखाएं होती हैं, जिससे यह सबसे सघन होता है। ये संरचनाएं कार्बन-हाइड्रोजन बंधों को स्पष्ट रूप से दर्शाती हैं, प्रत्येक कार्बन की चतुर्योग्यता को पूरा करती हैं, जबकि अलग-अलग आणविक व्यवस्था दिखाती हैं।
In simple words: For C6H14, there are several structural isomers, including n-hexane (straight chain), 2-methylpentane (one branch), and 2,2-dimethylbutane (two branches), all having the same molecular formula but different arrangements.

🎯 Exam Tip: Drawing isomers requires careful and systematic exploration of possible carbon skeleton arrangements (straight, branched, cyclic). Always verify that each carbon atom forms exactly four bonds and hydrogen atoms form one bond.

Try This! (Text Book Page No. 124)

Question 1. Light a bunsen burner. Open and close the air hole at the bottom of the burner by means of the movable ring around it. When do you get yellow sooty flame? When do you get blue flame?
Answer: When the air hole at the bottom of the burner is open, sufficient oxygen is mixed gaseous fuel for complete combustion and a clean blue flame is obtained. When the air hole is partially blocked by means of the movable ring around it, the air supply is limited which results in incomplete combustion. Hence, yellow sooty flame is produced.
In simple words: A Bunsen burner produces a clean blue flame when the air hole is open, allowing complete combustion. If the air hole is partially closed, limited oxygen leads to incomplete combustion and a yellow, sooty flame.

🎯 Exam Tip: Relate flame color and sootiness directly to the completeness of combustion, which depends on oxygen supply. Blue flame means complete, yellow/sooty means incomplete.

Question 1. The names of four fatty acids separated from vegetable oils are given in the table. Identify the number of carbon - carbon double bonds from their structure and molecular formula from the below fatty acids which one when reacts with iodine will make the colour of iodine disappear.
Answer:

Name	Molecular Formula	Number of C = C double bonds	Will it decolorise I2?
Stearic acid	C17H35COOH	0	no
Oleic acid	C17H33COOH	One double bond	yes
Plamitic acid	C15H31COOH	0	no
Linoleic acid	C17H31COOH	Two double bonds	yes

In simple words: Fatty acids with carbon-carbon double bonds (like Oleic and Linoleic acid) are unsaturated and will decolorize iodine, while saturated fatty acids (like Stearic and Palmitic acid) will not.

🎯 Exam Tip: Remember that the iodine test is used to detect unsaturation. Compounds with double or triple bonds will react with iodine, causing its color to disappear.

Use Your Brain Power! (Text Book Page No. 128)

Question 1. Explain by writing a reaction, what will happen when pieces of sodium metal are put in n-propyl alcohol.
Answer: n-Propyl alcohol reacts with pieces of sodium metal, sodium propoxide and hydrogen gas are obtained.
2CH3-CH2-CH2-OH + 2Na \( \implies \) 2CH3-CH2-CH2 - ONa + H2 ↑
n-propyl alcohol Sodium propoxide
In simple words: When sodium metal reacts with n-propyl alcohol, it produces sodium propoxide and releases hydrogen gas, characteristic of alcohol-metal reactions.

🎯 Exam Tip: Alkali metals react vigorously with alcohols to produce hydrogen gas. This reaction demonstrates the acidic nature of the hydrogen on the -OH group in alcohols.

Question 2. Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.
Answer: When n-butyl alcohol is heated with concentrated sulphuric acid, one molecule of water is removed from its molecule to form 1-butene.
CH3-CH2-CH2-CH2-OH \(\xrightarrow{\text{Conc. H2SO4, 170 °C}}\) CH3-CH2-CH=CH2 + H2O
n-butyl alcohol 1-butene
In simple words: Heating n-butyl alcohol with concentrated sulfuric acid leads to dehydration, removing a water molecule and forming 1-butene.

🎯 Exam Tip: Concentrated sulfuric acid acts as a dehydrating agent in alcohol reactions. Remember that dehydration of primary alcohols often leads to the formation of an alkene.

Use Your Brain Power! (Text Book Page No. 129)

Question 1. Which one of ethanoic acid and hydrochloric acid is stronger?
Answer: Hydrochloric acid is stronger acid.
In simple words: Hydrochloric acid is a much stronger acid than ethanoic acid.

🎯 Exam Tip: Distinguish between strong acids (like HCl) and weak acids (like ethanoic acid). Strong acids ionize completely in water, while weak acids only partially ionize.

Question 2. Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid?
Answer: pH paper is useful to distinguish between ethanoic acid and hydrochloric acid.
In simple words: pH paper can differentiate between ethanoic acid (weak acid) and hydrochloric acid (strong acid) by showing different pH values.

🎯 Exam Tip: Litmus paper only shows if a solution is acidic or basic. pH paper provides a more precise pH value, allowing differentiation between strong and weak acids.

Use Your Brain Power! (Text Book Page No. 130)

Question 1. Explain why does the lime water turns milky in the reaction of acetic acid with sodium carbonate.
Answer: In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.
2CH3COOH + Na2CO3 \( \implies \) 2CH3COONa + H2O + CO2↑
Acetic acid Sodium carbonate Sodium acetate
Ca(OH)2 + CO2(g) \( \implies \) CaCO3 ↓ + H2O
Lime water Carbon dioxide Calcium carbonate (milky)
In simple words: Acetic acid reacts with sodium carbonate to release carbon dioxide gas, which then reacts with lime water (calcium hydroxide) to form insoluble calcium carbonate, making the lime water milky.

🎯 Exam Tip: This reaction is a classic test for carbon dioxide. Remember the two steps: CO2 production and then its reaction with lime water to form a precipitate (milkiness).

Question 2. Explain the reaction that would take place when a piece of sodium metal is dropped in ethanoic acid.
Answer: When a piece of sodium metal is dropped in ethanoic acid, sodium acetate and hydrogen gas is formed.
2CH3COOH + 2Na \( \implies \) 2CH3COONa + H2 ↑
Ethanoic acid Sodium Sodium ethanoate
In simple words: Sodium metal reacts with ethanoic acid to produce sodium ethanoate and hydrogen gas.

🎯 Exam Tip: Similar to alcohols, carboxylic acids react with active metals like sodium to produce hydrogen gas, indicating the presence of an acidic hydrogen atom.

Question 3. Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Answer: Ethanol does not react with sodium bicarbonate, while ethanoic acid reacts with sodium bicarbonate to form carbon dioxide gas.
CH3COOH + NaHCO3 \( \implies \) CH3COONa + H2O + CO2↑
Ethanoic acid Sodium ethanoate
In simple words: To distinguish between ethanol and ethanoic acid, add sodium bicarbonate: ethanoic acid will react to produce carbon dioxide gas (fizzing), while ethanol will not.

🎯 Exam Tip: The reaction with sodium bicarbonate (or sodium carbonate) is a standard test for carboxylic acids, as it produces carbon dioxide bubbles, which are easily observable.

Use Your Brain Power! (Text Book Page No. 131)

Question 1. When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin? What do you think?
Answer: The functional group carboxylic acid (-COOH) is present in fat whereas the functioned group hydroxyl group (-OH) is present in glycerin.
In simple words: Fats are esters containing carboxylic acid functional groups, while glycerin (a product of fat saponification) contains hydroxyl (-OH) functional groups.

🎯 Exam Tip: Saponification (soap making) is a key reaction. Remember that fats are essentially triglycerides (esters) and that hydrolysis yields fatty acids (or their salts, soap) and glycerol (which has hydroxyl groups).

Can You Tell? (Text Book Page No. 131)

Question 1. What are the chemical names of the nutrients that we get from the foodstuff, namely, cereals, pulses and meat?
Answer: The nutrients that we get from the foodstuff, namely cereals, pulses and meat are alpha amino acids.
In simple words: Cereals and pulses provide carbohydrates and proteins, while meat primarily provides proteins, which are made of alpha amino acids.

🎯 Exam Tip: Connect common food sources to their primary nutritional components. Alpha amino acids are the building blocks of proteins, a vital nutrient.

Use your brain power!

Question 1. Structural formulae of some monomers are given below. Write the structural formula of the homopolymer formed from them.
CH3
CH3
(i) CH2 = C
(ii) CH2 = C
CH3
CH3

Answer:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पहली संरचना 2-मिथाइलप्रोपीन का मोनोमर है, जिसमें दो कार्बन परमाणु एक डबल बॉन्ड से जुड़े हैं और एक कार्बन परमाणु पर दो मिथाइल समूह लगे हैं। इसके बहुलीकरण से एक लंबी श्रृंखला बनती है जहाँ प्रत्येक पुनरावृत्त इकाई में कार्बन-कार्बन सिंगल बॉन्ड और एक मिथाइल समूह होता है।

(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दूसरी संरचना एक्रेलोनाइट्राइल का मोनोमर है, जिसमें दो कार्बन परमाणु एक डबल बॉन्ड से जुड़े हैं और एक कार्बन परमाणु पर एक साइनाइड (CN) समूह लगा है। इसके बहुलीकरण से एक पॉलीमर श्रृंखला बनती है जहाँ प्रत्येक पुनरावृत्त इकाई में कार्बन-कार्बन सिंगल बॉन्ड और एक साइनाइड (CN) समूह होता है।
In simple words: Monomers are small repeating units. Polymerization joins these units. The diagrams show how 2-methylpropene and acrylonitrile monomers link up to form their respective homopolymers.

🎯 Exam Tip: When drawing polymer structures, ensure the repeating unit is clearly enclosed in brackets with 'n' as a subscript, indicating repetition. Pay attention to the double bond breaking to form single bonds in the polymer chain.

Question 2. From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह पॉलीविनाइल एसीटेट (पॉलीमर) की संरचना है, जिसमें एक लंबी कार्बन श्रृंखला है। इस श्रृंखला में प्रत्येक कार्बन परमाणु पर एक हाइड्रोजन और दूसरे कार्बन परमाणु पर एक हाइड्रोजन और एक एसीटेट (C=O-CH3) समूह लगा होता है। 'n' इसकी पुनरावृत्ति इकाई को दर्शाता है।

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह विनाइल एसीटेट (मोनोमर) की संरचना है। इसमें एक डबल बॉन्ड से जुड़े दो कार्बन परमाणु होते हैं, जिसमें एक कार्बन परमाणु पर दो हाइड्रोजन और दूसरे कार्बन परमाणु पर एक हाइड्रोजन और एक एसीटेट (C=O-CH3) समूह लगा होता है।
In simple words: The polymer polyvinyl acetate is formed from the monomer vinyl acetate. To find the monomer, remove the 'n' brackets and convert the single bond in the repeating unit back into a double bond.

🎯 Exam Tip: Recognizing the repeating unit is key to deducing the monomer from a polymer structure. The monomer will always have a double bond that breaks to form the polymer chain.

Text Book Page No. 133

Question 1. Complete the following table by writing their Structural formulae and Molecular formulae.
Answer:
(Answer is given directly in bold letters.)

 

Straight chain of Carbon atoms	Structural formula	Molecular formula	Name
C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कार्बन परमाणु (C) से जुड़ा चार हाइड्रोजन परमाणुओं (H) की संरचना को दर्शाता है, जो मीथेन अणु बनाता है। प्रत्येक H-C बंधन एक एकल सहसंयोजक बंधन है।	CH4	Methane
C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो कार्बन परमाणुओं की संरचना को दर्शाता है, जो एक एकल बंधन से जुड़े हैं। प्रत्येक कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़ा है, जिससे इथेन अणु बनता है।	C2H6	Ethane
C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़ा है, जिससे प्रोपेन अणु बनता है।	C3H8	Propane
C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह चार कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे ब्यूटेन अणु बनता है।	C4H10	Butane
C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह पांच कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे पेंटेन अणु बनता है।	C5H12	Pentane
C-C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह छह कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे हेक्सेन अणु बनता है।	C6H14	Hexane
C-C-C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह सात कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे हेप्टेन अणु बनता है।	C7H16	Heptane
C-C-C-C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह आठ कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे ऑक्टेन अणु बनता है।	C8H18	Octane
C-C-C-C-C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह नौ कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे नोनैन अणु बनता है।	C9H20	Nonane
C-C-C-C-C-C-C-C-C-C	ℹ️ चित्र व्याख्या (Diagram Explanation): यह दस कार्बन परमाणुओं की एक सीधी श्रृंखला को दर्शाता है, जो एकल बंधनों से जुड़े हैं। टर्मिनल कार्बन परमाणु तीन हाइड्रोजन परमाणुओं से जुड़े हैं, और मध्य कार्बन परमाणु दो हाइड्रोजन परमाणुओं से जुड़े हैं, जिससे डेकेन अणु बनता है।	C10H22	Decane

In simple words: This table illustrates the homologous series of alkanes, showing how increasing the number of carbon atoms in a straight chain leads to different alkanes with predictable molecular and structural formulas.

🎯 Exam Tip: For alkanes, remember the general formula CnH2n+2. Practice drawing structural formulas for various chain lengths to ensure accuracy in representing single bonds and hydrogen atoms.

Fill in the gaps in the table:

a. Homologous series of alkanes.

 

Name	Molecular formula	Condensed Structural Formula	Number of carbon atoms	Number of -CH2- Units	Boiling Point °C
Methane	CH4	CH4	1	1	-162
Ethane	C2H6	CH3-CH3	2	2	-88.5
Propane	C3H8	CH3-CH2-CH3	3	3	-42
Butane	C4H10	CH3-CH2-CH2-CH3	4	4	0
Pentane	C5H12	CH3-CH2-CH2-CH2-CH3	5	5	36
Hexane	C6H14	CH3-CH2-CH2-CH2-CH2-CH3	6	6	69

b. Homologous series of alcohol.

Answer:

Name	Molecular formula	Condensed Structural Formula	Number of carbon atoms	Number of -CH2-Units	Boiling- Point °C
Methanol	CH4O	CH3OH	1	1	63
Ethanol	C2H6O	CH3-CH2OH	2	2	78
Propanol	C3H8O	CH3-CH2-CH2-OH	3	3	97
Butanol	C4H10O	CH3-CH2-CH2-CH2-OH	4	4	118

c. Homologous series of alkenes.

Answer:

Name	Molecular formula	Condensed Structural Formula	Number of carbon atoms	Number of -CH2- Units	Boiling Point °C
Ethyne	C2H4	H2C = CH2	2	0	-102
Propane	C3H6	CH3-CH = CH2	3	1	-48
1-Butene	C4H8	CH3-CH2-CH = CH2	4	2	-6.5
1-Pentane	C5H10	CH3-CH2-CH2-CH = CH2	5	3	+30

In simple words: These tables show homologous series for alkanes, alcohols, and alkenes, illustrating how their properties and formulas change predictably with an increasing number of carbon atoms.

🎯 Exam Tip: Understanding the general formula for each homologous series (alkanes: CnH2n+2, alkenes: CnH2n, alkynes: CnH2n-2, alcohols: CnH2n+1OH) is essential for quickly determining molecular formulas and identifying compounds.

Text Book Page No. 123

Question 1. Complete the table by writing the IUPAC names in the third column.
(Answer is directly given with underline.)

Answer:

Common name	Structural formula	IUPAC name
Ethylene	CH2 = CH2	Ethene
Acetylene	HC = CH	Ethyne
Acetic acid	CH3-COOH	Ethanoic acid
Methyl alcohol	CH3-OH	Methanol
Ethyl alcohol	CH3-CH2-OH	Ethanol

In simple words: This table maps common names and structural formulas to their systematic IUPAC names, which are standardized for chemical compounds.

🎯 Exam Tip: Memorize common names for simple organic compounds and practice converting them to IUPAC names, paying close attention to functional groups and parent chain length.

Use your brain power!

Question 1. By how many -CH2- (methylene) units do the formulae of the first two members of homologous series of alkanes, methane (CH4) and ethane (C2H6) differ? Similarly, by how many -CH2- units do the neighbouring members ethane (C2H6) and propane (C3H8) differ from each other?

Answer: The first two members of homologous series of alkanes, methane (CH4) and ethane (C2H6) differed by one -CH2- unit. Similarly, ethane (C2H6) and propane (C3H8) differed by -CH2- unit.
In simple words: Consecutive members in a homologous series always differ by one -CH2- unit.

🎯 Exam Tip: The constant difference of a -CH2- group between consecutive members is a defining characteristic of a homologous series, impacting their molecular mass and physical properties incrementally.

Question 2. How many methylene units are extra in the formula of the fourth member than the third member of the homologous series of alcohols?

Answer: There is only one, methylene unit extra in the formula of the fourth member and the third member of the homologous series of alcohols.
In simple words: The fourth member of any homologous series will have one more -CH2- unit than its third member.

🎯 Exam Tip: This question tests the understanding of the homologous series concept, where each successive member differs by a -CH2- group, irrespective of the specific functional group.

Question 3. How many methylene units are less in the formula of the second member than the third member of the homologous series of alkenes?

Answer: There is only one methylene unit less in the formula of the second member of and the third member of the homologous series of alkenes.
In simple words: The second member of a homologous series has one less -CH2- unit compared to the third member.

🎯 Exam Tip: This reinforces the understanding that the difference between adjacent members in any homologous series is consistently a single -CH2- unit.

 

Fill in the blanks and rewrite the completed statements:

Question 1. The organic compounds having double or triple bonds in them are termed as ............

Answer: The organic compounds having double or triple bonds in them are termed as unsaturated hydrocarbons.
In simple words: Organic compounds with double or triple bonds are called unsaturated hydrocarbons.

🎯 Exam Tip: Remember that "saturated" means only single bonds, while "unsaturated" indicates the presence of at least one double or triple bond.

Question 2. The general formula of alkane is ............

Answer: The general formula of alkane is CnH2n + 2.
In simple words: Alkanes follow the general formula CnH2n+2.

🎯 Exam Tip: Knowing the general formulas for alkanes, alkenes, and alkynes is fundamental for writing molecular formulas and understanding their classification.

Question 3. The compounds of homologous series have the same ............ group.

Answer: The compounds of homologous series have the same functional group.
In simple words: Members of a homologous series share the same functional group.

🎯 Exam Tip: The functional group is responsible for the characteristic chemical properties of a homologous series, even as the carbon chain length changes.

Question 4. A double bond is formed between carbon atoms by ............ pairs of electrons.

Answer: A double bond is formed between carbon atoms by two pairs of electrons.
In simple words: Two pairs of electrons are shared to form a double bond between carbon atoms.

🎯 Exam Tip: A single bond involves one shared pair, a double bond two shared pairs, and a triple bond three shared pairs of electrons.

Question 5. The compounds having different structural formulae having the same molecular formula is called ............

Answer: The compounds having different structural formulae having the same molecular formula is called structural isomerism.
In simple words: Isomerism describes compounds with the same molecular formula but different arrangements of atoms.

🎯 Exam Tip: Structural isomerism is a common phenomenon in organic chemistry; understanding it helps in predicting different possible structures for a given molecular formula.

Question 6. The functional group of ether is ............

Answer: The functional group of ether is -O-.
In simple words: The -O- group connecting two alkyl or aryl groups defines an ether.

🎯 Exam Tip: Identify common functional groups like -OH (alcohol), -CHO (aldehyde), -COOH (carboxylic acid), and -O- (ether) to classify organic compounds correctly.

 

Question 7. The general formula of alkene is ............

Answer: The general formula of alkene is CnH2n.
In simple words: Alkenes follow the formula CnH2n.

🎯 Exam Tip: Remember the differences in hydrogen count for alkanes, alkenes, and alkynes, which stem from the type of carbon-carbon bonds present.

Question 8. The bond between two atoms of nitrogen is a ............ bond.

Answer: The bond between two atoms of nitrogen is a triple bond.
In simple words: Nitrogen atoms form a triple bond in an N2 molecule.

🎯 Exam Tip: Nitrogen's high valency allows it to form a very strong triple bond, crucial for its stability in the atmosphere.

Question 9. Benzene ring is made up of ............ carbon atoms.

Answer: Benzene ring is made up of six carbon atoms.
In simple words: A benzene ring consists of six carbon atoms in a cyclic structure.

🎯 Exam Tip: Benzene's unique six-carbon ring with alternating single and double bonds (or delocalized electrons) is fundamental to aromatic chemistry.

Question 10. Due to ............., vegetable oil is converted into vanaspati ghee.

Answer: Due to hydrogenation, vegetable oil is converted into vanatspati ghee.
In simple words: Hydrogenation adds hydrogen to vegetable oil, changing it into vanaspati ghee.

🎯 Exam Tip: Hydrogenation is an important industrial process that converts unsaturated fats (oils) into saturated fats (ghee) using a catalyst.

Question 11. ............ control the heredity at molecular level.

Answer: Nucleic acids control the heredity at molecular level.
In simple words: Nucleic acids (DNA, RNA) are the molecules that carry genetic information.

🎯 Exam Tip: Nucleic acids are polymers of nucleotides and play a central role in genetics and protein synthesis.

Question 12. The regular repetition of a small unit is called ............

Answer: The regular repetition of a small unit is called polymer.
In simple words: A polymer is a large molecule made of many repeating smaller units called monomers.

🎯 Exam Tip: Understand the relationship between monomers (small units) and polymers (large chains formed by repeating monomers).

 

Question 13. The structural formula of polypropylene is ............

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पॉलीप्रोपाइलीन (पॉलीमर) की संरचना है, जिसमें एक लंबी कार्बन श्रृंखला होती है। इस श्रृंखला में प्रत्येक वैकल्पिक कार्बन परमाणु पर एक हाइड्रोजन और एक मिथाइल (CH3) समूह लगा होता है। 'n' इसकी पुनरावृत्ति इकाई को दर्शाता है।
In simple words: Polypropylene's structure consists of repeating units of propylene monomers, forming a long chain polymer.

🎯 Exam Tip: Polymer structures are shown with repeating units enclosed in brackets and an 'n' subscript. The monomer for polypropylene is propylene (propene), CH2=CH-CH3.

Question 14. The monomers of proteins are ............

Answer: The monomers of proteins are alpha amino acids.
In simple words: Proteins are built from smaller units called amino acids.

🎯 Exam Tip: Amino acids are the fundamental building blocks of proteins, linked together by peptide bonds.

Question 15. The monomer of cellulose is ............

Answer: The monomer of cellulose is glucose.
In simple words: Cellulose is a polysaccharide made up of glucose units.

🎯 Exam Tip: Glucose is a common monosaccharide and the monomer for many complex carbohydrates like starch and cellulose.

Question 16. ............ have sweet odour.

Answer: Esters have sweet odour.
In simple words: Esters are organic compounds known for their fruity and sweet smells.

🎯 Exam Tip: Esters are formed from the reaction of an alcohol and a carboxylic acid, and their characteristic aromas are often used in perfumes and flavorings.

Choose the correct alternative and rewrite the statement:

Question 1. The property of direct bonding between atoms of the same element to form a chain is called ............
(a) catenation
(b) isomerism
(c) dehydration
(d) polymerization

Answer: (a) catenation
In simple words: Catenation is the ability of an element to form bonds with itself, creating long chains or rings.

🎯 Exam Tip: Carbon's strong catenation property is the primary reason for the vast number and diversity of organic compounds.

Question 2. The molecular weight of two adjacent members in homologous series of an alkane differ by ............ units.
(a) 16
(b) 20
(c) 14
(d) 12

Answer: (c) 14
The molecular weight of two adjacent members in homologous series of an alkane differ by 14 units.
In simple words: Each -CH2- unit adds 14 atomic mass units (12 for C, 2 for H) to the molecular weight.

🎯 Exam Tip: This difference of 14 units (from a -CH2- group) is characteristic of all homologous series, not just alkanes.

Question 3. Consecutive members of a homologous series differ by ......... group.
(a) -CH
(b) -CH2
(C) -CH3
(d) -CH4

Answer: (b) -CH2
Consecutive members of a homologous series differ by CH2 group.
In simple words: The -CH2 group is the fundamental unit by which homologous series members differ.

🎯 Exam Tip: Understanding the -CH2- difference is crucial for predicting the properties and formulas of homologous series members.

Question 4. ............ is used to prepare carbon black.
(a) Methane
(b) Ethene
(c) Propane
(d) Butane

Answer: (a) Methane
Methane is used to prepare carbon black.
In simple words: Methane is cracked or incompletely burned to produce carbon black.

🎯 Exam Tip: Carbon black, a form of amorphous carbon, has various industrial uses, including as a pigment and a reinforcing filler in rubber products.

Question 5. ............ is the general formula of alkene.
(a) CnH2n
(b) CnH2n + 2
(c) CnH2n-2
(d) CnHn-2

Answer: (a) CnH2n
CnH2n is the general formula of alkene.
In simple words: Alkenes, having one double bond, follow the general formula CnH2n.

🎯 Exam Tip: Differentiate between the general formulas for alkanes (CnH2n+2), alkenes (CnH2n), and alkynes (CnH2n-2) as they indicate the degree of saturation.

 

Question 6. The reaction of methane with chlorine in the presence of sunlight is called ............
(a) pyrolysis
(b) an elimination reaction
(c) a substitution reaction
(d) an addition reaction

Answer: (c) a substitution reaction
The reaction of methane with chlorine in the presence of sunlight is called a substitution reaction.
In simple words: In this reaction, a chlorine atom replaces a hydrogen atom in methane.

🎯 Exam Tip: Substitution reactions are characteristic of saturated hydrocarbons (alkanes), where one atom or group is replaced by another.

Question 7. The general formula for alkynes is ............
(a) CnH2n
(b) CnH2n + 2
(c) CnH2n-2
(d) CnH2n-1

Answer: (c) CnH2n-2
The general formula for alkynes is CnH2n - 2
In simple words: Alkynes, with a triple bond, have the general formula CnH2n-2.

🎯 Exam Tip: The decreasing number of hydrogen atoms from alkanes to alkenes to alkynes reflects the presence of double and triple bonds.

Question 8. The reaction of ............ with ethanol is a fast reaction.
(a) calcium
(b) magnesium
(c) sodium
(d) aluminum

Answer: (c) sodium
The reaction of sodium with ethanol is a fast reaction.
In simple words: Sodium reacts vigorously with ethanol to produce sodium ethoxide and hydrogen gas.

🎯 Exam Tip: Alkali metals like sodium react readily with alcohols due to the acidic nature of the hydrogen atom attached to oxygen.

Question 9. Ethylene has ............ bond between two carbon atoms.
(a) a single
(b) a double
(c) a triple
(d) an ionic

Answer: (b) a double
Ethylene has a double bond between two carbon atoms.
In simple words: Ethylene (ethene) is an alkene, characterized by a carbon-carbon double bond.

🎯 Exam Tip: The presence of a double bond makes alkenes more reactive than alkanes, undergoing addition reactions.

 

Question 10. The saturated hydrocarbons are those in which carbon atom are linked by ............
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond

Answer: (a) a single bond
The saturated hydrocarbons are those in which carbon atom are linked by a single bond.
In simple words: Saturated hydrocarbons contain only carbon-carbon single bonds.

🎯 Exam Tip: The term "saturated" means that the carbon atoms are bonded to the maximum possible number of hydrogen atoms through single bonds.

Question 11. C7H16 is ............
(a) hexane
(b) octane
(c) methane
(d) heptane

Answer: (d) heptane
C7H16 is heptane.
In simple words: The formula C7H16 fits the general formula for alkanes (CnH2n+2) with n=7, which corresponds to heptane.

🎯 Exam Tip: Be able to match molecular formulas with the correct alkane names based on the number of carbon atoms.

Question 12.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कार्बन परमाणु (C) से डबल बॉन्ड द्वारा जुड़ा एक ऑक्सीजन परमाणु (O) और एकल बॉन्ड द्वारा जुड़ा एक हाइड्रॉक्सिल (OH) समूह दर्शाता है।
-C-OH is called ............
(a) carboxylic acid group
(b) aldehyde group
(c) ketonic group
(d) alcohol group

Answer: (a) carboxylic acid group
In simple words: The -COOH group, consisting of a carbonyl and a hydroxyl group, is characteristic of carboxylic acids.

🎯 Exam Tip: Visually identify and differentiate key functional groups; -COOH is distinctly the carboxylic acid group.

Question 13. The possible isomers for C5H12 are ............
(a) 2

(b) 4
(c) 1
(d) 3

Answer: (d) 3
The possible isomers for C5H12 are 3.
In simple words: For pentane (C5H12), there are three structural isomers: n-pentane, isopentane, and neopentane.

🎯 Exam Tip: Practice drawing all possible structural isomers for simple alkanes like C4H10 (2 isomers) and C5H12 (3 isomers) to master the concept of isomerism.

Question 14. ............ contains alcoholic functional group.
ℹ️ चित्र व्याख्या (Diagram Explanation): (a) एक कार्बन परमाणु से जुड़ा एक हाइड्रोजन, एक ऑक्सीजन, और एक हाइड्रोजन परमाणु को दर्शाता है, जिसमें ऑक्सीजन से एक हाइड्रोजन जुड़ा है। (b) एक कार्बन परमाणु से जुड़ा एक मिथाइल समूह (CH3) और एक हाइड्रोजन परमाणु दर्शाता है। (c) एक कार्बन परमाणु से जुड़ा एक मिथाइल समूह (CH3) और एक हाइड्रोजन परमाणु दर्शाता है। (d) (a), (b) और (c) तीनों विकल्प।
(a) H-C-O-H
H
(b) CH3-C-H
OH
(c) CH3-C-CH3
OH
(d) all of these

Answer: (d) all of these
In simple words: An alcoholic functional group is an -OH group attached to a carbon atom. All the given options (a, b, c) depict structures containing an -OH group on a carbon, therefore all contain an alcoholic functional group.

🎯 Exam Tip: The -OH group directly attached to an aliphatic carbon is the signature of an alcohol. Recognize its placement in various carbon skeletons.

Question 15. Oxygen molecule has ............ bond between two oxygen atoms.
(a) a double
(b) a single
(c) a triple
(d) an ionic

Answer: (a) a double
Oxygen molecule has a double bond between two oxygen atoms.
In simple words: In an O2 molecule, two oxygen atoms share two pairs of electrons, forming a double bond.

🎯 Exam Tip: Remember common diatomic molecules and the types of bonds they form (e.g., H2 single, N2 triple, O2 double).

Question 16. Some acetic acid is treated with solid NaHCO3. The resulting solution will be ............
(a) colourless
(b) blue
(c) green
(d) yellow

Answer: (a) colourless
Some acetic acid is treated with solid NaHCO3. The resulting solution will be colourless.
In simple words: The reaction between acetic acid and sodium bicarbonate produces sodium acetate, water, and carbon dioxide, all of which are colorless.

🎯 Exam Tip: This is a common test for carboxylic acids, producing colorless CO2 gas that can be observed as effervescence, and the resulting solution remains colorless.

 

Question 17. Ethanoic acid has a ......... odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like

Answer: (d) vinegar-like
Ethanoic acid has a vinegar-like odour.
In simple words: Ethanoic acid is commonly known as acetic acid, the main component of vinegar.

🎯 Exam Tip: Associating common names and properties (like odor) with IUPAC names helps in quick recall for exams.

Question 18. Acetic acid ............
(a) turns red litmus blue
(b) has pungent odour
(c) is red in colour
(d) is odourless

Answer: (b) has pungent odour
Acetic acid has pungent odour.
In simple words: Acetic acid has a sharp, strong smell.

🎯 Exam Tip: Acetic acid is a weak acid, turning blue litmus red, but its most distinct sensory property is its pungent, vinegar-like smell.

Question 19. When acetic acid reacts with sodium metal ............ gas is formed.
(a) oxygen
(b) hydrogen
(c) chlorine
(d) nitrogen

Answer: (b) hydrogen
When acetic acid reacts with sodium metal hydrogen gas is formed.
In simple words: Acetic acid, being an acid, reacts with reactive metals like sodium to release hydrogen gas.

🎯 Exam Tip: This is a characteristic reaction of acids with reactive metals, producing a salt and hydrogen gas.

Question 20. The molecular formula of acetic acid (ethanoic acid) is ............
(a) HCOOH
(b) CH3COOH
(C) C2H5COOH
(d) C3H7COOH

Answer: (b) CH3COOH

 

The molecular formula of acetic acid (ethanoic acid) is CH3COOH.
In simple words: Acetic acid has one methyl group (CH3) and one carboxylic acid group (COOH).

🎯 Exam Tip: Know the molecular formulas for simple carboxylic acids; methanoic acid (HCOOH), ethanoic acid (CH3COOH), propanoic acid (C2H5COOH).

Question 21. When sodium bicarbonate solution is added to dilute acetic acid ............
(a) a gas is evolved
(b) a solid settles at the bottom
(c) the mixture becomes warm
(d) the colour of the mixture becomes yellow

Answer: (a) a gas is evolved
When sodium bicarbonate solution is added to dilute acetic acid a gas-is evolved.
In simple words: Acetic acid reacts with sodium bicarbonate to produce carbon dioxide gas, which is observed as bubbles.

🎯 Exam Tip: The evolution of carbon dioxide gas is a common test for acids reacting with carbonates or bicarbonates.

Question 22. 2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in ............
(a) test tube A
(b) test tube B
(c) test tube C
(d) all the test tubes

Answer: (d) all the test tubes
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in all the test tubes.
In simple words: Ethanoic acid is miscible with water in all proportions, meaning it forms a clear solution regardless of the amount of water added within this range.

🎯 Exam Tip: The miscibility of ethanoic acid with water is due to its ability to form hydrogen bonds with water molecules, making it soluble in various concentrations.

Question 23. In the presence of acid catalyst, ethanoic acid reacts with ethanol and ............ ester is produced.
(a) ethanol
(b) ethanoic
(c) ethyl ethanoate
(d) ethyl ethanol (Practice Activity Sheet - 1)

Answer: (c) ethyl ethanoate
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ethyl ethanoate ester is produced.
In simple words: Ethanoic acid and ethanol react in the presence of an acid catalyst to form ethyl ethanoate and water.

🎯 Exam Tip: This is an esterification reaction, a reversible process where a carboxylic acid and an alcohol form an ester and water, typically catalyzed by a strong acid like H2SO4.

 

Question 24. The following structural formula belongs to which carbon compound?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक षटकोणीय रिंग संरचना दर्शाती है जिसमें छह कार्बन परमाणु और छह हाइड्रोजन परमाणु हैं। प्रत्येक कार्बन परमाणु एक हाइड्रोजन परमाणु से जुड़ा है और रिंग के भीतर वैकल्पिक एकल और दोहरा बंधन है, जो एक सुगंधित यौगिक को दर्शाता है।

(a) Camphor
(b) Benzene
(c) Starch
(d) Glucose
Answer: (b) Benzene
In simple words: The given structural formula with a hexagonal ring of carbon atoms and alternating single and double bonds is characteristic of benzene, an aromatic hydrocarbon.

🎯 Exam Tip: Recognizing common structural formulae like benzene is crucial for identifying organic compounds in exams.

 

Question 25. What type of reaction is shown below?
Sunlight
CH4 + Cl2 \( \longrightarrow \) CH3 - Cl + HCl
(a) Addition
(b) Substitution
(c) Decomposition
(d) Reduction
Answer: (b) substitution
In simple words: In this reaction, a hydrogen atom in methane is replaced by a chlorine atom, which is a classic example of a substitution reaction.

🎯 Exam Tip: Understand the definitions and typical examples of different types of organic reactions, especially substitution, addition, and oxidation-reduction.

 

Question 26. The carbon compound is used in daily life is ...........
(a) edible oil
(b) salt
(c) carbon dioxide
(d) baking soda
Answer: (a) edible oil
In simple words: Edible oil is a common carbon compound used daily for cooking and other purposes.

🎯 Exam Tip: Be familiar with everyday applications and examples of carbon compounds to answer questions related to their practical utility.

 

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

 

Question 1. Generally the melting and boiling points of carbon compounds are high.
Answer: False. (Generally the melting and boiling points of carbon compounds are low.)
In simple words: Carbon compounds typically have low melting and boiling points because they are held together by relatively weak intermolecular forces, not strong ionic bonds.

🎯 Exam Tip: Remember that organic compounds often have lower melting and boiling points compared to inorganic compounds due to their covalent nature and weaker intermolecular forces.

 

Question 2. Till now the number of known carbon compounds is about 10 million.
Answer: True.
In simple words: Due to carbon's unique ability to catenate and form bonds with many elements, a vast number of carbon compounds are known.

🎯 Exam Tip: The vast number of carbon compounds highlights the importance of organic chemistry and carbon's unique bonding properties.

 

Question 3. Unsaturated hydrocarbons are less reactive than saturated hydrocarbons.
Answer: False. (Unsaturated hydrocarbons are more reactive than saturated hydrocarbons.)
In simple words: Unsaturated hydrocarbons, with their double or triple bonds, are more reactive than saturated hydrocarbons because these multiple bonds can easily undergo addition reactions.

🎯 Exam Tip: Multiple bonds (double or triple) in unsaturated hydrocarbons are centers of reactivity, making them more prone to chemical reactions than single bonds in saturated hydrocarbons.

 

Question 4. Benzene is an aromatic compound.
Answer: True.
In simple words: Benzene is indeed classified as an aromatic compound due to its cyclic, planar structure and delocalized electron system.

🎯 Exam Tip: Recall the characteristics of aromatic compounds, such as the presence of a benzene ring or a similar delocalized pi-electron system, which distinguishes them.

 

Question 5. The carbon-carbon double and triple bonds are also recognised as functional groups.
Answer: True.
In simple words: Double and triple bonds between carbon atoms are considered functional groups because they impart specific chemical properties and reactivity to the molecule.

🎯 Exam Tip: Functional groups are atoms or groups of atoms that determine the characteristic chemical reactions of organic compounds, including multiple bonds.

 

Question 6. The general formula of alkyne is CnH2n.
Answer: False. (The general formula of alkyne is CnH2n - 2)
In simple words: The general formula for alkynes, which contain a carbon-carbon triple bond, is CnH2n - 2, representing two fewer hydrogen atoms than alkenes.

🎯 Exam Tip: Memorize the general formulae for alkanes (CnH2n+2), alkenes (CnH2n), and alkynes (CnH2n-2) to correctly identify and classify hydrocarbons.

 

Question 7. Ethanoic acid has a ......... odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like
Answer: (d) vinegar-like
In simple words: Ethanoic acid, also known as acetic acid, is the main component of vinegar and thus shares its characteristic sharp, vinegar-like odor.

🎯 Exam Tip: Associating common chemicals with their distinct physical properties like odor can help in identification and recall during exams.

 

Question 8. When vegetable oil and tincture iodine react, the color of iodine does not change.
Answer: False. (When vegetable oil and tincture iodine react, the colour of iodine changes.)
In simple words: Vegetable oils, being unsaturated, react with iodine in an addition reaction, causing the iodine color to disappear, indicating the presence of double bonds.

🎯 Exam Tip: The decolorization of iodine water is a standard test for unsaturation in organic compounds; remember that saturated compounds do not decolorize iodine.

 

Question 9. Saturated fats are healthy.
Answer: False. (Saturated fats are harmful to health.)
In simple words: Saturated fats are generally considered harmful to health because they can raise cholesterol levels and increase the risk of heart disease.

🎯 Exam Tip: Differentiate between saturated and unsaturated fats based on their health implications; unsaturated fats are generally healthier.

 

Question 10. Aqueous solution of ethanol is found to be neutral.
Answer: True.
In simple words: Ethanol is an alcohol and does not significantly ionize in water, resulting in a neutral aqueous solution.

🎯 Exam Tip: Alcohols are generally neutral compounds; distinguishing their pH properties from acids and bases is important.

 

Question 11. Denatured ethanol is used as industrial solvent.
Answer: True.
In simple words: Denatured ethanol is ethanol mixed with small amounts of poisonous substances to make it unfit for consumption, and it is widely used as a solvent in industries.

🎯 Exam Tip: Understand the purpose and composition of denatured alcohol, which is to make it non-potable for industrial use while avoiding high taxation on drinking alcohol.

 

Question 12. Vinegar is a 12-15 % aqueous solution of acetic acid.
Answer: False. (Vinegar is a 5-8 % aqueous solution of acetic acid.)
In simple words: Vinegar is typically a dilute solution of acetic acid, usually containing 5-8% acetic acid by volume.

🎯 Exam Tip: Know the common concentration ranges for household chemicals like vinegar; this detail can be tested in factual questions.

 

Question 13. The functional group of ethanoic acid is a carboxylic group.
Answer: True.
In simple words: Ethanoic acid, also known as acetic acid, contains the -COOH group, which is characteristic of carboxylic acids.

🎯 Exam Tip: Correctly identifying functional groups (e.g., -COOH for carboxylic acids, -OH for alcohols) is fundamental to organic chemistry nomenclature and reactions.

 

Question 14. Sodium hydroxide is used in the preparation of soap from fats and oils.
Answer: True.
In simple words: Sodium hydroxide is a key ingredient in saponification, the chemical process used to convert fats and oils into soap.

🎯 Exam Tip: Recall the process of saponification and the role of strong bases like sodium hydroxide in soap manufacturing.

 

Question 15. Rubber is a manmade macromolecule.
Answer: False. (Rubber is a natural macromolecule.)
In simple words: Natural rubber is a polymer derived from the latex of rubber trees, making it a natural macromolecule, not manmade.

🎯 Exam Tip: Distinguish between natural polymers (like rubber, cellulose) and synthetic polymers (like PVC, polystyrene).

 

Question 16. Polyvinyl chloride is used in the manufacture of P.V.C. pipes and bags.
Answer: True.
In simple words: Polyvinyl chloride (PVC) is a synthetic polymer widely used in the production of pipes, bags, and various other plastic products due to its versatility and durability.

🎯 Exam Tip: Be aware of the practical applications of common polymers and their corresponding full forms (e.g., PVC for Polyvinyl chloride).

 

Question 17. Polyethylene is a homopolymer.
Answer: True.
In simple words: Polyethylene is a homopolymer because it is formed by the repeated addition of identical ethylene monomer units.

🎯 Exam Tip: A homopolymer is formed from a single type of monomer, while a copolymer is formed from two or more different types of monomers.

 

Question 18. The chemical bonds in carbon compounds do not produce ions.
Answer: True.
In simple words: Carbon compounds primarily form covalent bonds by sharing electrons, rather than forming ions through electron transfer.

🎯 Exam Tip: Remember that organic compounds are generally covalent and do not dissociate into ions in solution, unlike many ionic inorganic compounds.

 

Find the odd one out:

 

Question 1. Propane, methane, ethene, pentane
Answer: Ethene. (Others are saturated hydrocarbons.)
In simple words: Ethene is an unsaturated hydrocarbon with a double bond, while propane, methane, and pentane are all saturated alkanes with only single bonds.

🎯 Exam Tip: Classify hydrocarbons based on their saturation (presence of single, double, or triple bonds) to easily identify the odd one out.

 

Question 2. Methane, butane, benzene, sodium chloride
Answer: Sodium chloride. (Others are organic compounds.)
In simple words: Sodium chloride is an inorganic ionic compound, whereas methane, butane, and benzene are all organic covalent compounds containing carbon.

🎯 Exam Tip: Distinguish between organic and inorganic compounds; organic compounds contain carbon, typically bonded to hydrogen, while inorganic compounds generally do not.

 

Question 3. CH4, C2H6, C3H8, CaCO3
Answer: CaCO3. (Others are organic compounds.)
In simple words: Calcium carbonate (CaCO3) is an inorganic compound, while methane (CH4), ethane (C2H6), and propane (C3H8) are all organic hydrocarbons.

🎯 Exam Tip: Always be mindful of the basic definition of organic vs. inorganic compounds to correctly categorize them.

 

Question 4. C2H2, C3H8, C2H6, CH4
Answer: C2H2. (Others are saturated hydrocarbons.)
In simple words: Ethyne (C2H2) is an alkyne with a triple bond and is unsaturated, while propane (C3H8), ethane (C2H6), and methane (CH4) are saturated alkanes.

🎯 Exam Tip: Pay attention to the molecular formula and the corresponding general formula (CnH2n+2, CnH2n, CnH2n-2) to determine saturation.

 

Question 5. C2H4, C4H10, C3H8, CH4
Answer: C2H4. (Others are saturated hydrocarbons.)
In simple words: Ethene (C2H4) is an alkene with a double bond and is unsaturated, whereas butane (C4H10), propane (C3H8), and methane (CH4) are saturated alkanes.

🎯 Exam Tip: The CnH2n general formula signals an alkene (unsaturated), distinguishing it from alkanes (CnH2n+2, saturated).

 

Question 6. Polyethylene, Polysaccharide, Polystyrene, Polypropylene
Answer: Polysaccharide (Others are manmade polymers.)
In simple words: Polysaccharide is a natural polymer (like starch or cellulose), while polyethylene, polystyrene, and polypropylene are all synthetic, manmade polymers.

🎯 Exam Tip: Recognize common examples of natural polymers (e.g., proteins, polysaccharides, natural rubber) versus synthetic polymers.

 

Question 7. -NH2, -COOH,-SO4, -Br
Answer: -SO4 (Others are functional groups.)
In simple words: -SO4 (sulfate ion) is an inorganic ion, while -NH2 (amine), -COOH (carboxylic acid), and -Br (bromo) are common organic functional groups.

🎯 Exam Tip: Familiarize yourself with common organic functional groups and their characteristic structures to differentiate them from inorganic ions.

 

Question 8. Methane, Ethane, Propene, Propane, Butane
Answer: Propene (Others are members of homologous series of alkanes.)
In simple words: Propene is an alkene with a double bond, making it an unsaturated hydrocarbon, while methane, ethane, propane, and butane are all saturated alkanes.

🎯 Exam Tip: The suffix "-ene" indicates an alkene (double bond), while "-ane" indicates an alkane (single bonds).

 

Match the columns:

 

Question 1.

Column I	Column II
(1) CH4	(a) CH2 = CH2
(2) Ethane	(b) CnH2n-2
(3) Alkene	(c) Methane
(4) Alkyne	(d) C2H6
	(e) C3H8

Answer:
(1) CH4 - Methane
(2) Ethane - C2H6
(3) Alkene - CH2 = CH2
(4) Alkyne - CnH2n - 2.
In simple words: This matching exercise pairs chemical formulas and classifications with their corresponding names or general formulas. Methane is CH4, Ethane is C2H6, Alkene refers to compounds with a C=C double bond (e.g., CH2=CH2), and Alkyne has the general formula CnH2n-2.

🎯 Exam Tip: A strong understanding of hydrocarbon nomenclature, molecular formulas, and general formulas for different homologous series is essential for accurately matching columns.

 

Question 2.

Column I	Column II
(1) Aromatic hydrocarbon	(a) Propyne
(2) Alkane	(b) Benzene
(3) Alkyne	(c) Saturated hydrocarbon
(4) Alkene	(d) CnH2n
	(e) Cn H2n-1 OH

Answer:
(1) Aromatic hydrocarbon - Benzene
(2) Alkane - Saturated hydrocarbon
(3) Alkyne - Propyne
(4) Alkene - CnH2n
In simple words: This match connects hydrocarbon classes with their examples or general characteristics. Benzene is a classic aromatic hydrocarbon, alkanes are saturated, propyne is an example of an alkyne, and alkenes have the general formula CnH2n.

🎯 Exam Tip: Familiarity with key examples and characteristic features (like general formulas or saturation levels) for each class of hydrocarbons is crucial for matching type questions.

 

Question 3.

Column I	Column II
(1) Cyclohexane	(a) CH3COOH
(2) Methanol	(b) CH3Cl
(3) Acetaldehyde	(c) CH2Cl2
(4) Ethanoic acid	(d) CH3OH
	(e) C6H12
	(f) CH3CHO

Answer:
(1) Cyclohexane - C6H12
(2) Methanol - CH3OH
(3) Acetaldehyde - CH3CHO
(4) Ethanoic acid - CH3COOH.
In simple words: This matching links organic compound names with their correct molecular formulas. Cyclohexane is C6H12, Methanol is CH3OH, Acetaldehyde is CH3CHO, and Ethanoic acid is CH3COOH.

🎯 Exam Tip: Knowing the common names and corresponding molecular formulas for basic organic compounds is essential for successful matching and problem-solving.

 

Question 4.

Column I	Column II
(1) - OH	(a) Amine
(2) - COOH	(b) Aldehyde
(3) - CHO	(c) Ketone
(4) \( \mathrm{\underset{\Vert}{C}-} \)	(d) Alcohol
	(e) Carboxylic acid

Answer:
(1) (-OH) - Alcohol
(2) (-COOH) - Carboxylic acid
(3) (-CHO) - Aldehyde
(4) (\( \mathrm{\underset{\Vert}{C}-} \)) - Ketone.
In simple words: This match connects functional group structures with their class names. -OH is an alcohol, -COOH is a carboxylic acid, -CHO is an aldehyde, and a carbonyl group (\( \mathrm{\underset{\Vert}{C}-} \)) within a chain is a ketone.

🎯 Exam Tip: Accurately identifying functional groups by their structural representation is a fundamental skill in organic chemistry nomenclature and understanding reactivity.

 

Question 5.

Column I	Column II
(1) Ethyne	(a) C2H6
(2) Ethene	(b) C2H2
(3) Ethane	(c) C3H6
(4) Propyne	(d) C2H4
	(e) C3H4

Answer:
(1) Ethyne - C2H2
(2) Ethene - C2H4
(3) Ethane - C2H6
(4) Propyne - C3H4.
In simple words: This match connects the names of hydrocarbons with their correct molecular formulas. Ethyne is C2H2, Ethene is C2H4, Ethane is C2H6, and Propyne is C3H4.

🎯 Exam Tip: Mastery of IUPAC nomenclature and corresponding molecular formulas for simple alkanes, alkenes, and alkynes is critical for success in organic chemistry.

 

Question 6.

Column I	Column II
(1) Cellulose	(a) P.V.C. pipes, bags
(2) R.N.A	(b) Blankets
(3) Polyacrylonitrile	(c) Wood
(4) Polyvinyl chloride	(d) Chromosomes of plants

Answer:
(1) Cellulose - Wood
(2) R.N.A. - Chromosomes of plants
(3) Polyacrylonitrile - Blankets
(4) Polyvinyl chloride - P.V.C. pipes, bags.
In simple words: This match connects various polymers with their primary applications or natural occurrences. Cellulose is found in wood, RNA is a component of chromosomes in plants, polyacrylonitrile is used for blankets, and polyvinyl chloride (PVC) is used for pipes and bags.

🎯 Exam Tip: Understanding the applications and biological roles of common natural and synthetic polymers is useful for both chemistry and biology questions.

 

Define the following:

 

Question 1. Define Alkane
Answer: Alkane: In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: Methane (CH4), Ethane (C2H6)
In simple words: Alkanes are a class of saturated hydrocarbons where all carbon atoms are linked by single bonds, ensuring each carbon satisfies its tetravalency.

🎯 Exam Tip: Remember that alkanes are saturated hydrocarbons and their general formula is CnH2n+2. They are generally less reactive due to single bonds.

 

Question 2. Define Alkene.
Answer: Alkene: The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example : Ethene (CH2 = CH2)
In simple words: Alkenes are hydrocarbons that contain at least one carbon-carbon double bond, making them unsaturated and more reactive than alkanes.

🎯 Exam Tip: Alkenes follow the general formula CnH2n. The presence of a double bond makes them susceptible to addition reactions.

 

Question 3. Define Alkyne.
Answer: Alkyne: The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes.
Example: Ethyne C2H2 (CH = CH).
In simple words: Alkynes are unsaturated hydrocarbons characterized by the presence of at least one carbon-carbon triple bond.

🎯 Exam Tip: Alkynes have the general formula CnH2n-2 and are highly reactive due to the triple bond, undergoing addition reactions similar to alkenes, but often in two steps.

 

Question 4. Define Addition reaction.
Answer: Addition reaction: When a carbon compound combines with another compound to form a product that contains all the atoms in both the reactants; it is called an addition reaction.
Nil
Vegetable oil + H2(g) \( \underset{\text{catalyst}}{\xrightarrow{\hspace{1.5cm}}} \) Vanaspati ghee
In simple words: An addition reaction occurs when atoms are added across a double or triple bond in an unsaturated compound, forming a single saturated product without losing any atoms.

🎯 Exam Tip: Addition reactions are characteristic of unsaturated compounds (alkenes and alkynes) and involve the breaking of pi bonds to form new sigma bonds.

 

Question 5. Define Substitution reaction.
Answer: Substitution reaction: The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms, is called substitution reaction.
Sunlight
CH4 + Cl2 \( \longrightarrow \) CH3Cl + HCl
Sunlight
CH3Cl + Cl2 \( \longrightarrow \) CH2Cl2 + HCl
Sunlight
CH2Cl2 + Cl2 \( \longrightarrow \) CHCl3 + HCl
In simple words: A substitution reaction is a chemical process where one atom or functional group in a molecule is replaced by another atom or functional group.

🎯 Exam Tip: Substitution reactions are typical for saturated hydrocarbons (alkanes) and aromatic compounds, often requiring specific conditions like UV light for alkanes.

 

Question 6. Define Esterification.
Answer: Esterification: A carboxylic acid reacts with an alcohol in presence of an acid catalyst, an ester is formed. The reaction is known as esterification.
Acid
CH3-COOH + CH3-CH2-OH \( \underset{\text{catalyst}}{\xrightarrow{\hspace{1.5cm}}} \) CH3COOC2H5 + H2O
In simple words: Esterification is the chemical reaction between a carboxylic acid and an alcohol, typically catalyzed by an acid, to produce an ester and water.

🎯 Exam Tip: Remember that esterification is a condensation reaction where water is eliminated, and esters are known for their characteristic fruity smells.

 

Question 7. Define Saponification.
Answer: Saponification: When an ester reacts with the alkali, i.e. sodium hydroxide, the corresponding alcohol and sodium salt of carboxylic acid are obtained. This reaction is called saponification reaction. It is used in the preparation of soap.
Ester + Sodium hydroxide \( \longrightarrow \) Sodium salt of carboxylic acid + Alcohol.
In simple words: Saponification is the hydrolysis of an ester, usually a fat or oil, using a strong base like sodium hydroxide to produce an alcohol and a carboxylate salt, which is soap.

🎯 Exam Tip: Saponification is essentially the reverse of esterification, specifically applied to fats and oils to produce soap and glycerol.

 

Question 8. Define Polymerization.
Answer: Polymerization: The reaction by which monomer molecules are converted into a polymer is called polymerization.
Polymerization
nCH2 = CH2 \( \underset{\text{Monomer}}{\xrightarrow{\hspace{1.5cm}}} \) (CH2-CH2)n
In simple words: Polymerization is the process where many small molecules, called monomers, link together to form a large chain-like molecule called a polymer.

🎯 Exam Tip: Distinguish between addition polymerization (monomers add without loss of atoms, like in polyethene) and condensation polymerization (monomers join with loss of small molecules like water).

 

Name the following:

 

Question 1. The higher homologue of hexane.
Answer: Heptane.
In simple words: The next member in the alkane homologous series after hexane is heptane, which has one more carbon atom.

🎯 Exam Tip: In a homologous series, each successive member differs by a -CH2- unit and follows a similar naming pattern (e.g., methane, ethane, propane, butane, pentane, hexane, heptane).

 

Question 2. The number of double bonds in benzene.
Answer: Three.
In simple words: Benzene has a six-membered carbon ring with three alternating double bonds.

🎯 Exam Tip: Remember that benzene's structure includes three double bonds, which are delocalized, giving it unique stability and aromatic properties.

 

Question 3. The functional group in ether and halogen.
Answer: Functional groups:
Ether: - O -
Halogen: - X (-Cl, -Br, -I).
In simple words: The ether functional group is an oxygen atom bonded to two alkyl or aryl groups, while halogens are represented by -X, standing for fluorine, chlorine, bromine, or iodine.

🎯 Exam Tip: Be able to identify and draw the structures of common functional groups as they define the chemical properties of organic compounds.

 

Question 4. Polymer of tetrafluoroethylene.
Answer: Teflon.
In simple words: Teflon is a synthetic polymer made from repeating units of tetrafluoroethylene, known for its non-stick properties.

🎯 Exam Tip: Connect monomers to their corresponding polymers and be aware of common uses (e.g., tetrafluoroethylene -> Teflon for non-stick coatings).

 

Question 5. The monomer of polysaccharide.
Answer: Glucose.
In simple words: Polysaccharides like starch or cellulose are large polymers made up of repeating glucose monomer units.

🎯 Exam Tip: Glucose is the fundamental building block (monomer) for many important biological polysaccharides.

 

Question 6. The Polymer of nucleotide.
Answer: D.N.A./R.N.A.
In simple words: DNA and RNA are nucleic acids, which are biological polymers made up of repeating nucleotide units.

🎯 Exam Tip: Nucleotides are the monomers for nucleic acids (DNA and RNA), which carry genetic information.

 

Question 7. The Monomer of rubber.
Answer: Isoprene.
In simple words: Natural rubber is a polymer formed from the repeated linking of isoprene monomer units.

🎯 Exam Tip: Know the monomer for natural rubber (isoprene) and understand its importance in the rubber industry.

 

Question 8. Two oxidising compounds.
Answer: Potassium permanganate, Potassium dichromate.
In simple words: Potassium permanganate and potassium dichromate are common strong oxidizing agents used in chemistry.

🎯 Exam Tip: Remember common oxidizing agents and reducing agents, as they are frequently involved in redox reactions in organic chemistry.

 

Question 9. IUPAC name of sodium acetate.
Answer: Sodium ethanoate.
In simple words: The common name sodium acetate corresponds to the IUPAC name sodium ethanoate.

🎯 Exam Tip: Be proficient in converting common names of organic compounds to their IUPAC names and vice versa, especially for carboxylic acid derivatives.

 

Question 10. The main component of natural gas.
Answer: Methane.
In simple words: Methane is the primary and most abundant component of natural gas, a fossil fuel.

🎯 Exam Tip: Methane (CH4) is the simplest alkane and is a major source of energy.

 

Question 11. Two isomers of butane.
Answer: n-butane and i-butane.
In simple words: Butane (C4H10) exists as two structural isomers: normal butane (a straight chain) and isobutane (a branched chain).

🎯 Exam Tip: Isomers have the same molecular formula but different structural arrangements; be able to draw and name simple isomers.

 

Question 12. A nomenclature system based on the structure of the compounds and it was accepted all over the world.
Answer: International Union of Pure and Applied Chemistry (IUPAC).
In simple words: The IUPAC system provides a standardized, universally accepted method for naming chemical compounds based on their structure.

🎯 Exam Tip: The IUPAC nomenclature system is essential for unambiguous communication in chemistry; understanding its rules is fundamental.

 

Answer the following questions in one sentence each:

 

Question 1. State the atomic number and electronic configuration of carbon.
Answer: The atomic number of carbon is 6 and the electronic configuration of carbon is (2, 4).
In simple words: Carbon has an atomic number of 6, meaning it has 6 protons and 6 electrons, with 2 electrons in the first shell and 4 in the outermost shell.

🎯 Exam Tip: Knowing the atomic number and electronic configuration of common elements like carbon is foundational for understanding their bonding behavior.

 

Question 2. State number of electrons in the outermost orbit of carbon and valency of carbon.
Answer: Four electrons are present in the outermost orbit of carbon and the valency of carbon is 4.
In simple words: Carbon has four valence electrons, which gives it a valency of 4, allowing it to form four covalent bonds.

🎯 Exam Tip: Carbon's valency of 4 (tetravalency) is the key reason for its ability to form a vast number of diverse organic compounds.

 

Question 3. What are hydrocarbons? Give one example.
Answer: The compounds containing only carbon and hydrogen are called hydrocarbons. These compounds are known as organic compounds. E.g. Methane, Ethane.
In simple words: Hydrocarbons are organic compounds composed solely of carbon and hydrogen atoms, such as methane.

🎯 Exam Tip: Hydrocarbons are the simplest organic compounds and serve as the backbone for more complex organic molecules containing functional groups.

 

Question 4. What is the molecular formula and structural of methane?
Answer: The molecular formula of methane is CH4.


ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कार्बन परमाणु (केंद्रीय बिंदु) को चार हाइड्रोजन परमाणुओं (ऊपर, नीचे, बाएँ, दाएँ) से एकल सहसंयोजक बंधों (हाइफ़न) के माध्यम से जुड़ा हुआ दिखाता है, जो मीथेन की चतुष्फलकीय संरचना को दर्शाता है।

In simple words: The molecular formula for methane is CH4, and its structural formula shows a central carbon atom bonded to four hydrogen atoms.

🎯 Exam Tip: Be able to write both the molecular and structural formulas for simple alkanes like methane, remembering carbon's tetravalency.

 

Question 5. How many atoms of carbon and hydrogen are present in methane?
Answer: The molecule of methane has one carbon atom and four hydrogen atoms.
In simple words: A methane molecule consists of one carbon atom and four hydrogen atoms.

🎯 Exam Tip: The molecular formula CH4 directly indicates the number of each type of atom present in a methane molecule.

 

Question 6. State the general formula of alkane.
Answer: The general formula of an alkane is CnH2n + 2.
In simple words: The general formula CnH2n + 2 represents any alkane, where 'n' is the number of carbon atoms.

🎯 Exam Tip: This general formula is crucial for determining the molecular formula of any alkane given its number of carbon atoms.

 

Question 7. Give two examples of alkanes.
Answer: Methane (CH4) and ethane (C2H6) are alkanes.
In simple words: Methane and ethane are common examples of alkanes, which are saturated hydrocarbons.

🎯 Exam Tip: Be familiar with the first few members of the alkane series and their corresponding molecular formulas.

 

Question 8. Give two examples of alkenes.
Answer: Ethene (CH2 = CH2) and propene (CH3 - CH = CH2) are alkenes.
In simple words: Ethene and propene are two examples of alkenes, characterized by the presence of a carbon-carbon double bond.

🎯 Exam Tip: Recognize the "-ene" suffix and the C=C double bond as key indicators for alkenes.

Question 9.Give two examples of alkynes.
Answer:Ethyne (HC = CH) and propyne (CH3-C = CH) are alkynes.
In simple words: Alkynes are hydrocarbons with a triple bond. Ethyne and propyne are two common examples of such compounds.

🎯 Exam Tip: Remembering the general formula for alkynes (\(C_nH_{2n-2}\)) and simple examples like ethyne and propyne is crucial for nomenclature and identifying hydrocarbon types.

Question 10.Write the name and molecular formula of a higher homologue of propane.
Answer:Butane (C4H10) is a higher homologue of propane.
In simple words: A higher homologue means the next member in the same series with one more -CH2 unit. For propane, that's butane.

🎯 Exam Tip: Understanding homologous series and how each member differs by a -CH2 unit is fundamental for predicting chemical properties and molecular formulas.

Question 11.Write the structure and molecular formula of ethane.
Answer:Structure of ethane:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथेन की संरचना को दर्शाता है, जिसमें दो कार्बन परमाणु एक एकल बंधन से जुड़े होते हैं और प्रत्येक कार्बन परमाणु से तीन हाइड्रोजन परमाणु एकल बंधन द्वारा जुड़े होते हैं। यह एक सीधा श्रृंखला वाला संतृप्त हाइड्रोकार्बन है। Molecular formula: C2H6
In simple words: Ethane has two carbon atoms linked by a single bond, with each carbon also bonded to three hydrogen atoms, giving it the formula \(C_2H_6\).

🎯 Exam Tip: When drawing structural formulas, ensure each carbon atom satisfies its tetravalency by forming four bonds, and hydrogen forms one bond.

Question 12.What is meant by catenation power?
Answer:Carbon has a unique ability to form strong covalent bonds with other carbon atoms, this result in formation of big molecules. This property of carbon is called catenation power.
In simple words: Catenation is carbon's special ability to form long chains or rings by bonding with itself, creating large and diverse molecules.

🎯 Exam Tip: Catenation is a key characteristic of carbon that explains the vast number of organic compounds. Mentioning "strong covalent bonds" and "big molecules" are important points for full marks.

Question 13.State the structural and molecular formula of benzene.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह बेंजीन की संरचना को दर्शाता है, जिसमें छह कार्बन परमाणु एक हेक्सागोनल रिंग बनाते हैं। प्रत्येक कार्बन परमाणु एक हाइड्रोजन परमाणु से जुड़ा होता है, और कार्बन-कार्बन बंधों में एकांतर एकल और दोहरा बंधन होते हैं, जो अनुनाद को दर्शाते हैं। Molecular formula of benzene: C6H6
In simple words: Benzene is a cyclic hydrocarbon with six carbon atoms forming a ring, each bonded to one hydrogen, and features alternating single and double bonds.

🎯 Exam Tip: Clearly state both the molecular formula (\(C_6H_6\)) and describe the cyclic, alternating bond structure for benzene. Understanding its aromatic nature is also helpful.

Question 14.Which functional groups are present in aldehyde and ketone?
Answer:The functional group -CHO is present in aldehyde and the functional group \( \text{O} \) \( \text{||} \) \( \text{-C-} \) is present in ketone.
In simple words: Aldehydes contain a carbonyl group bonded to a hydrogen atom and an alkyl group, while ketones have a carbonyl group bonded to two alkyl groups.

🎯 Exam Tip: Remember that both aldehydes and ketones contain the carbonyl group (\( \text{C=O} \)), but their attachments differentiate them: -CHO for aldehyde, and -CO- for ketone.

Question 15.Which functional group is present in \( \text{O} \) \( \text{||} \) CH3-C-OH?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह CH3-C=O-OH संरचना का प्रतिनिधित्व करता है, जिसे सामान्यतः कार्बोक्सिलिक एसिड समूह के रूप में जाना जाता है। इसमें एक मिथाइल समूह, एक कार्बोक्सिल समूह होता है जिसमें एक कार्बन परमाणु दोहरे बंधन से ऑक्सीजन से जुड़ा होता है और एक एकल बंधन से एक हाइड्रॉक्सिल समूह से जुड़ा होता है। Carboxylic group, \( \text{O} \) \( \text{||} \) \( \text{-C-OH} \) functional group is present in CH3-C-OH
In simple words: The functional group present is the carboxylic acid group, characterized by a carbon atom double-bonded to an oxygen and single-bonded to a hydroxyl (-OH) group.

🎯 Exam Tip: Identify the -COOH group as the carboxylic acid functional group. This group gives organic acids their acidic properties.

Question 16.Compare: The proportions of carbon atoms in ethanol (C2H5OH) and naphthalene (C10H8).
Answer:Ethanol contains two carbon atoms while naphthalene contains 10 carbon atoms. Ethanol is a saturated hydrocarbon and naphthalene is an unsaturated hydrocarbon.
In simple words: Ethanol has fewer carbon atoms and is saturated, while naphthalene has many more carbons in an unsaturated, ring structure.

🎯 Exam Tip: A good comparison highlights both the number of carbon atoms and the saturation/unsaturation status, as these fundamentally influence chemical behavior.

Question 17.What are the products of combustion of methane?
Answer:Carbon dioxide (CO2) and water (H2O) are the products of combustion of methane.
In simple words: When methane burns completely, it produces carbon dioxide and water.

🎯 Exam Tip: For complete combustion of any hydrocarbon, the products are always carbon dioxide and water. Incomplete combustion can produce carbon monoxide or soot.

Question 18.Which gas is evolved when ethanol reacts with sodium?
Answer:Hydrogen gas (H2) is evolved when ethanol reacts with sodium.
In simple words: When sodium reacts with ethanol, hydrogen gas is released.

🎯 Exam Tip: This reaction is characteristic of alcohols acting as weak acids. The evolution of hydrogen gas is a key indicator of this reaction.

Question 19.Compare: How is the transformation of ethanol into ethanoic acid on oxidation reaction?
Answer:The transformation of ethanol into ethanoic acid is an oxidation process, in which ethanol accepts oxygen.
In simple words: Ethanol converts to ethanoic acid through oxidation, meaning it gains oxygen.

🎯 Exam Tip: Recognize oxidation in organic chemistry as the gain of oxygen or loss of hydrogen. This specific transformation is a common example of alcohol oxidation to a carboxylic acid.

Question 20.Complete the following: \( \text{CH}_3\text{-CH}_2\text{-OH} \xrightarrow{(\text{O})} \text{alkaline KMnO}_4 \)
Answer:\( \text{CH}_3\text{-CH}_2\text{-OH} \xrightarrow{\text{(O) alkaline KMnO}_4} \text{CH}_3\text{-C-OH} \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथेनॉल के ऑक्सीकरण की अभिक्रिया को दर्शाता है। एथेनॉल (CH3-CH2-OH) क्षारीय पोटेशियम परमैंगनेट की उपस्थिति में ऑक्सीकृत होकर एथेनोइक एसिड (CH3-C(=O)-OH) बनाता है। यह एक अल्कोहल से कार्बोक्सिलिक एसिड में परिवर्तन है। Ethanol \( \longrightarrow \) Ethanoic acid
In simple words: Ethanol, upon oxidation with alkaline potassium permanganate, converts into ethanoic acid.

🎯 Exam Tip: Alkaline potassium permanganate (\( \text{KMnO}_4 \)) is a strong oxidizing agent that can convert primary alcohols directly to carboxylic acids.

Question 21.Which of the following hydrocarbons undergo addition reactions: C2H6, C3H8, C2H4, C3H8?
Answer:C2H4 (ethene) and C3H6 (propene) undergo addition reactions.
In simple words: Unsaturated hydrocarbons like ethene (\(C_2H_4\)) and propene (\(C_3H_6\)), which contain double or triple bonds, undergo addition reactions.

🎯 Exam Tip: Addition reactions are characteristic of unsaturated hydrocarbons (alkenes and alkynes) due to the presence of double or triple bonds, while saturated hydrocarbons (alkanes) typically undergo substitution reactions.

Question 22.How many covalent bonds are there in a molecule of cyclohexane?
Answer:A molecule of cyclohexane contains 18 covalent bonds.
In simple words: Cyclohexane (\(C_6H_{12}\)) has 18 covalent bonds: 6 carbon-carbon bonds in the ring and 12 carbon-hydrogen bonds.

🎯 Exam Tip: To count covalent bonds, sum up all C-C and C-H bonds. For cyclohexane (\(C_6H_{12}\)), there are 6 C-C bonds and 12 C-H bonds, totaling 18.

Question 23.Give the IUPAC name for CH3COOH.
Answer:The IUPAC name for CH3COOH is ethanoic acid.
In simple words: CH3COOH is commonly known as acetic acid, but its IUPAC name is ethanoic acid.

🎯 Exam Tip: For carboxylic acids, the IUPAC name is derived from the parent alkane by replacing '-e' with '-oic acid'. With two carbons, methane becomes methanoic, ethane becomes ethanoic.

Question 24.Write the IUPAC name of CH3COONa.
Answer:IUPAC name of CH3COONa is sodium ethanoate.
In simple words: CH3COONa is the sodium salt of ethanoic acid, thus named sodium ethanoate.

🎯 Exam Tip: Salts of carboxylic acids are named by replacing the '-oic acid' suffix with '-oate' and adding the name of the metal cation (e.g., sodium).

Question 25.What is meant by denatured alcohol?
Answer:Ethanol is the important commercial solvent. To prevent the misuse of this solvent, it is mixed with the poisonous methanol. Such ethanol is called denatured spirit.
In simple words: Denatured alcohol is ethanol made unsuitable for drinking by adding toxic substances like methanol, to avoid its consumption.

🎯 Exam Tip: The key purpose of denaturing alcohol is to prevent its consumption as an alcoholic beverage while still allowing its use as an industrial solvent, typically done by adding methanol or pyridine.

Question 26.What is meant by glacial acetic acid?
Answer:The melting point of pure acetic acid is 17 °C. Therefore, during winter in old countries acetic acid freezes at room temperature itself and looks like ice. Therefore it is named glacial acetic acid.
In simple words: Glacial acetic acid is pure ethanoic acid that freezes into an ice-like solid at or below 17 °C, hence its name.

🎯 Exam Tip: The term "glacial" refers to its ice-like appearance when frozen. Emphasize its purity and freezing behavior at a relatively high temperature compared to water.

Question 26.Which useful components of hydro-carbon are obtained by fractional distillation of crude oil?
Answer:Various useful components of hydrocarbon such as CNG, LPG, petrol (gasoline), rockel, diesel, engine oil, lubricant, etc. are obtained by separation of crude oil using fractional distillation.
In simple words: Fractional distillation of crude oil yields many useful products like petrol, diesel, LPG, and various oils.

🎯 Exam Tip: List several common products obtained from crude oil refining to show a comprehensive understanding of its importance. Fractional distillation separates these based on boiling points.

Question 28.Which functional groups are present in ester and amine?
Answer:Ester: -COO- Amine: -NH2-
In simple words: Esters have the -COO- group, and amines have the -NH2- group.

🎯 Exam Tip: Clearly distinguish between the carbonyl-oxygen-oxygen linkage for esters and the nitrogen-hydrogen linkage for amines.

Question 29.Give two examples of natural macromolecules.
Answer:Examples: Polysaccharide, protein and nucleic acid.
In simple words: Polysaccharides and proteins are examples of large, naturally occurring molecules.

🎯 Exam Tip: Natural macromolecules are large organic molecules found in living systems. Mentioning polysaccharides, proteins, or nucleic acids are good examples.

Question 30.Write the structure of polystyrene and give its uses.
Answer:Structure:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पॉलीस्टाइनिन की पुनरावृत्ति इकाई संरचना को दर्शाता है, जिसमें कार्बन परमाणुओं की एक मुख्य श्रृंखला होती है। प्रत्येक दूसरे कार्बन से एक फिनाइल समूह (एक बेंजीन रिंग) जुड़ा होता है। n का मतलब है कि यह इकाई कई बार दोहराई जाती है, जो एक बहुलक बनाती है। Polystyrene is used to make thermocoal articles.
In simple words: Polystyrene is a polymer used to make items like Styrofoam cups and insulation.

🎯 Exam Tip: When providing a polymer's structure, show the repeating monomer unit within brackets with 'n' as a subscript. Mentioning at least one practical use is also important.

Question 31.Write the name and the structure of monomer of polyacrylonitrile.
Answer:The name and structure of monomer: Acrylonitrile CH2 = CH - CN
In simple words: The monomer for polyacrylonitrile is acrylonitrile, which has a double bond and a cyanide group.

🎯 Exam Tip: For monomers, identify the double bond (or triple bond) that facilitates polymerization and any characteristic functional groups (like the nitrile group here).

Question 32.Write the name and the structure of monomer of teflon and its uses.
Answer:The name and structure of monomer: Tetrafluro ethylene CF2 = CF2 Teflon is used to make nonstick cookware.
In simple words: The monomer of Teflon is tetrafluoroethylene, and Teflon is widely used for nonstick coatings on pans.

🎯 Exam Tip: When asked for monomer and use, ensure both are provided. Tetrafluoroethylene's high fluorine content gives Teflon its non-stick and chemical-resistant properties.

Question 33.What is meant by copolymers?
Answer:The polymers formed from two or more monomers are called copolymers. Examples: Poly ethylene terephthalate.
In simple words: Copolymers are polymers made from two or more different types of monomer units.

🎯 Exam Tip: The key distinction for copolymers is the involvement of *multiple* different monomers, unlike homopolymers which use a single type of monomer.

Answer the following questions:

Question 1.How is hydrogen molecule formed?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हाइड्रोजन अणु के निर्माण को दर्शाता है। दो अलग-अलग हाइड्रोजन परमाणु, प्रत्येक में एक इलेक्ट्रॉन होता है, करीब आते हैं। वे अपने इलेक्ट्रॉनों को साझा करते हैं, एक एकल सहसंयोजक बंधन बनाते हैं, जिसके परिणामस्वरूप H2 अणु बनता है। 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.
In simple words: Two hydrogen atoms each share their single electron to form a single covalent bond, creating a stable hydrogen molecule (\(H_2\)).

🎯 Exam Tip: Explain the octet rule (or duet rule for hydrogen) and how sharing electrons helps atoms achieve a stable electronic configuration, forming a covalent bond.

Question 2.Describe the formation of oxygen molecule (O2).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑक्सीजन अणु के निर्माण को दर्शाता है। दो ऑक्सीजन परमाणु, प्रत्येक में बाहरी कोश में छह इलेक्ट्रॉन होते हैं, करीब आते हैं। वे दो इलेक्ट्रॉन युग्मों को साझा करते हैं, जिसके परिणामस्वरूप ऑक्सीजन परमाणुओं के बीच एक दोहरा सहसंयोजक बंधन बनता है, जिससे एक स्थिर O2 अणु बनता है। (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 (=).
In simple words: Two oxygen atoms each share two electrons to form a double covalent bond, achieving stability and forming an oxygen molecule (\(O_2\)).

🎯 Exam Tip: Detail the electronic configuration, the number of electrons needed for stability, and how sharing two pairs of electrons leads to a double covalent bond in \(O_2\).

Question 3.Describe the formation of nitrogen molecule.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह नाइट्रोजन अणु के निर्माण को दर्शाता है। दो नाइट्रोजन परमाणु, प्रत्येक के बाहरी कोश में पाँच इलेक्ट्रॉन होते हैं, करीब आते हैं। वे तीन इलेक्ट्रॉन युग्मों को साझा करते हैं, जिसके परिणामस्वरूप नाइट्रोजन परमाणुओं के बीच एक तिहरा सहसंयोजक बंधन बनता है, जिससे एक स्थिर N2 अणु बनता है। (1) The atomic number of nitrogen is 7. The electronic configuration of nitrogen is (2, 5). Nitrogen has 5 electrons in the outermost shell. (2) It requires three more electrons to complete the L shell and attain the configuration of neon (Ne). (3) Two nitrogen atoms come close together and share three pairs of electrons with each other, resulting in the formation of a triple bond. (4) Thus, two nitrogen atoms are bound with a triple bond (=) to form a nitrogen molecule.
In simple words: Two nitrogen atoms share three pairs of electrons to create a triple covalent bond, forming a stable nitrogen molecule (\(N_2\)).

🎯 Exam Tip: Emphasize the need for three shared pairs of electrons to satisfy the octet rule for nitrogen, resulting in a strong triple covalent bond.

Question 4.How is the methane molecule formed?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह मीथेन अणु के निर्माण को दर्शाता है। एक केंद्रीय कार्बन परमाणु अपने बाहरी कोश के चार इलेक्ट्रॉनों को चार हाइड्रोजन परमाणुओं के एक-एक इलेक्ट्रॉन के साथ साझा करता है। यह चार एकल सहसंयोजक बंधों के माध्यम से कार्बन से जुड़े चार हाइड्रोजन परमाणुओं के साथ एक टेट्राहेड्रल संरचना बनाता है। (1) The electronic configuration of carbon is (2, 4). Carbon has four electrons in the outermost shell, hence it is tetravalent. (2) The electronic configuration of hydrogen is 1, hence it is monovalent. (3) Carbon needs four electrons to complete the L shell and attain the configuration of neon (Ne). (4) Four atoms of hydrogen share 1 electron each with 4 electrons of carbon. (5) A single covalent bond is formed by sharing of two electrons.
In simple words: Methane is formed when one carbon atom shares its four valence electrons with four hydrogen atoms, creating four single covalent bonds.

🎯 Exam Tip: Highlight carbon's tetravalency and how it forms four single covalent bonds with four monovalent hydrogen atoms to achieve a stable octet, resulting in \(CH_4\).

Question 5.State the various compounds and its formulae formed by a single atom of carbon with monovalent hydrogen and chlorine.
Answer:

Compounds	Names
CH4	Methane
CH3Cl	Methyl chloride
CH2Cl2	Methylene dichloride
CHCl3	Methylene trichloride
CCl4	Carbon tetrachloride

In simple words: A single carbon atom can form various compounds by bonding with different combinations of hydrogen and chlorine atoms, from methane to carbon tetrachloride.

🎯 Exam Tip: This table demonstrates the concept of substitution reactions where hydrogen atoms are progressively replaced by halogen atoms, leading to a series of chlorinated methanes.

Question 6.Observe the straight chain hydrocarbons given below and answer the following questions:
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र A प्रोपेन की सीधी-श्रृंखला संरचना को दर्शाता है, जहाँ तीन कार्बन परमाणु एकल बंधों द्वारा जुड़े होते हैं और शेष संयोजकता हाइड्रोजन परमाणुओं द्वारा पूरी की जाती है। चित्र B प्रोपीन की सीधी-श्रृंखला संरचना को दर्शाता है, जिसमें तीन कार्बन परमाणु होते हैं, जिनमें से दो के बीच एक दोहरा बंधन होता है, और शेष संयोजकता हाइड्रोजन परमाणुओं द्वारा पूरी की जाती है। (i) A is a saturated hydrocarbon, B is an unsaturated hydrocarbon. (ii) A = Propane, B = Propene (iii) The chemical formula of A = C3H8 and number of -CH2 units are 3. The chemical formula of B = C3H6 and number of -CH2 unit is 1.
In simple words: Image A shows propane, a saturated hydrocarbon (\(C_3H_8\)), while Image B shows propene, an unsaturated hydrocarbon (\(C_3H_6\)), differing in bond type and -CH2 units.

🎯 Exam Tip: Differentiate saturated from unsaturated hydrocarbons based on the presence of only single bonds versus at least one double or triple bond. Count -CH2 units carefully based on the formula, not necessarily the chain length directly.

Question 7.Draw electron-dot and line structure of an ethane molecule.
Answer:The molecular formula of ethane is C2H6.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथेन के इलेक्ट्रॉन-डॉट और लाइन संरचना को दर्शाता है। इलेक्ट्रॉन-डॉट संरचना में, कार्बन परमाणुओं के बीच साझा इलेक्ट्रॉन युग्मों को डॉट्स से दिखाया गया है, जबकि लाइन संरचना में, सहसंयोजक बंधों को लाइनों से दर्शाया गया है। प्रत्येक कार्बन परमाणु दूसरे कार्बन परमाणु से और तीन हाइड्रोजन परमाणुओं से एकल बंधन से जुड़ा होता है।
In simple words: Ethane's line structure shows two carbons connected by a single line, with three lines branching off each carbon to hydrogens; its electron-dot structure shows shared dots for each bond.

🎯 Exam Tip: For electron-dot structures, ensure all valence electrons are shown and shared appropriately to satisfy the octet rule for each atom. For line structures, each line represents a pair of shared electrons.

Question 8.The molecular formula of propane is C3H8. From this draw its structural formula.
Answer:The structural formula of propane:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेन की सीधी-श्रृंखला संरचना को दर्शाता है, जिसमें तीन कार्बन परमाणु एक-दूसरे से एकल बंधों द्वारा जुड़े हुए हैं। प्रत्येक कार्बन परमाणु की शेष संयोजकता हाइड्रोजन परमाणुओं द्वारा पूरी की जाती है, जिससे प्रत्येक टर्मिनल कार्बन पर तीन हाइड्रोजन और मध्य कार्बन पर दो हाइड्रोजन होते हैं।
In simple words: The structural formula for propane shows three carbon atoms linked in a straight chain, with hydrogen atoms filling the remaining valencies around each carbon.

🎯 Exam Tip: When drawing structural formulas, ensure each carbon atom has four bonds and each hydrogen atom has one bond. For straight chain alkanes, the carbons form a zigzag pattern with hydrogens around them.

Question 9.Draw the structure and carbon skeleton for cyclohexane.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह साइक्लोहेक्सेन की पूर्ण अणु संरचना और उसके कार्बन कंकाल को दर्शाता है। पूर्ण संरचना में, छह कार्बन परमाणु एक रिंग में एकल बंधों द्वारा जुड़े होते हैं, और प्रत्येक कार्बन परमाणु से दो हाइड्रोजन परमाणु जुड़े होते हैं। कार्बन कंकाल केवल कार्बन रिंग को दर्शाता है, जहाँ प्रत्येक शीर्ष पर एक कार्बन परमाणु होता है।
In simple words: Cyclohexane's structure is a six-carbon ring with each carbon bonded to two hydrogens; its carbon skeleton just shows the six-membered carbon ring.

🎯 Exam Tip: For cyclic compounds, ensure the ring is closed and all valencies are satisfied. The carbon skeleton simplifies the structure by omitting hydrogen atoms, assuming they are present to complete valencies.

Question 10.Classify into saturated and unsaturated hydrocarbons: (1) Methane (2) Ethene (3) Ethane (4) Ethyne (5) Propene (6) Propyne (7) Butane (8) Cyclohexene (9) Cyclopentane (10) Heptane.
Answer:(i) Saturated hydrocarbons: (1) Methane (2) Ethane (3) Butane (4) Cyclopentane (5) Heptane. (ii) Unsaturated hydrocarbons: (1) Ethene (2) Ethyne (3) Propene (4) Propyne (5) Cyclohexene.
In simple words: Saturated hydrocarbons like methane and ethane have only single bonds, while unsaturated ones like ethene and ethyne contain double or triple bonds.

🎯 Exam Tip: Saturated hydrocarbons are alkanes and cycloalkanes (only C-C single bonds). Unsaturated hydrocarbons are alkenes, alkynes, and cycloalkenes (containing at least one C=C or C≡C bond).

Question 11.Classify into alkanes, alkenes and alkynes: (1) Ethane (2) Ethene (3) Methane (4) Propene (5) Ethyne (6) Propyne (7) Butane (8) Pentane.
Answer:Alkanes: (1) Ethane (2) Methane (3) Butane. (4) Pentane Alkenes: (1) Ethene (2) Propene Alkynes: (1) Ethyne (2) Propyne
In simple words: This classification groups hydrocarbons based on their carbon-carbon bonding: alkanes have single bonds, alkenes have double bonds, and alkynes have triple bonds.

🎯 Exam Tip: Remember the general formulas: alkanes \(C_nH_{2n+2}\), alkenes \(C_nH_{2n}\), and alkynes \(C_nH_{2n-2}\) to quickly classify these compounds.

Question 12.Classify into straight chain carbon compounds, branched chain carbon compounds and ring carbon compounds: (1) Propene (2) Butane (3) Iso-butane (4) Cyclopentane (5) Benzene (6) Isobutylene.
Answer:Straight chain carbon compounds: 1. Propene 2. Butane. Branched chain carbon compounds: 1. Iso-butane 2. Isobutylene. Ring carbon compounds: 1. Cyclopentane 2. Benzene.
In simple words: Hydrocarbons can be straight (like propene), branched (like iso-butane), or form a closed ring (like cyclopentane).

🎯 Exam Tip: Visualizing the carbon skeleton helps in classifying compounds into straight, branched, or cyclic structures. Straight chains are linear, branched chains have side groups, and ring compounds form closed loops.

Question 13.Draw chain and ring structures of organic compound having six carbon atoms in it.
Answer:Chain structures of an organic compound having six carbon atoms:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह छह कार्बन परमाणुओं वाले कार्बनिक यौगिकों की विभिन्न श्रृंखला संरचनाओं को दर्शाता है। इनमें सीधी-श्रृंखला वाला हेक्सेन, साथ ही विभिन्न शाखा-श्रृंखला वाले आइसोमर जैसे 2-मिथाइलपेंटेन और 2,3-डाइमिथाइलब्यूटेन शामिल हैं, जहाँ शाखाएँ मुख्य कार्बन श्रृंखला से जुड़ी होती हैं। Ring structures of an organic compound having six carbon atoms:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह छह कार्बन परमाणुओं वाले कार्बनिक यौगिकों की विभिन्न वलय संरचनाओं को दर्शाता है। इनमें साइक्लोहेक्सेन (एक एकल बंधित छह-सदस्यीय वलय) और बेंजीन (एक छह-सदस्यीय वलय जिसमें एकांतर एकल और दोहरे बंधन होते हैं) शामिल हैं।
In simple words: For six carbon atoms, chain structures can be straight or branched (like hexane isomers), while ring structures include cyclohexane and benzene.

🎯 Exam Tip: When drawing isomers, ensure each structure is unique and all carbon atoms maintain tetravalency. For rings, remember to show the cyclic arrangement.

Question 14.Explain the structure of benzene.
Answer:The molecular formula of benzene is C6H6. It is a cyclic unsaturated hydrocarbon. Benzene ring is made of six carbon atoms. In benzene, each carbon atom is linked to two other carbon atoms, on one side by a single bond and on the other side by a double bond, i.e. three alternate single bonds and double bonds in the six membered ring structure of benzene. The compound having this characteristic unit in their structure are called aromatic compounds.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह बेंजीन की संरचना को दर्शाता है, जिसमें छह कार्बन परमाणु एक हेक्सागोनल रिंग बनाते हैं। प्रत्येक कार्बन परमाणु एक हाइड्रोजन परमाणु से जुड़ा होता है, और कार्बन-कार्बन बंधों में एकांतर एकल और दोहरा बंधन होते हैं, जो अनुनाद को दर्शाते हैं।
In simple words: Benzene is a stable, cyclic unsaturated hydrocarbon (\(C_6H_6\)) with six carbon atoms in a ring, featuring alternating single and double bonds, making it an aromatic compound.

🎯 Exam Tip: Key points for benzene structure are: cyclic, six carbons, six hydrogens, alternating single and double bonds (delocalized electrons), and its classification as an aromatic compound.

Question 15.Draw the structures of isomers of pentane (C5H12).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पेंटेन (\(C_5H_{12}\)) के तीन संरचनात्मक आइसोमर को दर्शाता है। पहला n-पेंटेन है, एक सीधी-श्रृंखला वाला हाइड्रोकार्बन। दूसरा i-पेंटेन (आइसोपेंटेन) है, जिसमें एक शाखाबद्ध श्रृंखला है। तीसरा नियो-पेंटेन (2,2-डाइमिथाइलप्रोपेन) है, जिसमें एक केंद्रीय कार्बन से चार मिथाइल समूह जुड़े होते हैं।
In simple words: Pentane (\(C_5H_{12}\)) has three isomers: n-pentane (straight chain), isopentane (branched chain), and neopentane (more branched).

🎯 Exam Tip: Isomers have the same molecular formula but different structural arrangements. For pentane, ensure you can draw all three distinct structures (n-pentane, isopentane, neopentane) while maintaining carbon's tetravalency.

Question 16.Recognize the carbon chain type for each of the following:
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): पहला चित्र एक सीधी कार्बन श्रृंखला को दर्शाता है जहाँ कार्बन परमाणु एक सीधी रेखा में जुड़े होते हैं और हाइड्रोजन परमाणुओं द्वारा उनकी संयोजकता पूरी की जाती है। दूसरा चित्र एक शाखाबद्ध कार्बन श्रृंखला को दर्शाता है जहाँ एक केंद्रीय कार्बन परमाणु से तीन अन्य कार्बन परमाणु जुड़े होते हैं, और शेष संयोजकता हाइड्रोजन परमाणुओं द्वारा पूरी की जाती है।
In simple words: The first structure shows a straight carbon chain, while the second depicts a branched carbon chain.

🎯 Exam Tip: Straight chains have carbons linked sequentially without side branches. Branched chains have carbon atoms attached to the main chain at multiple points, forming "branches."

Question 17.What is meant by functional group? Give examples. (OR) Explain the term functional group with example.
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. Example: Methyl alcohol, acetic acid. In methane (CH4), when one hydrogen atom is replaced by an -OH group, methyl alcohol (CH3OH), is formed. The -OH is known as the alcoholic functional group. Similarly, from methane (CH4) when one hydrogen atom is replaced by -COOH group, acetic acid (CH3COOH) is formed. The -COOH group is known as the carboxylic acid functional group.
In simple words: A functional group is a specific atom or group of atoms within a molecule that determines its characteristic chemical properties, like the -OH in alcohols or -COOH in carboxylic acids.

🎯 Exam Tip: Define functional group as the site of chemical reactivity. Provide diverse examples like -OH (alcohol) and -COOH (carboxylic acid) to illustrate its impact on molecular properties.

Question 18.Define functional group and complete the following table:
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.

Functional group	Compound	Formula
-OH	Ethyl alcohol	C2H5OH
-CHO	Acetaldehyde	CH3CHO

In simple words: A functional group is an atom or group that dictates a molecule's chemical behavior, and examples include the -OH group in ethyl alcohol and the -CHO group in acetaldehyde.

🎯 Exam Tip: When defining functional groups, remember to emphasize their role in determining chemical properties. For the table, accurately match the functional group with its compound and molecular formula.

 

Question 19. What is meant by homologous series?
Answer:The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH2- (methylene) unit and their mass differ by 14 units.
The homologous series of straight chain alkanes can be represented by the general formula \(C_nH_{2n} + 2\). The members of this series are as follows:

Methane - \(CH_4\)	- These differ by - \(CH_2\) units
Ethane - \(C_2H_6\)
Ethane - \(C_2H_6\)	- These differ by - \(CH_2\) units
Propane - \(C_3H_8\)
Butane - \(C_4H_{10}\)	- These differ by - \(CH_2\) units
Pentane - \(C_5H_{12}\)

In simple words: A homologous series is a group of organic compounds with similar chemical properties, differing by a repeating -CH2- unit in their structure. This results in a gradual change in physical properties along the series.

🎯 Exam Tip: Understanding homologous series is crucial for predicting the properties of organic compounds based on their position in the series.

 

Question 20. State the four characteristics of homologous series.
Answer:Characteristics of Homologous series:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH2-) gets added
(b) molecular mass increases by 14 u (c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) While going in an increasing order of the length there is gradation in the physical properties i.e. the boiling and melting points.In simple words: Homologous series are characterized by a -CH2- unit difference between consecutive members, a mass increase of 14u, similar chemical properties due to the same functional group, and a general formula that applies to all members.

🎯 Exam Tip: Listing all four characteristics clearly with sub-points can earn full marks. Focus on molecular difference, mass change, chemical similarity, and physical property gradation.

 

Question 21. Write names of first four homologous series of alcohols:
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आरेख है जो अल्कोहल के सजातीय श्रृंखला को दर्शाने के लिए खाली स्थान प्रदान करता है, जहाँ प्रत्येक सदस्य में एक -OH कार्यात्मक समूह होता है। First four homologous series of alcohols are
1. Methanol \(CH_3\) - OH
2. Ethanol \(C_2H_5\) - OH
3. Propanol \(C_3H_7\) - OH
4. Butanol \(C_4H_9\) - OHIn simple words: The first four alcohols in a homologous series are Methanol, Ethanol, Propanol, and Butanol, each differing by a -CH2- unit and having an -OH functional group.

🎯 Exam Tip: Remember the 'meth', 'eth', 'prop', 'but' prefixes for 1, 2, 3, and 4 carbon atoms, respectively, and add '-ol' for alcohols.

 

Question 22. Describe the IUPAC rules of naming organic compounds.
Answer:IUPAC nomenclature system: International Union for Pure and Applied Chemistry (IUPAC) put forth a nomenclature system based on the structure of the compounds and it was accepted all over the world. There are three units in the IUPAC name of any carbon compound: parent, suffix and prefix. These are arranged in the name as follows:
Prefix-parent-suffix:
An IUPAC name is given to a compound on the basis of the name of its parent alkane. The name of the compound is constructed by attaching appropriate suffix and prefix to the name of the parent-alkane. The steps in the IUPAC nomenclature of straight chain compounds are as follows:
Step 1: Draw the structural formula of the straight chain compound and count the number of carbon atoms in it. The alkane with the same number of carbon atoms is the parent alkane of the concerned compound. Write the name of this alkane.
In case the carbon chain of concerned compound contains a double bond, change the ending of the parent name from 'ane' to 'ene'. If the carbon chain in the concerned compound contains a triple bond, change the ending of the parent name from 'ane' to 'yne'.

Sr. No.	Structural formula	Straight chain	Parent name
1.	\(CH_3-CH_2-CH_3\)	C-C-C	propane
2.	\(CH_3-CH_2-OH\)	C-C	ethane
3.	\(CH_3-CH_2-COOH\)	C-C-C	propane
4.	\(CH_3-CH_2-CH_2-CHO\)	C-C-C-C	butane
5.	\(CH_3-CH=CH_2\)	C-C = C	propene
6.	\(CH_3-C\equiv CH\)	C-C \( \equiv \) C	propyne

Step 2: If the structural formula contains a functional group, replace the last letter 'e' from the parent name by the condensed name of the functional group as the suffix. (Exception: The condensed name of the functional group 'halogen' is always attached as the prefix.)
Step 3: Number the carbon atoms in the carbon chain from one end to the other. Assign the number T to carbon in the functional group -CHO or -COOH, if present. Otherwise, the chain can be numbered in two directions. Accept that numbering which gives smaller number to the carbon carrying the functional group. In the final name, a digit (number) and a character (letter) should be separated by a small horizontal line.In simple words: IUPAC naming follows a prefix-parent-suffix structure. First, determine the parent alkane based on the carbon chain length. Then, modify the parent name's ending based on the presence of double/triple bonds ('ene' or 'yne') or functional groups (e.g., '-ol' for alcohol). Finally, number the carbon chain to give the lowest possible number to functional groups or multiple bonds.

🎯 Exam Tip: Practicing naming various compounds (alkanes, alkenes, alkynes, and those with different functional groups) is key to mastering IUPAC nomenclature.

 

Question 23. Write the IUPAC names of the following structural formulae.
a. \(CH_3 - CH_2 - CH = CH_2\)
Answer:Number of carbon atoms in the longest chain: 4
Parent alkane: Butene
Functional group: double bond
Assign the number:
⁴\(CH_3\)-³\(CH_2\)-²\(CH\)=¹\(CH_2\)
In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the double bond and the other c-atoms are numbered accordingly.
Parent suffix: But-1-ene
IUPAC name: But-1-ene
In simple words: The compound \(CH_3 - CH_2 - CH = CH_2\) has a four-carbon chain with a double bond at the first carbon, making its IUPAC name But-1-ene.

🎯 Exam Tip: For alkenes, always prioritize numbering the carbon chain from the end closest to the double bond to get the lowest possible position number for the double bond.

b. \(CH_3 - C \equiv C-H\)
Answer:Number of carbon atoms in the longest chain: 3
Parent alkane: Propyne
Functional group: triple bond
Parent suffix: Propyne
IUPAC name: Propyne
In simple words: The compound \(CH_3 - C \equiv C-H\) has a three-carbon chain with a triple bond, hence its IUPAC name is Propyne.

🎯 Exam Tip: For alkynes, the triple bond determines the 'yne' suffix. For a three-carbon alkyne, propyne is the only isomer without explicitly stating the position.

c. \(CH_3-CH(Cl)-CH_2-CH_2-CH_3\)
Answer:The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2
¹\(CH_3\)-²\(CH(Cl)\)-³\(CH_2\)-⁴\(CH_2\)-⁵\(CH_3\)
The carbon atom to which the -Cl atom is attached is numbered as \(C_2\) and the other C atoms are numbered accordingly.
Prefix parent: 2-Chloropentane
IUPAC name: 2-Chloropentane
In simple words: The given structure has a five-carbon chain (pentane) with a chlorine atom attached to the second carbon, thus it is named 2-Chloropentane.

🎯 Exam Tip: When a halogen is present, it acts as a prefix. Number the chain to give the halogen the lowest possible number.

d. \(CH_3 - CH_2 - CH_2 - Br\)
Answer:The number of carbon atoms in the longest chain: 3
Parent alkane : Propane Prefix functional group: Bromo
Assign the number:
³\(CH_3\)-²\(CH_2\)-¹\(CH_2\)-Br
The carbon atom to which the -Br atom is attached is numbered as \(C_1\) and the other C atoms are numbered accordingly.
Prefix parent: 1-Bromopropane
IUPAC name: 1-Bromopropane
In simple words: This compound has a three-carbon chain (propane) with a bromine atom on the first carbon, hence its IUPAC name is 1-Bromopropane.

🎯 Exam Tip: For haloalkanes, ensure the numbering starts from the end closest to the halogen substituent.

e. \(CH_3-CH_2-CH(OH)-CH_3\)
Answer:The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:
⁴\(CH_3\)-³\(CH_2\)-²\(CH(OH)\)-¹\(CH_3\)
The carbon atom to which the -OH group is attached is numbered as \(C_2\).
If the carbon chain of the compound contains a -OH group, then change the ending 'e' of the parent name, i.e. ,'e' of butane is replaced by 'ol' (ol for alcohol).
Parent suffix: Butan-2-ol
IUPAC name: Butan-2-ol
In simple words: The compound \(CH_3-CH_2-CH(OH)-CH_3\) is a four-carbon alcohol with the hydroxyl group on the second carbon, making its IUPAC name Butan-2-ol.

🎯 Exam Tip: When naming alcohols, identify the longest carbon chain containing the -OH group and number it to give the hydroxyl group the lowest possible position number.

f. \(CH_3-CH(NH_2)-CH_3\)
Answer:The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH2 (amine)
Assign the number:
¹\(CH_3\)-²\(CH(NH_2)\)-³\(CH_3\)
If the carbon chain of the compound contains a -NH2 group then change the ending of the parent name, i.e. 'e' of propane is replaced by 'amine'.
Parent suffix: 2-Propanamine
IUPAC name: 2-Propanamine
In simple words: This compound has a three-carbon chain (propane) with an amine group attached to the second carbon, so its IUPAC name is 2-Propanamine.

🎯 Exam Tip: For amines, the suffix is '-amine', and the position of the amino group is indicated by a number if necessary.

g. HCOOH
Answer:The number of carbon atoms: 1
Parent alkane: Methane
Functional group: -COOH (-oic cid)
If the carbon chain of the compound contains a -COOH group, then change the ending of the parent name, i.e. 'e' of methane is replaced by '-oic acid'.
Parent suffix: Methanoic acid
IUPAC name: Methanoic acid
In simple words: HCOOH is a one-carbon carboxylic acid, commonly known as formic acid, but its IUPAC name is Methanoic acid.

🎯 Exam Tip: For carboxylic acids, the carbon of the -COOH group is always \(C_1\) and is included in the parent chain naming.

h. \(CH_3 - CH_2 - CH_2 - CHO\)
Answer:The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CHO (al)
Assign the number: 1
Assign the number '1' to carbon in the functional group
⁴\(CH_3\)-³\(CH_2\)-²\(CH_2\)-¹\(CHO\)
If the carbon chain of the compound contains a -CHO group then change the ending of the parent name, i.e. 'e' of the butane is replaced by 'al'.
Parent suffix: Butanal
IUPAC name: Butanal
In simple words: The compound \(CH_3 - CH_2 - CH_2 - CHO\) is a four-carbon aldehyde, meaning its IUPAC name is Butanal.

🎯 Exam Tip: For aldehydes, the carbon of the -CHO group is always \(C_1\) and contributes to the parent chain length, with '-al' as the suffix.

i. \(CH_3-CH_2-CO-CH_2-CH_3\).
Answer:The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:
-CO- (one)
Assign the numbering:
¹\(CH_3\)-²\(CH_2\)-³\(CO\)-⁴\(CH_2\)-⁵\(CH_3\)
In the longest chain, the numbering of carbon atom starts from the carbon nearest to the functional group (both the numbering equivalent).
If the carbon chain of compound contains a
\[\begin{array}{l} \text{O} \\ \text{(}-C-) \\ \end{array}\] group, then change the ending of the parent name i.e. 'e' of pentane is replaced by 'one'.
Parent suffix: Pentan-3-one
IUPAC name: Pentan-3-one.
In simple words: This is a five-carbon ketone where the carbonyl group (CO) is on the third carbon, giving it the IUPAC name Pentan-3-one.

🎯 Exam Tip: For ketones, number the carbon chain to give the carbonyl carbon the lowest possible number, and use the suffix '-one'.

 

Question 24. What happens when methane is burnt in air? Write the balanced chemical equation for the same.
Answer:When methane burns in air, carbon dioxide and water are formed. The reaction is exothermic with release of large amount of heat and light.
\[CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(g)} + \text{Heat and light}\] Methane Oxygen Carbon dioxide
In simple words: Methane burning in air is a combustion reaction that produces carbon dioxide, water, and releases a significant amount of heat and light.

🎯 Exam Tip: For combustion reactions, remember that hydrocarbons react with oxygen to produce carbon dioxide and water. Balancing the equation correctly is essential.

 

Question 25. What happens when ethanol is burnt in air?
Answer:When ethanol is burnt in air, it burns with a clean blue flame, carbon dioxide and water are formed. In this reaction, release of large amount of heat and light takes place.
\[CH_3-CH_2-OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(g)} + \text{Heat and light}\] Ethanol
In simple words: When ethanol burns in air, it undergoes combustion to produce a clean blue flame, releasing carbon dioxide, water, heat, and light.

🎯 Exam Tip: Similar to methane, ethanol combustion yields \(CO_2\) and \(H_2O\). The key difference is the amount of oxygen required and the products' coefficients due to ethanol's extra carbon and oxygen atom.

 

Question 26. What happens when ethanol is treated with alkaline potassium permanganate? Write the balanced chemical equation for the same.
Answer:When ethanol is treated with alkaline potassium permanganate, ethanol gets oxidised by alkaline potassium permanganate to form ethanoic acid.
\[CH_3-CH_2-OH \underset{\text{alkaline }KMnO_4}{\xrightarrow{(O)}} CH_3-COOH\] Ethanol Ethanoic acid
In simple words: Alkaline potassium permanganate oxidizes ethanol to ethanoic acid, which is a key reaction in organic chemistry.

🎯 Exam Tip: Alkaline \(KMnO_4\) is a strong oxidizing agent. Remember that primary alcohols are oxidized to carboxylic acids under these conditions.

 

Question 27. What happens when vegetable oil is hydrogenated? Write the balanced chemical equation.
Answer:When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, vanaspati ghee (saturated compound) is formed.
\[ \begin{array}{ccc} \text{-C=C-} & + H_2 & \xrightarrow{Pt/Ni} & \text{-C-C-} \\ & & & \\ \text{Unsaturated compound} & & & \text{Saturated compound} \\ \text{(vegetable oil)} & & & \text{(vanaspati ghee)} \end{array} \]In simple words: Hydrogenation converts unsaturated vegetable oil into saturated vanaspati ghee by adding hydrogen across double bonds in the presence of a metal catalyst like nickel.

🎯 Exam Tip: This is an addition reaction (hydrogenation) where unsaturated fats are converted to saturated fats. The catalyst (Ni, Pt, or Pd) and heat are crucial conditions.

 

Question 28. What happens when chlorine is treated with methane? (OR) Describe the action of chlorine on methane. (OR) Write a note on chlorination of methane.
Answer:Methane reacts rapidly with chlorine in the presence of sunlight to form four products. In this reaction, chlorine atoms replace, one by one, all the hydrogen atoms in the methane.
\[CH_4 + Cl_2 \xrightarrow{Sunlight} CH_3Cl + HCl\] Methane Methyl chloride
\[CH_3Cl + Cl_2 \xrightarrow{Sunlight} CH_2Cl_2 + HCl\] Methylene dichloride
\[CH_2Cl_2 + Cl_2 \xrightarrow{Sunlight} CHCl_3 + HCl\] Chloroform
\[CHCl_3 + Cl_2 \xrightarrow{Sunlight} CCl_4 + HCl\] Carbon tetrachloride
The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms is called substitution reaction. Chlorination of methane is a substitution reaction.In simple words: When methane reacts with chlorine in sunlight, it undergoes a substitution reaction where hydrogen atoms are progressively replaced by chlorine atoms, forming methyl chloride, dichloromethane, trichloromethane (chloroform), and tetrachloromethane.

🎯 Exam Tip: Remember that chlorination of methane is a free radical substitution reaction that requires sunlight or UV light and can lead to multiple substitution products.

 

Question 29. What happens when ethanol is reacted with sodium?
Answer:When ethanol is reacted with sodium at room temperature, sodium ethoxide is formed and hydrogen gas is liberated.
\[2C_2H_5OH + 2Na \longrightarrow 2C_2H_5ONa + H_2 \uparrow\] Ethanol Sodium Sodium ethoxide Hydrogen
In simple words: Ethanol reacts with active metals like sodium to produce sodium ethoxide and hydrogen gas, similar to how water reacts with sodium.

🎯 Exam Tip: This reaction demonstrates the acidic nature of alcohols, as they release hydrogen gas when reacting with reactive metals like sodium.

 

Question 30. What happens when ethanol is heated at 170 °C with excess of conc. sulphuric acid?
Answer:When ethanol is heated at 170 °C with excess of conc. sulphuric acid, one molecule of water is removed from its molecule to form ethene (unsaturated compound).
\[CH_3-CH_2-OH \xrightarrow{\text{Conc. }H_2SO_4, 170^\circ C} CH_2 = CH_2 + H_2O\] Ethanol Ethene Water
In simple words: Heating ethanol with concentrated sulfuric acid at 170 °C causes dehydration, removing a water molecule to form ethene, an unsaturated compound.

🎯 Exam Tip: Concentrated sulfuric acid acts as a dehydrating agent in this reaction, converting an alcohol into an alkene. Remember the specific temperature and the role of the acid.

 

Question 31. What happens when ethanoic acid is treated with sodium hydroxide? Write the balanced equation for the same.
Answer:When ethanoic acid is treated with sodium hydroxide, neutralization takes place to form sodium acetate (sodium ethanoate) and water.
\[CH_3-COOH + NaOH \longrightarrow CH_3COONa + H_2O\] Ethanoic acid Sodium hydroxide Sodium ethanoate Water
In simple words: Ethanoic acid reacts with sodium hydroxide in a neutralization reaction to form sodium ethanoate (a salt) and water.

🎯 Exam Tip: This is a classic acid-base neutralization reaction. Ethanoic acid (acetic acid) is a weak acid reacting with a strong base (sodium hydroxide).

 

Question 32. What happens when ethanoic acid is treated with sodium carbonate?
Answer:When ethanoic acid is treated with sodium carbonate, sodium ethanoate, carbon dioxide and water is formed.
\[2CH_3COOH + Na_2CO_3 \longrightarrow 2CH_3COONa + H_2O + CO_2 \uparrow\] Acetic acid Sodium carbonate Sodium acetate
In simple words: Ethanoic acid reacts with sodium carbonate to produce sodium ethanoate, water, and carbon dioxide gas.

🎯 Exam Tip: The evolution of carbon dioxide gas is a characteristic test for carboxylic acids reacting with carbonates. Ensure the equation is balanced.

 

Question 33. What happens when ethanoic acid is treated with sodium bicarbonate?
Answer:When ethanoic acid is treated with sodium bicarbonate, sodium ethanoate, water and carbon dioxide is formed.
\[CH_3-COOH + NaHCO_3 \longrightarrow CH_3COONa + H_2O + CO_2 \uparrow\] Ethanoic acid Sodium ethanoate
In simple words: Ethanoic acid reacts with sodium bicarbonate to yield sodium ethanoate, water, and carbon dioxide gas, which causes effervescence.

🎯 Exam Tip: This reaction is a common laboratory test for carboxylic acids, as it produces effervescence (bubbles of \(CO_2\)) when bicarbonate is added.

 

Question 34. What happens when ethanoic acid is treated with ethanol? Write the balanced equation for the same.
Answer:When ethanoic acid is treated with ethanol in the presence of an acid catalyst, an ester, i.e., ethyl ethanoate is formed.
\[CH_3COOH + CH_3-CH_2-OH \xrightarrow{\text{Acid Catalyst}} CH_3COO-CH_2-CH_3 + H_2O\] Ethanoic acid Ethanol Ethyl ethanoate Water
In simple words: Ethanoic acid reacts with ethanol in the presence of an acid catalyst to form ethyl ethanoate and water, a process called esterification.

🎯 Exam Tip: Esterification is a condensation reaction between a carboxylic acid and an alcohol, producing an ester and water. It often requires an acid catalyst like concentrated sulfuric acid.

 

Question 35. What happens when ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst?
Answer:When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).
\[nCH_2 = CH_2 \xrightarrow{\text{Heat, high pressure, Catalyst}} (CH_2-CH_2)_n\] Ethylene Polythene or Polyethylene
(\(n\) = 100 to 1000)
CH2-CH2-CH2-CH2-CH2.... etc.
OR
\[ \begin{array}{cccccc} H & H & H & H & H & H \\ | & | & | & | & | & | \\ \text{-C-C-C-C-C-C-} \\ | & | & | & | & | & | \\ H & H & H & H & H & H \end{array} \]In simple words: Ethylene gas undergoes polymerization under high pressure and temperature with a catalyst to form polyethylene, a long-chain plastic.

🎯 Exam Tip: This is an example of addition polymerization, where many monomer units (ethylene) join together to form a large polymer (polyethylene) without the loss of any atoms.

 

Question 36. State the physical properties of ethyl alcohol ethanol.
Answer:1. Ethanol is a colourless liquid and it is soluble in water in all proportions and has pleasant odour.
2. The boiling point of ethanol is 78 °C and the freezing point is -114 °C.
3. It is combustible and burns with a blue flame.
4. An aqueous solution of ethanol is neutral to litmus paper.In simple words: Ethanol is a colorless, sweet-smelling liquid that is miscible with water, has a boiling point of 78 °C, burns with a blue flame, and is neutral to litmus.

🎯 Exam Tip: When describing physical properties, always include appearance, odor, solubility, and key transition temperatures like boiling and freezing points.

 

Question 37. State the properties of ethanoic acid.
Answer:1. Ethanoic acid is a colourless liquid with boiling point 118 °C and melting point 17 °C. It has a pungent odour.
2. Its aqueous solution is acidic and turns blue litmus red.
3. A 5-8% aqueous solution of acetic acid is used as vinegar.
4. It is a weak acid.In simple words: Ethanoic acid is a pungent, colorless liquid with a boiling point of 118 °C, which forms acidic solutions that turn blue litmus red, and is the active component of vinegar.

🎯 Exam Tip: For ethanoic acid, remember its characteristic pungent odor, its acidic nature (turns blue litmus red), and its common name and use as vinegar.

Write Short Notes:

 

Question 1. Catenation power.
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.
(2) Carbon shows catenation. Two or more carbon atoms can share their valence electrons and bond with each other. Thus, carbon chains can be straight or branched or closed chain ring structure forming large molecules. 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.
(3) Hence, carbon atoms can form an unlimited number of compounds.
ℹ️ चित्र व्याख्या (Diagram Explanation): कार्बन परमाणुओं की सीधी श्रृंखला (C-C-C-C-C-C), शाखित श्रृंखला (बीच में एक C के साथ लंबी C-C श्रृंखला) और चक्रीय संरचनाएं (C-C-C-C-C-C रिंग) दिखाई गई हैं, जो कार्बन की कैटेनेशन शक्ति को दर्शाती हैं।In simple words: Catenation is carbon's unique ability to form strong covalent bonds with other carbon atoms, creating long straight, branched, or ring-shaped chains and leading to a vast number of diverse compounds.

🎯 Exam Tip: Define catenation, explain how carbon forms various structures (straight, branched, rings), and emphasize the strength and stability of C-C bonds as the reason for this property.

 

Question 2. Characteristics of Carbon.
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.In simple words: Carbon's main characteristic is catenation, allowing it to form strong covalent bonds with other carbon atoms in various structures like straight, branched, or ring chains, leading to a vast array of compounds.

🎯 Exam Tip: Focus on catenation and the stability of carbon bonds. Mentioning the different types of chains (open, closed, straight, branched) demonstrates a comprehensive understanding.

 

Question 3. Functional group.
Answer:(1) The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of the length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called functional groups.
All organic compounds are derivatives of hydrocarbons. The derivatives are formed by replacing one or more H-atom/atoms of hydrocarbon by some other hetero atom or groups of atoms containing hetero atoms. After replacement, a new compound is formed which has properties different from the parent hydrocarbon.
Examples: For methane, if one hydrogen atom is replaced by an -OH group, then a compound is methyl alcohol (\(CH_3OH\)). The -OH group is known as the alcoholic functional group.
Functional group is organic compound:
1. Alcohol: -OH (hydroxy group)
2. Aldehyde: H-C=O
3. Ketones: -CO-
4. Carboxylic acid : -COOH
In simple words: A functional group is a specific atom or group of atoms within a molecule that dictates its characteristic chemical reactions, regardless of the carbon chain's size.

🎯 Exam Tip: Define functional groups clearly, explain their role in determining chemical properties, and provide at least 2-3 examples with their names and structures.

 

Question 4. Homologous series.
Answer:The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH2- (methylene) unit and their mass differ by 14 units.
The homologous series of straight chain alkanes can be represented by the general formula \(C_nH_{2n} + 2\)
The members of this series are as follows:

Methane - \(CH_4\)	- These differ by - \(CH_2\) units
Ethane - \(C_2H_6\)
Ethane - \(C_2H_6\)	- These differ by - \(CH_2\) units
Propane - \(C_3H_8\)
Butane - \(C_4H_{10}\)	- These differ by - \(CH_2\) units
Pentane - \(C_5H_{12}\)

Characteristics:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH2-) gets added
(b) molecular mass increases by 14 u
(c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) while going in an increasing order of the length there is gradation in the physical properties i.e., the boiling and melting points.In simple words: A homologous series is a family of organic compounds with similar chemical properties, the same general formula, and consecutive members differing by a -CH2- unit, showing a gradual change in physical properties.

🎯 Exam Tip: Clearly define homologous series, state the common difference between members (-CH2- unit and 14u mass), and list the characteristics, including similar chemical properties and gradual change in physical properties.

Give Scientific Reasons:

Question 1. Carbon atoms are capable of forming an unlimited number of compounds.
Answer:1. Carbon has the property of catenation. Two or more carbon atoms can share some of their valence electrons to form (single, double and triple) bonds.
2. The straight chains or branched chains or rings may have different shapes and sizes. This results in formation of many compounds. Hence, carbon atoms are capable of forming an unlimited number of compounds.
In simple words: Carbon's ability to form long chains, branched structures, and rings, along with its capacity for single, double, and triple bonds, allows it to create a vast number of diverse compounds.

🎯 Exam Tip: Focus on 'catenation' and 'tetravalency' as key reasons for carbon's diverse compound formation for maximum points.

Question 2. Ethylene is an unsaturated hydrocarbon.
Answer:(1) Ethylene (CH2 = CH2) contains a double bond between carbon atoms.
(2) Thus, the valencies of the two carbon atoms are not fully satisfied by single covalent bonds. Hence, ethylene is an unsaturated hydrocarbon.
In simple words: Ethylene has a double bond between its carbon atoms, meaning not all carbon valencies are satisfied by single bonds, making it an unsaturated hydrocarbon.

🎯 Exam Tip: Clearly state the presence of a double bond and relate it to the unsatisfied valencies for a complete answer.

Question 3. Naphthalene burns with a yellow flame.
Answer:(1) Naphthalene is an unsaturated compound. In unsaturated hydrocarbon the proportion of carbon is larger than that of saturated hydrocarbon. As a result, some unburnt carbon particles are also formed during combustion of unsaturated compounds.
(2) In the flame. these unburnt hot carbon particles emit yellow light and therefore the flame appears yellow. Hence, naphthalene burns with a yellow flame.
In simple words: Naphthalene is an unsaturated hydrocarbon with a high carbon-to-hydrogen ratio, leading to incomplete combustion and the emission of yellow light from unburnt carbon particles.

🎯 Exam Tip: Emphasize the connection between unsaturated hydrocarbons, incomplete combustion, and the production of a sooty, yellow flame.

Question 4. The colour of iodine disappears in the reaction between vegetable oil and iodine.
Answer:(1) Vegetable oils (unsaturated compound) contains a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) The addition reaction of vegetable oil with iodine takes place instantaneously at room temperature. The colour of iodine disappears in this reaction. This iodine test indicates the presence of a multiple bond in vegetable oil.
In simple words: Vegetable oils, being unsaturated, have multiple bonds that react with iodine in an addition reaction, causing the iodine's color to disappear and indicating unsaturation.

🎯 Exam Tip: Remember that iodine test is a common method to detect unsaturation in organic compounds; link color change to the addition reaction.

Question 5. The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee.
Answer:(1) The molecules of vegetable oil contain long and unsaturated carbon chains. These unsaturated hydrocarbons contain a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, the addition reaction takes place, vanaspati ghee (saturated compound) is formed.
In simple words: Vegetable oil, being unsaturated, undergoes an addition reaction with hydrogen in the presence of a nickel catalyst, converting its multiple bonds to single bonds and forming saturated vanaspati ghee.

🎯 Exam Tip: Highlight the role of nickel catalyst and the conversion of unsaturated multiple bonds to saturated single bonds in hydrogenation reactions.

Distinguish Between The Following:

Question 1. Saturated hydrocarbons and Unsaturated hydrocarbons.
Answer:Saturated hydrocarbons:
1. In saturated hydrocarbons, the carbon atoms are linked to each other only by single covalent bonds.
2. They contain only a single bond.
3. They are chemically less reactive.
4. Substitution reaction is a characteristic property of these hydrocarbons.
5. Their general formula is CnH2n + 2.
Unsaturated hydrocarbons:
1. In unsaturated hydrocarbons, the valencies of carbon atoms are not fully satisfied by single covalent bonds.
2. They contain carbon to carbon double or triple bonds.
3. They are chemically more reactive.
4. Addition reaction is a characteristic property of these hydrocarbons.
5. Their general formula is CnH2n or CnH2n - 2
In simple words: Saturated hydrocarbons have only single carbon-carbon bonds (CnH2n+2) and are less reactive, while unsaturated hydrocarbons have double or triple carbon-carbon bonds (CnH2n or CnH2n-2) and are more reactive.

🎯 Exam Tip: For distinguishing questions, use clear, concise points of comparison and include general formulas and typical reactions (substitution vs. addition).

Question 2. Open chain hydrocarbons and closed chain hydrocarbons.
Answer:Open chain hydrocarbons:
1. A hydrocarbon in which the chain of carbon atoms is not cyclic is called an open chain hydrocarbon.
2. All aliphatic hydrocarbons contain open chains.
Closed chain hydrocarbons:
1. A hydrocarbon in which the chain of carbon atoms is present in a cyclic form or ring form is called a closed chain hydrocarbon.
2. All aromatic hydrocarbons contain closed chains.
In simple words: Open chain hydrocarbons have carbon atoms arranged in a non-cyclic, linear structure, while closed chain hydrocarbons have their carbon atoms forming a ring or cyclic structure.

🎯 Exam Tip: Define each type clearly, focusing on the arrangement of carbon atoms (linear vs. cyclic) and mentioning common examples like aliphatic for open and aromatic for closed chains.

Question 3. Alkane and Alkene.
Answer:Alkane
1. Alkanes in which the carbon atoms are linked to each other only by single bonds.
2. The general formula of an alkane is CnH2n + 2
3. They are chemically less reactive.
Alkene:
1. Alkenes in which carbon atoms are linked to each other by double bonds.
2. The general formula of an alkene is CnH2n.
3. They are chemically more reactive.
In simple words: Alkanes are saturated hydrocarbons with only single bonds (CnH2n+2) and are less reactive, whereas alkenes are unsaturated hydrocarbons with at least one double bond (CnH2n) and are more reactive.

🎯 Exam Tip: Distinguish based on bond type (single vs. double), general formula, and chemical reactivity (less reactive vs. more reactive).

Project:

Question 1. Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.