Selina Concise Solutions for ICSE Class 10 Chemistry Chapter 12 Organic Chemistry

Exercise 12(A)

Question. Write the IUPAC name of the following:
(a) \( CH_3 - C(CH_3)_2 - CH_3 \)
(b) \( CH_3 - CH(CH_3) - CH_2 - CH_3 \)
(c) \( H - C = C - H \) with \( H \) attached to each carbon
(d) \( CH_3 - C(CH_3)_2 - CH_2 - CH_2 - CH_3 \)
(e) \( CH_3 - C \equiv C - CH_2 - CH_3 \)
(f) \( CH_3 - C(CH_3) \equiv C - H \)
(g) \( CH_3 - CH(Cl) - CH(Cl) - CH_2 - CH_3 \)
(h) \( CH_3 - CH(CH_3) - CH_2 - CH_2 - CH(CH_3)_2 \)
(i) \( CH_3 - CH(CH_3) - CH_2 - CH_3 \)
(j) \( CH_3 - C \equiv C - CH_2 - CH_2 - CH_2 - CH_3 \)
(k) \( CH_3 - C(CH_3)_2 - CH_2 - CH_2 - CH_2 - CHO \)
(l) \( CH_3 - CH(OH) - CH_2 - CH_2 - CH_3 \)
(m) \( CH_3 - CH_2 - CH(CH_3) - CH_2 - COOH \)
(n) \( CH_3 - C(CH_3)(Br) - CH_2 - CH_3 \)
(o) \( CH_3 - CH_2 - CH(CH_3) - CH_2 - Br \)
Answer:
(a) 2,2-dimethylpropane
(b) 2-methyl butane
(c) Prop-1-ene
(d) 2,2-dimethyl pentane
(e) Pent-2-yne
(f) 3-methyl but-1-yne
(g) 2,3-dichloropentane
(h) 3-methylheptane
(i) 2-methyl butane
(j) Hept-2-yne
(k) 2,2-dimethyl hexanal
(l) Pentan-2-ol
(m) 4-methylpentanoic acid
(n) 2-bromo2-methyl butane
(o) 1-bromo3-methyl butane
In simple words: IUPAC naming follows rules where we find the longest carbon chain, number it to give functional groups the lowest numbers, and name substituents alphabetically with their positions.

📝 Teacher's Note: Have students practice drawing structures from names first, then work backwards to naming. Use colored pencils to highlight the main chain versus side chains to make the pattern clearer.

🎯 Exam Tip: Always identify the functional group first (it gets priority in numbering), then find the longest chain containing that group. Write substituents alphabetically, not in order of position numbers.

Question. Write the structure of the following compounds:
(a) Prop-1-ene
(b) 2,3-dimethyl butane
(c) 2-methyl propane
(d) 3-hexene
(e) prop-1-yne
(f) 2-methylprop-1-ene
(g) Alcohol with molecular formula \( C_4H_{10}O \)
Answer:
(a) Prop-1-ene: \( CH_3 - CH = CH_2 \)
(b) 2,3-dimethylbutane: \( CH_3 - CH(CH_3) - CH(CH_3) - CH_3 \)
(c) 2-methylpropane: \( CH_3 - CH(CH_3) - CH_3 \)
(d) 3-hexene: \( CH_3 - CH_2 - CH = CH - CH_2 - CH_3 \)
(e) Prop-1-yne: \( CH_3 - C \equiv CH \)
(f) 2-methylprop-1-ene: \( CH_3 - C(CH_3) = CH_2 \)
(g) Alcohol with molecular formula \( C_4H_{10}O \): \( CH_3 - CH_2 - CH_2 - CH_2 - OH \)
In simple words: When drawing structures from names, start with the main carbon chain, then add double/triple bonds at the right positions, and finally attach all the side groups.

📝 Teacher's Note: Emphasize that students should always count carbons in the main chain first, then place the functional group, then add substituents. Common mistakes include miscounting carbons or placing groups on wrong carbons.

🎯 Exam Tip: After drawing each structure, quickly count the carbons and hydrogens to verify your molecular formula matches what's expected. This catches most structural errors.

Question. Choose the correct answer:
(a) \( C_5H_{11} \) is an
(i) alkane (ii) alkene (iii) alkyne (iv) alkyl group
(b) A hydrocarbon of the general \( C_nH_{2n} \) is
(i) \( C_{15}H_{30} \) (ii) \( C_{12}H_{26} \)
(iii) \( C_8H_{20} \) (iv) \( C_6H_{14} \)
(c) A hydrocarbon with molecular mass 72 is
(i) an alkane (ii) an alkene (iii) an alkyne
(d) The total number of different carbon chains that four carbon atoms form in alkane is
(i) 5 (ii) 4 (iii) 3 (iv) 2
(e) \( CH_3 - CH_2 - OH \) and \( CH_3 - O - CH_3 \) are
(i) position isomers (ii) chain isomers
(iii) homologous (iv) functional-group isomers
(f) The IUPAC name of the compound \( CH_3 - CH_2 - CH_2 - CH(CH_3) - CH_2 - CH_3 \) is
(i) 3-trimethylhexane (ii) 3-methyl hexane (iii) 4-methyl hexane
Answer:
(a) (iv) alkyl group - \( C_nH_{2n+1} \) is the formula for alkyl groups
(b) (i) \( C_{15}H_{30} \) - follows the alkene formula \( C_nH_{2n} \)
(c) (ii) an alkene - using \( C_nH_{2n} \), we get n = 24, giving molecular mass 72
(d) (iv) 2 - four carbon atoms can form butane (straight chain) and methylpropane (branched chain)
(e) (iv) functional-group isomers - alcohol and ether have same molecular formula but different functional groups
(f) (ii) 3-methyl hexane - longest chain is 6 carbons with methyl group at position 3
In simple words: These questions test understanding of hydrocarbon formulas, isomerism types, and basic naming rules. Each hydrocarbon family has its own formula pattern.

📝 Teacher's Note: Use the "formula game" - give students a molecular formula and have them determine if it's alkane, alkene, or alkyne before attempting to draw structures. This builds pattern recognition.

🎯 Exam Tip: Memorize the three key formulas: alkanes \( C_nH_{2n+2} \), alkenes \( C_nH_{2n} \), alkynes \( C_nH_{2n-2} \). Most multiple choice questions can be solved using these formulas.

Question. Fill in the blanks:
(a) Propane and ethane are ____________
(b) A saturated hydrocarbon does not participate in a/an _________ reaction.
(c) Succeeding members of a homologous series differ by ______________
(d) As the molecular masses of hydrocarbons increase, their boiling points ________ and melting point ____________
(e) \( C_{25}H_{52} \) and \( C_{50}H_{102} \) belong to _______________ homologous series.
(f) CO is an__________ Compound.
(g) The chemical properties of an organic compound are largely decided by the ___________ and the physical properties of an organic compound are largely decided by the ___________
(h) CHO is the functional group of an __________.
(i) The root in the IUPAC name of an organic compound depends upon the number of carbon atoms in _____________
(j) But-1-ene and but-2-ene are examples of ___________ isomerism.
Answer:
(a) Propane and ethane are homologues
(b) A saturated hydrocarbon does not participate in a/an addition reaction
(c) Succeeding members of a homologous series differ by \( CH_2 \)
(d) As the molecular masses of hydrocarbons increase, their boiling points increase and melting point increase
(e) \( C_{25}H_{52} \) and \( C_{50}H_{102} \) belong to the same homologous series
(f) CO is an organic Compound
(g) The chemical properties of an organic compound are largely decided by the functional group and the physical properties of an organic compound are largely decided by the functional group
(h) CHO is the functional group of an aldehyde
(i) The root in the IUPAC name of an organic compound depends upon the number of carbon atoms in the longest chain
(j) But-1-ene and but-2-ene are examples of position isomerism
In simple words: These fill-in-the-blanks cover key concepts about homologous series (compounds that differ by CH₂), functional groups that determine properties, and types of isomerism in organic chemistry.

📝 Teacher's Note: Create a concept map showing how molecular size affects physical properties. Students often confuse the different types of isomerism, so use clear examples and visual aids.

🎯 Exam Tip: For isomerism questions, always identify what's different between the compounds - if it's the position of a functional group, it's positional; if it's different functional groups, it's functional isomerism.

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

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

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

Exercise 12B

Question 1. State the sources of Alkanes.
Answer: Sources of alkane: The principal sources of alkanes are Natural gas and petroleum.
In simple words: Alkanes come mainly from natural gas (which we use for cooking) and petroleum (crude oil that we get from underground).

📝 Teacher's Note: Connect this to students' daily life by mentioning cooking gas, petrol, and diesel. Show petroleum products around the classroom to make it tangible.

🎯 Exam Tip: Always mention both sources - "natural gas and petroleum" - as the question asks for sources (plural).

Question 2. Methane is a greenhouse gas comment.
Answer: Methane is a primary constituent of natural gas. It absorbs outgoing heat radiation from the earth, and thus contributes to the green house effect and so it is considered as a green house gas.
In simple words: Methane traps heat from the Earth like a blanket, making our planet warmer, which is why it's called a greenhouse gas.

📝 Teacher's Note: Use the analogy of a greenhouse or blanket to explain how methane traps heat. Discuss global warming briefly to make it relevant.

🎯 Exam Tip: Mention the key phrase "absorbs outgoing heat radiation" and connect it to the greenhouse effect for full marks.

Question 3. Give the general formula of alkanes.
Answer: The general formula of alkane is: \( C_nH_{2n+2} \)
In simple words: For any alkane, if you have n carbon atoms, you will have (2n+2) hydrogen atoms - this formula always works.

📝 Teacher's Note: Verify with examples - CH4 (n=1, H=4=2×1+2), C2H6 (n=2, H=6=2×2+2). Let students practice with more examples.

🎯 Exam Tip: Write the formula clearly with proper subscripts and remember the "+2" part - it's commonly forgotten in exams.

Question 4. Draw the structures of isomers of:
(a) butane (b) pentane
Write the IUPAC and common names of these isomers
Answer: (a) The structures of isomers of butane are:
(i)
H H H H
| | | |
H-C-C-C-C-H
| | | |
H H H H
Common name: n-Butane
IUPAC name: Butane

(ii)
H H H
| | |
H-C-C-C-H
| | |
H H H
|
H-C-H
|
H
Common name: iso butane
IUPAC name: 2-methyl propane

(b) The structures of isomers of Pentane are:
(i)
H H H H H
| | | | |
H-C-C-C-C-C-H
| | | | |
H H H H H
Common name: n-pentane
IUPAC name: Pentane

(ii)
H H H H
| | | |
H-C-C-C-C-H
| | | |
H CH3 H H
Common name: iso pentane
IUPAC name: 2-methyl butane

(iii)
CH3
|
H3C-C-CH3
|
CH3
Common name: neo pentane
IUPAC name: 2,2-dimethyl propane
In simple words: Isomers are like different ways to arrange the same building blocks - same number of carbon and hydrogen atoms but arranged differently.

📝 Teacher's Note: Use molecular model kits to show how atoms can be rearranged. Emphasize that isomers have the same molecular formula but different structures.

🎯 Exam Tip: Always draw clear structural formulas and provide both common and IUPAC names when asked. Count carbons carefully for numbering.

Question 5. Write the:
(a) molecular formula
(b) electron dot formula and
(c) structural formula of methane and ethane.
Answer: For methane:
(a) Molecular formula is CH4
(b) Electron dot formula
[Electron dot structure showing C with 4 dots around it, each paired with H]
(c) Structural formula
H
|
H-C-H
|
H

For ethane:
(a) Molecular formula is: C2H6
(b) Electron dot formula:
[Electron dot structure showing two C atoms connected, each with appropriate H atoms]
(c) Structural Formula:
H H
| |
H-C-C-H
| |
H H
In simple words: These formulas show the same molecule in different ways - molecular formula shows what atoms are there, electron dot shows shared electrons, and structural formula shows how atoms connect.

📝 Teacher's Note: Start with electron dot structures to show covalent bonding, then simplify to structural formulas. Emphasize that each line represents a shared pair of electrons.

🎯 Exam Tip: In electron dot formulas, show all valence electrons clearly. In structural formulas, make sure each carbon has 4 bonds and each hydrogen has 1 bond.

Question 6. How is:
(a) methane and
(b) ethane prepared in the laboratory?
Answer: (a) Laboratory preparation of methane:
When the mixture of sodium ethanoate and soda lime is taken in a hard glass test tube and heated, the gas evolved is methane. It is collected by downward displacement of water.
CH3COONa + NaOH \( \xrightarrow{CaO, 300°C} \) Na2CO3 + CH4

(b) Laboratory preparation of ethane:
When the mixture of sodium propionate and soda lime is taken in the boiling tube and heated the ethane gas is evolved. It is also collected by downward displacement of water.
C2H5COONa + NaOH \( \xrightarrow{CaO, 300°C} \) Na2CO3 + C2H6
In simple words: We heat sodium salts with soda lime (a base) at high temperature to produce these gases, which are lighter than water so we collect them by pushing water down.

📝 Teacher's Note: Explain the decarboxylation reaction concept - removing CO2 from carboxylate salts. Show why gases are collected by water displacement (they're lighter).

🎯 Exam Tip: Always mention the exact reagents (sodium ethanoate/propionate + soda lime), conditions (CaO, 300°C), and collection method (downward displacement of water).

Question 7. How are methane and ethane prepared from methyl iodide and ethyl bromide?
Answer: When methyl iodide is reduced by nascent hydrogen at ordinary room temperature then methane is formed.
CH3I + 2[H] → CH4 + HI

When bromoethane is reduced by nascent hydrogen at ordinary room temperature then ethane is produced.
C2H5Br + 2[H] → C2H6 + HBr
In simple words: We use fresh hydrogen atoms (nascent hydrogen) to replace the halogen atoms (iodine or bromine) with hydrogen, forming the alkane.

📝 Teacher's Note: Explain what nascent hydrogen means - newly formed, highly reactive hydrogen atoms. Connect this to reduction reactions (adding hydrogen).

🎯 Exam Tip: Write [H] for nascent hydrogen and show the halogen being replaced by hydrogen in the balanced equation.

Question 8. What is a substitution reaction?
Give the reaction of chlorine with ethane and name the product formed.
Answer: A reaction in which one atom of a molecule is replaced by another atom (or group of atoms) is called a substitution reaction.
When ethane reacts with chlorine:
C2H6 + Cl2 → C2H5Cl + HCl (Chloroethane)
C2H5Cl + Cl2 → C2H4Cl2 + HCl (Dichloroethane)
C2H4Cl2 + Cl2 → C2H3Cl3 + HCl (Trichloroethane)
C2H3Cl3 + Cl2 → C2H2Cl4 + HCl (Tetrachloroethane)
C2H2Cl4 + Cl2 → C2HCl5 + HCl (Pentachloroethane)
C2HCl5 + Cl2 → C2Cl6 + HCl (Hexachloroethane)
In simple words: In substitution, we replace hydrogen atoms one by one with chlorine atoms, like swapping players in a team but keeping the same team size.

📝 Teacher's Note: Use the analogy of swapping letters in a word. Show that the reaction continues until all hydrogens are replaced if excess chlorine is present.

🎯 Exam Tip: Define substitution clearly first, then show the step-wise chlorination with proper names for each product.

Question 9. Name the compounds formed when methane burns in:
(a) sufficient air, (b) insufficient air,
Give a balanced equation
Answer: (a) Sufficient air: When methane burns in sufficient air, then carbon dioxide and water vapors are formed.
CH4 + 2O2 → CO2 + 2H2O

(b) Insufficient air: When methane burns in insufficient air, then carbon monoxide and water is formed.
2CH4 + 3O2 → 2CO + 4H2O
In simple words: With enough oxygen, methane burns completely to make carbon dioxide and water. With less oxygen, it burns incompletely to make poisonous carbon monoxide.

📝 Teacher's Note: Emphasize the danger of incomplete combustion producing CO. Use examples like gas stoves in poorly ventilated rooms.

🎯 Exam Tip: Remember that sufficient oxygen gives complete combustion (CO2), insufficient oxygen gives incomplete combustion (CO). Balance equations carefully.

Question 10. Write the names and the formula of the products formed when:
(a) methane (b) ethane
Reacts with: (i) chlorine (ii) bromine
Write the chemical equations
Answer: (a)
(i) When methane reacts with chlorine in the presence of sunlight or UV light, it undergoes substitution reaction to form Tetrachloromethane.
CH4 + Cl2 \( \xrightarrow{hν} \) CH3Cl + HCl (Chloromethane)
CH3Cl + Cl2 \( \xrightarrow{hν} \) CH2Cl2 + HCl (Dichloromethane)
CH2Cl2 + Cl2 \( \xrightarrow{hν} \) CHCl3 + HCl (Trichloromethane)
CHCl3 + Cl2 \( \xrightarrow{hν} \) CCl4 + HCl (Tetrachloromethane)

(ii) When methane reacts with bromine in the presence of sunlight or UV light:
CH4 + Br2 \( \xrightarrow{hν} \) CH3Br + HBr (Bromomethane)
CH3Br + Br2 \( \xrightarrow{hν} \) CH2Br2 + HBr (Dibromomethane)
CH2Br2 + Br2 \( \xrightarrow{hν} \) CHBr3 + HBr (Tribromomethane)
CHBr3 + Br2 \( \xrightarrow{hν} \) CBr4 + HBr (Tetrabromomethane)

(b) When ethane reacts with chlorine:
C2H6 + Cl2 \( \xrightarrow{hν} \) C2H5Cl + HCl (Chloroethane)
And this continues similarly producing multiple chlorinated products.
In simple words: Halogens like chlorine and bromine replace hydrogen atoms one by one when light energy is provided, creating a series of different compounds.

📝 Teacher's Note: Emphasize the need for sunlight or UV light (hν) for these reactions. Show that bromine reacts similarly to chlorine but forms brominated products.

🎯 Exam Tip: Always show the light condition (hν) above the arrow and name each product correctly using the appropriate prefix (mono-, di-, tri-, tetra-).

Question. When it reacts with bromine it forms Tetrabromomethane
\( \mathrm{CH_4} + \mathrm{Br_2} \rightarrow \mathrm{CH_3Br} + \mathrm{HCl} \)
\( \mathrm{CH_3Br} + \mathrm{Br_2} \rightarrow \mathrm{CH_2Br_2} + \mathrm{HCl} \)
Dibromomethane
\( \mathrm{CH_2Br_2} + \mathrm{Br_2} \rightarrow \mathrm{CHBr_3} + \mathrm{HCl} \)
Tribromo methane
\( \mathrm{CHBr_3} + \mathrm{Br_2} \rightarrow \mathrm{CBr_4} + \mathrm{HCl} \)
Tetrabromomethane
Answer: (ii) When methane reacts with bromine in stages, it forms different brominated compounds. First methane forms bromomethane, then dibromomethane, then tribromomethane, and finally tetrabromomethane through successive substitution reactions.
In simple words: When bromine is added to methane step by step, it replaces hydrogen atoms one by one until all four hydrogens are replaced by bromine atoms.

📝 Teacher's Note: Use models or diagrams to show how each hydrogen atom gets replaced by bromine in sequence. Emphasize that this is a substitution reaction where like replaces like.

🎯 Exam Tip: Always show the step-wise mechanism for halogenation reactions. Write the names of intermediate compounds to get full marks.

Question. When ethane reacts with chlorine it forms hexachoroethane.
\( \mathrm{C_2H_6} + \mathrm{Cl_2} \rightarrow \mathrm{C_2H_5Cl} + \mathrm{HCl} \)
Chloroethane
\( \mathrm{C_2H_5Cl} + \mathrm{Cl_2} \rightarrow \mathrm{C_2H_4Cl_2} + \mathrm{HCl} \)
Dichloroethane
\( \mathrm{C_2H_4Cl_2} + \mathrm{Cl_2} \rightarrow \mathrm{C_2H_3Cl_3} + \mathrm{HCl} \)
Trichloroethane
\( \mathrm{C_2H_3Cl_3} + \mathrm{Cl_2} \rightarrow \mathrm{C_2H_2Cl_4} + \mathrm{HCl} \)
Tetrachloroethane
\( \mathrm{C_2H_2Cl_4} + \mathrm{Cl_2} \rightarrow \mathrm{C_2HCl_5} + \mathrm{HCl} \)
Pentachloroethane
\( \mathrm{C_2HCl_5} + \mathrm{Cl_2} \rightarrow \mathrm{C_2Cl_6} + \mathrm{HCl} \)
Hexachloroethane
Answer: (i) When ethane reacts with chlorine in stages, it forms different chlorinated compounds. Starting from ethane, each step replaces one hydrogen atom with chlorine until all six hydrogens are replaced, forming hexachloroethane.
In simple words: Ethane has 6 hydrogen atoms, and chlorine replaces them one by one in six steps, making the final product with 6 chlorine atoms.

📝 Teacher's Note: Point out that ethane has 6 hydrogen atoms (unlike methane's 4), so it takes 6 steps to complete the substitution. Use counting to help students remember.

🎯 Exam Tip: Count the hydrogen atoms in the starting compound to predict how many substitution steps are needed. Always name each intermediate compound.

Question. When ethane reacts with bromine it forms Hexabromoethane
\( \mathrm{C_2H_6} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_5Br} + \mathrm{HBr} \)
Bromoethane
\( \mathrm{C_2H_5Br} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_4Br_2} + \mathrm{HBr} \)
Dibromoethane
\( \mathrm{C_2H_4Br_2} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_3Br_3} + \mathrm{HBr} \)
Tribromoethane
\( \mathrm{C_2H_3Br_3} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_2Br_4} + \mathrm{HBr} \)
Tetrabromoethane
\( \mathrm{C_2H_2Br_4} + \mathrm{Br_2} \rightarrow \mathrm{C_2HBr_5} + \mathrm{HBr} \)
Pentabromoethane
\( \mathrm{C_2HBr_5} + \mathrm{Br_2} \rightarrow \mathrm{C_2Br_6} + \mathrm{HBr} \)
HexaBromoethane
Answer: (ii) When ethane reacts with bromine in successive steps, all six hydrogen atoms are gradually replaced by bromine atoms, forming hexabromoethane as the final product.
In simple words: Just like with chlorine, bromine also replaces hydrogen atoms in ethane one by one until all 6 hydrogens become bromines.

📝 Teacher's Note: Compare this with the chlorination of ethane to show the pattern. Both halogens follow the same substitution mechanism.

🎯 Exam Tip: The pattern is the same for all halogens - only the halogen and hydrogen halide change in the equations.

Question 11. Name the compound prepared from: (a) sodium propionate, (b) methyl iodide and (c) ethyl bromide Write a balanced equation for the same
Answer: (a) Ethane is prepared from sodium propionate.
\( \mathrm{C_2H_5COONa} + \mathrm{NaOH} \xrightarrow{\mathrm{CaO, 300°C}} \mathrm{Na_2CO_3} + \mathrm{C_2H_6} \)
(b) Methane is prepared from methyl iodide.
\( \mathrm{CH_3I} + 2[\mathrm{H}] \rightarrow \mathrm{CH_4} + \mathrm{HI} \)
(c) Ethane is prepared from ethyl bromide.
\( \mathrm{C_2H_5Br} + 2[\mathrm{H}] \rightarrow \mathrm{C_2H_6} + \mathrm{HBr} \)
In simple words: These are different ways to make alkanes - from salt-like compounds using heat, or from halogen compounds using hydrogen.

📝 Teacher's Note: Explain that [H] represents nascent hydrogen, which is very reactive hydrogen. Show how decarboxylation removes CO₂ group while reduction removes halogen.

🎯 Exam Tip: Remember the conditions: sodium salts need heat and base, while halides need reducing agents. Write [H] for nascent hydrogen in reduction reactions.

Question 12. What is pyrolysis or cracking? Explain with example.
Answer: The decomposition of a compound by heat in the absence of air is called Pyrolysis. When pyrolysis occurs in alkanes, the process is termed cracking. For example: Alkanes on heating under high temperature or in the presence of a catalyst in absence of air broken down into lower alkanes, alkenes and hydrogen.
\( 2\mathrm{CH_4} \xrightarrow{1500°C} \mathrm{HC≡CH} + 3\mathrm{H_2} \)
In simple words: Pyrolysis is like breaking big molecules into smaller pieces using very high heat without any air present.

📝 Teacher's Note: Emphasize the "absence of air" condition - this prevents combustion and allows decomposition. Use the analogy of breaking a long chain into shorter pieces.

🎯 Exam Tip: Always mention "absence of air" and "high temperature" as key conditions. Show both the starting material breaking down and the products formed.

Question 13. Convert: (a) Methane into chloroform (b) sodium acetate into methane (c) Methyl iodide into ethane (d) Aluminum carbide into methane
Answer: (a) Methane into chloroform
\( \mathrm{CH_4} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3Cl} + \mathrm{HCl} \)
\( \mathrm{CH_3Cl} + \mathrm{Cl_2} \rightarrow \mathrm{CH_2Cl_2} + \mathrm{HCl} \)
\( \mathrm{CH_2Cl_2} + \mathrm{Cl_2} \rightarrow \mathrm{CHCl_3} + \mathrm{HCl} \)
(b) Sodium acetate into methane
\( \mathrm{CH_3COONa} + \mathrm{NaOH} \xrightarrow{\mathrm{CaO, 300°C}} \mathrm{Na_2CO_3} + \mathrm{CH_4} \)
(c) Methyl iodide into ethane
\( 2\mathrm{CH_3I} + 2\mathrm{Na} \xrightarrow{\mathrm{dry\ ether}} \mathrm{CH_3-CH_3} + 2\mathrm{NaI} \)
(d) Aluminium carbide into methane
\( \mathrm{Al_4C_3} + 12\mathrm{H_2O} \rightarrow 3\mathrm{CH_4} + 4\mathrm{Al(OH)_3} \)
In simple words: These are different chemical tricks to make one compound from another - some use heat, some use metals, and some use water.

📝 Teacher's Note: Show students that there are multiple pathways to make the same product. Emphasize the specific conditions needed for each reaction type.

🎯 Exam Tip: Learn the reaction conditions by heart - dry ether for Wurtz reaction, CaO/heat for decarboxylation, water for hydrolysis. Don't forget intermediate steps for halogenation.

Question 14. Give three uses of: (a) methane (b) ethane
Answer: (a) Methane: Three uses of methane are:
(i) Methane is a source of carbon monoxide and hydrogen
(ii) It is used in the preparation of ethyne, methanal, chloromethane, carbon tetrachloride.
(iii) It is employed as a domestic fuel.
(b) Ethane: Three uses of ethane are:
(i) It is used in the preparation of ethene, ethanol, and ethanol.
(ii) It forms ethyl chloride, which is used to make tetraethyllead.
(iii) It is also a good fuel.
In simple words: Both methane and ethane are useful as fuels for cooking and heating, and also as starting materials to make many other useful chemicals.

📝 Teacher's Note: Connect these uses to everyday life - mention LPG, natural gas, and industrial chemicals. Show how one simple molecule leads to many products.

🎯 Exam Tip: Remember the pattern: fuel use + industrial raw material + specific chemical synthesis. Give concrete examples of the chemicals that can be made.

Question 15. Under what conditions does ethane get converted to: (a) ethyl alcohol (b) acetaldehyde (c) acetic acid
Answer: (a) When a mixture of ethane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.
\( 2\mathrm{C_2H_6} + \mathrm{O_2} \xrightarrow{120\ atm, Cu\ tubes, 475K} 2\mathrm{C_2H_5OH} \)
(b) When mixture of ethane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Acetaldehyde.
\( \mathrm{C_2H_6} + \mathrm{O_2} \xrightarrow{MoO} \mathrm{CH_3CHO} + \mathrm{H_2O} \)
(c) Ethanol formed from ethane gets oxidized to acetic acid.
\( 2\mathrm{C_2H_6} + \mathrm{O_2} \xrightarrow{120\ atm, Cu\ tubes, 475K} 2\mathrm{C_2H_5OH} \)
\( \mathrm{C_2H_5OH} + \mathrm{O_2} \xrightarrow{Pt, 300°C} \mathrm{CH_3COOH} + \mathrm{H_2O} \)
In simple words: Different catalysts and conditions help ethane pick up oxygen to form alcohol, then aldehyde, then acid - like a step-by-step addition of oxygen.

📝 Teacher's Note: Emphasize the role of different catalysts (Cu, MoO, Pt) and how they control which product is formed. Show the oxidation sequence.

🎯 Exam Tip: Remember specific conditions for each product: high pressure + Cu for alcohol, MoO for aldehyde, Pt for acid. Write temperatures and pressures exactly.

Question 16. Give the inter-relationship of methane, methyl alcohol, formaldehyde and formic acid with conditions.
Answer: (a) Methane to methyl alcohol: When a mixture of methane and oxygen is compressed to about 120atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.
\( 2\mathrm{CH_4} + \mathrm{O_2} \xrightarrow{120\ atm, Cu\ tubes, 475K} 2\mathrm{CH_3OH} \)
(b) Methane to formaldehyde: When mixture of methane and oxygen is passed through heated molybdenum oxide, the mixture is oxidized to Formaldehyde.
\( \mathrm{CH_4} + \mathrm{O_2} \xrightarrow{MoO} \mathrm{HCHO} + \mathrm{H_2O} \)
(c) Methane to Formic acid: When a manganese based catalyst is used methane is oxidized to formic acid.
\( 2\mathrm{CH_4} + 3\mathrm{O_2} \xrightarrow{Mn\ compound} 2\mathrm{HCOOH} + 2\mathrm{H_2O} \)
In simple words: Methane can be turned into alcohol, aldehyde, or acid by using different catalysts - like having different tools for different jobs.

📝 Teacher's Note: Draw a flowchart showing methane at the center with arrows to the three products. Highlight how catalyst choice determines the product.

🎯 Exam Tip: Associate each catalyst with its product: Cu → alcohol, MoO → aldehyde, Mn → acid. These are key industrial processes.

Exercise 12(C)

Question 1. Write: (a) molecular formula, (b) electron dot formula and (c) structural formula of ethane (ethylene)
Answer: (a) The molecular formula of ethene is \( \mathrm{C_2H_4} \)
(b) Electron dot formula of ethene is: [electron dot structure showing shared electrons between carbon atoms]
(c) Structural formula of ethene: [structural formula showing double bond \( \mathrm{C=C} \)]
In simple words: Ethene has 2 carbon atoms and 4 hydrogen atoms, with a double bond between the carbons that makes it different from ethane.

📝 Teacher's Note: Emphasize the difference between ethane (single bond) and ethene (double bond). Use models to show electron sharing in the double bond.

🎯 Exam Tip: Always draw the electron dot structure clearly showing shared electron pairs. Remember ethene = C₂H₄ with double bond.

Question 2. The molecules of alkene family are represented by a general formula \( \mathrm{C_nH_{2n}} \). Answer the following: (a) What do n and 2n signify? (b) what is the name of alkene when n = 4? (c) What is the molecular formula of alkene when n = 4? (d) what is the molecular formula of the alkene if there are ten H atoms in it? (e) what is the structural formula of the third member of the alkene family? (f) write the molecular formula of lower and higher homologous of an alkene which contains four carbon atoms.
Answer: (a) n signifies the number of carbon atoms and 2n signifies the number of hydrogen atoms.
(b) The name of alkene when n = 4 is Butene.
(c) The molecular formula of alkene when n = 4 is \( \mathrm{C_4H_8} \).
(d) The molecular formula of alkene when there are 10 H atom in it \( \mathrm{C_5H_{10}} \).
(e) The structural formula of the third member of alkene is H-C-C-C-H with double bond
(f) Lower homologous of alkene which contain four carbons is \( \mathrm{C_3H_6} \). Higher homologous of alkene which contain four carbons is \( \mathrm{C_5H_{10}} \).
In simple words: The formula CₙH₂ₙ is like a pattern - it tells us for any number of carbons (n), we will have exactly double that number of hydrogens (2n).

📝 Teacher's Note: Use the general formula to predict any alkene's formula. Show students how to count backwards from hydrogen atoms to find carbon atoms.

🎯 Exam Tip: For part (d), if there are 10 H atoms, then 2n = 10, so n = 5. Always work systematically through the general formula.

Question 3. Discuss isomers in double bond compounds taking example of butane. Draw their structures and write IUPAC names.
Answer: The isomers of Butene are:
(i) \( \mathrm{CH_3-CH_2-CH=CH_2} \), But-1-ene
(ii) \( \mathrm{CH_3-CH=CH-CH_3} \), But-2-ene
(iii) \( \mathrm{CH_2=C(CH_3)-CH_3} \), 2-methyl propene
In simple words: Even with the same number of atoms, we can arrange them differently - like rearranging furniture in the same room to get different layouts.

📝 Teacher's Note: Show students how the double bond can be in different positions (1-ene vs 2-ene) or how branches can form (2-methyl propene). Use molecular models if available.

🎯 Exam Tip: Always number from the end that gives the double bond the lowest number. Remember branched isomers have different names entirely.

Question 4. Give a balanced equation for the lab. Preparation of ethylene. How is the gas collected?
Answer: Balanced Equation of ethylene:
\( \mathrm{CH_3-CH_2OH} + \mathrm{H_2SO_4} \rightarrow \mathrm{CH_3-CH_2HSO_4} + \mathrm{H_2O} \)
\( \mathrm{CH_3-CH_2HSO_4} \xrightarrow{\text{excess } \mathrm{H_2SO_4}, 160°C} \mathrm{CH_2=CH_2} \)
The gas is collected by downward displacement of water.
In simple words: Ethylene is made by heating ethyl alcohol with concentrated sulfuric acid, and the gas bubbles are collected in a water-filled container.

📝 Teacher's Note: Explain that sulfuric acid acts as a dehydrating agent, removing water from alcohol. Show the setup for gas collection over water.

🎯 Exam Tip: Write both steps of the reaction and mention the temperature (160°C). Always state the collection method for gas preparation questions.

Question 5. How is ethane prepared by: (a) dehydrohalogenation reaction (b) dehydration reaction? Give equations and name the products formed.
Answer: (a) Dehydrohalogenation reaction: [Answer continues but is cut off in the visible text]
In simple words: Both methods remove something from a molecule - dehydrohalogenation removes hydrogen and halogen, while dehydration removes water.

📝 Teacher's Note: Emphasize the prefixes: "de-hydro-halogenation" means removing H and halogen, "de-hydration" means removing water. Show how both create double bonds.

🎯 Exam Tip: Learn the word meanings: dehydrohalogenation = removal of HX (hydrogen halide), dehydration = removal of H₂O. Both form alkenes from alkanes/alcohols.

Question 6. Give the conditions and the main products formed by hydrogenation of ethylene.
Answer: When ethene and hydrogen are passed over finely divided catalyst such as platinum or palladium at ordinary temperature or nickel at 200°C, the two atoms of hydrogen molecule are added to the unsaturated molecule, which thus becomes a saturated one.
\( C_2H_4 + H_2 \xrightarrow{Ni, 200°C} C_2H_6 \)
In simple words: Ethene gas is mixed with hydrogen gas and passed over a nickel catalyst at 200°C, which adds hydrogen atoms to make ethane gas.

📝 Teacher's Note: Demonstrate with molecular models showing how the double bond "opens up" to accept two hydrogen atoms. Emphasize that this is an addition reaction, not a substitution.

🎯 Exam Tip: Always mention the catalyst (Ni/Pt/Pd) and temperature conditions - these are crucial for full marks in hydrogenation questions.

Question 7. Ethylene when reacts with halogens (chlorine and bromine) form saturated products. Name them and write balanced equations.
Answer: Chlorine and bromine are added to the double bond of ethene to form saturated ethylene chloride and ethylene bromide respectively.
\( CH_2 = CH_2 + Cl_2 \rightarrow CH_2(Cl) - CH_2(Cl) \)
1,2-dichloro ethane
\( CH_2 = CH_2 + Br_2 \rightarrow CH_2(Br) - CH_2(Br) \)
1,2-dibromo ethane
In simple words: When ethene reacts with chlorine or bromine, the halogen molecules break and attach to both carbon atoms, forming saturated compounds.

📝 Teacher's Note: Show students that this is an addition reaction where the double bond breaks and each carbon gets one halogen atom. Use the orange bromine water test as a practical example.

🎯 Exam Tip: Remember the naming: 1,2-dichloro ethane (not just "ethylene chloride") - the IUPAC names score more marks than common names.

Question 8. How is ethanol converted into ethene using (i) solid dehydrating agent (ii) hot conc. H₂SO₄? Give only balanced equations
Answer:
(i) Solid dehydrating agent:
\( C_2H_5OH \xrightarrow{Al_2O_3, 300°C} C_2H_4 + H_2O \)
Ethene
(ii) Hot conc. H₂SO₄:
\( C_2H_5OH \xrightarrow{Conc. H_2SO_4, 160-170°C} C_2H_4 + H_2SO_4 \)
In simple words: Both methods remove water molecules from ethanol to form ethene - one uses aluminum oxide powder, the other uses concentrated sulfuric acid at specific temperatures.

📝 Teacher's Note: Emphasize that dehydration means "removal of water" - students often confuse this with other reactions. Show how OH and H are removed as H₂O.

🎯 Exam Tip: Always write the exact temperature and reagent conditions - Al₂O₃ at 300°C and conc. H₂SO₄ at 160-170°C are specific requirements.

Question 9. Write the following properties of ethene: (a) Physical state (b) Odour (c) Density as compared to air (d) Solubility
Answer:
(a) Physical state: Ethene is a colourless and inflammable gas.
(b) Odour: It has faint sweetish odour.
(c) Density as compared to air: It has density less than one hence it is lighter than air.
(d) Solubility: It is sparingly soluble in water but highly soluble in organic solvents like alcohol, ether and chloroform.
In simple words: Ethene is a colorless gas that smells slightly sweet, floats in air because it's lighter, and doesn't mix well with water but mixes easily with alcohol-like liquids.

📝 Teacher's Note: Connect the "lighter than air" property to why ethene can be collected by downward displacement of air. Use everyday examples like how oil doesn't mix with water to explain solubility.

🎯 Exam Tip: For density questions, always compare with air and mention "lighter than air" or "heavier than air" - this comparison is what examiners expect.

Question 10. How would you convert: (a) ethene into 1, 2-dibromoethane? (b) ethene into ethyl bromide?
Answer:
(a) Ethene into 1, 2-dibromoethane: Ethene reacts with bromine at room temperature to form saturated ethylene chloride.
\( CH_2 = CH_2 + Br_2 \rightarrow CH_2(Br) - CH_2(Br) \)
1,2-dibromo ethane
(b) Ethene into ethyl bromide: When ethene is treated with HBr bromoethane is formed.
\( CH_2 = CH_2 + HBr \rightarrow CH_3 - CH_2Br \)
Ethyl bromide
In simple words: In the first reaction, both carbons get bromine atoms, but in the second reaction, only one carbon gets bromine while the other gets hydrogen.

📝 Teacher's Note: Highlight the difference between Br₂ (gives symmetrical products) and HBr (gives unsymmetrical products following Markovnikov's rule). Draw the electron movement if students are ready.

🎯 Exam Tip: For HBr addition, remember that H goes to the carbon with more hydrogens already attached - this follows Markovnikov's rule.

Question 11. Give balanced equations when: (a) ethene is burnt in excess of oxygen (b) ethene reacts with chlorine (c) ethene combines with hydrogen chloride (d) a mixture of ethene and hydrogen is passed over nickel at 200° C.
Answer:
(a) \( C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O + heat \)
(b) \( CH_2 = CH_2 + Cl_2 \rightarrow CH_2(Cl) - CH_2(Cl) \)
(c) \( CH_2 = CH_2 + HCl \rightarrow CH_3 - CH_2-Cl \)
(d) \( C_2H_4 + H_2 \xrightarrow{Ni, 200°C} C_2H_6 \)
In simple words: These show four different reactions - complete burning with oxygen, adding chlorine molecules, adding hydrogen chloride, and adding hydrogen gas with a catalyst.

📝 Teacher's Note: For combustion, remind students to balance oxygen carefully - ethene needs 3 oxygen molecules. For addition reactions, show how the double bond always breaks to accept new atoms.

🎯 Exam Tip: In combustion reactions, always include "heat" as a product and ensure the equation is balanced properly - count atoms on both sides.

Question 12. Give the formula and name of A, B C and D in the following equations:
(a) \( CH_4 \xrightarrow{Cl_2, -HCl} A \xrightarrow{Cl_2, -HCl} B \xrightarrow{Cl_2, -HCl} C \xrightarrow{Cl_2, -HCl} D \)
(b) \( C_2H_2 \xrightarrow{H_2} A \xrightarrow{H_2} B \xrightarrow{Br_2, -HBr} C \xrightarrow{Br_2, -HBr} D \)
(c) \( C_2H_4 + Cl_2 \rightarrow A \)
(d) \( C_2H_4 + B \xrightarrow{Ni, 200°C} C_2H_6 \)
Answer:
(a) \( CH_4 \xrightarrow{Cl_2, -HCl} CH_3Cl \xrightarrow{Cl_2, -HCl} CH_2Cl_2 \xrightarrow{Cl_2, -HCl} CHCl_3 \xrightarrow{Cl_2, -HCl} CCl_4 \)
A = monochloromethane
B = dichloromethane
C = Trichloromethane
D = Tetrachloromethane
(b) \( C_2H_2 \xrightarrow{H_2} C_2H_4 \xrightarrow{H_2} C_2H_6 \xrightarrow{Br_2, -HBr} C_2H_5Br \xrightarrow{Br_2, -HBr} C_2H_4Br \)
A= Ethene
B = ethane
C = bromoethane
D = dibromoethane
(c) \( C_2H_4 + Cl_2 \rightarrow C_2H_4Cl_2 \)
A = 1,2-dichloro ethane
(d) \( C_2H_4 + H_2 \xrightarrow{Ni, 200°C} C_2H_6 \)
B = hydrogen
In simple words: These are step-by-step reactions where simple molecules get more complex by adding atoms one at a time, like building with blocks.

📝 Teacher's Note: Show students the pattern in successive chlorination - each step replaces one more H with Cl. For hydrogenation sequences, emphasize how triple bonds become double bonds, then single bonds.

🎯 Exam Tip: For systematic naming, count the number of substituents carefully - mono (1), di (2), tri (3), tetra (4) are crucial prefixes.

Question 13. Write the name and formula of the product formed in each case below:
(a) \( C_2H_4 + Cl_2 \rightarrow \) …………….
(b) \( C_2H_5Br + KOH (alc.) \rightarrow \) ……….
(c) \( H_2C = CH_2 \xrightarrow{alk. KMnO_4} \) ………………
(d) \( H_2C = CH_2 + HBr \rightarrow \) …………….
(e) \( H_2C = CH_2 + O_3 \rightarrow \) …………….
Answer:
(a) \( C_2H_4 + Cl_2 \rightarrow CH_2(Cl) - CH_2(Cl) \)
1,2- dichloro ethane
(b) \( C_2H_5Br + KOH (alc.) \rightarrow C_2H_4 + KBr + H_2O \)
Ethane
(c) \( CH_2 = CH_2 \xrightarrow{alk. KMnO_4} CH_2(OH) - CH_2(OH) \)
1,2- Ethanediol
(d) \( CH_2 = CH_2 + HBr \rightarrow CH_3-CH_2Cl \)
Chloroethane
(e) \( CH_2 = CH_2 + O_3 \rightarrow \) [Ozonide formation]
In simple words: These reactions show different ways to modify ethene - adding halogens, removing atoms to form alkenes, adding oxygen groups, or forming special oxygen compounds.

📝 Teacher's Note: For reaction (b), emphasize that alcoholic KOH causes elimination (removes HBr), while aqueous KOH would cause substitution. The KMnO₄ reaction is an oxidation that adds OH groups.

🎯 Exam Tip: Remember the difference between alcoholic and aqueous KOH - alcoholic removes (elimination), aqueous substitutes. This distinction often appears in exams.

Question 14. What do you observe when ethylene is passed through alkaline KMnO₄ solution?
Answer: When ethylene is passed through alkaline KMnO₄ solution 1, 2-Ethanediol is formed. The Purple color of KMnO₄ decolorizes.
\( CH_2 = CH_2 + H - O - H + [O] \xrightarrow{Cold alkaline KMnO_4 solution} CH_2(OH) - CH_2(OH) \)
In simple words: The purple permanganate solution loses its color and becomes colorless because ethene uses up the oxygen from it to form a compound with two OH groups.

📝 Teacher's Note: This is an excellent visual test for unsaturation. Demonstrate the color change and explain that the purple color disappearing means the double bond has reacted.

🎯 Exam Tip: Always mention both the product formed (1,2-ethanediol) AND the color change (purple to colorless) for complete marks in observation questions.

Question 15. Name three compounds formed by ethylene and give the use of these compounds.
Answer: Three compounds formed by ethylene are:
Polythene
Ethanol
Epoxyethane
Uses of above compounds:
Polythene is used as carry bags.
Ethanol is used as a starting material for other products, mainly cosmetics and toiletry preparation.
Epoxyethane is used in the manufacture of detergents.
In simple words: Ethylene can be turned into plastic bags, alcohol for cosmetics, and special chemicals for making soaps and detergents.

📝 Teacher's Note: Connect these to everyday items students use - polythene bags, hand sanitizers (ethanol), and detergents. This makes the chemistry relevant to their daily lives.

🎯 Exam Tip: For "uses" questions, be specific - don't just say "industrial use" but mention exact applications like "carry bags" or "cosmetics manufacture".

Exercise 12 D

Question 1. What are the sources for alkynes? Give the general formula of alkynes.
Answer: Natural gas and Petroleum are sources for alkynes.
The general formula of alkynes are:
\( C_nH_{2n-2} \)
In simple words: Alkynes come from natural gas and oil, and their chemical formula follows the pattern where hydrogen atoms are always 2 less than double the carbon atoms.

📝 Teacher's Note: Compare the general formulas of alkanes (CₙH₂ₙ₊₂), alkenes (CₙH₂ₙ), and alkynes (CₙH₂ₙ₋₂) to show the pattern of decreasing hydrogen atoms with increasing unsaturation.

🎯 Exam Tip: Remember the formula pattern: alkanes (+2), alkenes (0), alkynes (-2) compared to 2n. This helps in identifying compound types quickly.

Question 2. Give an example of isomers shown by triple bond hydrocarbon (alkynes) and write its IUPAC name.
Answer: Butyne is an example, its isomers are:
H-C-C≡C-H and H-C≡C-C-H (with appropriate hydrogen atoms attached)
IUPAC name: But-2-yne But-1-yne
In simple words: The same four carbon atoms can be arranged with the triple bond either in the middle (but-2-yne) or at the end (but-1-yne) of the chain.

📝 Teacher's Note: Use molecular models to show how the position of the triple bond creates different compounds. Emphasize that the numbers indicate where the triple bond starts.

🎯 Exam Tip: For IUPAC naming of alkynes, the number indicates the position of the triple bond - count from the end that gives the lowest number.

Question 3. How is acetylene prepared in the laboratory? (a) draw a diagram (b) Give an equation (c) How is pure dry gas collected?
Answer:
(a) Diagram of acetylene preparation: [Laboratory apparatus diagram shown]
(b) \( CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 \)
(c) The pure dry gas is collected by downward displacement of water, since it is insoluble in water.
In simple words: Calcium carbide is mixed with water to produce acetylene gas, which bubbles up and is collected in a jar placed upside down over water.

📝 Teacher's Note: Emphasize safety precautions as this reaction is vigorous and acetylene is highly inflammable. Show students why downward displacement works - the gas is less dense than water.

🎯 Exam Tip: For gas collection methods, always mention the reason - "downward displacement of water because acetylene is insoluble in water" gives complete explanation.

Question 4. Give the method of preparation of ethyne by 1, 2− dibromoethene.
Answer: When 1,2-dibromoethane is boiled with alcoholic potassium hydroxide, ethyne is formed.
\( CH_2Br-CH_2Br + 2KOH(alc.) \rightarrow C_2H_2 + 2KBr + 2H_2O \)
In simple words: The alcoholic KOH removes both bromine atoms and two hydrogen atoms from dibromoethane, creating a triple bond to form ethyne.

📝 Teacher's Note: This is a double elimination reaction. Show how two molecules of HBr are eliminated to form the triple bond. Compare with single elimination that forms double bonds.

🎯 Exam Tip: For elimination reactions forming alkynes, always specify "alcoholic KOH" and mention that it's a double elimination (removes 2 HBr molecules).

Question 5. Classify the following compounds as alkanes, alkenes and alkynes.
\( C_3H_4 \):-
\( C_3H_8 \):-
\( C_5H_8 \):-
\( C_3H_6 \):-
Answer: The following compounds can be classified as:
\( C_3H_4 \):- Alkynes
\( C_3H_8 \):- Alkanes
\( C_5H_8 \):- Alkynes
\( C_3H_6 \):- Alkenes
In simple words: To classify these compounds, we look at the hydrogen to carbon ratio. Alkanes have the most hydrogen atoms, alkenes have fewer, and alkynes have the least because they have double or triple bonds.

📝 Teacher's Note: Teach students the general formulas: alkanes (\( C_nH_{2n+2} \)), alkenes (\( C_nH_{2n} \)), and alkynes (\( C_nH_{2n-2} \)). Have them practice identifying the type by counting carbons and hydrogens.

🎯 Exam Tip: Memorize the general formulas and always check the H:C ratio. Count carbons first, then see if hydrogens match alkane, alkene, or alkyne formula.

Question 6. Give a chemical test to distinguish between
(a) saturated and unsaturated compounds
(b) ethane and ethene
(c) ethene (ethylene) and ethyne (acetylene)
Answer: Chemical test to distinguish:
(b) Ethane and ethene:

(c) Ethene and ethyne:

In simple words: Unsaturated compounds (alkenes and alkynes) react with bromine water and make it colorless, while saturated compounds (alkanes) don't react. Alkynes also form colored precipitates with special metal solutions.

📝 Teacher's Note: Demonstrate these tests with actual chemicals if possible. Students remember visual color changes better than theory. Emphasize that only unsaturated compounds decolorize bromine water.

🎯 Exam Tip: Always mention the color changes clearly - "reddish brown to colorless" for bromine test, "purple to colorless" for permanganate test. These specific observations get full marks.

Question 7. Name the products formed and write an equation when ethyne is added to the following in an inert solvent:
(a) chlorine
(b) bromine
(c) iodine
(d) hydrogen
(e) excess of hydrochloric acid
Answer: (a) Ethyne in an inert solvent of carbon tetrachloride adds chlorine to change into 1,2-dichloro ethene with carbon-carbon double bond, and then to an 1,1,2,2-tetrachloro ethane with carbon-carbon single bond.
\( C_2H_2 \xrightarrow{Cl_2} C_2H_2Cl_2 \xrightarrow{Cl_2} C_2H_2Cl_4 \)
1,2-dichloro ethene → 1,1,2,2-tetrachloro ethane
(b) Ethyne in an inert solvent of carbon tetrachloride adds bromine to change into 1,2-dibromo ethene and then to 1,1,2,2-tetrabromo ethane.
\( C_2H_2 \xrightarrow{Br_2} C_2H_2Br_2 \xrightarrow{Br_2} C_2H_2Br_4 \)
(c) Iodine reacts slowly in the presence of alcohol to form di-iodo ethene
\( CH \equiv CH + I_2 \rightarrow ICH = CHI \)
1,2-di-iodoethene
(d) In the presence of nickel, platinum or palladium ethyne change to ethene and then to ethane.
\( CH \equiv CH \xrightarrow{H_2/Ni} CH_2 = CH_2 \xrightarrow{H_2/Ni} CH_3 - CH_3 \)
In simple words: Ethyne (triple bond) can add different chemicals step by step. First it becomes a double bond compound, then a single bond compound. It's like filling up empty spaces in the molecule.

📝 Teacher's Note: Use molecular models to show how triple bonds become double bonds, then single bonds. Students understand addition reactions better when they see the 3D structure changes.

🎯 Exam Tip: Write both steps of the reaction clearly with intermediate products. Don't skip the intermediate compound - examiners expect to see the complete pathway.

Question 8. Name the hydrocarbon which;
(a) is a tetrahedral molecule
(b) is a planar molecule
(c) is a linear molecule
(d) forms a red precipitate with ammoniacal solution of copper (I) chloride
(e) is known as paraffin
(f) is known as olefin
Answer: (a) The hydrocarbon which is tetrahedral is Methane.
(b) The hydrocarbon which is planar molecule is ethene.
(c) The hydrocarbon which is a linear molecule is Ethyne.
(d) The hydrocarbon which forms a red precipitate with ammoniacal solution of copper chloride is acetylene.
(e) Alkanes are also called as paraffin.
(f) Alkenes are also called olefin.
In simple words: Different hydrocarbons have different shapes. Methane looks like a pyramid, ethene is flat like a sheet of paper, and ethyne is straight like a stick. Each shape comes from how the atoms are bonded together.

📝 Teacher's Note: Use ball-and-stick models to show molecular geometry. Connect the shapes to bond types - single bonds allow rotation (tetrahedral), double bonds are rigid and flat (planar), triple bonds are straight (linear).

🎯 Exam Tip: Remember the connection: alkanes = tetrahedral = paraffin, alkenes = planar = olefin, alkynes = linear = acetylene test. This helps you answer multiple parts quickly.

Exercise 12E

Question 1.
(a) what are alcohols? State their sources
(b) give general formulae of monohydric alcohol
Answer: (a) Alcohols are the hydroxyl derivatives of alkanes and are formed by replacing one or more hydrogen atoms of the alkane with OH group. Methanol is obtained from destructive distillation of wood while ethanol is obtained from fermentation of sugar.
(b) General formula of monohydric alcohol: \( C_nH_{2n+1}OH \)
In simple words: Alcohols are like alkanes but with an -OH group attached. Think of them as modified hydrocarbons where one hydrogen is replaced by a hydroxyl group. We get them from wood or sugar fermentation.

📝 Teacher's Note: Connect alcohols to students' everyday experience - hand sanitizer, rubbing alcohol. Explain that ethanol in drinks comes from natural fermentation, while methanol is poisonous and comes from wood.

🎯 Exam Tip: Always write the -OH group clearly in the general formula. Remember: monohydric = one -OH group, so it's \( C_nH_{2n+1}OH \).

Question 2. Give the
(a) Dot diagram
(b) Abbreviated formula
(c) Structure of second member of the alcohol group.
Answer: (a) Dot diagram
[Molecular structure with electron dots shown]
(b) Abbreviated formula
\( C_2H_5OH \)
(c) Structure:
H H
| |
H-C-C-O-H
| |
H H
In simple words: Ethanol has two carbon atoms connected in a chain, with hydrogen atoms filling up the remaining bonds and an -OH group at one end. It's the alcohol found in drinks.

📝 Teacher's Note: Start with the carbon skeleton, then add hydrogens to satisfy valency of 4 for each carbon. Show how the -OH group replaces one hydrogen from the alkane structure.

🎯 Exam Tip: Draw structures systematically: carbon skeleton first, then add hydrogens, finally place the -OH group. Count all atoms to verify your formula matches \( C_2H_5OH \).

Question 3. State the method of preparation of ethanol:
(a) by hydrolysis of ethane,
(b) by hydrolysis of alkyl halide
Answer: (a) By hydrolysis of ethene: When concentrated sulphuric acid is added to ethene at a temperature of 80°C and pressure of 30 atm. ethyl hydrogen sulphate is produced. Ethyl hydrogen sulphate on hydrolysis with boiling water gives ethanol.
\( C_2H_4 + H_2SO_4 \xrightarrow{80°C, 30atm} C_2H_5HSO_4 \)
\( C_2H_5HSO_4 + H_2O \rightarrow C_2H_5OH + H_2SO_4 \)
(b) By hydrolysis of alkyl halide: Alcohols can be prepared by the hydrolysis of alkyl halide with a hot dilute alkali.
\( C_2H_5Cl + KOH \xrightarrow{boil} C_2H_5OH + KCl \)
In simple words: We can make ethanol by adding water to ethene using acid, or by treating ethyl chloride with base. Both methods basically add an -OH group to replace something else.

📝 Teacher's Note: Emphasize the two-step process in method (a) - first acid addition, then hydrolysis. Students often miss the intermediate compound. Show how both methods follow SN2 mechanism.

🎯 Exam Tip: Write reaction conditions clearly (temperature, pressure, "boil"). Include the intermediate product in method (a) - don't jump directly from ethene to ethanol.

Question 4. How is ethanol prepared by fermentation?
Answer: Ethanol is prepared by the fermentation of sugar by the enzymes invertase and zymase.
\( C_{12}H_{22}O_{11} + H_2O \xrightarrow{Invertase\ Fermentation} C_6H_{12}O_6 + C_6H_{12}O_6 \)
Glucose Fructose
\( C_6H_{12}O_6 \xrightarrow{Zymase(yeast)} 2C_2H_5OH + 2CO_2 \)
Ethanol
In simple words: Yeast breaks down sugar into alcohol and carbon dioxide. First the complex sugar splits into simple sugars, then these simple sugars turn into alcohol. This is how wine and beer are made naturally.

📝 Teacher's Note: Connect this to real-life brewing and wine-making. Explain that yeast is a living organism that "eats" sugar and "produces" alcohol as waste. This makes the biochemical process more relatable.

🎯 Exam Tip: Write both enzyme names (invertase and zymase) and show the two-step process. Remember the 2:1 ratio - one glucose molecule gives two ethanol molecules.

Question 5. Give the lab. Prepared of:
(a) ethyl alcohol
(b) methyl alcohol
Answer: (a) Ethyl alcohol:
Ethyl chloride reacts with aqueous potassium hydroxide to form ethyl alcohol.
\( C_2H_5Cl + KOH \xrightarrow{boil} C_2H_5OH + KCl \)
(b) Methyl alcohol:
Methyl bromide reacts with aqueous potassium hydroxide to form methyl alcohol.
\( CH_3Br + KOH \xrightarrow{boil} CH_3OH \)
In simple words: In the lab, we make alcohols by heating alkyl halides (compounds with Cl, Br, I) with strong base like KOH. The base kicks out the halogen and replaces it with -OH.

📝 Teacher's Note: Explain this as a substitution reaction where the halide ion is a good leaving group. Students should understand that we need heat and aqueous conditions for this reaction to work.

🎯 Exam Tip: Always write "boil" as the reaction condition and include the salt product (KCl or KBr). This shows complete understanding of the reaction.

Question 6.
(a) how do the boiling point and melting point change in the homologous series of alcohols?
(b) Name the product formed when ethanol reacts with acetic acid. Give an equation.
(c) What is the name given to this type of reaction?
Answer: (a) The melting and boiling point of the successive members of the homologous series of alcohols increase with the increase in molecular mass.
(b) When ethanol reacts with acetic acid ethyl acetate is formed.
\( C_2H_5OH + CH_3COOH \xrightarrow{Conc. H_2SO_4} CH_3COOC_2H_5 + H_2O \)
(c) This reaction is known as esterification reaction.
In simple words: Longer alcohol molecules have higher boiling points because they stick together more strongly. When alcohol meets acid, they combine to form an ester (like making artificial fruit flavors) and water.

📝 Teacher's Note: Explain intermolecular hydrogen bonding in alcohols for the boiling point trend. For esterification, mention that esters often have fruity smells - this helps students remember the reaction.

🎯 Exam Tip: Write "Conc. \( H_2SO_4 \)" as catalyst in esterification. Remember that water is always a product when alcohol and carboxylic acid react.

Question 7. What is the effect ethanol on human body.
Answer: Ethanol affects that part of the brain which controls our muscular movements and then gives temporary relief from tiredness. But it damages the liver and kidney too.
In simple words: Alcohol makes people feel relaxed by affecting the brain, but it's actually harmful to important organs like the liver and kidneys. The relaxation is temporary but the damage can be permanent.

📝 Teacher's Note: Use this as a health education moment. Explain how the liver processes alcohol and why excessive drinking causes liver damage. Connect to real-world consequences of alcohol abuse.

🎯 Exam Tip: Mention both the temporary effect (brain/tiredness relief) and the harmful effects (liver and kidney damage) for complete answer.

Question 8. How are the following obtained
(a) absolute alcohol
(b) spurious alcohol
(c) methylated spirit?
Answer: (a) Absolute alcohol: Absolute alcohol may be obtained by distilling moist alcohol with benzene. The mixture of water and benzene distills off and anhydrous alcohol is left behind.
(b) Spurious alcohol: It is made by improper distillation. It contains large portions of methanol in a mixture of alcohols.
(c) Methylated spirit: Methylated spirit or denatured alcohol is ethyl alcohol with 5% methyl alcohol, a coloured dye and some pyridine.
In simple words: Absolute alcohol is completely pure alcohol with no water. Spurious alcohol is dangerous because it has poisonous methanol. Methylated spirit is ethanol mixed with poisonous substances to prevent people from drinking it.

📝 Teacher's Note: Warn students about the dangers of spurious alcohol - it causes blindness and death. Explain that methylated spirit is deliberately made undrinkable for industrial use.

🎯 Exam Tip: Mention the specific percentages and additives for methylated spirit (5% methyl alcohol, dye, pyridine). This shows detailed knowledge.

Question 9. Name the products formed and give appropriate chemical equations for the following:
(a) sodium reacting with ethyl alcohol
(b) Ethanol oxidized by acidified potassium dichromate
Answer: (a) Sodium reacting with ethyl alcohol:
\( 2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \)
Products: Sodium ethoxide and hydrogen gas
(b) Ethanol oxidized by acidified potassium dichromate:
\( C_2H_5OH + K_2Cr_2O_7 + H_2SO_4 \rightarrow CH_3COOH + Cr_2(SO_4)_3 + K_2SO_4 + H_2O \)
Product: Acetic acid (ethanoic acid)
In simple words: When sodium metal touches alcohol, it releases hydrogen gas and forms sodium ethoxide. When alcohol is oxidized with acidified dichromate, it turns into acetic acid (vinegar).

📝 Teacher's Note: Demonstrate the sodium-alcohol reaction if possible - students remember the vigorous hydrogen gas evolution. Connect the oxidation to how wine turns to vinegar over time.

🎯 Exam Tip: Balance chemical equations completely. For oxidation, remember that primary alcohols go to carboxylic acids, not just aldehydes.

Question 9.
(a) Sodium reacting with ethyl alcohol:
\( 2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \)
When sodium reacts with ethyl alcohol hydrogen is evolved with formation of sodium ethoxide.
(b) Ethanol oxidized by \( K_2Cr_2O_7 \):
\( C_2H_5OH \xrightarrow{[O]} CH_3CHO + H_2O \xrightarrow{[O]} CH_3COOH \)
Alcohols gets oxidized and get converted into ethanal and then into acetic acid.
Answer: (a) When sodium reacts with ethyl alcohol, hydrogen gas is released and sodium ethoxide forms. (b) When ethanol is oxidized by potassium dichromate, it first becomes ethanal (acetaldehyde) and then further oxidizes to form acetic acid.
In simple words: Sodium pulls hydrogen from alcohol to make hydrogen gas, while strong oxidizing agents convert alcohol step by step into acetic acid (vinegar).

📝 Teacher's Note: Demonstrate the hydrogen gas evolution with sodium and ethanol — students can see the bubbling. For oxidation, use color changes of dichromate (orange to green) as a visual indicator of the reaction progress.

🎯 Exam Tip: Always show both steps of oxidation (alcohol → aldehyde → acid) and mention the oxidizing agent. Write balanced equations with proper arrows and conditions.

Question 10. Give the trivial (common) names and the IUPAC names of the following:
(a) \( C_3H_6 \) (b) \( C_2H_4 \) (c) \( C_2H_2 \) (d) \( CH_3OH \) (e) \( C_2H_5OH \)
Answer:

In simple words: Each compound has an old common name that people still use and a newer IUPAC name that follows international rules — like how people have nicknames and official names.

📝 Teacher's Note: Emphasize that common names are still widely used in industry and daily life, while IUPAC names follow systematic rules. Use memory tricks like "acetylene for welding" to help students remember.

🎯 Exam Tip: Learn both names for each compound. Questions often ask for one type specifically, and using the wrong naming system loses marks even if the structure is correct.

Question 11. Ethanol can be oxidized to ethanoic acid. Write the equation and name the oxidizing agent.
Answer: \( C_2H_5OH \xrightarrow{[O]} CH_3CHO + H_2O \xrightarrow{[O]} CH_3COOH \)
The oxidizing agents that can be used are potassium dichromate and potassium permanganate.
In simple words: Ethanol (alcohol) gets converted to acetic acid (vinegar) when treated with strong oxidizing chemicals in two steps.

📝 Teacher's Note: Show that this is a two-step process with aldehyde as intermediate. Mention that this is how vinegar can be made from wine or other alcoholic beverages through bacterial oxidation.

🎯 Exam Tip: Always show both steps of oxidation and name both possible oxidizing agents. Don't forget to write [O] above the arrows to show oxidation.

Question 12. Complete and balanced the following equations. State the conditions wherever necessary.
(a) \( CH \equiv CH + H_2 \rightarrow \ldots + H_2 \rightarrow \ldots \)
(b) \( C_2H_4 + Cl_2 \rightarrow \ldots \)
(c) \( C_2H_4 + HCl \rightarrow \ldots \)
(d) \( CaC_2 + H_2O \rightarrow \ldots \)
(e) \( C_2H_2 + Br_2 \rightarrow \ldots \)
(f) \( C_2H_5OH \xrightarrow{[o] K_2Cr_2O_7} \ldots \)
Answer:
(a) \( CH \equiv CH + H_2 \xrightarrow{Ni, 200°C} CH_2 = CH_2 + H_2 \xrightarrow{Ni, 200°C} CH_3-CH_3 \)
(b) \( C_2H_4 + Cl_2 \rightarrow CH_2(Cl)-CH_2(Cl) \)
(c) \( C_2H_4 + HCl \rightarrow CH_3-CH_2Cl \)
(d) \( CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2 \)
(e) \( C_2H_2 + Br_2 \rightarrow H(Br)C = C(Br)H \)
(f) \( C_2H_5OH \xrightarrow{[o] K_2Cr_2O_7} CH_3CHO \)
In simple words: These reactions show how hydrocarbons and alcohols react with different chemicals to form new compounds — like adding pieces to a puzzle.

📝 Teacher's Note: Emphasize the difference between addition reactions (adding across double/triple bonds) and oxidation reactions. Use models to show how atoms are added or removed.

🎯 Exam Tip: Always include reaction conditions (temperature, catalyst) when given. For addition reactions, remember that hydrogen adds across multiple bonds in stages.

Question 13. Name an organic compound which is:
(a) used for illuminating country houses
(b) Used for making a household plastic material
(c) Called 'wood spirit'
(d) Poisonous and contains OH group
(e) Consumed as a drink
(f) Made from water gas
Answer:
(a) Used for illuminating country houses: Ethyne
(b) Used for making a household plastic material: ethyne
(c) Called 'wood spirit': Methanol
(d) Poisonous: Methanol
(e) Consumed as a drink: Ethanol
(f) Made from water gas: Methanol
In simple words: Different organic compounds have specific uses — ethyne for lighting and plastics, methanol as a poisonous wood spirit, and ethanol as drinking alcohol.

📝 Teacher's Note: Discuss why methanol is called "wood spirit" (originally made from wood distillation) and emphasize the difference between toxic methanol and safe ethanol for consumption.

🎯 Exam Tip: Remember that ethyne (acetylene) is used for both lighting and plastic making, while methanol and ethanol are both alcohols but only ethanol is safe to drink.

Exercise 12F

Question 1. What are carboxylic acids? Give their general formula
Answer: An organic compound containing the carboxyl group (COOH) is known as carboxylic acid. The general formula: \( C_nH_{2n+1}COOH \)
In simple words: Carboxylic acids are organic compounds that contain a special -COOH group that makes them acidic, like the acetic acid in vinegar.

📝 Teacher's Note: Draw the carboxyl group structure clearly showing both the C=O and C-OH parts. Explain that this combination gives the acidic properties.

🎯 Exam Tip: Always write the general formula correctly with the COOH group separate from the alkyl part. Don't forget the subscript in C_n H_{2n+1}.

Question 2. Write the common name, IUPAC name and formula of one monocarboxylic acid and one dicarboxylic acid
Answer:
Monocarboxylic acid:
Formula: HCOOH
Common name: Formic acid
IUPAC name: Methanoic acid
Dicarboxylic acid:
Formula: COOH-COOH
Common name: Oxalic acid
IUPAC name: Ethane-di-oic acid
In simple words: Monocarboxylic acids have one -COOH group (like formic acid from ant stings), while dicarboxylic acids have two -COOH groups (like oxalic acid found in spinach).

📝 Teacher's Note: Mention that formic acid is found in ant stings and stinging nettles, while oxalic acid is found in many green vegetables. This helps students remember the compounds.

🎯 Exam Tip: For dicarboxylic acids, write "di-oic" in the IUPAC name. The formula HOOC-COOH can also be written as (COOH)_2.

Question 3. Write the names of:
(a) First three members of carboxylic acids series.
(b) Three compounds that can be oxidized directly or in stages to produce acetic acid.
Answer:
(a) First three members of carboxylic acids are:
Methanoic acid
Ethanoic acid
Propanoic acid
(b) Three compounds that can be oxidized directly or in stages to produce acetic acid are:
Ethanol
Acetylene
Ethanal
In simple words: The carboxylic acid series starts with methanoic, ethanoic, and propanoic acids, while ethanol, acetylene, and ethanal can all be converted to acetic acid through chemical reactions.

📝 Teacher's Note: Show the progression in carbon atoms (C1, C2, C3) for the acid series. For part (b), draw the oxidation pathways showing how each compound leads to acetic acid.

🎯 Exam Tip: Remember the systematic naming: meth- (1C), eth- (2C), prop- (3C) + -anoic acid. For oxidation, all three compounds eventually form the same product: acetic acid.

Question 4. Vinegar in greyish in colour with a particular taste Explain.
Answer: Vinegar commonly called Sirka is a dilute solution of acetic acid. The presence of colouring matter gives it a greyish colour while the presence of some other organic acids and organic compounds impart it the usual taste and flavour.
In simple words: Vinegar looks greyish and tastes different from pure acetic acid because it contains other natural substances that add color and flavor.

📝 Teacher's Note: Compare commercial vinegar with pure acetic acid solution to show the difference. Explain that vinegar is made from natural fermentation, which adds impurities that enhance taste.

🎯 Exam Tip: Mention both the color (due to impurities) and taste (due to other organic compounds) in your answer. Don't just say it's acetic acid.

Question 5. Give the structural formulae and IUPAC name of acetic acid. What is glacial acetic acid?
Answer: Structural formula of acetic acid:
H
|
H-C-C-OH
| ||
H O
IUPAC name of acetic acid is: Ethanoic acid
Glacial acetic acid is the pure form of acetic acid. It does not contain water.
In simple words: Acetic acid has a methyl group attached to a carboxyl group, its IUPAC name is ethanoic acid, and glacial acetic acid is 100% pure without any water mixed in.

📝 Teacher's Note: Draw the structure clearly showing the tetrahedral arrangement around carbons. Explain why it's called "glacial" (crystallizes like ice at room temperature when pure).

🎯 Exam Tip: Draw the structural formula properly with all bonds shown. Remember that glacial = pure = no water, which is why it can freeze at room temperature.

Question 6. Complete:
(a) Vinegar is prepared by the bacterial oxidation of ...
(b) The organic acid present in vinegar is ...
(c) The next higher homologue of ethanoic acid is ...
Answer:
(a) Ethanol
(b) Acetic acid
(c) Propanoic acid
In simple words: Vinegar comes from ethanol being oxidized by bacteria, contains acetic acid as the main component, and the next acid in the series is propanoic acid.

📝 Teacher's Note: Explain the bacterial fermentation process that converts alcohol to vinegar. Show the homologous series pattern where each member differs by CH2.

🎯 Exam Tip: Remember that bacterial oxidation is slower but more complete than chemical oxidation. Higher homologue means adding one more CH2 group.

Question 7. How is acetic prepared from
(a) ethanol (b) acetylene?
Answer:
(a) It is prepared in the lab by the oxidation of ethanol with acidified potassium dichromate.
\( C_2H_5OH \xrightarrow{[O]} CH_3CHO \xrightarrow{[O]} CH_3COOH \)
(b) Acetylene is first converted to acetaldehyde by passing through 40% \( H_2SO_4 \) at 60°C in the presence of 1% \( HgSO_4 \).
The acetaldehyde is then oxidised to acetic acid in the presence of catalyst manganous acetate at 70°C.
\( C_2H_2 + H_2O \xrightarrow{H_2SO_4(dil), HgSO_4} CH_3CHO \)
\( CH_3CHO + O \xrightarrow{Catalyst} 2CH_3COOH \)
In simple words: Ethanol is directly oxidized to acetic acid in two steps, while acetylene first needs to be converted to acetaldehyde and then oxidized to acetic acid.

📝 Teacher's Note: Emphasize that both pathways go through acetaldehyde as an intermediate. The acetylene route is more complex and requires specific catalysts and conditions.

🎯 Exam Tip: Always show the intermediate step (acetaldehyde) in both methods. Include reaction conditions for the acetylene method as they are crucial for the reaction to occur.

Question 8. What do you notice when acetic acid reacts with
(a) litmus (b) metals (c) alkalies (d) alcohol
Answer:
(a) When acetic acid reacts with litmus it turns blue litmus red.
(b) When acetic acid reacts with metals hydrogen is evolved.
\( 2CH_3COOH + Zn \rightarrow (CH_3COO)_2Zn + H_2 \)
(c) When acetic acid reacts with alkalies it forms salt
\( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \)
(d) Acetic acid reacts with alcohols forming esters
\( CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O \)
In simple words: Acetic acid shows typical acid behavior - turns litmus red, releases hydrogen with metals, neutralizes bases to form salts, and combines with alcohols to make sweet-smelling esters.

📝 Teacher's Note: Demonstrate the litmus test and hydrogen gas evolution. Explain that esterification produces pleasant fruity smells that students might recognize in perfumes or flavorings.

🎯 Exam Tip: Show balanced equations for each reaction type. Remember that esterification requires an acid catalyst and produces water as a byproduct.

Question 9. Acetic acid is a typical acid. Write one equation in each case for its reactions with
(a) a metal (b) a base / alkali
(c) a carbonate (d) a bicarbonate
Answer:
(a) \( 2CH_3COOH + Zn \rightarrow (CH_3COO)_2Zn + H_2 \)
(b) \( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \)
(c) \( 2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2 \)
(d) \( CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \)
In simple words: Acetic acid behaves like other acids - it reacts with metals to give hydrogen, neutralizes bases, and reacts with carbonates to produce carbon dioxide gas.

📝 Teacher's Note: Highlight the gas evolution in reactions (c) and (d) - this is a good test for acids. Show that carbonates and bicarbonates both produce CO2 but in different ratios.

🎯 Exam Tip: Balance equations carefully, especially with carbonates where you need 2 moles of acid. Gas evolution (H2 or CO2) is a key identifying feature of these reactions.

Question 10. Name:
(a) compound formed when acetic acid and ethanol react together
(b) reducing agent used to convert acetic acid into ethanol
(c) substance used to change acetic acid to acetic anhydride.
Answer:
(a) When acetic acid and ethanol react it results in the formation of ethyl acetate.
(b) Lithum aluminium hydride (\( LiAlH_4 \)) is used to convert acetic acid to ethanol.
(c) Phosphorous pentoxide (\( P_2O_5 \)) is heated along with acetic acid to form acetic anhydride.
In simple words: Acetic acid and ethanol make ethyl acetate (an ester), lithium aluminium hydride can reverse acid to alcohol, and phosphorous pentoxide removes water to form anhydride.

📝 Teacher's Note: Explain that ethyl acetate has a fruity smell and is used in nail polish remover. LiAlH4 is a powerful reducing agent, and P2O5 is a dehydrating agent.

🎯 Exam Tip: Learn the specific names of these reagents. Ethyl acetate is an ester, LiAlH4 is a reducing agent, and P2O5 removes water molecules.

Question 11. Give two tests to show that \( CH_3COOH \) is acidic in nature.
Answer: Test to show that \( CH_3COOH \) is acidic are:
When litmus test is done, it turns blue litmus red.
In simple words: To prove acetic acid is acidic, you can use litmus paper (turns red) or react it with metals to see hydrogen gas bubbles.

📝 Teacher's Note: The answer seems incomplete - add a second test like reaction with zinc to evolve hydrogen gas or neutralization with sodium hydroxide. Demonstrate both tests practically.

🎯 Exam Tip: Always give two complete tests as asked. Common acid tests include litmus paper, metal reaction (H2 evolution), or neutralization with base.

Question 12. What do you observe when acetic acid is added to:
(a) sodium bicarbonate
(b) ethyl alcohol in the presence of sulphuric acid
(c) neutral FeCl₃ solution?
Answer:
(a) When acetic acid is added to sodium bicarbonate, carbon dioxide is liberated.
\( CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \)
(b) When acetic acid is added to ethyl alcohol in presence of sulphuric acid, ester (ethyl acetate) is formed.
\( CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O \)
(c) When acetic acid is added to neutral FeCl₃, wine red color is produced.
In simple words: Acetic acid reacts differently with different substances - it bubbles with baking soda, makes a sweet-smelling compound with alcohol when heated with acid, and turns iron chloride solution wine-red.

📝 Teacher's Note: Demonstrate these reactions live in class - the CO₂ bubbling, the fruity smell of ester formation, and the dramatic color change with FeCl₃. Students remember visual demonstrations better than theory.

🎯 Exam Tip: For esterification, always mention the catalyst (H₂SO₄) and for the FeCl₃ test, specify "wine red" color - these are key marking points.

Miscellaneous

Question. Draw structural formula for each of the following compounds:
(a) isomer of n-butane
(b) vinegar
(c) 2-propanol
(d) ethanal
(e) acetone
(f) diethyl ether
What is used to describe these compounds taken together?
Answer:
(a) Ethane:
H H
| |
H-C-C-H
| |
H H

(b) Vinegar:
H
|
H-C-COOH
|
H

(c) Marsh gas (methane):
H
|
H-C-H
|
H

(d) Ethanal:
H O
| ||
H-C-C
| |
H H

(e) Acetone:
O
||
H₃C-C-CH₃

These compounds are called organic compounds.
In simple words: All these different molecules contain carbon atoms and are called organic compounds - they make up most of the chemicals found in living things.

📝 Teacher's Note: Use molecular model kits to show 3D structures. Emphasize that organic compounds always contain carbon and are the building blocks of life.

🎯 Exam Tip: Draw structural formulas clearly with correct bond angles. Always conclude with "These are organic compounds" when asked to classify them together.

Question. What is the special feature of the structure of:
(i) C₂H₄ (ii) C₂H₂
What type of reaction is common to both these compounds? Why methane does not undergo this type of reaction?
Answer:
(a)
(i) H₂C=CH₂ (contains double bond)
(ii) HC≡CH (contains triple bond)
They both are unsaturated compounds. The structure (i) contains double bond where as structure (ii) contains triple bond.
(b) Both the compounds undergo addition reactions. Methane does not undergo addition reactions because it is a saturated hydrocarbon with only single bonds.
In simple words: These molecules have extra bonds (double or triple) that can break and add new atoms, unlike methane which has all single bonds and is "full up".

📝 Teacher's Note: Compare to a bus with empty seats (unsaturated) versus a full bus (saturated). Empty seats can accommodate more passengers, just like multiple bonds can add more atoms.

🎯 Exam Tip: Always mention "unsaturated" for double/triple bonds and "addition reactions" - these are key terms examiners look for.

Question. Give the names and structural formula of:
(a) saturated hydrocarbon (b) unsaturated hydrocarbon
Which type of reaction will they undergo?
Answer:

The saturated hydrocarbons undergo substitution reactions whereas unsaturated hydrocarbons undergo addition reactions.

In simple words: Saturated hydrocarbons swap atoms (substitution) while unsaturated ones add new atoms at their double or triple bonds (addition).

📝 Teacher's Note: Use the analogy of a crowded room (saturated) where people can only swap places versus a room with empty chairs (unsaturated) where new people can sit down.

🎯 Exam Tip: Remember the pattern: saturated → substitution, unsaturated → addition. This is a frequently tested concept.

Question. Write an equation for the laboratory preparation of (i) an unsaturated hydrocarbon from calcium carbide. (ii) an alcohol from ethyl bromide.
What would you see, when ethyne is bubbled through a solution of bromine in carbon tetrachloride?
Name the addition product formed between ethene and water.
Answer:
(a) \( CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 \)
(b) When bromine in carbon tetrachloride is added to ethyne, the orange colour of the bromine disappears due to the formation of the colourless ethylene bromide.
(c) Water reacts with ethene to form ethanol.
\( CH_2=CH_2 + H_2O \xrightarrow{H^+} C_2H_5OH \)
In simple words: Calcium carbide makes acetylene gas with water, bromine solution loses its orange color when mixed with acetylene, and ethene plus water makes alcohol.

📝 Teacher's Note: Demonstrate the bromine test - the color change is dramatic and helps students remember this important test for unsaturation.

🎯 Exam Tip: For the bromine test, mention both the color change (orange to colorless) and the reason (formation of addition product).

Question. Choosing only words from the following list, write down appropriate words to fill in the blanks from (a) to (e) given below.
Addition, carbohydrates, CₙH₂ₙ₋₂, CₙH₂ₙ, CₙH₂ₙ₊₂, electrochemical homologous, hydrocarbon, saturated, substitution, unsaturated.
The alkanes form an (a) ___________ series with the general formula (b) __________ The alkanes are (c)____________ (d)__________ which generally undergo (e) ________ reactions.
Answer:
The alkanes form an (a) Homologous series with the general formula (b) CₙH₂ₙ₊₂. The alkanes are (c) saturated (d) hydrocarbon which generally undergo (e) substitution reactions.
In simple words: Alkanes are a family of similar compounds with the same pattern, all single bonds, and they replace atoms rather than add new ones.

📝 Teacher's Note: Explain homologous series using the analogy of family members with similar features but different sizes.

🎯 Exam Tip: Remember CₙH₂ₙ₊₂ for alkanes - each carbon has the maximum number of hydrogens possible (saturated).

Question. Ethanol can be converted into ethene which can be changed into ethane. Choose the correct word or phrase from the brackets to complete the following sentences.
(a) The conversion of ethanol into ethene is an example of ___________
(b) Converting ethanol into ethene requires the use of ___________
(c) The conversion of ethene into ethane is an example of ___________
(d) The catalyst used in the conversion of ethene into ethane is commonly ___________
Answer:
(a) The conversion of ethanol into ethene is an example of Dehydration.
(b) Converting ethanol into ethene requires the use of Conc. H₂SO₄.
(c) The conversion of ethene into ethane is an example of hydrogenation.
(d) The catalyst used in the conversion of ethene into ethane is commonly nickel.
In simple words: Alcohol loses water to become ethene (dehydration), then ethene gains hydrogen to become ethane (hydrogenation) using nickel as a helper.

📝 Teacher's Note: Emphasize the opposite nature of dehydration (removing water) and hydrogenation (adding hydrogen). Use the reversibility concept.

🎯 Exam Tip: Remember: concentrated H₂SO₄ for dehydration and nickel catalyst for hydrogenation - these specific conditions are often tested.

Question. Give reasons:
(a) ethyne is more reactive than ethene
(b) Ethene is more reactive than ethane
(c) Hydrocarbons are excellent fuels
Answer:
(a) Ethyne is a highly reactive compound than ethene because of the presence of a triple bond between its two carbon atoms.
(b) Ethene is a highly reactive compound than ethane because of the presence of a double bond between its two carbon atoms.
(c) Hydrocarbons such as alkanes undergo combustion reactions with oxygen to produce carbon dioxide and water vapour. Alkanes are flammable which makes them excellent fuels. Methane for example is the principal component of natural gas.
\( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)
In simple words: More bonds between carbons mean more reactivity - triple bonds are most reactive, double bonds are medium, single bonds are least reactive. Hydrocarbons burn easily and release lots of energy.

📝 Teacher's Note: Use the analogy of a tightly wound spring (triple bond) versus a loose spring (single bond) - the tighter spring has more stored energy and reacts more readily.

🎯 Exam Tip: Always link reactivity to bond type and mention the energy released during combustion for the fuel question.

Question 1(2004). Write balanced equation when ethane is burnt in air.
Answer:
\( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \)
In simple words: Ethane burns in oxygen to make carbon dioxide and water, releasing energy as heat and light.

📝 Teacher's Note: Emphasize balancing equations step by step - count atoms on both sides to ensure conservation of mass.

🎯 Exam Tip: Always check your balanced equation by counting each type of atom on both sides - this prevents simple calculation errors.

Question 2(2004). Write the equation for the preparation of ethylene from ethyl alcohol
Write the general formula for a saturated hydrocarbon and give one example of a saturated hydrocarbon with its structural formula
Name a compound which will give acetylene gas when treated with water.
Answer:
(a) \( C_2H_5OH \xrightarrow[170°C]{Conc.H_2SO_4} CH_2=CH_2 + H_2O \)
(b) General formula of saturated hydrocarbon is: CₙH₂ₙ₊₂
Example: CH₄
H
|
H-C-H
|
H
(c) Calcium carbide reacts with water to give acetylene gas.
In simple words: Alcohol heated with strong acid loses water to form ethylene, saturated hydrocarbons follow the pattern CₙH₂ₙ₊₂, and calcium carbide makes acetylene when water is added.

📝 Teacher's Note: Show the temperature and catalyst requirements for dehydration - these conditions are crucial for the reaction to occur.

🎯 Exam Tip: Include temperature (170°C) and catalyst (conc. H₂SO₄) in dehydration equations - these details fetch extra marks.

Question 1(2005). Draw the structural formula of a compound with two carbon atoms in each of the following cases
(a) An alkane with a carbon to carbon single bond
Answer:
(a) Ethane:
H H
| |
H-C-C-H
| |
H H
In simple words: Ethane has two carbon atoms connected by a single bond, with each carbon having as many hydrogens as possible.

📝 Teacher's Note: Emphasize that in alkanes, carbon forms four bonds and hydrogen forms one bond - this helps students draw correct structures.

🎯 Exam Tip: Show all bonds clearly in structural formulas and ensure carbon has exactly four bonds in total.

Question. Write the structural formula for:
(b) An alcohol containing two carbon atoms
(c) An unsaturated hydrocarbon with a carbon to carbon triple bond.
Answer:
(a) An alkane is ethane
H H
| |
H-C-C-H
| |
H H
Ethane
(b) The alcohol is ethanol
H H
| |
H-C-C-O-H
| |
H H
(c) An unsaturated hydrocarbon is ethyne
H-C≡C-H
In simple words: These are basic organic compounds - ethane is a simple gas, ethanol is the alcohol in drinks, and ethyne is used for welding torches.

📝 Teacher's Note: Use molecular model kits to show the 3D structures. Emphasize how the functional groups (-OH, triple bond) change the properties of the base hydrocarbon chain.

🎯 Exam Tip: Always show all hydrogen atoms clearly in structural formulas. Missing hydrogens will cost marks even if the carbon skeleton is correct.

Question. Ethane, Ethene, ethanoic acid, Ethyne, Ethanol
From the box, name
(a) The compound with -OH as the part of its structure.
(b) The compound with -COOH as the part of its structure.
(c) Homologue of Homologous series with general formula \( C_nH_{2n} \).
Answer:
(a) Ethanol
(b) Ethanoic acid
(c) Ethene
In simple words: Look for the functional groups - OH means alcohol (ethanol), COOH means acid (ethanoic acid), and the formula \( C_nH_{2n} \) is for alkenes (ethene).

📝 Teacher's Note: Make students memorize the general formulas for alkanes (\( C_nH_{2n+2} \)), alkenes (\( C_nH_{2n} \)), and alkynes (\( C_nH_{2n-2} \)). Use simple examples to reinforce.

🎯 Exam Tip: When identifying functional groups, look for specific atom arrangements: -OH (alcohol), -COOH (carboxylic acid), C=C (alkene), C≡C (alkyne).

Question. Write the equations for the following lab. Preparations:
(a) Ethane from sodium propionate,
(b) Ethene from Iodoethane
(c) ethyne from calcium carbide
(d) Methanol from Iodomethane.
Answer:
(a) Ethane from sodium propionate
\( C_2H_5COONa + NaOH \xrightarrow{CaO, 300°C} Na_2CO_3 + C_2H_6 \)
(b) Ethene from iodoethane
\( C_2H_5I + KOH(alcoholic) → C_2H_4 + KI + H_2O \)
(c) Ethyne from calcium carbide
\( CaC_2 + 2H_2O → Ca(OH)_2 + C_2H_2 \)
(d) Methanol from iodomethane
\( CH_3I + NaOH → CH_3OH + NaI \)
In simple words: These are standard preparation reactions - decarboxylation removes CO₂, elimination removes HI, hydrolysis adds water, and nucleophilic substitution replaces iodine with OH.

📝 Teacher's Note: Emphasize reaction conditions (temperature, catalysts). Show students how to balance equations step by step. Relate to industrial processes where possible.

🎯 Exam Tip: Always include reaction conditions (temperature, catalyst) above the arrow. Balance equations carefully - unbalanced equations lose marks.

Question. Give the correct IUPAC name and the functional group for each of the compounds whose structural formulae are given below:
(a) H-C-C-C-H with H, H, O arrangement
(b) H-C-C-C-OH with proper hydrogen arrangement
Answer:
(a) IUPAC name: Propanal
Functional group: -CHO
(b) IUPAC name: Propanol
Functional group: -OH
In simple words: Propanal has the aldehyde group (-CHO) which smells strong, while propanol has the alcohol group (-OH) which makes it a type of alcohol.

📝 Teacher's Note: Teach students to identify the longest carbon chain first, then locate the functional group. Practice with different structural representations.

🎯 Exam Tip: For IUPAC naming, always find the longest carbon chain containing the functional group. Number from the end closest to the functional group.

Question. (a) write the equation for the preparation of carbon tetrachloride from methane .
(b) draw the structural formula of ethyne
(c) How is the structure of alkynes different from that of alkenes?
Answer:
(a) Preparation of carbon tetrachloride from methane:
\( CH_4 + Cl_2 \xrightarrow{Diffused\ sunlight\ or\ 600K} CH_3Cl + HCl \)
\( CH_3Cl + Cl_2 → CH_2Cl_2 + HCl \)
\( CH_2Cl_2 + Cl_2 → CHCl_3 + HCl \)
\( CHCl_3 + Cl_2 → CCl_4 + HCl \)
(b) Structural formula of ethyne:
H-C≡C-H
(c) Alkynes contain triple bond where as alkenes contain double bond.
In simple words: Methane reacts with chlorine step by step, replacing one hydrogen at a time. Alkynes have three lines between carbons (≡) while alkenes have two lines (=).

📝 Teacher's Note: Show this as a step-wise halogenation. Use models to demonstrate the difference between double and triple bonds. Explain why multiple products form.

🎯 Exam Tip: Show all four steps of chlorination clearly. Remember that this is a free radical substitution requiring UV light or heat.

Question. Fill in the blanks with the correct words from the brackets:
Alkenes are the (a) ___________ series of (b) ___________ hydrocarbons. They differ from alkanes due to presence of (c) ___________ bonds. Alkenes mainly undergo (d) ___________ reactions.
Answer:
Alkenes are the (a) homologous series of (b) unsaturated hydrocarbons. They differ from alkanes due to presence of (c) double bonds. Alkenes mainly undergo (d) addition reactions.
In simple words: Alkenes are a family of compounds with double bonds that can easily add other atoms across the double bond, unlike alkanes which only have single bonds.

📝 Teacher's Note: Connect the presence of double bonds to the ability to undergo addition reactions. Give examples like adding hydrogen, halogens, or water across the double bond.

🎯 Exam Tip: Remember that unsaturated means having double or triple bonds. Alkenes undergo addition reactions because the double bond can break and add new atoms.

Question. (a) draw the structural formulae of the two isomers of Butane. Give the correct IUPAC name of each isomer.
(b) State one use of acetylene.
Answer:
(a) Structural formulae of isomers of Butane are:
H-C-C-C-C-H (with all hydrogens shown)
Butane

H-C-C-C-H with CH₃ branch on second carbon
2-methyl propane

(b) Use of acetylene: For Oxy-acetylene welding at very high temperatures.
In simple words: Butane has two forms - straight chain and branched chain with the same number of atoms but different arrangements. Acetylene burns very hot for cutting and joining metals.

📝 Teacher's Note: Use molecular models to show how the same atoms can be arranged differently. Explain how branching affects boiling points and other properties.

🎯 Exam Tip: For structural isomers, draw all bonds and atoms clearly. IUPAC names must reflect the longest chain and branch positions correctly.

Question. Give the IUPAC names of the following compounds numbered (i) to (v). The IUPAC names of the compounds on the left are to guide you for giving the correct IUPAC names of the compounds on the right.
Answer:
(i) Propyne
(ii) Pentan-3-ol
(iii) 2-methylpropane
(iv) Ethanoic acid
(v) 1,2-dichloroethane
In simple words: These names follow IUPAC rules - count the longest carbon chain, identify functional groups, and number to give the lowest possible numbers to important groups.

📝 Teacher's Note: Use the left examples as templates to guide students. Emphasize systematic approach: find longest chain, identify functional group, number appropriately.

🎯 Exam Tip: Always start by identifying the functional group, then find the longest carbon chain containing it. Number from the end that gives the functional group the lowest number.

Question. Copy and complete the following table which relates to three homologous series of hydrocarbons:

Answer: The homologous series of hydrocarbons are:

In simple words: This table shows three families of hydrocarbons - alkanes have single bonds and do substitution, alkenes have double bonds and do addition, alkynes have triple bonds and also do addition reactions.

📝 Teacher's Note: Start with methane and ethane examples to show saturated bonds, then demonstrate ethene and ethyne to show unsaturated bonds. Use molecular models if available to make the bond differences visual.

🎯 Exam Tip: Remember the pattern: alkanes end in -ane (single bonds), alkenes end in -ene (double bonds), alkynes end in -yne (triple bonds). This helps identify the series quickly.

Question. Name the organic compound prepared by each of the following reactions:
(i) \( C_2H_5COONa + NaOH \)
(ii) \( CH_3I + 2[H] \)
(iii) \( C_2H_5Br + KOH \) (alcoholic solution)
(iv) \( CO + 2H_2 \) (zinc oxide catalyst)
(v) \( CaC_2 + 2H_2O \)

Answer:
(i) \( C_2H_5COONa + NaOH \)
\( \implies CaO, 300°C \)
\( \implies Na_2CO_3 + C_2H_6 \) (Ethane)
(ii) \( CH_3I + 2[H] \)
\( \implies CH_4 + HI \) (Methane)
(iii) \( C_2H_5Br + KOH \)
\( \implies C_2H_4 + KBr + H_2O \) (Ethene)
(iv) \( CO + 2H_2 \)
\( \implies CH_3OH \) (Methanol)
(v) \( CaC_2 + 2H_2O \)
\( \implies Ca(OH)_2 + C_2H_2 \) (Ethyne)
In simple words: These are different chemical reactions that produce simple organic compounds like methane, ethane, ethene, methanol, and ethyne from various starting materials.

📝 Teacher's Note: Emphasize the conditions needed for each reaction (temperature, catalyst, alcoholic vs aqueous solutions). Show how different reaction conditions lead to different products.

🎯 Exam Tip: Learn the specific conditions for each reaction - mentioning "alcoholic KOH" vs "aqueous KOH" can make the difference between full marks and partial marks.

Question. Write the equations for the following reactions:
(i) calcium carbide and water
(ii) ethene and water (steam)
(iii) Bromoethane and an aqueous solution of sodium hydroxide.

Answer:
(i) Calcium carbide and water:
\( CaC_2 + 2H_2O \)
\( \implies Ca(OH)_2 + C_2H_2 \)
(ii) Ethene and water:
\( CH_2 = CH_2 + H_2O \)
\( \implies^{H^+} C_2H_5OH \)
(iii) Bromoethane and aqueous solution of sodium hydroxide
\( C_2H_5Br + NaOH \)
\( \implies C_2H_5OH + NaBr \)
In simple words: These reactions show how calcium carbide makes acetylene gas, ethene adds water to form alcohol, and bromoethane reacts with base to form ethanol.

📝 Teacher's Note: Point out that reaction (ii) needs an acid catalyst (H+) and that aqueous NaOH gives substitution while alcoholic KOH gives elimination. These distinctions are crucial.

🎯 Exam Tip: Always show the catalyst or conditions above/below the arrow. Missing these details costs marks even if the products are correct.

Question. Distinguish between the saturated hydrocarbon ethane and the unsaturated hydrocarbon ethene by drawing their structural formulae.

Answer:

In simple words: Ethane has all single bonds (saturated) like a completely filled container, while ethene has a double bond (unsaturated) with room for more atoms to attach.

📝 Teacher's Note: Use the analogy of a train with single vs double tracks. Single bonds are like single tracks (substitution - replace passengers), double bonds are like having extra space (addition - add more passengers).

🎯 Exam Tip: Draw clear structural formulas showing all bonds. Mention saturation/unsaturation AND the type of reaction each undergoes for complete answers.

Question. Addition reactions and substitution reactions are types of organic reactions. Which type of reaction is shown by:
(i) ethane
(ii) ethene?

Answer:
(i) Ethane undergoes substitution reaction.
(ii) Ethene undergoes addition reactions.
In simple words: Ethane (single bonds) can only replace hydrogen atoms, while ethene (double bonds) can add new atoms across the double bond.

📝 Teacher's Note: Demonstrate with examples: ethane + Cl₂ → C₂H₅Cl + HCl (substitution), ethene + Cl₂ → C₂H₄Cl₂ (addition). Show how the bond count changes.

🎯 Exam Tip: Remember: saturated = substitution, unsaturated = addition. This simple rule works for most hydrocarbon reactions.

Question. (i) write the equation for the complete combustion of ethane (ii) Using appropriate catalysts, ethane can be oxidized to an alcohol, an aldehyde and an acid. Name the alcohol, aldehyde and acid formed when ethane is oxidized.

Answer:
(i) \( 2C_2H_6 + 7O_2 \)
\( \implies 4CO_2 + 6H_2O \)
(ii) Ethane can be oxidized as follows:
When a mixture of ethane and oxygen in the ratio 9:1 by volume is compressed to about 120 atm pressure and passed over copper tubes at 475K, ethyl alcohol is formed.
\( 2C_2H_6 + O_2 \)
\( \implies^{120 atm, Cu tubes, 475K} 2C_2H_5OH \)
When a mixture of ethane and oxygen is passed through heated MoO, the mixture is oxidized to ethanal.
\( C_2H_6 + O_2 \)
\( \implies^{MoO} CH_3CHO + H_2O \)
When a manganese based catalyst is used 100°C, ethane can be oxidized to ethanoic acid.
\( 2C_2H_6 + 3O_2 \)
\( \implies^{Mn Compound} 2CH_3COOH + 2H_2O \)
In simple words: Ethane burns completely to make carbon dioxide and water. With controlled oxidation using different catalysts, we can make ethyl alcohol, ethanal (aldehyde), and ethanoic acid from ethane.

📝 Teacher's Note: Emphasize that complete combustion always produces CO₂ and H₂O, while controlled oxidation with specific catalysts gives useful products. The conditions determine what product forms.

🎯 Exam Tip: For combustion equations, balance oxygen last. For oxidation products, remember the sequence: alcohol → aldehyde → acid (increasing oxidation level).

Question. (i) Why is pure acetic acid known as glacial acetic acid? (ii) what type of compound is formed by the reaction between acetic acid and an alcohol?

Answer:
(i) Pure acetic acid on cooling forms crystalline mass resembling ice and for this reason it is called glacial acetic acid.
(ii) When acetic acid reacts with alcohol, ester is formed.
\( CH_3COOH + C_2H_5OH \)
\( \implies^{Conc. H_2SO_4} CH_3COOC_2H_5 + H_2O \)
In simple words: Pure acetic acid looks like ice crystals when it freezes, so it's called "glacial." When acids and alcohols react, they form esters which often have pleasant fruity smells.

📝 Teacher's Note: Show students that "glacial" means ice-like appearance. Esterification is important in making artificial flavors and fragrances - connect to everyday examples like banana essence.

🎯 Exam Tip: Always mention that esterification requires concentrated sulfuric acid as catalyst and that water is eliminated. This shows understanding of the mechanism.

Question. What do you understand by organic chemistry?

Answer: Organic chemistry may be defined as the chemistry of hydrocarbons and its derivatives.
In simple words: Organic chemistry is the study of carbon-containing compounds - basically all the chemicals found in living things and many man-made materials.

📝 Teacher's Note: Start with examples students know - like sugar, plastic, petrol. Show that all contain carbon and hydrogen as the basic building blocks, then add other elements.

🎯 Exam Tip: Define it as "chemistry of carbon compounds" or "chemistry of hydrocarbons and their derivatives" - both are acceptable and score full marks.

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

Answer: Vital Force Theory is a theory made by the Scientist Berzelius in 1809 which assumed that organic compounds are only formed in living cells and it is impossible to prepare them in laboratories. It was discarded because Friedrich Wohler showed that it was possible to obtain an organic compound (urea) in the laboratory.
In simple words: People used to think only living things could make organic chemicals, but Wohler proved this wrong by making urea (found in urine) artificially in his lab.

📝 Teacher's Note: Emphasize the historical importance - this discovery opened the door to synthetic chemistry. Connect to modern examples like synthetic drugs and plastics that are now made in factories.

🎯 Exam Tip: Mention Berzelius (proposed the theory), Wohler (disproved it), and urea (the compound that disproved it). These three key points ensure full marks.

Question. Name a few sources of organic chemistry

Answer: Few sources of organic compounds are:
Plants
Animals
Coal
Petroleum
Wood
In simple words: Organic compounds come from living things (plants, animals), fossil fuels (coal, petroleum), and materials that were once alive (wood).

📝 Teacher's Note: Connect each source to everyday examples - plants give sugar and cotton, animals give proteins and fats, petroleum gives petrol and plastics, coal gives dyes and medicines.

🎯 Exam Tip: List at least 4-5 sources. Include both natural (plants, animals) and fossil (coal, petroleum) sources to show comprehensive understanding.

Question. Give the various applications of organic chemistry

Answer: The various applications of organic chemistry is:
It is used in the production of soaps, shampoos, powders and perfumes.
Various fuels like natural gas, petroleum are also organic compounds.
The fabrics that we use to make various dresses are also made from organic compounds.
In simple words: Organic chemistry is everywhere - in our toiletries, clothes, fuels, medicines, and even the food we eat. It makes modern life possible.

📝 Teacher's Note: Ask students to look around the classroom and identify organic materials - clothing, plastic chairs, paper, etc. This makes the subject relevant and interesting.

🎯 Exam Tip: Categorize applications: personal care (soaps, cosmetics), energy (fuels), textiles (fabrics), medicine (drugs), and food industry (preservatives, flavors).

Question. Carbon shows some unique properties, name them

Answer: The unique properties shown by carbon are:
Tetravalency of carbon
Catenation
Isomerism
In simple words: Carbon is special because it can form four bonds (tetravalency), link with other carbon atoms in chains (catenation), and arrange in different ways (isomerism).

📝 Teacher's Note: Use building blocks analogy - carbon is like a LEGO block that can connect in 4 directions, link together in chains, and be arranged in countless different structures.

🎯 Exam Tip: These three properties explain why carbon can form millions of compounds. Mention all three for complete marks.

Question. Explain the following: (a) Tetravalency (b) Catenation

Answer:
(a) Tetravalency: Carbon can neither lose nor gain electrons to attain octet. Thus it shares four electrons with other atoms. This characteristics of carbon by virtue of which it forms four covalent bonds, is called Tetravalency of carbon.
(b) Catenation: The property of self-linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation. Carbon-carbon bond is strong so carbon can combine with other carbon atoms to form chains or rings and can involve single, double and triple bonds.
In simple words: Tetravalency means carbon always makes 4 bonds. Catenation means carbon atoms can join together like beads in a necklace to form long chains or rings.

📝 Teacher's Note: Draw simple structures showing CH₄ (tetravalency) and ethane, propane, butane (catenation). Show how the chain length increases while keeping the 4-bond rule for carbon.

🎯 Exam Tip: For tetravalency, mention "4 covalent bonds." For catenation, mention "chains, rings, single/double/triple bonds" to show complete understanding.

Question. Write any four properties of organic compounds that distinguish them from inorganic compounds.

Answer: Four properties of organic compound that distinguish them from inorganic compounds are:
1. Organic compounds generally have lower melting and boiling points compared to inorganic compounds
2. Most organic compounds are insoluble in water but soluble in organic solvents
3. Organic compounds are generally combustible while inorganic compounds are not
4. Organic reactions are generally slower and require catalysts while inorganic reactions are faster
In simple words: Organic compounds usually melt/boil easily, don't dissolve in water, can burn, and react slowly compared to inorganic compounds.

📝 Teacher's Note: Demonstrate with examples - sugar vs salt (melting point), oil vs salt (water solubility), paper vs metal (combustibility), to make these differences concrete and memorable.

🎯 Exam Tip: Compare properties directly - "organic compounds have... while inorganic compounds have..." This shows clear understanding of the distinction.

(i) Presence of carbon.
(ii) Solubility in the organic solvents.
(iii) Forming of covalent bonds.
(iv) Having low melting and boiling points.

Question 7. Why are organic compounds studies as a separate branch of chemistry?
Answer: Due to the unique nature of carbon atom, it gives rise to formation of large number of compounds. Thus this demands a separate branch of chemistry.

📝 Teacher's Note: Emphasize that carbon's unique properties like tetravalency and catenation make it special among all elements. Compare it to how cricket needs separate rules from football because it's so different.

🎯 Exam Tip: Always mention "unique nature of carbon" and "large number of compounds" - these are key phrases examiners look for in this answer.

Question 8. What are hydrocarbons? Compare saturated and unsaturated hydrocarbons?
Answer: Hydrocarbons are compounds that are made up of only carbon and hydrogen.
Comparison of saturated and unsaturated hydrocarbons:

📝 Teacher's Note: Use the analogy of a single lane road vs multi-lane highway to explain bonding differences. Multiple bonds provide more "traffic" (electrons) for reactions.

🎯 Exam Tip: Create a clear table format in your answer and mention all three key differences: bonding, reactivity, and reaction type.

Question 9. Give reason for the existence of the large number of organic compounds.
Answer: Due to presence of unique properties of carbon like Tetravalency, catenation and Isomerism large number of organic compounds are formed.

📝 Teacher's Note: Explain each property separately with simple examples. Show how tetravalency allows 4 bonds, catenation allows chains, and isomerism allows different arrangements.

🎯 Exam Tip: Remember the three magic words: "Tetravalency, catenation, isomerism" - write all three to get full marks.

Question 10. Give at least one example in each case to show structure of isomers of:
(a) single bond compound
(b) double bond compound
(c) triple bond compound
Answer:
(a) Single Bond compound: For example: In pentane
\( CH_3 - CH_2 - CH_2 - CH_2 - CH_3 \) (n-pentane)
and iso-pentane with branched structure

(b) Double bond compound: For example: In pentene
\( CH_2 = CH - CH_2 - CH_2 - CH_3 \) (1-pentene)
\( CH_3 - CH = CH - CH_2 - CH_3 \) (2-pentene)
and isopentene with different branching

(c) Triple bond compound: In case of Hexyne with different positions of triple bond

📝 Teacher's Note: Draw structures on the board to show how the same atoms can be arranged differently. Use building blocks analogy - same blocks, different structures.

🎯 Exam Tip: Always draw the structural formulas clearly and label each isomer properly with its name for maximum marks.

Question 11. Name a compound of each type and draw the figure.
(a) Cyclic compound with single bond
(b) Cyclic compound with triple bond.
Answer:
(a) Cyclic compound with single bond: cyclopentane
Structure: [Five-membered ring with all single bonds]

(b) Cyclic compound with triple bond: cyclopentyne
Structure: [Five-membered ring with one triple bond]

📝 Teacher's Note: Explain that cyclic means "ring-shaped" like a circle. Show how atoms connect in a closed loop rather than a straight chain.

🎯 Exam Tip: Draw clear ring structures and indicate the type of bonds (single/triple) clearly in your diagrams.

Question 12. Give the name of one member of each of the following:
(a) saturated hydrocarbons
(b) unsaturated hydrocarbons
Answer:
The member of each of the following is:
(a) Saturated Hydrocarbon: Hexane (\( C_6H_{14} \))
(b) Unsaturated Hydrocarbon: Hexene (\( C_6H_{12} \))

📝 Teacher's Note: Point out the pattern in molecular formulas - saturated has maximum hydrogen atoms, unsaturated has fewer due to multiple bonds.

🎯 Exam Tip: Include the molecular formula in brackets after the compound name to show you understand the difference completely.

Question 13. Define substitution and addition reaction.
Answer:
Substitution reaction: A reaction in which one atom of a molecule is replaced by another atom (or group of atoms) is called a substitution reaction.

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

📝 Teacher's Note: Use the analogy of replacing a player in a team (substitution) vs adding more players to the team (addition) to make concepts clear.

🎯 Exam Tip: Mention "replacement" for substitution and "addition to double/triple bonds" for addition reaction - these are key distinguishing features.

Question 14. Define or explain chain isomerism and position isomerism with examples in each case.
Answer:
Chain isomerism: Chain isomerism arises due to the difference in arrangement of C atoms in the chain. For example, there are two isomers of butane, \( C_4H_{10} \). In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.

\( CH_3 - CH_2 - CH_2 - CH_3 \) (straight chain)
and branched structure with \( CH_3 \) group attached to middle carbon

Position isomerism: It is due to the difference in position of functional groups. For example, there are two structural isomers with the molecular formula \( C_3H_7Br \). In one of them, the bromine atom is on the end of the chain, whereas in the other it is attached in the middle.

\( CH_3 - CH_2 - CH_2Br \) (1-bromopropane)
\( CH_3 - CHBr - CH_3 \) (2-bromopropane)

📝 Teacher's Note: Use the analogy of different ways to arrange the same furniture in a room (chain isomerism) vs placing the same decoration in different positions (position isomerism).

🎯 Exam Tip: Always provide clear examples with proper naming and show the structural difference visually for maximum marks.

Question 15. (a) Define the term isomerism. State two main causes of isomerism.
(b) draw the chain isomers of hexane (\( C_6H_{12} \))
(c) Draw position isomers of butane (\( C_4H_8 \))
Answer:
(a) Isomerism: Compounds having the same molecular formula but different structural formula are known as isomers and the phenomenon as isomerism.

Two main causes of isomerism are:
- Difference in mode of linking of atoms.
- Difference in the arrangement of atoms or groups in space.

(b) [Chain isomers of hexane with different branching patterns]

(c) \( CH_2=CHCH_2CH_3 \) (1-butene)
\( H_3C-CH=CHCH_3 \) (2-butene)

📝 Teacher's Note: Emphasize that same atoms can be connected differently - like using same Lego blocks to build different structures. The molecular formula stays same but arrangement changes.

🎯 Exam Tip: State the definition clearly, mention both causes, and draw all possible structures systematically to avoid missing any isomers.

Question 16. Define a functional group and give the structural formula of the following:
(a) Ketone, (b) alcohols (c) aldehydes.
Answer:
A functional group is an atom or a group of atoms that defines the structure (or the properties of a particular family) of organic compounds.

The structural formula of:
(a) Halides: R-X
Example: \( H_3C-Cl \) (chloromethane)

(b) Alcohols: R-OH
Example: \( CH_3-OH \) (methanol)

📝 Teacher's Note: Explain that functional groups are like family names - they give compounds their characteristic properties and reactions, just like how surnames connect family members.

🎯 Exam Tip: Always write the general formula (R-X, R-OH) first, then give a specific example with proper naming for complete marks.

Question 17. Identify the functional groups of the following:
(a) \( CH_3OH \) (b) \( HCHO \) (c) \( CH_3COOH \)
Answer: The functional group present in the following compounds are:
(a) \( CH_3OH \) :- Alcohol
(b) \( HCHO \):- Aldehyde
(c) \( CH_3COOH \):- Carboxyl
In simple words: Each organic compound has a special part that decides how it behaves chemically - this is called the functional group. Like alcohol has -OH group, aldehyde has -CHO group, and carboxylic acid has -COOH group.

📝 Teacher's Note: Use everyday examples - alcohol in sanitizer (methanol), aldehyde in formaldehyde (preservative), and carboxyl in vinegar (acetic acid). This makes functional groups relatable and memorable.

🎯 Exam Tip: Always write the full name of functional groups - "Alcohol" not just "OH", "Aldehyde" not just "CHO", "Carboxyl" not just "COOH" to get full marks.

Question 18. What will be the formula and structure of benzene?
Answer: Formula of benzene: \( C_6H_6 \)
Structure of benzene:
A hexagonal ring with 6 carbon atoms, each bonded to one hydrogen atom, with alternating single and double bonds between carbon atoms.
In simple words: Benzene is like a perfect hexagon made of 6 carbon atoms with 6 hydrogen atoms attached, forming a very stable ring structure.

📝 Teacher's Note: Draw the benzene ring on the board and emphasize the alternating double bonds. Mention that benzene is the simplest aromatic compound and forms the basis for many important chemicals.

🎯 Exam Tip: Always draw the complete structural formula showing all carbon and hydrogen atoms clearly. Don't forget to mention both molecular formula \( C_6H_6 \) and the ring structure.

Question 19. Which part of an organic compound determines
(i) physical properties (ii) chemical properties?
Answer:
(i) Physical properties: The alkyl group determines the physical properties.
(ii) Chemical properties: The functional group is responsible for the chemical properties.
In simple words: Think of organic compounds as having two parts - the alkyl group (like the body) controls things like boiling point and solubility, while the functional group (like the head) controls how it reacts with other chemicals.

📝 Teacher's Note: Use the analogy of a person - the body (alkyl group) determines height and weight (physical properties), while the brain (functional group) determines behavior (chemical properties).

🎯 Exam Tip: Remember the clear distinction: alkyl group = physical properties, functional group = chemical properties. Don't mix them up.

Question 20. Name the alkyl radical and the functional group of the following organic compounds:
(a) \( CH_3OH \) (b) \( C_2H_5OH \) (c) \( C_3H_7CHO \) (d) \( C_4H_9COOH \)
Answer: The alkyl radical and the functional group are:

In simple words: Each compound has two parts - the alkyl part (methyl, ethyl, etc.) which is like the carbon chain, and the functional group (-OH, -CHO, -COOH) which gives it special properties.

📝 Teacher's Note: Teach students to identify the functional group first, then count the carbon atoms in the remaining part to name the alkyl radical. Use the pattern: 1C = methyl, 2C = ethyl, 3C = propyl, 4C = butyl.

🎯 Exam Tip: Present answers in tabular format for clarity. Always identify the functional group correctly first, then count carbons for the alkyl radical name.

Question 21.
(a) What is an alkyl group?
(b) Give the names of any three alkyl radicals. How are they formed?
Answer:
(a) An alkyl group is obtained by removing one atom of hydrogen from an alkane molecule. Alkyl group is named by replacing the suffix 'ane' of the alkane with the suffix -yl.
(b) The name of three alkyl radicals are:
Methyl
Ethyl
Propyl
They are formed by removing 1 hydrogen from an alkane.
\( CH_4 \rightarrow -CH_3 + H^+ \)
Methyl
\( CH_3 - CH_3 \rightarrow CH_3 - CH_2^- + H^+ \)
Ethyl
\( CH_3 - CH_2 - CH_3 \rightarrow CH_3 - CH_2 - CH_2^- + H^+ \)
Propyl
In simple words: An alkyl group is what remains when you remove one hydrogen atom from an alkane. It's like taking away one hydrogen from methane to get methyl, or from ethane to get ethyl.

📝 Teacher's Note: Use physical models or drawings to show how removing one hydrogen creates a "free hand" that can bond with functional groups. Emphasize the naming pattern: methane → methyl, ethane → ethyl.

🎯 Exam Tip: Always show the formation reaction clearly with proper chemical equations. Remember to change 'ane' to 'yl' for naming alkyl radicals.

Question 22. Give the names and the structural formula of the first three members of the homologous series of alkanes.
Answer: The names and the structural formula of first three members of the homologous series of alkane are:
(i) methane (\( CH_4 \))
H
|
H-C-H
|
H

(ii) \( C_2H_6 \) Ethane
H H
| |
H-C-C-H
| |
H H
ethane (\( C_2H_6 \))

(iii) \( C_3H_8 \) Propane
H H H
| | |
H-C-C-C-H
| | |
H H H
propane (\( C_3H_8 \))
In simple words: Alkanes are like a family of compounds that start with methane (1 carbon), then ethane (2 carbons), then propane (3 carbons) - each new member just adds one more \( CH_2 \) group.

📝 Teacher's Note: Start with methane and show how each successive alkane adds a \( CH_2 \) unit. Use molecular models if available to demonstrate the 3D structure and tetrahedral geometry around each carbon.

🎯 Exam Tip: Draw complete structural formulas showing all bonds and atoms. Write both molecular formula and structural formula for full marks. Remember the pattern: \( CH_4, C_2H_6, C_3H_8 \).

Question 23.
(a) What is homologous series?
(b) What is the difference in the molecular formula of any two adjacent homologues:
(i) in terms of molecular mass,
(ii) in terms of number and kind of atoms per molecule?
Answer:
(a) A homologous series is a group of organic compounds having a similar structure and similar chemical properties in which the successive compounds differ by a \( CH_2 \) group.
(b) The difference in molecular formula of any two adjacent homologues is
(i) It differs by 14 a.m.u in terms of molecular mass.
(ii) It differs by three atoms. The kind of atoms it differs is one carbon and two hydrogen.
In simple words: A homologous series is like steps on a ladder - each step (compound) is similar to the next but has one extra \( CH_2 \) unit, making it 14 units heavier with 1 more carbon and 2 more hydrogen atoms.

📝 Teacher's Note: Use the alkane series as the main example. Show students how methane → ethane → propane follows the pattern. Calculate molecular masses to demonstrate the 14 a.m.u difference (C=12, H=1, so \( CH_2 \) = 14).

🎯 Exam Tip: Remember the key numbers: difference of \( CH_2 \) group, 14 a.m.u in mass, and 3 atoms (1 carbon + 2 hydrogen). These are frequently asked in exams.