Answer

Lassaigne’s Test – Principle

Lassaigne fusion involves converting heteroatoms (such as nitrogen, sulfur, and halogens) in an organic compound into water-soluble sodium salts, which enables their qualitative detection. The typical conversions are:

• NaCN for carbon and nitrogen (C + N)
• Na₂S for sulfur (S)
• NaX (where X = Cl, Br, I) for halogens

These salts are then tested in aqueous solution to identify the corresponding elements.

Note: The CuO–carbon reduction is a metallurgical reaction and not related to the sodium fusion method. Therefore, it is excluded from Lassaigne’s test.

Answer

First-Order Kinetics Calculation

For a first-order reaction, the time t required for the concentration of a reactant to change from [R]₀ to [R] is given by the integrated rate law:

t = (2.303 / k) × log([R]₀ / [R])

Given:

• Rate constant, k = 0.03 s⁻¹
• Initial concentration, [R]₀ = 7.2 mol·L⁻¹
• Final concentration, [R] = 0.9 mol·L⁻¹

Step 1: Calculate the ratio

[R]₀ / [R] = 7.2 / 0.9 = 8

Step 2: Calculate the logarithm

log(8) = log(2³) = 3 × log(2) = 3 × 0.301 ≈ 0.903

Step 3: Plug into the equation

t = (2.303 / 0.03) × 0.903 ≈ 76.77 × 0.903 ≈ 69.3 s

Conclusion: The time required for the concentration to drop from 7.2 mol·L⁻¹ to 0.9 mol·L⁻¹ is approximately 69.3 seconds.

Answer

Thermodynamic Behavior and Equilibrium Shift in NO Formation

The reaction is strongly endothermic (ΔH = +180.7 kJ·mol⁻¹), so raising the temperature shifts equilibrium toward NO.

Le Chatelier’s Principle also predicts that increasing the concentration of either reactant (N₂ or O₂) drives the equilibrium forward to re-establish balance.

Lower temperature would favor the exothermic reverse step, resulting in less formation of NO.

Answer

Molecular Property Analysis

• Dipole Moments:
  H₂O (1.85 D) > NH₃ (1.47 D) > CHCl₃ (≈1.04 D)
  ✔ Statement A is true.
• Lone-Pair Counts:
  XeF₂ (3) > XeF₄ (2) > XeO₃ (1)
  ✘ Statement B is false.
• Bond Lengths:
  O–H (96 pm) < C–H (110 pm) < N–O (136 pm)
  ✘ Statement C is incorrect.
• Bond Enthalpies:
  N≡N (946 kJ·mol⁻¹) > O=O (498 kJ·mol⁻¹) > H–H (436 kJ·mol⁻¹)
  ✔ Statement D is true.

Answer

Cyclic Ether Skeletons for C₄H₈O

Cyclic ethers are ring structures containing an oxygen atom. For the molecular formula C₄H₈O, several ring-based isomers are possible. Below is a detailed description of each structure and its stereochemical variants:

Answer

Application of Hess’s Law to Calculate Enthalpy of Formation

Consider the precipitation reaction:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

This reaction forms solid barium sulfate from its aqueous ions. The associated enthalpy change is the enthalpy of crystallization:

ΔH° = ΔH°_crys. = −4.5 kcal mol⁻¹

Using Hess’s Law

We want to find the standard enthalpy of formation for Ba²⁺(aq):

ΔH°_f[BaSO₄(s)] = ΔH°_f[Ba²⁺(aq)] + ΔH°_f[SO₄²⁻(aq)] + ΔH°_crys.

Substituting the known values:

• ΔH°_f[BaSO₄(s)] = −349 kcal mol⁻¹
• ΔH°_f[SO₄²⁻(aq)] = −216 kcal mol⁻¹
• ΔH°_crys. = −4.5 kcal mol⁻¹

−349 = ΔH°_f[Ba²⁺(aq)] − 220.5

⇒ ΔH°_f[Ba²⁺(aq)] = −128.5 kcal mol⁻¹

Conclusion

The standard enthalpy of formation of Ba²⁺(aq) is:

ΔH°_f[Ba²⁺(aq)] = −128.5 kcal mol⁻¹

This example illustrates how Hess’s Law is applied to calculate unknown thermodynamic values using known enthalpy data.

Answer

SN2 Reaction and the Role of Leaving Groups

The rate of an SN2 reaction depends significantly on the tendency of the leaving group to depart.

• 1-Iodobutane reacts faster than 1-chlorobutane in SN2 reactions.

Reason: Superior Leaving Group Ability of Iodide

• Iodide ion (I⁻) is a larger, more polarizable anion compared to chloride (Cl⁻).
• Its size and polarizability allow it to disperse negative charge more effectively, making it a more stable ion in solution.
• As a result, iodide is a significantly superior leaving group than chloride.

Conclusion

In SN2 mechanisms, the stability and charge distribution of the leaving group are critical. Since iodide ion disperses charge better, it enhances reaction rate, making 1-iodobutane more reactive than 1-chlorobutane.

Answer

Selective Reduction of Esters to Aldehydes

Reaction Overview:

Reagent Specificity:

• DIBAL-H (Diisobutylaluminium hydride): Selectively reduces esters to aldehydes under controlled conditions. It stops at the aldehyde stage, avoiding further reduction to alcohols.
• LiAlH₄ (Lithium Aluminium Hydride): Too strong; reduces esters all the way to primary alcohols.
• NaBH₄ (Sodium Borohydride): Too mild; ineffective for esters.
• Rosenmund Reduction (H₂/Pd-BaSO₄): Works only on acid chlorides, not esters.

Conclusion:

To reduce esters to aldehydes effectively and selectively:

• Use DIBAL-H at low temperatures (typically –78°C), followed by aqueous work-up.
• Avoid LiAlH₄ (over-reduction) and NaBH₄ (no reduction).
• Rosenmund reduction is suitable for acid chlorides only.

Answer

Composition and Properties of Natural Honey

• Natural honey contains equal amounts of D-glucose and D-fructose.
• D-fructose is a ketohexose, which means it is a six-carbon sugar with a ketone group (a keto-sugar).
• In aqueous solution, D-fructose undergoes cyclization to form:
  • Fructofuranose (5-membered ring)
  • Fructopyranose (6-membered ring)
• These cyclic forms exist as anomeric pairs:
  • α (alpha) anomer
  • β (beta) anomer
• The specific rotation of D-fructose is –92°, making it laevorotatory (rotates plane-polarized light to the left).

Key Terms

• Ketohexose: A monosaccharide with six carbon atoms and a ketone group.
• Anomers: Isomers that differ in configuration around the anomeric carbon in a cyclic sugar.
• Laevorotatory: A compound that rotates plane-polarized light counterclockwise (to the left).

Answer

🧪 pH Scale and Substance Classification

pH = -log[H⁺]