Essay Questions

Iron Reaction Pathway

By tracking the reactions visually:

• Compound (B) is a black oxide that reacts with dilute H₂SO₄. It must be FeO (Iron II Oxide).
• Reacting FeO with dilute H₂SO₄ gives Compound (C), which is FeSO₄.
• Heating FeSO₄ strongly yields Compound (D), which is Fe₂O₃.
• Element (A) is elemental Iron (Fe), which also reacts with dil. H₂SO₄ to give FeSO₄.

(a) Process Names:
- Process (1) converts FeO to Fe₂O₃. This is an Oxidation reaction.
- Process (2) converts Fe₂O₃ down to elemental Fe. This is a Reduction reaction.

(b) Chemical Formulas:
- Compound (C): FeSO₄
- Compound (D): Fe₂O₃

Synthesis of Oxalic Acid

Let's map out this sequence from the end backward:

• Compound (F) is a dibasic acid produced from oxidation. It is Oxalic acid.
• It is prepared by the complete oxidation of (E), which must be Ethylene glycol.
• Ethylene glycol is prepared by oxidizing (D) with H₂O₂. Thus, (D) is Ethene.
• Ethene is prepared by dehydrating (C), so (C) is Ethanol.
• Ethanol is produced by the alcoholic fermentation of (B), which is Glucose.
• Glucose comes from the hydrolysis of (A), which is Sucrose.

(1) Highest pH Value

Solution C (0.02M). In a weak acid, dilution decreases the hydronium ion concentration [H₃O⁺], which consequently increases the pH value.

(2) Highest Fluoride Ion Concentration

Solution A (0.4M). Higher initial analytical concentration results in a larger total concentration of dissociated F^- ions at equilibrium, because [F-] = √(Ka × Ca).

(3) K_a Value of Solution B

6.7 × 10^-4. The acid dissociation constant (K_a) is temperature-dependent and remains constant for the same acid at room temperature.

(4) Finding Solution with Dissociation Percentage = 4.9%

This matches Solution B.

Step-by-Step Pathway

• A: Formic acid (CH₂O₂)
• B: Methanol (CH₃OH)
• C: Methyl formate (HCOOCH₃)
• D: Formamide (HCONH₂)

Answers:

1- Common Name for Compound (A): Formic acid

2- Structural Formula of the ester formed (Methyl salicylate / Oil of Wintergreen):

3- IUPAC Name for Compound (C): Methyl methanoate

4- Structural Formula of Compound (D):

(a) Anode and Cathode of the Lithium Battery:

• The Li⁺ cell is connected to the lead accumulator for charging (it acts as an electrolytic cell).
• Electrode (Y) → connected to negative (−) terminal A → Cathode (Reduction): Li⁺ + e⁻ → Li
• Electrode (X) → connected to positive (+) terminal B → Anode (Oxidation)

(b) Spontaneous Reactions:

• The Lead Accumulator is the source of electric current → spontaneous (galvanic) reactions occur inside it.
• The lithium cell is being charged (non-spontaneous/electrolytic process).

Step 1 — Identify (F): Catalytic hydration of propyne:

F = Acetone (Propanone)

Step 2 — Identify (A): Preservative for frozen fruit = Benzoic acid (C₆H₅COOH)

Step 3 — Identify (B):

B = Sodium benzoate (C₆H₅COONa)

Step 4 — Identify (D): Since (F = Acetone) converts to (D) by process (E), (D) must be Propan-2-ol (Isopropyl alcohol):

D = Propan-2-ol

(1) Structural Formulas:

• (B) Sodium benzoate: C₆H₅–COO⁻Na⁺
• (D) Propan-2-ol: CH₃–CH(OH)–CH₃

(2) Chemical Processes:

• Process (E): Reduction (Catalytic Hydrogenation) — Acetone → Propan-2-ol
• Process (C): Conversion of Sodium benzoate (B) → Propan-2-ol (D) involves multiple steps; overall the process can be classified as dry distillation then halogenation then hydrolysis.

1. Concentrated Sulfuric Acid (H₂SO₄): Will cause effervescence (CO₂) with A, brown fumes (NO₂) with B, acidic fumes (HCl) with D, and no reaction with C.
2. FeSO₄ (Solution C): Forms a greenish-white precipitate of Fe(OH)₂ which does not dissolve in excess NaOH. (Al³⁺ in B forms a precipitate that dissolves in excess).
3. Ca(HCO₃)₂ (Solution A): Reacting with dil. H₂SO₄ yields CO₂ gas and sparingly soluble CaSO₄ (precipitate).
4. KCl (Solution D): Potassium (K⁺) salts are highly soluble; detection usually requires a physical flame test (violet flame) rather than standard chemical precipitation.

Chemistry Breakdown for 46:
Sucrose (A) hydrolysis yields Glucose and Fructose (isomers). Fermentation of Glucose (B) yields Ethanol(D).
Ethanol (D) + H₂SO₄ at 180°C → Ethene (Y).
Ethanol (D) + H₂SO₄ at 140°C → Diethyl ether (X).

Analysis

Salt X reacts with HCl to release gas Y. Gas Y passes into ZCl₂ solution → forms black precipitate W.

Black precipitate + ZCl₂ suggests metal sulfides (PbS, CuS are black).

So gas Y = H₂S, and salt X is a sulfide (e.g., Na₂S, FeS).

Answer (1):

(a) Salt ZCl₂: PbCl₂ (or CuCl₂)

(b) Black precipitate W: PbS (or CuS)

Answer (2): Two Other Nitrate Salts

• Pb(NO₃)₂ — Pb²⁺ + H₂S → PbS↓ (black)

• Cu(NO₃)₂ — Cu²⁺ + H₂S → CuS↓ (black)

Part (a): 1 mol compound + 2 mol Br₂ → saturated

Requires a triple bond: D: CH≡CH (acetylene)

CH≡CH + 2Br₂ → CHBr₂-CHBr₂

IUPAC name: 1,1,2,2-Tetrabromoethane

Part (b): HBr + compound (Markovnikov)

Using C: CH₃-CH=CH₂ (propene)

CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (H adds to C with more H; Br to C with less H)

IUPAC name: 2-Bromopropane

Part (c): 2 mol compound + 2 mol Cl₂ → two mono-chloro isomers

Using F: Toluene with Cl₂/FeCl₃ (substitution on ring, ortho/para directing by -CH₃)

Products (two isomers):

• 2-Chloromethyl benzene (ortho)

• 4-Chloromethyl benzene (para)

Or alternatively named: 2-Chloro toluene and 4-Chloro toluene

(a) Chemical formulas of C and D:

• (A) = Fe₂O₃ (red oxide of iron)
• (B) = Fe (iron metal, from reduction of Fe₂O₃ at 800°C)
• (C) = FeCl₃ (iron reacts with X = Cl₂ gas/Δ → FeCl₃)
• (D) = Fe₂(SO₄)₃ (Fe₂O₃ + conc. H₂SO₄ → Fe₂(SO₄)₃ + H₂O)
• (Z) = NaOH (added to D to give Fe(OH)₃)

(b) Intermetallic alloy for (B) = Fe: Steel (Fe + C) or Stainless steel (Fe + Cr + Ni)

(c) Name of (C) when X_(g) replaced by dilute HCl_(aq):
Fe + 2HCl_(dilute) → FeCl₂ + H₂↑ → Compound = Iron(II) chloride (Ferrous chloride)

Identification:

• (Y) = Simplest tertiary alcohol = 2-methylpropan-2-ol [(CH₃)₃COH]
• (X) = Alkene that gives Y by catalytic hydration = 2-methylpropene [(CH₃)₂C=CH₂]
• (W) = HCl (hydrogen chloride)
• (Z) = Product = 2-chloro-2-methylpropane [(CH₃)₃CCl]

IUPAC Names:

• X: 2-methylpropene (isobutylene)
• Z: 2-chloro-2-methylpropane (tert-butyl chloride)

Structural formulas of saturated isomers of X (C₄H₈):

Analysis

• Salt (X): FeS — Iron(II) sulfide
• Salt (Y): FeCl₂ — Iron(II) chloride
• Precipitate (A): CuS — Copper(II) sulfide (black)
• Precipitate (B): Fe(OH)₂ — Iron(II) hydroxide (white, turns greenish in air)

Identifying the Compounds

Answers

• 1. Excess HBr to Z (CH≡CH) → CH₃CHBr₂ (1,1-dibromoethane). Its isomer = 1,2-dibromoethane.
• 2. Reduction of X (CH₃COOH) → Ethanol.
• 3. Y (Sodium acetate): CH₃COONa   |   L (Ethanal): CH₃CHO

Identification: A=FeCO₃, C=FeO, D=Fe₂O₃, E=FeSO₄, F=Fe₂(SO₄)₃, B=Fe(OH)₃.

(1) Compounds giving Fe₂O₃ on heating: Iron(III) hydroxide and hydrated iron(III) oxide.

(2) Fe₂O₃ + conc. HCl → Iron(III) chloride (ferric chloride).

(3) Formula of F = Fe₂(SO₄)₃.

• Y = acetic acid (rayon) = CH₃COOH
• X = ethanol = C₂H₅OH
• Z = methyl acetate = CH₃COOCH₃
• L = ethyl methanoate = HCOOC₂H₅

Z and L are isomers, both C₃H₆O₂.