Organic Chemistry

Complete Hydrogenation of Hydrocarbons

Let's decipher the identity of each hydrocarbon based on the clues:

• Compound (A): Kept in a smaller ratio in hot countries to manage LPG cylinder pressure. This is Propane (C₃H₈).
• Compound (B): Has 3 carbons but 2 fewer H atoms. This is Propene (C₃H₆).
• Compound (C): Has 3 carbons but 2 fewer H atoms than B. This is Propyne (C₃H₄).

The Chemistry: Completely hydrogenating any 3-carbon unsaturated hydrocarbon maxes out its hydrogen bonds, turning it back into the fully saturated alkane. Thus, all three turn into/remain Propane.

Synthesis of Methane

First, identify the compounds (all have 2 carbons): A is Ethane, B is Ethanol (1 OH group), C is Ethylene glycol (2 OH groups). The target is a saturated hydrocarbon with a lower boiling point than Ethane, which must be Methane (CH₄).

Synthesis Pathway from Ethanol (B):

1. Complete Oxidation: Ethanol is oxidized to Acetic acid (CH₃COOH).
2. Neutralization (Add NaOH): Acetic acid reacts with NaOH to form Sodium acetate (CH₃COONa).
3. Dry Distillation: Heating Sodium acetate with soda lime cleaves the molecule, producing Methane gas.

Ester Isomers

Let's deduce the reactants and the product:

• The Acid: An acid where the number of carbons equals the number of carboxyl groups is Formic acid (HCOOH) (1 C, 1 COOH).
• The Alcohol: The alcohol referenced with a low freezing point around -110°C is Ethanol (C₂H₅OH).
• The Ester: Esterification produces Ethyl formate (HCOOC₂H₅), with a total molecular formula of C₃H₆O₂.

Finding Isomers: How many other esters share the formula C₃H₆O₂? There is only exactly one: Methyl acetate (CH₃COOCH₃).

Reactions of Lactic Acid

The clue "causes muscle contraction/fatigue" immediately identifies acid (A) as Lactic acid (CH₃CH(OH)COOH).

1. Step 1: Reacting lactic acid with NaOH neutralizes it to form Sodium lactate (B).
2. Step 2 (Process C): Heating sodium lactate with soda lime (Dry Distillation) cleaves the carboxylate group, converting the remaining structure into Ethanol (D). (C₂H₆O).
3. Step 3: Complete oxidation of ethanol yields Ethanoic/Acetic acid (E). (C₂H₄O₂).

Preparation of Ethylene Glycol (Antifreeze)

The target antifreeze compound is Ethylene glycol. The journey from cane sugar involves four distinct steps:

1. Hydrolysis: Cane sugar (Sucrose) is broken down into simple sugars (glucose/fructose).
2. Alcoholic Fermentation: Yeast enzymes convert glucose into Ethanol.
3. Dehydration (180°C): Heating ethanol with concentrated H₂SO₄ at exactly 180°C eliminates water to form Ethene.
4. Oxidation (H₂O₂ / Bayer's): Oxidizing ethene adds two hydroxyl (-OH) groups, resulting in Ethylene glycol.

Identifying Non-Carboxylic Organic Acids

Let's map out the identities of these specific compounds:

• Acid A: A weak solid acid with a distinct hospital smell is Phenol (also historically called carbolic acid).
• Acid B: Benzene + mineral acid = Benzenesulfonic acid (sulfonation).
• Acid C: Used to treat burns (and as an explosive) is Picric acid (2,4,6-trinitrophenol).

To prepare B from A: You must first reduce Phenol (A) back to Benzene by heating it with Zinc dust. Then, perform a substitution reaction by adding concentrated sulfuric acid to get Benzenesulfonic acid (B).

Combustion Stoichiometry

Let's work backward from the combustion products to find the molecular formula:

• 3 mol CO₂: Contains exactly 3 Carbon atoms.
• 3 mol H₂O: Contains exactly 6 Hydrogen atoms (3 × 2).

This means the original hydrocarbon must have the formula C₃H₆. Among the choices, Propane is C₃H₈, and Propanol is C₃H₈O. Only Cyclopropane (and propene) fits the formula C₃H₆.

Roles of Core Reagents

Both NaOH and H₂SO₄ play definitive roles in these organic preparations:

• Paraffins (Alkanes): Prepared via dry distillation of sodium salts. The role of NaOH (in soda lime) is to cleave and remove the carboxylate group (-COONa) as sodium carbonate (Na₂CO₃).
• Olefins (Alkenes): Prepared by dehydrating alcohols. Concentrated H₂SO₄ acts as a powerful dehydrating agent, eliminating water from the molecule to form the double bond. (Note: "illuminating" is a known translational typo for eliminating/removing).

Functional Group Analysis

By visually breaking down the molecule, we find three key functional groups that govern its reactivity:

• 1. Carboxyl Group (-COOH): This acidic group reacts strongly with NaHCO₃, releasing CO₂ gas, causing visible effervescence that clouds lime water.
• 2. Phenolic Group (-OH on ring): Phenols famously react with Iron (III) chloride (FeCl₃) to produce a highly characteristic purple complex.
• 3. Ester Linkage (-O-CO-CH₂-NH₂): Acidic hydrolysis severs this ester bond, separating the phenol ring and isolating the amino acid Glycine (NH₂-CH₂-COOH).

Preparation of Carbon Black

The simplest alkyne is ethyne (acetylene). Its catalytic hydration yields Acetaldehyde. The target substance (used heavily in printing inks) is Carbon black.

The Multi-Step Synthesis:

1. Oxidation: Acetaldehyde oxidizes to form Acetic acid.
2. Neutralization: Adding NaOH to acetic acid forms Sodium acetate.
3. Dry Distillation: Heating sodium acetate with soda lime produces Methane gas.
4. Thermal Cracking: Heating methane to 1000°C in the absolute absence of air decomposes it into Hydrogen gas and solid Carbon black.

Selective Precipitation & Amphoterism

To precipitate exactly one cation, we need a reagent that has distinct, highly specific interactions with the metals:

• Aluminum (Al³⁺): It initially precipitates as Al(OH)₃, but because it is amphoteric, adding an abundance of NaOH causes it to re-dissolve completely into soluble Sodium aluminate.
• Calcium (Ca²⁺): Calcium hydroxide Ca(OH)₂ is moderately soluble and does not precipitate effectively at these typical testing concentrations.
• Iron (Fe²⁺): Reacts to form Iron (II) hydroxide, Fe(OH)₂, a stable, insoluble precipitate that does not dissolve in excess base.

Multi-Step Synthesis (Backward Analysis)

To find starting material X, we trace the reactions backwards from the final product:

1. Complete Hydrogenation: 1,2-dimethylcyclohexane is formed by fully hydrogenating the aromatic ring of 1,2-dimethylbenzene (o-xylene).
2. Catalytic Reforming: O-xylene (an 8-carbon aromatic) comes from reforming an 8-carbon aliphatic chain with a 6-carbon straight core: 3,4-dimethylhexane.
3. Dry Distillation: This process drops one carbon atom. So, the precursor was a 9-carbon salt: Sodium 3,4-dimethylheptanoate.
4. Neutralization & Oxidation: The salt came from neutralizing an acid, which was produced by completely oxidizing a primary alcohol. This alcohol must be 3,4-dimethyl-1-heptanol.

Preparation of Sodium Benzoate

The renowned organic food preservative that inhibits fungal growth is Sodium benzoate.

The Industrial Pathway:

1. Catalytic Reforming: Normal hexane is heated with a Pt catalyst to form Benzene.
2. Alkylation: Benzene reacts with methyl chloride (Friedel-Crafts) to form Toluene.
3. Oxidation: Toluene is oxidized using V₂O₅ at 400°C to yield Benzoic acid.
4. Neutralization: Benzoic acid is neutralized with NaOH to produce the water-soluble salt Sodium benzoate.

Oxidation of Butanol Isomers

Let's map the isomers of C₄H₁₀O based on their structural methyl count and oxidize them:

• (A) Two methyl groups: This is isobutanol (2-methyl-1-propanol). Complete oxidation yields 2-methylpropanoic acid.
• (B) Three methyl groups: This is tert-butanol (2-methyl-2-propanol). Because the carbon bonded to the OH group has no attached hydrogen atoms, it strongly resists oxidation under normal conditions (No reaction).
• (C) One methyl group: This is the straight-chain 1-butanol. Oxidation seamlessly yields Butanoic acid.

Oxidation and Dehydration Constraints

Let's filter the choices based on the two distinct chemical requirements:

• Constraint 1 (Must Oxidize): The compound must be a primary or secondary alcohol. Tertiary alcohols do not oxidize. This instantly eliminates choices (a) and (c).
• Constraint 2 (Does NOT give 2-methyl-1-butene):
  Choice (d) directly dehydrates to form 2-methyl-1-butene.
  Choice (b) 2,2-Dimethyl-1-propanol (Neopentyl alcohol) is a primary alcohol that oxidizes, but it lacks beta-hydrogens (the adjacent carbon is bonded only to methyl groups). Thus, it cannot undergo standard E2 dehydration to yield simple alkenes without highly complex carbocation rearrangements.

Structural Formula Analysis

IUPAC Rule Verification

The correct IUPAC name for d is 4,4-dimethyl-1-pentyne

Drawing the Isomers

To have an ethyl branch (–CH₂CH₃), the longest chain must be at least 5 carbons (pentene) to accommodate the branch at a non-terminal carbon:

1. 1-pentene:
   CH2=CH–CH2–CH2–CH3
2. 2-pentene:
   CH3–CH=CH–CH2–CH3
3. 2-methyl-1-butene:
   CH2=C(CH3)–CH2–CH3

These are the only 3 possible alkene structures containing an ethyl branch.

Nucleophilic Substitution Reaction

The starting material is CH3–CH(CH3)–CH(Br)–CH3. Alkaline hydrolysis substitutes the halogen (Br) with a hydroxyl group (OH):

Numbering from the right gives the -OH group the lowest locant (2), so the IUPAC name is 3-methyl-2-butanol.

Markovnikov's Addition Analysis

• For X (2-methyl-1-butene): H adds to CH₂, and Br adds to the tertiary carbon:
  CH3–CH2–C(CH3)=CH2 + HBr → CH3–CH2–C(Br)(CH3)–CH3 (2-bromo-2-methylbutane).
• For Y (2-methyl-2-butene): H adds to CH, and Br adds to the tertiary carbon:
  CH3–CH=C(CH3)2 + HBr → CH3–CH2–C(Br)(CH3)2 (2-bromo-2-methylbutane).

Synthesis Route

Goal: Convert Methane (CH₄) to Benzoic acid (C₆H₅COOH):

1. Strong heating and rapid cooling: Methane is converted to ethyne (C₂H₂) at 1500°C.
2. Polymerization: Ethyne undergoes cyclic polymerization to form Benzene (C₆H₆).
3. Alkylation: Benzene reacts with CH₃Cl to form Toluene (C₆H₅CH₃).
4. Oxidation: Toluene is oxidized using V₂O₅ catalyst to Benzoic acid.

• Starting compound: C₇H₆O₂ is phenyl formate.
• Target compound: C₆H₆O is phenol.
• Reaction: Acid hydrolysis splits the ester.
• Products: It yields phenol directly.

💡 Why Other Options Fail

• Basic hydrolysis: Forms a phenoxide salt.
• Dehydration: Removes water instead.
• Neutralization: Acid-base reaction only.

Step-by-Step Synthesis

Starting material is Phenol (C₆H₆O):

1. Heating with Zinc: Reduces phenol to benzene.
2. Alkylation: Friedel-Crafts alkylation to form toluene.
3. Friedel-Crafts reaction: A second alkylation yields 1,2-dimethylbenzene (o-xylene).
4. Hydrogenation: Saturates the benzene ring into 1,2-dimethylcyclohexane.

Dacron Synthesis

Process 1 :

Process 2 :

Process 3 :

Step 1: Identify Compound A & B

• Compound A: Phenol (carbolic acid) — reacts with HNO₃ (mineral acid) to form picric acid, but not with HCl.
• Compound B: Formaldehyde (HCHO) — oxidizes to formic acid, used in perfumes.

Step 2: Polymerization Product

Phenol reacts with formaldehyde in an acidic or basic medium to form Bakelite, a thermosetting polymer widely used as an insulator in electrical equipment.

Alloy Classification

• Intermetallic Alloys: Formed by metals chemically combining to produce compounds that do not follow normal chemical valency rules (e.g., Cementite Fe₃C, Gold-lead Au₂Pb, Duralumin Al-Cu).
• Substitution Alloys: Standard physical mixtures where atoms substitute for each other in the crystal lattice. Brass (Cu + Zn) is a substitution alloy.

Identification and Analysis

• Compound 1: Aliphatic Alcohol (e.g., Ethanol) — formula C_nH_2n+2O.
• Compound 2: Carboxylic Acid — formula C_nH_2nO₂ (reacts with both alcohols and NaOH).
• Compound 3: Phenol — acidic but cannot be prepared by simply oxidizing an aliphatic alcohol.

Step 1: Identify the Alcohol

Alcohols resistant to oxidation are tertiary alcohols (such as 2-methyl-2-propanol).

Step 2: Conversion Pathway

Reaction Analysis

1. X (Acetylene / Ethyne): H-C≡C-H (undergoes addition in 2 steps).
2. Y (Acetaldehyde / Ethanal): CH3CHO formed via catalytic hydration of X.
3. Z (Ethanol): CH3CH2OH formed by reduction of Y (primary alcohol).
4. W (Ethene): formed by dehydration of Z.

Esterification of Lactic Acid

The bifunctional molecule is lactic acid [CH₃CH(OH)COOH]:

• Reaction 1 (Acid function): reacts at –COOH with methanol to yield CH₃CH(OH)COOCH₃.
• Reaction 2 (Alcohol function): reacts at –OH with acetic acid to yield CH₃CH(OCOCH₃)COOH.

C₄H₈ isomers include:

Comprehensive Chemical Analysis of C₄H₈ Isomers

The molecular formula C₄H₈ represents an unsaturated hydrocarbon with one double bond (alkene) or a saturated cyclic compound (cycloalkane). There are exactly 5 structural isomers for this formula:

1. Open-Chain Isomers (Alkenes = 3 Isomers)

1. 1-Butene: CH₂=CH-CH₂-CH₃ (Asymmetric alkene)
2. 2-Butene: CH₃-CH=CH-CH₃ (Symmetric alkene)
3. 2-Methylpropene: CH₂=C(CH₃)₂ (Asymmetric alkene)

2. Closed-Chain Isomers (Cycloalkanes = 2 Isomers)

1. Cyclobutane
2. Methylcyclopropane

---

Column-by-Column Evaluation

Column X (Isomers obeying Markownikoff's rule)

• Markownikoff's rule applies when adding an asymmetric reagent (like HX or H₂O) to an asymmetric alkene.
• Among the isomers, 1-Butene and 2-Methylpropene are asymmetric because the double-bonded carbons hold unequal counts of hydrogen atoms.
• Count (X) = 2

Column Y (Isomers that decolorize alkaline potassium permanganate)

• Decolorizing purple alkaline KMnO₄ (Baeyer's Test) is a signature oxidation reaction for all carbon-carbon double bonds (alkenes).
• All 3 open-chain alkene isomers contain this double bond and successfully undergo this reaction.
• Count (Y) = 3

Column Z (Isomers that give by catalytic hydration alcohol difficult to be oxidized)

• Catalytic hydration (addition of water) of 2-Methylpropene follows Markownikoff's rule and produces a tertiary alcohol (2-methylpropan-2-ol):


CH₂=C(CH₃)₂ + H₂O →^H⁺ CH₃-C(OH)(CH₃)₂


• Tertiary alcohols lack a hydrogen atom attached directly to the carbinol carbon (C-OH), making them highly resistant to standard chemical oxidation. 2-Methylpropene is the only isomer that gives this result.
• Count (Z) = 1

Column W (Closed chain isomers)

• The ring structural configurations corresponding to the formula C₄H₈ are Cyclobutane and Methylcyclopropane.
• Count (W) = 2

---

Detailed Step-by-Step Chemical Conversion

To solve this problem, we must first identify the chemical substances mentioned:

• Simplest Alkyne: Ethyne (Acetylene, C₂H₂).
• Simplest Hydrocarbon: Methane (CH₄).

The correct industrial and laboratory pathway to convert ethyne into methane follows these four consecutive steps:

1. Catalytic Hydration

• Adding water to ethyne (C₂H₂) in the presence of an acid catalyst (H₂SO₄ / HgSO₄ at 60°C) produces an unstable vinyl alcohol intermediate, which quickly rearranges into Acetaldehyde (Ethanal, CH₃CHO).
• The Reaction Box:


C₂H₂ + H₂O →^{H₂SO₄ / HgSO₄} CH₃CHO

2. Complete Oxidation

• Subjecting acetaldehyde to complete chemical oxidation using acidified potassium permanganate or dichromate converts the aldehyde group into a carboxylic acid, yielding Acetic acid (Ethanoic acid, CH₃COOH).
• The Reaction Box:


CH₃CHO + [O] →^{KMnO₄ / H⁺} CH₃COOH

3. Neutralization

• Reacting acetic acid with a strong base like sodium hydroxide (NaOH) neutralizes the acid to form water and the organic salt Sodium acetate (CH₃COONa).
• The Reaction Box:


CH₃COOH + NaOH → CH₃COONa + H₂O

4. Dry Distillation

• Heating solid sodium acetate with soda-lime (NaOH + CaO) removes a molecule of carbon dioxide via decarboxylation, successfully yielding the target simplest hydrocarbon, Methane (CH₄).
• The Reaction Box:


CH₃COONa + NaOH →^{CaO / Δ} CH₄ ↑ + Na₂CO₃

Why the Other Options Are Incorrect

• (a) & (b): They suggest "reduction" instead of neutralization as the third step. Reducing acetic acid would lead back to alcohols or aldehydes rather than setting up the salt needed for methane extraction.
• (c): Suggests "Bayer's oxidation" as the initial step, which is an alkene-specific test utilizing cold alkaline potassium permanganate that cannot process triple-bonded alkynes into aldehydes.

Detailed Step-by-Step Analysis

1. Analyzing Organic Substance X (Violet Color Test)

• The Chemical Rule: Iron(III) chloride solution (FeCl₃) is the standard functional group reagent used to detect the presence of a phenolic OH group (a hydroxyl group directly linked to a benzene ring core).
• Identification: Carbolic acid is the common industrial name for Phenol (C₆H₅OH). When FeCl₃ is added to phenol, it coordinates to form a highly distinctive violet-colored complex solution.

2. Analyzing Organic Substance Y (Catalyst During Preparation)

• The Chemical Rule: Electrophilic aromatic substitution of benzene requires a Lewis acid catalyst to polarize the halogen molecule and generate the strong electrophile necessary to break aromaticity.
• Identification: In the preparation of a Halo benzene (such as chlorobenzene, C₆H₅Cl) via the direct halogenation of benzene, anhydrous Iron(III) chloride (FeCl₃) serves as the mandatory catalyst.
• The Reaction Box:


C₆H₆ (Benzene) + Cl₂ →^{FeCl₃ / UV} C₆H₅Cl (Halo benzene) + HCl

3. Why the Other Options Are Incorrect

• (A): While salicylic acid gives a violet color with FeCl₃, carbolic acid (phenol) is prepared industrially via the hydrolysis of chlorobenzene with NaOH at high temperature and pressure, not using FeCl₃ as a catalyst.
• (C): Sulphuric acid (H₂SO₄) is a strong mineral inorganic acid, not an organic compound, and is not prepared using an iron catalyst.
• (D): This choice completely reverses the roles; halo benzene does not react with FeCl₃ to yield a violet color.

Detailed Step-by-Step Analysis

1. Identifying the Reaction Type (Friedel-Crafts Alkylation)

• The Reactants: The reaction combines Benzene (C₆H₆) with Benzyl chloride (C₆H₅CH₂Cl).
• The Catalyst: Anhydrous Aluminium chloride (AlCl₃) acts as a classic Lewis acid catalyst.
• The Mechanism: This setup triggers a Friedel-Crafts alkylation reaction. The AlCl₃ catalyst polarizes and removes the chloride ion from benzyl chloride, generating a highly reactive carbocation intermediate (C₆H₅CH₂⁺). This carbocation then attacks the aromatic benzene ring.

---

2. Determining Products A and B

• When the benzyl carbocation (C₆H₅CH₂⁺) substitutes a hydrogen atom (H) on the benzene ring, the two aromatic groups link together via the central methylene bridge:
• Product A: Diphenylmethane (C₆H₅CH₂C₆H₅).
• The displaced hydrogen atom from the benzene ring immediately combines with the released chloride ion to form the stable inorganic byproduct:
• Product B: Hydrogen chloride gas (HCl).
• The Full Reaction Box:


C₆H₅CH₂Cl + C₆H₆ →^AlCl₃ C₆H₅CH₂C₆H₅ + HCl



---

3. Why the Other Options Are Incorrect

• (a) & (b): These options claim that hydrogen gas (H₂) is evolved, ignoring the fact that the chlorine atom from benzyl chloride must form a byproduct.
• (d): Claims that chlorine gas (Cl₂) is released. Free halogens are not produced in alkylation processes; instead, the halogen is eliminated as a halide salt or hydrogen halide.

• Carbolic acid = Phenol (C₆H₅OH)
• Acid (B) = Nitric acid (HNO₃)
• Phenol + conc. HNO₃ in presence of conc. H₂SO₄ → 2,4,6-trinitrophenol = Picric acid

• Propyl formate = HCOOC₃H₇. Acidic hydrolysis gives: Formic acid (HCOOH) + Propanol.
• Compound X = Formic acid (HCOOH) — formula CH₂O₂.
• (a) HCOOH has no structural isomers ✓
• (b) CH₂O₂ ✓
• (c) Formic acid does not react with halogenated acids (HCl, HBr) — that's a property of alcohols, not carboxylic acids ✗
• (d) Formic acid is secreted by ants and some insects ✓

• An alcohol that cannot be prepared by catalytic hydration = Methanol (CH₃OH), because catalytic hydration requires an alkene, and there's no C₁ alkene.
• (a) Can be prepared by the general method (halide + KOH) — technically yes for methyl halide ✓
• (b) Methanol (CH₃OH) has NO structural isomers — it's the simplest alcohol ✗
• (c) Methanol is toxic and causes blindness/death ✓
• (d) Methanol reacts with acids to form esters (e.g., methyl salicylate) ✓

• A = C₂H₆O = Ethanol (C₂H₅OH) — used as a disinfectant/antiseptic in medicine.
• B = C₇H₆O₃ = Salicylic acid — used to make aspirin ,dacron and other medical products.
• Both are used in medical fields. ✓
• Ethanol does not react with NaOH (neutral alcohol), so (b) is wrong.

Detailed Step-by-Step Analysis

The Core Chemical Rule: Acidified potassium permanganate (KMnO₄ / H⁺) is a powerful purple oxidizing agent. It undergoes decolorization (turns from purple to colorless) only when it reacts with substances that can be easily oxidized.

---

1. Why Option (c) is the Correct Exception

• Structure Analysis: 2-bromo-2-propanol is a halogenated derivative of a tertiary alcohol (Cyclic or structural formula: (CH₃)₂C(Br)(OH)).
• Oxidation Behavior: The central carbon atom holding the hydroxyl (-OH) group is attached to three other carbon chains and a bromine atom. It contains zero alpha-hydrogen atoms (H-C-OH).
• Conclusion: Because tertiary configurations lack alpha-hydrogens, they are highly stable against chemical oxidation under normal conditions. Therefore, it does not decolorize the purple KMnO₄ solution.

---

2. Why the Other Options Do Decolorize KMnO₄

• (a) Hydroxy cyclohexane (Cyclohexanol): This is a secondary alcohol. The carbon bonded to the -OH group possesses one alpha-hydrogen, allowing it to be easily oxidized to cyclohexanone, which decolorizes the solution.
• (b) Isobutyl alcohol (2-methylpropan-1-ol): This is a primary alcohol (R-CH₂OH). It contains two alpha-hydrogens and readily oxidizes first to an aldehyde and then to a carboxylic acid, decolorizing the solution.
• (d) 2-hydroxy propanoic acid (Lactic acid): The alpha-carbon contains a secondary alcohol group (CH₃-CH(OH)-COOH). It has one alpha-hydrogen and is smoothly oxidized into pyruvic acid (CH₃-CO-COOH), decolorizing the solution.

1. A (Sodium pentanoate) + NaOH/CaO/Δ → Dry distillation → B = Butane (C₄H₁₀)
2. Butane + Cl₂ → C = 1-chlorobutane (C₄H₉Cl)
3. 1-chlorobutane + KOH (aq) → D = 1-butanol (C₄H₉OH)
4. 1-butanol + Oxidation → E = Butanoic acid (C₃H₇COOH)

• Compound A appears in both schemes. From Scheme 2: A + V₂O₅/3O₂ → simplest aromatic acid (benzoic acid, C₆H₅COOH). The substance oxidized by V₂O₅ to give benzoic acid is Toluene (C₆H₅CH₃). So A = Toluene.
• Process Y: CₙH₂ₙ₊₂ → Toluene = Catalytic reforming (converting aliphatic hydrocarbons to aromatic ones). ✓
• From Scheme 1: RX + excess KOH/Δ → Toluene (A). Then Toluene → simplest aliphatic acid (formic acid HCOOH).
• Process X: Toluene → Formic acid. This would be complete oxidation. ✓

🧪 Chemical Explanation

• Compound (A) is Lactic Acid (C₃H₆O₃): Contains a secondary alcohol group (-CHOH-) which is easily oxidizable.
• Compound (B) is Citric Acid (C₆H₈O₇): Contains a tertiary alcohol group (-COH) which is non-oxidizable.
• Compound (C) is Carbolic Acid / Phenol (C₆H₆O): Acts as a weak organic acid but contains zero alcohol groups.

💡 Why Other Options Fail

• Option b: Reverses the correct identities of compounds A and B.
• Option c: Incorrectly lists Phenol (C₆H₆O) as compound A.
• Option d: Compound B here is Salicylic Acid, which does not fit the criteria.

• A = (CH)₆ = C₆H₆ = Benzene
• B = (CH₂)₆ = C₆H₁₂ = Cyclohexane
• (c): Benzene can be produced by reduction of carbolic acid (phenol): C₆H₅OH + Zn dust → C₆H₆ + ZnO ✓
• Cyclohexane is produced by hydrogenation of benzene: C₆H₆ + 3H₂ → C₆H₁₂ ✓

The correct option is a) (A) : Propanoic acid , (B) : Propanol , (C) : Propene.

Detailed Step-by-Step Analysis

1. Structural Logic Breakdown

• B → C (Dehydration): Dehydration of an alcohol removes water to give an alkene (unsaturated hydrocarbon). Thus, B must be an alcohol and C must be an alkene.
• A → B (Reduction): Reduction of compound A gives alcohol B. Since B is an alcohol, A must be a carboxylic acid.
• C → D (Oxidation): Oxidation of an alkene.

2. Verifying Option (a) Pathways

• Reduction Step (A → B): Propanoic acid (CH₃CH₂COOH) is reduced using hydrogen gas over a copper chromate catalyst to produce 1-Propanol (CH₃CH₂CH₂OH).


CH₃CH₂COOH + 2H₂ →^{CuCrO₄ / Δ} CH₃CH₂CH₂OH + H₂O


• Dehydration Step (B → C): Heating 1-Propanol with concentrated sulfuric acid (H₂SO₄) at 180°C eliminates water to form Propene (CH₃CH=CH₂), which is an unsaturated aliphatic hydrocarbon.


CH₃CH₂CH₂OH →^{conc. H₂SO₄ / 180°C} CH₃CH=CH₂ + H₂O

3. Why the Other Options Are Incorrect

• b) & d): These options state that Ethyl alcohol undergoes "reduction" to form aldehydes or acids, which is chemically impossible since primary alcohols can only be oxidized, not reduced, to these forms.
• c): Identifies C as Propane. Propane is a saturated alkane, directly contradicting the explicit condition that C is an unsaturated aliphatic hydrocarbon.

Compound X Definition
• Compound X: Ethyne (Acetylene, C₂H₂) reacts explosively with chlorine with a flame, producing carbon soot and HCl. It belongs to the alkyne family, which obeys the general formula C_nH_2n-2.
Compound Y Definition
• Compound Y: Gammaxane (C₆H₆Cl₆), used as an insecticide, is an alicyclic hydrocarbon derivative, not aromatic, which perfectly obeys the molecular relation C_nH_nCl_n.

• Methyl groups (-CH₃): 4 (two on the chain ends, two as side chains).
• Methylene groups (-CH₂-): 0 (only -CH- tertiary carbons are present).

1. Diethyl ether: CH₃-CH₂-O-CH₂-CH₃
2. Methyl propyl ether: CH₃-O-CH₂-CH₂-CH₃
3. Methyl isopropyl ether: CH₃-O-CH(CH₃)₂

1. Neutralization: React heptanoic acid with NaOH to obtain sodium heptanoate (C₆H₁₃COONa).
2. Dry Distillation: Heat sodium heptanoate with soda-lime to yield hexane (C₆H₁₄).
3. Catalytic Reforming: Pass hexane over a hot platinum catalyst to undergo cyclization and dehydrogenation, yielding benzene (C₆H₆).

• Acid Y (Citric acid): It is a tricarboxylic aliphatic acid containing 3 carboxyl groups (which react with 3 moles of NaOH) and 1 tertiary hydroxyl group (which reacts only with sodium metal, not NaOH). Citric acid is widely used as a preservative for frozen fruits.

1. Perform catalytic reforming of hexane to prepare benzene, which is then chlorinated to chlorobenzene.
2. Perform alkaline hydrolysis of chlorobenzene to produce phenol.
3. React phenol with formaldehyde via condensation polymerization to yield Bakelite.

🧪 Chemical Explanation

• Acid (A) Identification: Carbolic acid (Phenol) acts as a very weak acid but does not react with sodium bicarbonate, meaning it fails the acidity test.
• Reduction Behavior: Carbolic acid lacks a carbonyl or carboxyl group, so it cannot be reduced by hydrogen over a copper(II) chromate catalyst.
• Acid (B) Identification: Salicylic acid is a bifunctional organic acid used directly to manufacture aspirin (acetylsalicylic acid), which treats headaches.

💡 Why Other Options Fail

• Option a: Acetic acid is a carboxylic acid that passes the acidity test by reacting vigorously with sodium bicarbonate.
• Option c: Benzoic acid passes the acidity test, and it is not the primary compound used for making aspirin.
• Option d: Oxidizing para-methyl toluene produces terephthalic acid, not salicylic acid.

1. Alkylation: Alkylating benzene (C₆H₆) yields toluene, C₆H₅CH₃ (X).
2. Oxidation: Oxidizing toluene yields benzoic acid, C₆H₅COOH (Y).
3. Esterification: Reacting benzoic acid with ethanol (C₂H₅OH) produces ethyl benzoate (Z).

• Compound X: Acetic acid reacts with methanol (CH₃OH) to produce methyl acetate, which has a fruity odor.
• Compound Y: 2-propanol is a secondary alcohol that oxidizes easily, causing the violet color of acidified KMnO₄ to disappear.
• Compound Z: Carbolic acid (phenol) reacts with bromine water to form a white precipitate of 2,4,6-tribromophenol.

1. Alkaline Hydrolysis: Heat the ester with NaOH to obtain 1-propanol and sodium propanoate.
2. Dehydration: Heat 1-propanol with concentrated H_2SO4 at 180°C to obtain propene.
3. Catalytic Hydrogenation: React propene with H₂ gas over a nickel catalyst to obtain propane.

• Why Option (a) fails to produce Fe₂O₃: Reducing Fe_3O4 at high temperatures (>700^°C) yields pure Iron (Fe). Adding diluted H_2SO4 to Iron produces Iron(II) sulfate, which reacts with an alkali to form Iron(II) hydroxide. Heating Fe(OH)₂ above 200^°C yields Iron(II) oxide (FeO), not Fe_2O3:
  Fe + H_2SO4(dil) → FeSO₄ + H₂
  FeSO₄ + 2NaOH → Fe(OH)₂ ↓ + Na_2SO4
  Fe(OH)₂ xrightarrowΔ >200^°C FeO + H_2O

💡 Why Other Options Successfully Produce Fe₂O₃

• Option b: Passing hot air over red-hot iron produces magnetic iron oxide (Fe_3O4), which can then be completely oxidized to form Iron(III) oxide:
  3Fe + 2O₂ xrightarrowΔ Fe_3O4
  2Fe_3O4 + (1)/(2)O₂ xrightarrowΔ 3Fe_2O3
• Option c: Leaving FeSO₄ in air oxidizes it to an Iron(III) salt. Adding an alkali precipitates Iron(III) hydroxide, which easily thermally decomposes above 200^°C to give Iron(III) oxide:
  Fe³⁺ + 3OH^- → Fe(OH)₃ ↓
  2Fe(OH)₃ xrightarrowΔ >200^°C Fe_2O3 + 3H_2O
• Option d: Reacting iron with oxalic acid yields Iron(II) oxalate. Heating it in the absence of air yields FeO, which is then oxidized in the final step to form Iron(III) oxide:
  Fe(COO)₂ xrightarrowΔ (no air) FeO + CO + CO₂
  2FeO + (1)/(2)O₂ xrightarrowΔ Fe_2O3

• 1. Neutralization: Reacting octanoic acid with NaOH produces sodium octanoate (C₇H₁₅COONa).
• 2. Dry Distillation: Heating sodium octanoate with soda lime produces Heptane (C₇H₁₆).
• 3. Catalytic Reforming: Passing heptane over heated Pt reshapes it into an aromatic ring, yielding Toluene (C₆H₅CH₃).
• 4. Oxidation: Oxidizing toluene with oxygen (V₂O₅ cat.) yields Benzoic acid (C₆H₅COOH).

Step 1: Identify Compound X

Dry distillation of sodium acetate with soda lime produces Methane:

Chlorinating Methane by replacing three hydrogen atoms yields Chloroform (CHCl₃):

Properties of X: Chloroform is a substituted alkane. It was historically used as an anesthetic.

Step 2: Identify Compound Y

Dehydration of 2-chloroethanol (Cl-CH₂-CH₂-OH) at 180°C removes H and OH to form a double bond:

Properties of Y: The product is Vinyl Chloride (chloroethene). It is an alkene derivative used as the monomer to make PVC (Polyvinyl Chloride), which is used in water hoses and pipes. This makes option (d) true regarding compound Y,

Step 1: Determine the molecular formula.
Combustion of a hydrocarbon C_xH_y produces x moles of CO₂ and y/2 moles of H₂O.
If 5 moles of CO₂ are produced, x = 5 (Carbon atoms = 5).
If 5 moles of H₂O are produced, y/2 = 5 → y = 10 (Hydrogen atoms = 10).
The required molecular formula is C₅H₁₀.

Step 2: Check the options.

Chemical Analysis: The formula C₄H₈O corresponds to an aldehyde or a ketone.

• X oxidizes (turns dichromate green): X must be an aldehyde (e.g., butanal or 2-methylpropanal). Option C correctly identifies X as 2-methyl propanal (C₄H₈O).
• Y does not oxidize: Y must be a ketone (e.g., butanone). Option D suggests Y is 3-methyl-2-butanone, but that compound has 5 carbons (C₅H₁₀O), not 4, .

Step 1: Identify the Monomer

The repeating unit of the polymer is —CH₂—CH(C₂H₅)—. To find the monomer, we restore the double bond between the two carbon atoms of the backbone.

Step 2: Evaluate the Reactions of 1-Butene

1-Butene is an unsymmetrical alkene, so its addition reactions follow Markovnikov's Rule (the hydrogen adds to the carbon with more hydrogens in case of addition of asymmetric reagents only not all addition reactions).

Identify Compounds:

• A (C₂H₄O): Acetaldehyde (Ethanal)
• B (C₂H₆O): Ethanol (Ethyl alcohol)
• C (C₂H₄): Ethene (Ethylene)

Analyze Oxidation:
Oxidation of Acetaldehyde (A) yields Acetic acid (Ethanoic acid).
Oxidation of Ethanol (B) yields Acetaldehyde, then Acetic acid.
Acetic acid is a well-known raw material used in the chemical industry for the manufacture of dyes, plastics, and insecticides.

The compound with the formula C₇H₇NO is Benzamide (C₆H₅CONH₂). Count the atoms: C: 6+1=7, H: 5+2=7, N: 1, O: 1.

Amides not ester are prepared by the ammonolysis of esters (reacting an ester with ammonia).

Reacting Methyl benzoate (Option A) with ammonia perfectly yields Benzamide and Methanol.

Match formulas to functional groups:

• X (CH₂O): Formaldehyde (Methanal). This is an Aldehyde. (Ketones require at least 3 carbons).
• Y (C₂H₆O): Ethanol (CH₃CH₂OH) or Dimethyl ether. Ethanol is a Primary monohydric alcohol.
• Z (C₂H₆O₂): Ethylene glycol (HO-CH₂-CH₂-OH). This is a Primary dihydric alcohol.

Option B aligns perfectly with these structures.

Step 1: Identify C
Compound C does not react with NaOH, but it reacts with A to form a flavor (an ester). This means C is an alcohol. C₂H₆O is Ethanol.

Step 2: Identify A and B based on NaOH consumption

• A and B must be acids to react with NaOH.
• C₃H₆O₃ (Lactic acid): Contains one -COOH group and one aliphatic -OH group. Aliphatic -OH does not react with NaOH. So, it consumes 1 mole of NaOH.
• C₇H₆O₃ (Salicylic acid): Contains one -COOH group and one phenolic -OH group. Phenols DO react with NaOH. So, it consumes 2 moles of NaOH.

Since B consumes double the quantity of A, B is Salicylic acid (C₇H₆O₃) and A is Lactic acid (C₃H₆O₃). Heating A (Lactic acid) with C (Ethanol) produces Ethyl lactate, an ester used as a flavoring agent.

The boiling point of organic compounds with hydroxyl (-OH) groups is primarily determined by the extent of hydrogen bonding. More -OH groups mean stronger intermolecular forces.

Therefore, the correct decreasing order is A > B > C.

The compound C₆H₅OOCCH₃ is an ester. Specifically, it is formed from the acetate (ethanoate) radical CH₃COO- and the phenyl radical -C₆H₅. Its common name is phenyl acetate.

Acidic hydrolysis of an ester breaks it back down into its constituent carboxylic acid and alcohol/phenol:

The IUPAC name for CH₃COOH is Ethanoic acid. The IUPAC accepted name for C₆H₅OH is Phenol. (While "acetic acid" and "hydroxybenzene" are valid names, "ethanoic acid" and "phenol" perfectly fit standard IUPAC multiple-choice criteria here).

Step 1: Identify X and Y.
1. Identify elements X and Y: * The question states that X and Y are two consecutive transition elements in the first transition series, and each has a compound in its highest oxidation state that acts as an oxidizing agent. * Chromium (Cr, Z = 24): Its highest oxidation state is +6, found in potassium dichromate (K_2Cr2O7), which is a well-known strong oxidizing agent. * Manganese (Mn, Z = 25): Its highest oxidation state is +7, found in potassium permanganate (KMnO₄), which is also a powerful oxidizing agent. * Therefore, X and Y are Cr and Mn.

Step 2: Evaluate the uses.
If we let Y = Cr(V2O5) and Mn = Mn (or vice versa):

• (a) Iron (Fe) is the catalyst for ammonia (Haber process), not Cr or Mn.
• (b) Vanadium (V), not Chromium (Cr), is used in car springs., .
• Alloys of Nickel with Chromium or steel exhibit acid resistance, which does not match the description of the element following Cr (Manganese).
• The elements between which Cr (Z=24) and Mn (Z=25) are located are Vanadium (V, Z=23) and Iron (Fe, Z=26). * An alloy of Vanadium (V) and Iron (Fe) with carbon forms vanadium steel, which is characterized by high hardness and high elasticity (used in making car springs).

A "longer and weaker" O-H bond means the hydrogen is more easily lost as an H⁺ ion, indicating stronger acidity. In organic chemistry, phenols are more acidic than aliphatic alcohols because the benzene ring withdraws electron density, weakening the O-H bond.

Therefore, (A) is a Phenol and (B) is an Alcohol.

• Phenols (A) are acidic enough to react with strong bases like NaOH to form sodium phenoxide.
• Alcohols (B) are practically neutral and do not react with NaOH.

Step 1: Trace the reactions.

• X: Dry distillation of Sodium Decanoate (C₉H₁₉COONa) yields an alkane with one less carbon: Nonane (C₉H₂₀).
• Y & Z: Thermal cracking of Nonane yields a smaller alkane and an alkene. Z is the "simplest alkene" (Ethene, C₂H₄). So Y is Heptane (C₇H₁₆).
• W: Catalytic reforming of Heptane yields Toluene (methylbenzene, C₆H₅CH₃), an aromatic hydrocarbon.
• M: Hydrogenation of Toluene (adding H₂ to break the aromatic ring) yields Methylcyclohexane.

Step 2: Evaluate compound properties.

M (Methylcyclohexane) is a saturated cyclic hydrocarbon (but NOT cyclohexane exactly, making B incorrect).
W (Toluene) oxidizes with KMnO₄ to form Benzoic acid (an aromatic acid).

Option A perfectly describes the families of these compounds.

• Acid A: Alkaline hydrolysis of 1,1,1-trichloroethane (CH₃CCl₃ + 3NaOH) temporarily forms a tri-alcohol CH₃C(OH)₃, which is highly unstable. It instantly loses a water molecule to become Acetic acid (CH₃COOH). Formula: C₂H₄O₂.
• Acid B: Fermentation of milk sugars (lactose) yields Lactic acid. Formula: C₃H₆O₃.
• Acid C: The substance used for rheumatic pain is Methyl salicylate (oil of wintergreen). It is formed by reacting Methanol with Salicylic acid. Formula: C₇H₆O₃.

Salicylic Acid (X): Contains 1 Carboxyl group (-COOH) and 1 Phenolic group (-OH).
- NaOH reacts with BOTH (acids and phenols). Needs 2 moles.
- Na metal reacts with BOTH. Needs 2 moles.

Lactic Acid (Y): Contains 1 Carboxyl group (-COOH) and 1 Aliphatic Alcohol group (-OH).
- NaOH reacts ONLY with the acid group. Needs 1 mole.
- Na metal reacts with BOTH acid and alcohol groups. Needs 2 moles.

Reaction 1

CH₂Br-CF₃ + Cl₂ → CHClBr-CF₃ (Halothane)

Halothane is a hydrocarbon derivative (contains halogens) used as a safe anesthetic ✓

Reaction 2

C₂H₂ + H₂O → CH₃CHO (acetaldehyde, Y) — stable tautomer

Oxidation: CH₃CHO → CH₃COOH (acetic acid, Z)

Step 1: Alkylation Products of Toluene

Methyl group (-CH₃) in toluene is ortho/para directing. So methylation produces two main isomers:

• 1,2-Dimethyl benzene (ortho-xylene)

• 1,4-Dimethyl benzene (para-xylene)

Step 2: Identify Isomer X (Used in Artificial Heart Valves)

Artificial heart valves are made from PET (polyethylene terephthalate), a polymer derived from terephthalic acid.

Terephthalic acid is produced by oxidation of para-xylene:

Therefore: X = para-xylene (1,4-dimethyl benzene)

Step 3: Identify Isomer Y

The other isomer is Y = ortho-xylene (1,2-dimethyl benzene)

Oxidation of ortho-xylene produces phthalic acid (1,2-benzene dicarboxylic acid):

Hence statement (c) is correct: oxidation of Y produces phthalic acid ✓

Why Not (d)?

X gives terephthalic acid (not phthalic acid). Phthalic acid is derived from Y, not X.

Dry Distillation Rule

R-COONa + NaOH —(soda lime)→ R-H + Na₂CO₃

The carboxyl carbon is removed; resulting alkane has one fewer carbon.

Analysis of Each Option

Only isobutane is both branched and gaseous (alkanes with ≤4 carbons are gases).

Catalytic Hydration Reaction

3-methyl-1-butyne: (CH₃)₂CH-C≡CH

Hydration (Markovnikov, with Hg²⁺ catalyst):

(CH₃)₂CH-C≡CH + H₂O → (CH₃)₂CH-CO-CH₃ (3-methyl-2-butanone, a ketone)

General Formula

Ketones have general formula CₙH₂ₙO

Reaction with KMnO₄

Ketones do not decolorize acidified KMnO₄ — they resist oxidation (no α-H bonded to C=O that can be easily oxidized in this conditions).

Step 1: Dry Distillation

C₆H₄(C₂H₅)(COONa) + NaOH → C₆H₅-C₂H₅ (ethylbenzene, X) + Na₂CO₃

Step 2: Chlorination with FeCl₃

Ethyl group (-C₂H₅) is an alkyl group → ortho/para directing

Product: mixture of 2-chloro ethylbenzene (ortho) and 4-chloro ethylbenzene (para)

Identify X

Aromatic liquid with bp 80°C → Benzene (C₆H₆)

Reactions

Y: C₆H₆ + 3Cl₂ —UV→ C₆H₆Cl₆ (benzene hexachloride / lindane)

This is an aliphatic derivative (saturated, no aromatic ring) ✓

Z: C₆H₆ + Cl₂ —FeCl₃→ C₆H₅Cl (chlorobenzene)

This is an aromatic derivative (not hydrocarbon — contains Cl).

Identifying the Acids

X: Salt = NaOOC-COONa

→ Acid = HOOC-COOH = Oxalic acid ✓

Y: Salt = C₆H₄(COONa)₂

→ Acid = C₆H₄(COOH)₂ = Phthalic acid ✓

Both are dicarboxylic acids as stated in the problem.

Structural Analysis

Structure: HOOC-C(CH₃)(C₂H₅)-CH₂-C₂H₅ Wait, looking again: HOOC-C(CH₃)(C₂H₅)-CH(C₂H₅)...

From figure: COOH at top, central C has CH₃ and C₂H₅ branches, CH₂ branch leading to another C₂H₅

Longest chain containing COOH: COOH-C-CH₂-CH₂-CH₃ = 5 carbons → pentanoic acid

Substituent Naming

At C2: methyl and ethyl groups → alphabetical: ethyl first

Final name: 2-Ethyl-2-methyl pentanoic acid ✓

Step 1: Find the Simplest Alcohol Containing an Ethyl Branch

To have an ethyl group as a branch in the simplest possible alcohol, the main chain (containing -OH) should be as short as possible while still treating ethyl as a substituent.

The simplest alcohol satisfying this is 2-Ethyl-2-propanol:

Step 2: Derive Compound X (Alkyl Halide)

Alkaline hydrolysis replaces -Cl with -OH:

R-Cl + NaOH(aq) → R-OH + NaCl

To get 2-ethyl-2-propanol, the corresponding alkyl chloride is obtained by replacing -OH with -Cl:

Step 3: Why Not Option (b)?

1-Chloro-2-ethyl butane would give 2-ethyl-1-butanol — this contains 4 carbons in the main chain, which is NOT the simplest alcohol with an ethyl branch.

The simplest is 2-ethyl-2-propanol (only 3 carbons in main chain), so X = 2-Chloro-2-ethyl propane ✓

Reaction Sequence

Step 1 (dehydration): CH₃-CH₂-CH₂-OH → CH₃-CH=CH₂ (propene) + H₂O

Step 2 (hydration, Markovnikov): CH₃-CH=CH₂ + H₂O → CH₃-CHOH-CH₃

X = 2-propanol (isopropanol), a secondary alcohol

Properties of 2-Propanol

• 2° alcohol oxidizes (with KMnO₄) → propanone (acetone) ✓

• 2-propanol decolorizes KMnO₄ as it gets oxidized

Option (c) is correct.

Identifying Compounds

Evaluating Option (b)

Both Fe (A) and Fe (D) react with concentrated H₂SO₄:

Fe + conc. H₂SO₄ → mixture of FeSO₄ and Fe₂(SO₄)₃ ✓

Identifying Compounds

B: C₂H₄O₂ = CH₃COOH (acetic acid)

C: C₄H₈O₂ — gives amide of B on ammonolysis → ester of acetic acid: CH₃COOC₂H₅ (ethyl acetate)

A: C₃H₆O₃ = CH₃-CHOH-COOH (lactic acid) — has both -OH and -COOH

Why Option (b) is FALSE

Statement says "A does not react with B" — but lactic acid (A) does react with acetic acid (B)!

The -OH of lactic acid can be esterified by the -COOH of acetic acid → ester formation.

Therefore, statement (b) is false (the exception).

Step-by-Step Analysis

Step 1: CH₃COOH + NaOH → CH₃COONa

CH₃COONa + NaOH —(soda lime/Δ)→ CH₄ (Y) + Na₂CO₃

Step 2: CH₄ + Cl₂ —400°C→ CH₃Cl (Z) + HCl

Step 3: Simplest aromatic = benzene (C₆H₆)

C₆H₆ + CH₃Cl —AlCl₃→ C₆H₅-CH₃ = Toluene (W)

Step 4: C₆H₅-CH₃ + [O]/V₂O₅ → C₆H₅-COOH = Benzoic acid (M)

Sequence of Compounds

Compound: Propyl propanoate (CH₃CH₂COOCH₂CH₂CH₃)

Step 1: Alkaline Hydrolysis

CH₃CH₂COOCH₂CH₂CH₃ + NaOH → CH₃CH₂COONa + CH₃CH₂CH₂OH (1-propanol)

Step 2: Dehydration of Propanol

CH₃CH₂CH₂OH —H₂SO₄→ CH₃-CH=CH₂ (propene) + H₂O

Step 3: Catalytic Hydrogenation

CH₃-CH=CH₂ + H₂ —Ni→ CH₃-CH₂-CH₃ (propane) ✓

Propanol → Propene (CH₂=CHCH₃, C₃H₆) = compound X.

Isomer of C₃H₆: Cyclopropane — saturated (single bonds only) but chemically active due to ring strain (opens easily).

C_nH_2n−2BrCl: degree of unsaturation = 1 (one double bond) + Br and Cl substituting 2H. For n = 4: C₃H₄BrCl → 1-bromo-3-chloropropene.

Step 1: Understand the Target Formula

C_nH_2n+2O = general formula for saturated monohydric alcohols (R–OH).

Step 2: Identify Reducible Functional Groups

Functional groups that can be reduced to alcohols:

Step 3: Count Carboxylic Acid Isomers of C₄H₈O₂

Only carboxylic acids reduce to a single alcohol product matching C_nH_2n+2O:

Step 4: Why Esters Don't Count

Esters produce two different alcohols upon reduction (e.g., ethyl ethanoate → ethanol + ethanol, but methyl propanoate → methanol + propanol). The question asks for isomers that reduce to compounds (singular type) with formula C_nH_2n+2O, meaning a single alcohol product.

Step 5: Total Count

Only 2 carboxylic acid isomers reduce to a single alcohol product → Answer: (b) 2.

Dry distillation (with NaOH/CaO) gives alkane with one less carbon. Sodium methanoate (HCOONa) has only 1 carbon, producing H₂ gas, not an alkane.

CaC₂ + H₂O → Acetylene (C₂H₂)

3 C₂H₂ → Benzene (C₆H₆) by trimerisation

C₆H₆ + 3H₂ → Cyclohexane (C₆H₁₂) by hydrogenation (C_nH_2n, 6 CH groups → 6 methyl equivalent positions)

Step 1: Identify the Insecticide

C_nH_nCl_n with n = 6: C₆H₆Cl₆ = Gammexane (BHC — Benzene Hexachloride), a well-known insecticide.

Step 2: Trace the Synthesis from Hexane (C₆H₁)

Step 3: Why UV Light, Not Catalyst?

Step 4: Conclusion

Correct sequence: Catalytic reforming → Halogenation with UV → option (c).

Acid A = Salicylic acid (for aspirin). Oxidation of o-methylphenol (o-cresol) → Salicylic acid.

Acid B = Formic acid (HCOOH). Oxidation of methanol → formic acid. (Formic acid polymer: used in plastics industry context in some curricula.)

X = Glycerol (C₃H₈O₃) | Y = Fatty acid (e.g. hexanoic acid, n≥4)

Glycerol + 3 fatty acids → Triglyceride (fat/oil) = compound Z.

Triglycerides are the raw material for soap manufacturing (saponification).

X = Alcohol (higher BP due to H-bonding) | Y = Ether (isomer, lower BP).

2 R–OH + H₂SO₄ (140°C) → R–O–R + H₂O (intermolecular dehydration → ether).

At 180°C: intramolecular dehydration → alkene (eliminates water), not ether.

• A = Lactic acid (C₃H₆O₃): secondary –OH → oxidised to pyruvic acid (–COOH + C=O).
• B = Citric acid (C₆H₈O₇): tertiary –OH → cannot be oxidised by KMnO₄.
• C = Salicylic acid (C₇H₆O₃): phenolic –OH gives purple/violet colour with FeCl₃.

Step 1: Identify the Target Precipitate

Reddish-brown precipitate = Fe(OH)₃ (iron(III) hydroxide).

Step 2: Analyze Fe₃O₄ Composition

Fe₃O₄ = FeO·Fe₂O₃ = contains both Fe²⁺ and Fe³⁺ ions.

Step 3: Verify Option (a)

Step 4: Why "After a Period" is Key

Step 5: Why Other Options Fail

(b) Oxidation first would work, but the question asks for the simplest correct sequence. Option (a) achieves the same result more directly by allowing air oxidation during the waiting period.

(c) Reduction would convert Fe³⁺ to Fe²⁺, making it harder to get Fe(OH)₃.

(d) Dilute HCl produces FeCl₂ and FeCl₃, but without the waiting period, Fe(OH)₂ (green) would also form.

Ethanol (oxidation) → Acetic acid (CH₃COOH)

Benzoic acid (reduction) → Benzyl alcohol (C₆H₅CH₂OH)

Esterification: CH₃COOH + C₆H₅CH₂OH → C₆H₅CH₂–O–CO–CH₃ (benzyl acetate)

Coal mine gas = Methane (CH₄)

CH₄ + Cl₂ (UV) → CH₃Cl (Halogenation)

C₆H₆ + CH₃Cl (AlCl₃) → Toluene (Friedel-Crafts alkylation)

Toluene + 3 HNO₃/H₂SO₄ → 2,4,6-trinitrotoluene (TNT)

Step 1: Analyze the Scheme

The central compound is C_nH_2n (alkene, one double bond).

Step 2: Identify Process (1)

C_nH_2n + H₂O → C_nH_2n+2O

This is catalytic hydration (addition of water across the double bond using acid catalyst).

Product: Monohydroxy alcohol (one –OH group).

Step 3: Identify Process (2)

C_nH_2n + [O] + H₂O → C_nH_2n+2O₂

This is oxidation with cold, dilute KMnO₄ (Baeyer's reagent).

Step 4: Match with Options

Answer: (b) (2): Oxidation → dihydroxy alcohol.

Step 1: Identify the Target Compound

C₇H₆O₂: Degree of unsaturation = (2×7 + 2 − 6)/2 = 5

This corresponds to benzoic acid (C₆H₅COOH): benzene ring (4) + carboxyl group (1) = 5 ✓

Step 2: Identify Coal Tar Product

Fractional distillation of coal tar yields: benzene, toluene, xylene, phenol, naphthalene, etc.

The starting material for this synthesis is benzaldehyde (C₆H₅CHO) or a related compound from coal tar.

Step 3: Verify Sequence (b): Reduction → Alkylation → Oxidation

Step 4: Alternative Interpretation

Starting from phenol (C₆H₅OH) from coal tar:

Step 5: Why Other Options Fail

Answer: (b) Reduction → Alkylation → Oxidation.

The correct option is d).

X (calcium formate) needs 2 mol formic acid + 1 mol Ca(OH)₂; Y (calcium oxalate) needs 1 mol oxalic acid + 1 mol Ca(OH)₂. ✓

The correct option is b).

Decolourising Br₂/CCl₄ requires an addition reaction (C=C or C≡C). The compound that does NOT is aromatic benzene.

• (b) ✓ Catalytic reforming of hexane (C₆H₁₄ → C₆H₆) gives benzene. Without an FeBr₃ catalyst benzene undergoes substitution, not addition → does NOT decolourise Br₂/CCl₄.
• (a) ✗ CaC₂ + H₂O → acetylene (C≡C) → decolourises.
• (c) ✗ Propanol + conc. H₂SO₄ @180°C → propene (C=C) → decolourises.
• (d) ✗ Dry distillation of CH₃COONa → CH₄; strong heating + rapid cooling → acetylene (C₂H₂) → decolourises.

The correct option is c).

The full structure is:

The longest carbon chain that contains the double bond is only 5 carbons (pent-2-ene), since the C₃H₇ group is actually an isopropyl branch arrangement:

• Double bond at C2 → 2-pentene
• Phenyl group at C2
• Methyl groups at C3 and C4

Name → 3,4-dimethyl-2-phenyl-2-pentene (option c). The "hexene" reading in (d) is wrong because the longest chain through the C=C is 5 carbons, not 6. ✓

The correct option is c).

X = Benzoic acid (C₆H₅COOH). With methanol → Y = Methyl benzoate (C₆H₅COOCH₃), formula C₈H₈O₂. Phenyl acetate (CH₃COOC₆H₅) has the same formula C₈H₈O₂, so it is an isomer. ✓

The correct option is b).

• X (C₃H₅Cl) → 4 isomers: cis-1-chloropropene, trans-1-chloropropene, 2-chloropropene, 3-chloropropene (allyl chloride).
• Y (C₃H₇Cl) → 2 isomers: hydrogenation removes the C=C giving a saturated chloropropane → 1-chloropropane and 2-chloropropane only (degree of unsaturation = 0, so no ring possible).
• Z (C₃H₈O) → 2 isomers: alkaline hydrolysis gives the alcohols 1-propanol and 2-propanol.

Correction note: "cyclopropyl chloride" has formula C₃H₅Cl (not C₃H₇Cl), so it is NOT an isomer of Y — hence Y = 2, not 3.

The correct option is b).

X = 1-propanol → Y = Propene (C₃H₆) → Z = Cyclopropane (isomer of propene). Cyclopropane (MW=42) burns faster than cyclobutane (MW=56). ✓

The correct option is d).

• X = C₂H₆O₂ = Ethylene glycol (HOCH₂CH₂OH) — a di-alcohol.
• Y = C₂H₄O₂ = Acetic acid (CH₃COOH) — a carboxylic acid.

An alcohol + a carboxylic acid → ester, so (d) is correct. ✓

• (a) ✗ Glycol (alcohol) does not react with NaOH.
• (c) ✗ Acetic acid is NOT readily oxidised by KMnO₄.

Note: C₂H₄O₂ is acetic acid (CH₃COOH); formic acid is CH₂O₂ — a common mix-up.

The correct option is c).

A: 3 mol CO₂ → 3 carbons; substitution → alkane → propane (gas). B: addition in one step → alkene (liquid). ✓

The correct option is b).

• A: Toluene + HNO₃ → −CH₃ is an ortho/para director → 4-Nitrotoluene (para major).
• B: Benzoic acid + HNO₃ → −COOH is a meta director → 3-Nitrobenzoic acid.

The correct option is a).

• X = C₂F₄: Tetrafluoroethylene (alkene derivative) → monomer of Teflon (cooking utensils).
• Y = C₂H₄O₂: Acetic acid (aliphatic acid) → synthetic silk (cellulose acetate).
• Z = CCl₂F₂: Freon-12 (alkane derivative) → refrigerant/cooling agent.

The correct option is a).

D = CO (carbon monoxide) is the reducing agent in iron extraction. ✓

The correct option is c).

X gives ethylene glycol — a primary alcohol and the monomer of PEG. ✓

The correct option is b).

X = sodium propanoate; W = ethanol (primary alcohol). ✓

The correct option is a).

X = Propene. With H₂O (Markovnikov) → 2-propanol (secondary carbinol). With H₂O₂ → propylene glycol containing a secondary carbinol group. ✓

The correct option is b).

• X = C₇H₈ (Toluene): No reactive group → no reaction.
• Y = C₆H₆O (Phenol): Weak acid → reacts with NaOH ✓
• Z = C₇H₆O₂ (Benzoic acid): Carboxylic acid → reacts with NaOH ✓
• L = C₆H₁₄O₆ (polyol): Alcohol → no reaction with NaOH.

HBr on 2-methyl-2-butene (Markovnikov) → 2-bromo-2-methylbutane. So X = (b).

Propyl acetate = acetic acid + 1-propanol by esterification.

Y = glycerol; Z = nitroglycerin, used in treatment of heart problems (angina).

Simplest tertiary alcohol = tert-butanol. Dehydrate 2-methyl-1-propanol → 2-methylpropene → catalytic hydration (Markovnikov) → (CH₃)₃C-OH.

Oxidation of –CH₂OH → –COOH (carboxylic acid), which reacts with NaOH and Na₂CO₃.

❸ uses glucose (non-electrolyte); ❹ uses Na₂SO₄ forming insoluble PbSO₄/BaSO₄ → both block ion flow.

B must be asymmetric → 1-butene (not symmetrical 2-butene). So A=butyl hydrogen sulphate, B=1-butene, C=butane.

Excess Cl₂/FeCl₃ → 2,4,6-trichlorotoluene. 3H₂ hydrogenates ring → methyl cyclohexane.

Acetylene → benzene (cyclic polymerization) → alkylation → sulfonation → neutralization → detergent.

B burns → 2CO₂+2H₂O → C₂H₄ = ethene. C (aromatic, for explosives) means A is long-chain; D is a larger fragment → liquid. So (d).