Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 11f Carboxylic Acid

Carboxylic Acid

Solution 1:
Answer: Structure formula of Ethanoic acid:
H   O
|    ||
H-C-C-O-H
|
H

Molecular formula of Ethanoic acid: \( CH_3COOH \)

📝 Teacher's Note: Draw the structural formula step by step on the board, emphasizing the carboxyl group (-COOH) as the functional group. Students often confuse this with alcohols, so highlight the double-bonded oxygen.

🎯 Exam Tip: Always draw the structural formula clearly showing all bonds and write the molecular formula separately for full marks.

Solution 2:
Answer: Vinegar: Vinegar is 4 to 6% acetic acid (Ethanoic Acid). Glacial acetic acid: Pure acetic acid is called glacial acetic acid.

📝 Teacher's Note: Bring a bottle of vinegar to class and explain that household vinegar is diluted acetic acid. The term "glacial" comes from its ice-like appearance when pure.

🎯 Exam Tip: Remember the percentage range for vinegar (4-6%) and that "glacial" means pure - this distinction often appears in exams.

Solution 3:
Answer: Three physical properties of acetic acid:

• State: Liquid
• Odour: Pleasant smell- smell of vinegar
• Taste: Sour taste

📝 Teacher's Note: Let students smell vinegar safely to connect the theory with real experience. Emphasize that acids generally have sour taste but should never be tasted in lab.

🎯 Exam Tip: List exactly three properties as asked - state, odour, and taste are the most commonly expected answers.

Solution 4:
Answer: Uses of acetic acid:

• As a solvent for gums, resin, cellulose etc
• As a laboratory reagent
• In medicines
• As a vinegar for table purpose and for manufacturing pickles
• For making rubber, rayon, plastic, varnishes etc.
• For the manufacture of dyes, perfumes, pigments.

📝 Teacher's Note: Connect these uses to everyday items students know - pickles, medicines, plastics. This makes chemistry more relatable and memorable.

🎯 Exam Tip: Group the uses into categories: household (vinegar, pickles), industrial (plastics, dyes), and laboratory uses for easier recall.

Solution 5:
Answer: Acetic acid is the main constituent of vinegar.

📝 Teacher's Note: This is a simple fact but important - reinforce that vinegar is essentially diluted acetic acid with water and trace compounds.

🎯 Exam Tip: Keep the answer simple and direct - "main constituent" is the key phrase examiners look for.

Solution 6:
Answer: Oxidation of ethyl alcohol gives acetic acid.
\( C_2H_5OH + [O] \xrightarrow{K_2Cr_2O_7/H_2SO_4} CH_3CHO + H_2O \)
\( CH_3CHO + [O] \xrightarrow{K_2Cr_2O_7/H_2SO_4} CH_3COOH \)

📝 Teacher's Note: Explain this as a two-step process: alcohol → aldehyde → acid. Use the analogy of climbing stairs - each step requires oxygen.

🎯 Exam Tip: Always write both equations and include the catalyst conditions (K₂Cr₂O₇/H₂SO₄) for full marks.

Solution 7:
Answer: Acetic acid turns blue litmus red. It proves that it is acidic in nature.

📝 Teacher's Note: Demonstrate this test in class with litmus paper. Remind students that all acids turn blue litmus red - this is a universal property.

🎯 Exam Tip: State both the observation (blue litmus turns red) AND the conclusion (acidic nature) for complete marks.

Solution 8:
Answer: Boiling Point: 118°C

📝 Teacher's Note: Compare this to water's boiling point (100°C) to help students remember. The higher boiling point is due to hydrogen bonding in acetic acid.

🎯 Exam Tip: Remember 118°C exactly - numerical values are often tested and need to be precise.

Solution 9:
Answer: Pure acetic acid is called Glacial acetic acid because it forms an ice-like solid when cooled.

📝 Teacher's Note: The freezing point of pure acetic acid is 16.6°C, so it can freeze on a cold day. This is why it's called "glacial" - like ice or glacier.

🎯 Exam Tip: The key phrase is "ice-like solid when cooled" - this explains the term "glacial" clearly.

Solution 10:
Answer: The first four members of aliphatic carboxylic acid are:

• Methanoic acid
• Ethanoic acid
• Propanoic acid
• Butanoic acid

📝 Teacher's Note: Teach the pattern: meth- (1 carbon), eth- (2 carbons), prop- (3 carbons), but- (4 carbons). The ending is always -anoic acid for carboxylic acids.

🎯 Exam Tip: Learn the sequence in order - questions often ask for "first four" or "first three" members of the series.

Solution 11:
Answer: i) Ethene to Acetic acid
\( C_2H_4 + H_2 \xrightarrow{Pt} C_2H_6 \)
\( C_2H_6 + O_2 \xrightarrow{500°C} CH_3CHO + H_2O \)
\( CH_3CHO + [O] \xrightarrow{K_2Cr_2O_7/Conc.H_2SO_4} CH_3COOH \)

ii) Ethane to acetic acid
\( C_2H_6 + O_2 \xrightarrow{500°C} CH_3CHO + H_2O \)
\( CH_3CHO + [O] \xrightarrow{K_2Cr_2O_7/Conc.H_2SO_4} CH_3COOH \)

📝 Teacher's Note: Emphasize that route (i) requires an extra step - hydrogenation of ethene to ethane first. Route (ii) is more direct from ethane.

🎯 Exam Tip: Always include reaction conditions (temperature, catalysts) in chemical equations for full marks.

Solution 1999-1:
Answer: (i) Structural formula are as under:
(a) Ethane
H  H
|   |
H-C-C-H
|   |
H  H

(b) Vinegar
H   O
|    ||
H-C-C
|    O-H
H

(c) Marsh Gas
H
|
H-C-H
|
H

(ii) The three compounds taken together are known as Organic compounds.

📝 Teacher's Note: Draw these structures clearly on the board. Explain that marsh gas is methane (CH₄), vinegar contains acetic acid, and these are all organic because they contain carbon.

🎯 Exam Tip: Draw structural formulas showing all bonds clearly - don't use condensed formulas unless specifically asked.

Solution 1999-2:
Answer: (i). (a) The special feature of the structure of \( C_2H_2 \) is that there is presence of triple bond in the molecule.

(b) The special feature of the structure of \( C_2H_4 \) is that there is presence of double bond.

(ii) Addition reaction is common to both of these compounds.

📝 Teacher's Note: Draw the structures showing triple bond in acetylene and double bond in ethene. Explain that unsaturated compounds (with multiple bonds) undergo addition reactions.

🎯 Exam Tip: Clearly state "triple bond" for C₂H₂ and "double bond" for C₂H₄ - these are key identifying features.

Solution 2000-1:
Answer: (i) The name of saturated Hydrocarbon is called Alkane and the formula is \( C_nH_{2n+2} \) where n=1,2,3.......
(ii) The name of unsaturated hydrocarbon with double bond is called alkene and the formula is \( C_nH_{2n} \), where n=1,2,3,4......

📝 Teacher's Note: Write the general formulas on the board and show how they differ by H₂. Alkanes are "saturated" because they have maximum hydrogen atoms possible.

🎯 Exam Tip: Remember the formulas: Alkanes = CₙH₂ₙ₊₂, Alkenes = CₙH₂ₙ - these are frequently tested.

Solution 2000-2:
Answer: A saturated hydrocarbon will undergo Substitution reactions, whereas the typical reaction of an unsaturated hydrocarbon is Addition.

📝 Teacher's Note: Explain that saturated compounds have all single bonds, so atoms can only be substituted. Unsaturated compounds have multiple bonds that can "add" atoms across the bond.

🎯 Exam Tip: Learn the pattern: Saturated = Substitution, Unsaturated = Addition. This is a fundamental concept in organic chemistry.

Solution 2000-2:
Answer: (i) \( CaC_2 + 2H_2O → C_2H_2 + Ca(OH)_2 \)

(ii) Special feature of the structure of Ethyne is that there is presence of triple bond.
(iii) When Ethyne is bubbled through a solution of bromine in carbon tetrachloride, the orange colour of bromine disappears due to formation of colourless product.
(iv) Ethyl alcohol is formed when the addition reaction takes place between ethene and water.
\( CH_2 = CH_2 + H_2O \xrightarrow{H^+} CH_3 - CH_2OH \)

📝 Teacher's Note: Demonstrate the bromine test if possible - it's a classic test for unsaturation. The color change from orange to colorless is very striking and memorable.

🎯 Exam Tip: For the bromine test, mention both the color change (orange disappears) AND the reason (addition reaction forms colorless product).

Solution 2001-1:
Answer: When ethene is bubbled through a solution of bromine in tetrachloromethane, the orange colour of bromine disappears due to the formation of colourless ethylene bromide.
\( CH_2 = CH_2 + Br_2 → CH_2Br - CH_2Br \)

📝 Teacher's Note: This is the standard test for alkenes. Emphasize safety when discussing bromine - it's toxic and should only be handled by teachers in well-ventilated areas.

🎯 Exam Tip: Always mention the color change (orange to colorless) and name the product formed (ethylene bromide) for complete marks.

Solution 2001-2:
Answer: The alkanes form a homologous series with general formula \( C_nH_{2n-2} \). The alkanes are saturated, which generally undergo substitution reactions.

📝 Teacher's Note: There seems to be an error in the original - alkanes have formula CₙH₂ₙ₊₂, not CₙH₂ₙ₋₂. Make sure to correct this when teaching.

🎯 Exam Tip: Double-check formulas in your notes - alkanes are CₙH₂ₙ₊₂, alkenes are CₙH₂ₙ, and alkynes are CₙH₂ₙ₋₂.

Solution 2001-3:
Answer: (i) The conversion of ethanol to ethene is an example of dehydration (dehydration, dehydrogenation)
(ii) Converting ethanol to ethene requires the use of concentrated sulphuric acid (Concentrated hydrochloric acid, concentrated nitric acid and concentrated sulphuric acid).
(iii) The conversion of ethene to ethane is an example of hydrogenation (hydration, hydrogenation).
(iv) The catalyst used in the conversion of ethene to ethane is commonly nickel (iron, cobalt, nickel).

📝 Teacher's Note: Explain that dehydration means "removal of water" and hydrogenation means "addition of hydrogen". Use word roots to help students remember.

🎯 Exam Tip: Learn the specific catalysts: concentrated H₂SO₄ for dehydration, nickel for hydrogenation - these are standard industrial processes.

Solution 2001-4:
Answer: \( CaC_2 + 2H_2O → C_2H_2 + Ca(OH)_2 \)

📝 Teacher's Note: This is the industrial preparation of acetylene. Calcium carbide was historically very important for welding and lighting before electricity became widespread.

🎯 Exam Tip: Balance the equation carefully - 2 moles of water are needed for 1 mole of calcium carbide.

Solution 2002-1:

Answer:
(i) Substitution reaction takes between ethane and chlorine to form monochloroethane. This reaction is called chlorination.
(ii) Addition reaction takes place between ethene and chlorine and it is called halogenations.
(iii) (a) Structural formula of Ethene:
H-C=C-H (with H atoms attached to each carbon)
(b) Ethene can react with chlorine because there is presence of double bond which can result in the addition reaction.

📝 Teacher's Note: Use molecular models to show students how the double bond in ethene "opens up" to allow chlorine atoms to attach. Compare this with ethane's single bonds which require substitution.

🎯 Exam Tip: Always mention the type of reaction (addition vs substitution) and relate it to the bond structure - double bonds undergo addition, single bonds undergo substitution.

Solution 2002-2:

Answer:
(i) Ethene
(ii) Methane
(iii) Ethene
(iv) Methane
(v) Ethyne, Ethene

📝 Teacher's Note: Have students practice identifying hydrocarbons by their reactions - unsaturated hydrocarbons decolorize bromine water, while saturated ones don't.

🎯 Exam Tip: Remember the order: methane (saturated) < ethene (one double bond) < ethyne (triple bond) in terms of reactivity with halogens.

Solution 2002-3:

Answer:
(i) Ethane from sodium propionate:
CH₃CH₂COONa + NaOH → CH₄ - CH₃ + Na₂CO₃
(ii) Ethene from Ethanol:
C₂H₅OH → CH₂ = CH₂ + H₂O (at 170°C with conc. H₂SO₄)
(iii) Ethyne from Calcium carbide:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂
(iv) Ethanoic acid from ethane:
2C₂H₆ + O₂ → 2C₂H₅OH (at 500°C)
C₂H₅OH → CH₃CHO → CH₃COOH (using K₂Cr₂O₇/H₂SO₄)

📝 Teacher's Note: Emphasize the conditions for each reaction - temperature, catalysts, and reagents are crucial for these industrial processes.

🎯 Exam Tip: Write balanced chemical equations with proper conditions. Remember decarboxylation removes CO₂ to give one carbon less in the product.

Solution 2003-1:

Answer:
(i) Sodium propionate is heated with soda lime to obtain ethane gas in the laboratory.
(ii) CH₃CH₂COONa + NaOH → CH₄ - CH₃ + Na₂CO₃
(iii) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O + heat
(iv) Al₂O₃ can be used instead of sulphuric acid to prepare ethylene by dehydration of alcohol.
(v) Bromine solution can be used to distinguish between ethane and ethene.
(vi) Ethylene reacts with chlorine to form a product called as 1,2-dichloroethane. This reaction is known as Halogenation.
CH₂ = CH₂ + Cl₂ → CH₂Cl - CH₂Cl
1,2-dichloroethane

📝 Teacher's Note: Demonstrate the bromine water test - ethene will decolorize orange bromine water while ethane will not. This is a key distinguishing test.

🎯 Exam Tip: For preparation methods, always mention the specific reagent (soda lime = NaOH + CaO) and the type of reaction occurring.

Solution 2004-1:

Answer: Ethane is burnt in air
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O + heat
(a) C₂H₅OH → CH₂ = CH₂ + H₂O (using conc. H₂SO₄ at 170°C)
(b) General formula of saturated hydrocarbon: CₙH₂ₙ₊₂. For Example: Methane.
Structural Formula of methane is: CH₄ (tetrahedral structure with C-H bonds)
(c) Calcium carbide will react with water to give acetylene gas.

📝 Teacher's Note: Show students how to derive the general formula by looking at the hydrogen to carbon ratio in alkanes - each carbon can hold maximum 4 bonds.

🎯 Exam Tip: For combustion reactions, balance oxygen last. Remember complete combustion always gives CO₂ and H₂O as products.

Solution 2005-1:

Answer:
(a) (i) Ethane - structural formula showing two carbon atoms connected by single bond with hydrogen atoms
(ii) Ethanol - structural formula showing CH₃CH₂OH with OH functional group
(iii) Ethyne - structural formula showing H-C≡C-H with triple bond
(b) (i) Ethanol
(ii) Ethanoic acid
(iii) Ethene

📝 Teacher's Note: Use structural formulas to help students visualize the difference between functional groups - alcohol (-OH), carboxylic acid (-COOH), and alkenes (C=C).

🎯 Exam Tip: Draw structural formulas clearly showing all bonds and functional groups. This helps identify the compound type and predict its reactions.

Solution 2005-2:

Answer:
(i) Ethane from sodium propionate:
CH₃CH₂COONa + NaOH → CH₃ - CH₃ + Na₂CO₃
(ii) Ethene from iodoethane:
CH₃ - CH₂I + KOH → CH₃ = CH₂ + KI + H₂O
(iii) Ethyne from calcium Carbide:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

📝 Teacher's Note: Point out that dehydrohalogenation (removal of HX) from alkyl halides is another way to form alkenes, besides dehydration of alcohols.

🎯 Exam Tip: Learn multiple preparation methods for each hydrocarbon - exams often ask for alternative methods of preparation.

Solution 2005-3:

Answer: Carbon has the unique property of combining any number of carbon atoms to form straight chains, branched chains and rings of different sizes. This property is called catenation.

📝 Teacher's Note: Show examples of different carbon arrangements - linear (alkanes), branched (iso-alkanes), and cyclic (cycloalkanes) to illustrate catenation.

🎯 Exam Tip: Define catenation clearly as the ability of carbon to form bonds with other carbon atoms. This explains why organic compounds are so numerous.

Solution 2006-1:

Answer:
(i) IUPAC Name: Propanal
Functional group: Aldehyde
(i) IUPAC Name: Propan-1-ol
Functional group: Alcohol

📝 Teacher's Note: Teach students to identify functional groups first, then apply IUPAC naming rules. Aldehydes end in -al, alcohols end in -ol.

🎯 Exam Tip: For IUPAC naming, number the carbon chain to give the functional group the lowest possible number. Always state both the name and functional group.

Solution 2006-2:

Answer:
(i) CH₄ + 4Cl₂ → CCl₄ + 4HCl
(ii) H - C ≡ C - H
(iii) Alkynes contain triple bond between carbon atoms where as alkenes contain double bond.

📝 Teacher's Note: Emphasize the difference in bond types - single bonds in alkanes, double in alkenes, triple in alkynes. This affects their chemical behavior.

🎯 Exam Tip: For substitution reactions, remember that all hydrogen atoms can be replaced step by step. Write the final product when excess halogen is used.

Solution 2006-3:

Answer:
(i) Homologous
(ii) Unsaturated
(iii) Double
(iv) Addition

📝 Teacher's Note: Explain that homologous series have the same functional group but differ by CH₂ units. This gives them similar chemical properties.

🎯 Exam Tip: Remember key terms: homologous series, saturated/unsaturated, addition/substitution reactions. These concepts appear frequently in organic chemistry questions.

Solution 2007-1:

Answer:
(i) Propyne
(ii) Pentan-3-ol
(iii) 2-methylpropane
(iv) Ethanoic acid
(v) 1,2-dichloroethane

📝 Teacher's Note: Practice IUPAC naming by breaking down the structure - find longest chain, number it, identify substituents and functional groups.

🎯 Exam Tip: For naming, always find the longest carbon chain first, then number to give functional groups or substituents the lowest numbers.

Solution 2007-2:

Answer:

📝 Teacher's Note: Use this table to show the systematic relationship between structure and properties in hydrocarbons. The general formulas are key to identifying compound types.

🎯 Exam Tip: Memorize the general formulas - they help you identify whether a compound is alkane, alkene, or alkyne from its molecular formula.

Solution 2008-1:

Answer: (d) Addition

📝 Teacher's Note: Emphasize that unsaturated hydrocarbons (with double or triple bonds) undergo addition reactions because the multiple bonds can "open up" to add atoms.

🎯 Exam Tip: Quick rule: saturated compounds (single bonds only) undergo substitution; unsaturated compounds (multiple bonds) undergo addition reactions.

Solution 2008-2:

Answer:

(i) Ethane
\( C_2H_5COONa + NaOH \rightarrow C_2H_6 + Na_2CO_3 \)

(ii) Methane
\( CH_3I + 2H^- \rightarrow CH_4 + HI \)

(iii) Alkenes
\( C_2H_5Br + KOH(alcoholic solution) \rightarrow C_2H_4 + KBr + H_2O \)

(iv) Ethyne
\( CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2 \)

In simple words: These are preparation methods for different hydrocarbons - each one uses a specific chemical reaction to make alkanes, alkenes, or alkynes from simpler starting materials.

📝 Teacher's Note: Emphasize the reaction conditions for each preparation method - students often confuse alcoholic KOH (for elimination) with aqueous KOH (for substitution).

🎯 Exam Tip: Always write balanced chemical equations and mention specific reagents and conditions for each preparation method to score full marks.

Solution 2008-3:

Answer:

(i) \( C_2H_5Br + KOH \rightarrow C_2H_5OH + KBr \)
(ii) \( CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2 \)
(iii) \( CH_2 = CH_2 + H_2O \xrightarrow{H^+} CH_2 - CH_2 - OH \)

In simple words: These equations show how to prepare ethanol from ethyl bromide, ethyne from calcium carbide, and ethanol from ethene using different chemical reactions.

📝 Teacher's Note: Point out that equation (i) requires aqueous KOH for substitution, while alcoholic KOH would give elimination to form ethene instead.

🎯 Exam Tip: For hydration of alkenes, always mention the catalyst (H⁺) to show you understand the mechanism requires acid catalysis.

Solution 2008-4:

Answer:

(a)

(b) (i) Ethane shows Substitution Reaction.
(ii) Ethene shows Addition Reaction.

(c) (i) \( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O + Heat \)

(ii) Ethane → Alcohol
\( 2C_2H_6 + O_2 \xrightarrow{Cu/200°C} 2C_2H_5OH \)
The alcohol is ethanol.

Ethane → Aldehyde
\( C_2H_6 + O_2 \xrightarrow{MoO/350°C} CH_3CHO + H_2O \)

The Aldehyde formed is Ethanol.

Ethane → Acid
\( C_2H_6 + O_2 \xrightarrow{MoO/350°C} CH_3CHO + H_2O \)
\( CH_3CHO + [O] \xrightarrow{K_2Cr_2O_7/conc. H_2SO_4} CH_3COOH \)

The acid formed is Ethanoic acid.

In simple words: Ethane has single bonds so it undergoes substitution reactions, while ethene has double bonds so it undergoes addition reactions. Both can be oxidized to form different products.

📝 Teacher's Note: Use molecular models to show the difference in bonding between ethane and ethene - this visual aid helps students understand why their reactions differ.

🎯 Exam Tip: Always specify reaction conditions (temperature, catalyst) and write "The aldehyde formed is acetaldehyde" to show you know the product names.

Solution 2009-1:

Answer:

(i) (b) Statement is wrong. They can undergo addition as well as substitution reaction.
(ii) Acetic acid contains four hydrogen atoms in it.

In simple words: Alkenes can do both addition reactions (at the double bond) and substitution reactions (at other positions). Acetic acid (CH₃COOH) has exactly four hydrogen atoms.

📝 Teacher's Note: Draw the structure of acetic acid and count the hydrogen atoms with students - this reinforces molecular formula understanding.

🎯 Exam Tip: For alkene reactions, remember they are most famous for addition but can also do substitution under specific conditions - mention both for complete answers.

Solution 2009-2:

Answer:

\( CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O \)

In simple words: This shows the esterification reaction where acetic acid and ethanol combine to form ethyl acetate (an ester) and water.

📝 Teacher's Note: Emphasize that this is a reversible reaction and requires an acid catalyst - students often forget to mention the equilibrium nature.

🎯 Exam Tip: Always use the double arrow (⇌) for esterification reactions to show it's reversible, and mention that concentrated H₂SO₄ acts as a catalyst.

Solution 2009-3:

Answer:

\( C_5H_{10} \) is odd one out as it is an alkene whereas rests of organic compounds are Alkanes.

In simple words: C₅H₁₀ follows the alkene formula CₙH₂ₙ (with a double bond), while the others follow the alkane formula CₙH₂ₙ₊₂ (only single bonds).

📝 Teacher's Note: Teach students to quickly identify compound types by checking if the formula fits CₙH₂ₙ₊₂ (alkane), CₙH₂ₙ (alkene), or CₙH₂ₙ₋₂ (alkyne).

🎯 Exam Tip: When identifying the odd compound, always state the general formula it follows and explain why it's different from the others.

Solution 2009-4:

Answer:

Structural Formula of carbon tetrachloride:

Cl
|
Cl—C—Cl
|
Cl

The bond is Covalent Bond.

In simple words: Carbon tetrachloride has one carbon atom bonded to four chlorine atoms through covalent bonds, where electrons are shared between atoms.

📝 Teacher's Note: Explain that carbon forms four covalent bonds because it has four valence electrons to share, making it very stable.

🎯 Exam Tip: Always draw the structural formula clearly showing all bonds and state the bond type to demonstrate complete understanding.

Solution 2009-5:

Answer:

(a) \( CH_3COONa + NaOH \xrightarrow{CaO} CH_4 + Na_2CO_3 \)
(b) \( CH_2 = CH_2 + Cl_2 \rightarrow CH_2Cl - CH_2Cl \)
(c) \( CH_3Br - CH_3Br + 2KOH(alc.) \rightarrow CH \equiv CH + 2KBr + 2H_2O \)

In simple words: These show preparation of methane from sodium acetate, addition of chlorine to ethene, and elimination reaction to form acetylene from dibromoethane.

📝 Teacher's Note: Highlight that reaction (c) needs alcoholic KOH and high temperature for double elimination - this distinguishes it from single elimination reactions.

🎯 Exam Tip: For decarboxylation reactions like (a), always mention CaO as the heating agent along with the temperature conditions.

Solution 2009-6:

Answer:

(a) Ethyl chloride on hydrolysis with dilute alkali gives ethyl alcohol.
\( C_2H_5Cl + KOH \rightarrow C_2H_5OH + KCl \)

(b) Ethyl chloride by treating with alcoholic KOH gives Ethene.
\( CH_3 - CH_2Cl + KOH \rightarrow CH_2 = CH_2 + KBr + H_2O \)

(c) Ethene adds molecule of water in presence of mineral acids to form Ethyl alcohol.
\( CH_2 = CH_2 + H_2O \xrightarrow{H^+} CH_3 - CH_2 - OH \)

(d) When concentrated sulphuric acid is added to ethyl alcohol, it causes dehydration to give ethene. Ethene reacts with hydrogen in presence of Ni to give ethane.
\( C_2H_5OH \xrightarrow{Conc.H_2SO_4/170°C} CH_2 = CH_2 + H_2O \)
\( CH_2 = CH_2 + H_2 \xrightarrow{Ni/300°C} CH_3 - CH_3 \)

In simple words: These reactions show different ways to convert between ethyl chloride, ethanol, ethene, and ethane using various reagents and conditions.

📝 Teacher's Note: Emphasize the difference between aqueous KOH (substitution) and alcoholic KOH (elimination) - this is a common source of confusion.

🎯 Exam Tip: Always specify reaction conditions like temperature and catalyst type - these details often carry marks in organic chemistry questions.

Solution 2009-7:

Answer:

(a) Compounds having the same molecular formula, but different structural formula are called isomers and the phenomenon is called isomerism.

(b) IUPAC name of branched chain isomer of \( C_4H_{10} \) is 2-methyl propane.

In simple words: Isomers are like twins with the same number of atoms but arranged differently. Butane has two isomers - straight chain butane and branched 2-methylpropane.

📝 Teacher's Note: Use ball-and-stick models to show how the same atoms can be arranged in different ways - this makes the concept of isomerism very clear to students.

🎯 Exam Tip: When naming branched alkanes, always identify the longest carbon chain first, then name and number the branches correctly according to IUPAC rules.