Chemistry class 12 Full Syllabus Mock Test 5

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CBSE Class 12 Chemistry: FULL SYLLABUS MOCK TEST - 5

Time Allowed: 3 Hours Maximum Marks: 70

General Instructions:

1. This question paper contains 33 questions. All questions are compulsory.
2. Section A: Q. No. 1 to 16 are objective type questions carrying 1 mark each.
3. Section B: Q. No. 17 to 21 are very short answer questions carrying 2 marks each.
4. Section C: Q. No. 22 to 28 are short answer questions carrying 3 marks each.
5. Section D: Q. No. 29 to 30 are case-based questions carrying 4 marks each.
6. Section E: Q. No. 31 to 33 are long answer questions carrying 5 marks each.
7. Use of calculators is not allowed. Log tables may be used.

Section A (1 Mark Each)

[1] 1. Which of the following conditions is perfectly satisfied by an ideal solution?

• ΔH_mix > 0
• ΔV_mix < 0
• ΔS_mix = 0
• ΔH_mix = 0

[1] 2. How many Coulombs of charge are required for the reduction of 1 mole of MnO₄^- to Mn²⁺?

• 96500 C
• 193000 C
• 482500 C
• 289500 C

[1] 3. The plot of concentration of the reactant versus time for a reaction is a straight line with a negative slope. The reaction follows:

• Zero-order kinetics
• First-order kinetics
• Second-order kinetics
• Pseudo-first-order kinetics

[1] 4. Which of the following elements has the maximum number of unpaired d-electrons in its ground state?

• Iron (Fe)
• Chromium (Cr)
• Zinc (Zn)
• Copper (Cu)

[1] 5. In the complex [Co(en)₃]³⁺, the coordination number and oxidation state of Cobalt are respectively:

• 3 and +3
• 6 and +3
• 6 and +6
• 3 and 0

[1] 6. Chloroform, when slowly exposed to air and light, gets oxidized to form a highly poisonous gas known as:

• Mustard gas
• Phosgene
• Chloropicrin
• Tear gas

[1] 7. Phenol, when treated with bromine in CS₂ at low temperature, mainly yields:

• m-Bromophenol
• p-Bromophenol
• 2,4,6-Tribromophenol
• o-Bromophenol and p-Bromophenol mixture

[1] 8. The reaction of benzaldehyde with concentrated NaOH solution gives benzyl alcohol and sodium benzoate. This reaction is known as:

• Aldol condensation
• Cannizzaro reaction
• Cross-aldol condensation
• Rosenmund reduction

[1] 9. Primary amines on treating with nitrous acid (HNO₂) generally yield:

• Secondary amines
• Alcohols and Nitrogen gas
• Alkenes
• Amides

[1] 10. The nitrogenous base Thymine is found specifically in:

• RNA only
• DNA only
• Both DNA and RNA
• Proteins

[1] 11. Which of the following is NOT an essential amino acid?

• Valine
• Leucine
• Glycine
• Histidine

[1] 12. What is the product formed when an alkyl halide reacts with Silver Nitrite (AgNO₂)?

• Alkyl nitrite
• Nitroalkane
• Alkene
• Alcohol

Directions for Q13 to Q16: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:

• Both A and R are true and R is the correct explanation of A.
• Both A and R are true but R is not the correct explanation of A.
• A is true but R is false.
• A is false but R is true.

[1] 13. Assertion (A): The boiling points of isomeric haloalkanes decrease with branching.
Reason (R): Branching makes the molecule more spherical, resulting in a decrease in surface area and weaker van der Waals forces.

[1] 14. Assertion (A): Separation of Zirconium (Zr) and Hafnium (Hf) is extremely difficult.
Reason (R): Zirconium and Hafnium have nearly identical atomic radii due to the Lanthanoid contraction.

[1] 15. Assertion (A): In the electrochemical cell Zn | Zn²⁺ || Cu²⁺ | Cu, the mass of the Zinc electrode decreases over time.
Reason (R): Zinc undergoes reduction at the anode.

[1] 16. Assertion (A): Glucose forms a pentaacetate derivative on reacting with acetic anhydride.
Reason (R): The formation of pentaacetate confirms the presence of five hydroxyl (-OH) groups in glucose.

Section B (2 Marks Each)

[2] 17. State Kohlrausch's law of independent migration of ions. Write its mathematical expression for Barium Chloride (BaCl₂).

[2] 18. What is meant by the 'rate constant' (k) of a reaction? State the factors on which the rate constant of a reaction depends.

[2] 19. Write the IUPAC name and indicate the coordination number of the central metal ion in the complex: [Co(NH₃)₄(H₂O)Cl]Cl₂.

[2] 20. Predict the major products of the following reactions:
(i) CH₃CH=CH₂ + HBr →
(ii) CH₃CH₂Br + AgCN →

[2] 21. Differentiate between fibrous proteins and globular proteins on the basis of their structure and solubility. Give one example of each.

Section C (3 Marks Each)

[3] 22. A 5% (by mass) solution of cane sugar (Molar mass = 342 g mol^-1) in water has a freezing point of 271 K. Calculate the freezing point of a 5% (by mass) solution of glucose (Molar mass = 180 g mol^-1) in water. (Freezing point of pure water is 273.15 K).

[3] 23. Explain the following observations:
(i) Transition metals and their many compounds act as good catalysts.
(ii) Cr²⁺ is highly reducing in nature while Mn³⁺ is strongly oxidizing, despite both having a d⁴ configuration.
(iii) The enthalpies of atomization of transition metals are high.

[3] 24. How are the following conversions carried out?
(i) Phenol to Picric acid
(ii) Propene to Propan-1-ol
(iii) Anisole to p-Bromoanisole

[3] 25. Give chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Phenol and Benzoic acid
(iii) Primary (1°) and Secondary (2°) amines

[3] 26. Using Valence Bond Theory, discuss the hybridization, geometry, and magnetic behaviour of the complex [FeF₆]^3-. (Atomic number of Fe = 26)

[3] 27. Write the chemical equations for the following name reactions:
(i) Reimer-Tiemann reaction
(ii) Hoffmann bromamide degradation reaction
(iii) Sandmeyer's reaction

[3] 28. Define the following terms related to biomolecules:
(i) Zwitterion
(ii) Peptide linkage
(iii) Nucleotide

Section D: Case-Based (4 Marks Each)

[4] 29. Read the passage and answer the questions:

Liquid-liquid solutions can be classified as ideal and non-ideal on the basis of Raoult's law. Ideal solutions obey Raoult's law over the entire range of concentration, whereas non-ideal solutions deviate from it. A solution showing a large positive deviation forms a minimum boiling azeotrope at a specific composition, while a solution showing a large negative deviation forms a maximum boiling azeotrope. An azeotrope is a binary mixture having the same composition in the liquid and vapour phases and boiling at a constant temperature.

(a) What are the values of ΔH_mix and ΔV_mix for an ideal solution? [1 Mark]
(b) Give one example of a liquid mixture that forms a maximum boiling azeotrope. [1 Mark]
(c) Why does a mixture of chloroform and acetone show a negative deviation from Raoult's law? Explain with reference to intermolecular forces. [2 Marks]

[4] 30. Read the passage and answer the questions:

Standard electrode potentials are used to determine the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K_c) of a cell reaction. The Nernst equation allows the calculation of the cell potential under non-standard conditions. The relationship between the free energy change and the EMF of a cell provides insight into the spontaneity of the electrochemical reaction. If the cell potential is positive, the reaction is spontaneous.

(a) State the relationship between standard cell potential (E°_cell) and the standard Gibbs free energy change (ΔG°). [1 Mark]
(b) What is the value of E_cell when an electrochemical cell reaches equilibrium? [1 Mark]
(c) A cell is constructed using Zn/Zn²⁺ and Cu/Cu²⁺ half-cells. Calculate the standard Gibbs free energy change (ΔG°) for the reaction. (Given E°_Zn²⁺/Zn = -0.76 V, E°_Cu²⁺/Cu = +0.34 V, F = 96500 C mol^-1). [2 Marks]

Section E: Long Answer (5 Marks Each)

[5] 31. (a) Derive the integrated rate equation for a first-order reaction. [2 Marks]
(b) A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period (t_1/2).
(Given: log 10 = 1, log 7 = 0.845) [3 Marks]

[5] 32. An aromatic compound 'A' (C₇H₆O₂) on heating with aqueous ammonia forms compound 'B'. Compound 'B' on heating with Br₂ and KOH forms compound 'C' having the molecular formula C₆H₇N. Compound 'C' on reacting with NaNO₂ and HCl at 0-5°C forms an unstable compound 'D'. Compound 'D' on treatment with Cu₂Cl₂ and HCl yields compound 'E'.
(a) Identify the organic compounds A, B, C, D, and E. [2.5 Marks]
(b) Write the complete chemical equations for the following sequence of reactions:
    (i) Conversion of B to C.
    (ii) Conversion of C to D.
    (iii) Conversion of D to E. [2.5 Marks]

[5] 33. (a) Explain the mechanism of the base-catalyzed nucleophilic addition of Hydrogen Cyanide (HCN) to a carbonyl group. [2 Marks]
(b) Write the chemical equations for the following transformations:
    (i) Aldol condensation of Ethanal.
    (ii) Cannizzaro reaction of Formaldehyde.
    (iii) Clemmensen reduction of Propanone. [3 Marks]

ChemCA.in

SOLUTIONS & MARKING SCHEME

1. Answer: (d) ΔH_mix = 0

Reasoning: For an ideal solution, there is no change in enthalpy (ΔH_mix = 0) and no change in volume (ΔV_mix = 0) upon mixing, because the solute-solvent interactions are identical to the solute-solute and solvent-solvent interactions.

2. Answer: (c) 482500 C

Reasoning: The oxidation state of Mn in MnO₄^- is +7. To reduce it to Mn²⁺, 5 moles of electrons are required per mole of MnO₄^-. Charge = nF = 5 × 96500 C = 482500 C.

3. Answer: (a) Zero-order kinetics

Reasoning: For a zero-order reaction, the integrated rate equation is [R] = -kt + [R]₀. A plot of [R] versus time (t) yields a straight line with a negative slope (-k) and an intercept of [R]₀.

4. Answer: (b) Chromium (Cr)

Reasoning: Chromium (Z=24) has an exceptional ground state electronic configuration of [Ar] 3d⁵ 4s¹. It possesses a total of 6 unpaired electrons (5 in 3d, 1 in 4s), which is the maximum among the given 3d series elements.

5. Answer: (b) 6 and +3

Reasoning: Ethylenediamine (en) is a neutral, bidentate ligand. Three 'en' molecules form 6 coordinate bonds, so the coordination number is 6. Since the complex carries a +3 charge and 'en' is neutral, the oxidation state of Cobalt is +3.

6. Answer: (b) Phosgene

Reasoning: Chloroform (CHCl₃) is slowly oxidized by oxygen in the presence of light to form an extremely poisonous gas called carbonyl chloride (COCl₂), commonly known as Phosgene.

7. Answer: (d) o-Bromophenol and p-Bromophenol mixture

Reasoning: When phenol reacts with bromine in a non-polar solvent like Carbon disulphide (CS₂) at low temperature, mono-bromination occurs due to decreased polarization, yielding a mixture of ortho and para-bromophenol (para is the major product).

8. Answer: (b) Cannizzaro reaction

Reasoning: Benzaldehyde lacks an alpha-hydrogen atom. Upon treatment with concentrated alkali, it undergoes a self-oxidation-reduction (disproportionation) reaction to form benzyl alcohol (reduction product) and sodium benzoate (oxidation product). This is the Cannizzaro reaction.

9. Answer: (b) Alcohols and Nitrogen gas

Reasoning: Aliphatic primary amines react with nitrous acid (HNO₂, generated in situ from NaNO₂ + HCl) to form highly unstable aliphatic diazonium salts. These immediately decompose in water to yield primary alcohols, accompanied by the brisk evolution of nitrogen gas.

10. Answer: (b) DNA only

Reasoning: The four bases in DNA are Adenine, Guanine, Cytosine, and Thymine. In RNA, Thymine is replaced by Uracil. Therefore, Thymine is found exclusively in DNA.

11. Answer: (c) Glycine

Reasoning: Non-essential amino acids can be synthesized by the human body. Glycine is the simplest amino acid and can be readily synthesized in the body, making it a non-essential amino acid. Valine, Leucine, and Histidine must be obtained through diet.

12. Answer: (b) Nitroalkane

Reasoning: The nitrite ion (NO₂^-) is an ambidentate nucleophile. Silver nitrite (AgNO₂) is predominantly covalent, so the attack occurs through the lone pair on the Nitrogen atom, resulting in the formation of a nitroalkane (R-NO₂). (KNO₂ would yield an alkyl nitrite, R-O-N=O).

13. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: Boiling points decrease with increased branching because the molecule assumes a more compact, spherical shape. This minimizes the surface area of contact between molecules, weakening the intermolecular van der Waals dispersion forces.

14. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: Zirconium (Period 5) and Hafnium (Period 6) lie in the same group. Due to the poor shielding of the completely filled 4f orbitals (Lanthanoid Contraction), the expected size increase down the group is canceled out, making their sizes almost identical and their separation difficult.

15. Answer: (c) A is true but R is false.

Reasoning: The mass of the Zinc electrode does decrease because Zinc metal dissolves into the solution as Zn²⁺ ions. However, this process is Oxidation (loss of electrons at the anode), not reduction.

16. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: When glucose reacts with acetic anhydride, it undergoes acetylation to form glucose pentaacetate. This reaction specifically targets hydroxyl groups, confirming that there are exactly five -OH groups attached to different carbon atoms in the glucose molecule.

17. Solution:

Kohlrausch's Law: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual limiting molar conductivities of its constituent cations and anions.
Mathematical expression for BaCl₂:
Λ°_m(BaCl₂) = λ°(Ba²⁺) + 2λ°(Cl^-)

18. Solution:

Rate Constant (k): It is defined as the rate of the reaction when the concentration of all reactants is unity (1 mol/L). It is a measure of the reaction's speed.
Factors: The rate constant of a reaction depends primarily on the Temperature (increases with T) and the presence of a Catalyst. It is independent of the initial concentration of reactants.

19. Solution:

IUPAC Name: Tetraammineaquachloridocobalt(III) chloride.
(Alphabetical order of ligands: ammine, aqua, chlorido. Co is in +3 state: x + 4(0) + 0 - 1 = +2 ⇒ x = +3).
Coordination Number: 6 (Four NH₃, one H₂O, one Cl^-).

20. Solution:

(i) CH₃CH=CH₂ + HBr → CH₃-CH(Br)-CH₃ (2-Bromopropane). [Markovnikov's addition].
(ii) CH₃CH₂Br + AgCN → CH₃CH₂-NC (Ethyl isocyanide) + AgBr. [AgCN is covalent, attack via Nitrogen].

21. Solution:

Fibrous Proteins: Polypeptide chains run parallel to each other to form fiber-like structures. They are generally insoluble in water. Example: Keratin (hair, wool) or Myosin.
Globular Proteins: Polypeptide chains fold around each other to give a spherical, compact shape. They are generally soluble in water. Example: Insulin or Albumin.

22. Solution:

Given: For cane sugar (W₂ = 5g, M₂ = 342 g mol^-1, W₁ = 95g). Freezing point = 271 K. ΔT_f = 273.15 - 271 = 2.15 K.
Using ΔT_f = K_f × (W₂ / M₂) × (1000 / W₁), we find K_f.
2.15 = K_f × (5 / 342) × (1000 / 95)
K_f = (2.15 × 342 × 95) / 5000 = 69853.5 / 5000 = 13.97 K kg mol^-1.

For Glucose (W₂ = 5g, M₂ = 180 g mol^-1, W₁ = 95g).
ΔT_f' = K_f × (5 / 180) × (1000 / 95)
ΔT_f' = 13.97 × (5000 / 17100) = 13.97 × 0.292 = 4.08 K.
Freezing point of glucose solution = 273.15 - 4.08 = 269.07 K.

23. Solution:

(i) Transition metals possess multiple oxidation states (forming unstable intermediates) and offer a large surface area for reactants to adsorb, lowering the activation energy.
(ii) Cr²⁺ (3d⁴) is reducing because it oxidizes to Cr³⁺ (3d³), which has a highly stable exactly half-filled t_2g orbital set in an octahedral field. Mn³⁺ (3d⁴) is oxidizing because it reduces to Mn²⁺ (3d⁵), which is a highly stable exactly half-filled d-subshell.
(iii) Transition metals have a large number of unpaired d-electrons, resulting in exceptionally strong interatomic metallic bonding, which requires a vast amount of energy to break the metal lattice.

24. Solution:

(i) Phenol to Picric acid: Reaction with concentrated acids.
Phenol + Conc. HNO₃ / Conc. H₂SO₄ → 2,4,6-Trinitrophenol (Picric acid) + 3H₂O.
(ii) Propene to Propan-1-ol: Anti-Markovnikov addition via Hydroboration-Oxidation.
CH₃CH=CH₂ + (BH₃)₂ → (CH₃CH₂CH₂)₃B
(CH₃CH₂CH₂)₃B + 3H₂O₂ / OH^- → 3 CH₃CH₂CH₂OH + B(OH)₃.
(iii) Anisole to p-Bromoanisole: Electrophilic bromination.
C₆H₅OCH₃ + Br₂ in ethanoic acid → p-Bromoanisole (Major) + o-Bromoanisole (Minor).

25. Solution:

(i) Propanal and Propanone: Tollens' test. Propanal gives a silver mirror on warming with Tollens' reagent; propanone does not.
(ii) Phenol and Benzoic acid: NaHCO₃ test. Benzoic acid gives brisk effervescence of CO₂ gas with aqueous NaHCO₃; phenol does not.
(iii) 1° and 2° amines: Carbylamine test. 1° amine produces a foul-smelling isocyanide when heated with CHCl₃ and alc. KOH; 2° amine does not.

26. Solution:

Complex: [FeF₆]^3-. Oxidation state of Fe = +3.
Fe³⁺ configuration = [Ar] 3d⁵.
F^- is a weak field ligand, so pairing of electrons does not occur against Hund's rule.
The 5 electrons occupy the five 3d orbitals singly: ↑ | ↑ | ↑ | ↑ | ↑ .
It utilizes the outer 4d orbitals (one 4s, three 4p, two 4d) for bonding, resulting in sp³d² hybridization.
Geometry: Octahedral.
Magnetic Behaviour: Highly Paramagnetic (High Spin complex due to 5 unpaired electrons).

27. Solution:

(i) Reimer-Tiemann reaction: Phenol reacts with CHCl₃ + aq. NaOH at 340 K to form Salicylaldehyde.
(ii) Hoffmann bromamide degradation: Amide reacts with Br₂ and aqueous/alcoholic NaOH to yield a primary amine with one carbon less. (R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O).
(iii) Sandmeyer's reaction: Benzene diazonium chloride reacts with Cu₂Cl₂/HCl to form Chlorobenzene.

28. Solution:

(i) Zwitterion: A dipolar, neutral ion formed in aqueous solution when the carboxyl group (-COOH) of an amino acid donates a proton to its own amino group (-NH₂).
(ii) Peptide linkage: An amide bond (-CO-NH-) formed by the condensation of the carboxyl group of one amino acid with the amino group of another, eliminating water.
(iii) Nucleotide: The fundamental repeating unit of nucleic acids (DNA/RNA), consisting of a nitrogenous base, a pentose sugar, and a phosphate group.

29. Solution (Case-Based Solutions):

(a) For an ideal solution, the enthalpy of mixing (ΔH_mix = 0) and the volume of mixing (ΔV_mix = 0).
(b) Maximum boiling azeotropes are formed by solutions showing large negative deviations from Raoult's law. Example: A mixture of 68% Nitric acid (HNO₃) and 32% Water by mass.
(c) Chloroform and acetone show negative deviation because the chloroform molecule forms a new, strong intermolecular hydrogen bond with the oxygen atom of the acetone molecule. Since the A-B interactions are stronger than the A-A and B-B interactions, the escaping tendency of the molecules decreases, lowering the vapour pressure below that expected from Raoult's law.

30. Solution (Case-Based Electrochemistry):

(a) Standard Gibbs free energy change (ΔG°) is related to the standard cell potential (E°_cell) by the equation: ΔG° = -nFE°_cell.
(b) When an electrochemical cell reaches equilibrium, the reaction quotient equals the equilibrium constant, and the cell can do no more electrical work. Therefore, E_cell = 0 V.
(c) Numerical Calculation:
Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu. (n = 2).
E°_cell = E°_Cathode - E°_Anode = 0.34 V - (-0.76 V) = 1.10 V.
ΔG° = -nFE°_cell = -2 × 96500 C mol^-1 × 1.10 V = -212300 J mol^-1.
ΔG° = -212.3 kJ mol^-1.

31. Solution:

(a) Derivation for First Order:
For reaction R → Products, Rate = -d[R]/dt = k[R]¹.
Rearranging: d[R] / [R] = -k dt. Integrating both sides:
∫ (1/[R]) d[R] = -k ∫ dt ⇒ ln[R] = -kt + I.
At t=0, [R] = [R]₀, so I = ln[R]₀.
ln[R] = -kt + ln[R]₀ ⇒ kt = ln([R]₀/[R]).
Converting to log base 10: k = (2.303 / t) log([R]₀ / [R]).

(b) Numerical:
Given 30% decomposition. [R]₀ = 100, [R] = 70. t = 40 min.
k = (2.303 / t) log([R]₀ / [R]) = (2.303 / 40) log(100 / 70)
k = (2.303 / 40) × [log(10) - log(7)] = (2.303 / 40) × (1 - 0.845)
k = (2.303 / 40) × 0.155 = 0.00892 min^-1.
t_1/2 = 0.693 / k = 0.693 / 0.00892 = 77.7 minutes.

32. Solution:

(a) Identification:
- A (C₇H₆O₂) on heating with NH₃ forms B. This indicates A is an aromatic acid, and B is an amide. A is Benzoic Acid (C₆H₅COOH). B is Benzamide (C₆H₅CONH₂).
- B (Amide) + Br₂/KOH is Hoffmann degradation, forming C (a 1° amine). C is Aniline (C₆H₅NH₂).
- C (Aniline) + NaNO₂/HCl (0-5°C) is diazotization, forming unstable D. D is Benzene diazonium chloride (C₆H₅N₂⁺Cl^-).
- D + Cu₂Cl₂/HCl is Sandmeyer's reaction, forming E. E is Chlorobenzene (C₆H₅Cl).

(b) Chemical Equations:
(i) B to C: C₆H₅CONH₂ + Br₂ + 4KOH → C₆H₅NH₂ + K₂CO₃ + 2KBr + 2H₂O.
(ii) C to D: C₆H₅NH₂ + NaNO₂ + 2HCl (0-5°C) → C₆H₅N₂⁺Cl^- + NaCl + 2H₂O.
(iii) D to E: C₆H₅N₂⁺Cl^- + Cu₂Cl₂/HCl → C₆H₅Cl + N₂↑.

33. Solution:

(a) Nucleophilic addition of HCN:
Reaction with pure HCN is very slow. It is catalyzed by a base, which removes a proton from HCN to generate the much stronger cyanide nucleophile (CN^-).
Step 1: HCN + OH^- ⇔ H₂O + CN^- (Nucleophile)
Step 2: The strong nucleophile (CN^-) attacks the electrophilic carbonyl carbon to form a tetrahedral alkoxide intermediate.
>C=O + CN^- ⇔ >C(O^-)CN
Step 3: The intermediate rapidly abstracts a proton from water or HCN to form cyanohydrin.
>C(O^-)CN + H₂O ⇔ >C(OH)CN (Cyanohydrin) + OH^-.

(b) Chemical Equations:
(i) Aldol Condensation: 2 CH₃CHO →(dil. NaOH) CH₃CH(OH)CH₂CHO →(Δ, -H₂O) CH₃CH=CHCHO.
(ii) Cannizzaro reaction: 2 HCHO + Conc. NaOH → CH₃OH (Methanol) + HCOONa (Sod. formate).
(iii) Clemmensen reduction: CH₃COCH₃ + 4[H] →(Zn-Hg / Conc. HCl) CH₃CH₂CH₃ (Propane) + H₂O.

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