FYUG Even Semester Exam, 2025
UNIT-I
1. (a) What is inorganic benzene? Write its structure. 2
Borazine (B3N3H6) is known as inorganic benzene because its structure is isoelectronic and isostructural with benzene
. It consists of a hexagonal ring of alternating Boron and Nitrogen atoms with one Hydrogen atom attached to each.1. (b) Discuss briefly the catenation property of carbon. 2
Catenation is the ability of an element to form covalent bonds with other atoms of the same element, resulting in long chains or rings
. Carbon exhibits the highest catenation property due to its small size and high C-C bond energy, allowing for the vast diversity of organic compounds.1. (c) Write two structural differences between diamond and graphite. 2
| Feature | Diamond | Graphite |
|---|---|---|
| Hybridization | sp³ | sp² |
| Structure | 3D tetrahedral network | 2D hexagonal layered sheets |
2. (a) What happens when... 10
(i) Diborane is hydrolyzed: It reacts with water to form Boric acid and Hydrogen gas
.B2H6 + 6H2O → 2H3BO3 + 6H2
(ii) Diborane is treated with NH3:
- At low temperatures: Forms an unsymmetrical adduct [H2B(NH3)2]+ [BH4]- .
- On heating (1:2 ratio): Forms Borazine (B3N3H6) .
(iii) Calcium carbide is treated with water: It produces Ethyne (Acetylene) and Calcium hydroxide
.CaC2 + 2H2O → C2H2 + Ca(OH)2
(iv) Excess CO2 is passed through Ca(OH)2: The initial milkiness (CaCO3) disappears due to the formation of soluble Calcium bicarbonate
.CaCO3 + H2O + CO2 → Ca(HCO3)2
(v) BF3 is treated with NaBH4 in diglyme: It produces Diborane gas
.3NaBH4 + 4BF3 → 2B2H6 + 3NaBF4
2. (b) (ii) Discuss the bonding in B2H6. 3
Diborane (B2H6) features unique 3-center-2-electron (3c-2e) bonds, often called "banana bonds"
.- There are four terminal B-H bonds which are normal 2-center-2-electron bonds .
- The two bridging Hydrogen atoms are bonded to both Boron atoms, where two electrons are shared over three centers (B-H-B) .
UNIT-II
3. (a) What are open system and closed system? 2
- Open System: A system that can exchange both matter and energy with its surroundings .
- Closed System: A system that can exchange energy (heat or work) but not matter with its surroundings .
3. (c) State the first law of thermodynamics. Write its mathematical form. 2
The First Law of Thermodynamics states that energy can neither be created nor destroyed, only transformed from one form to another
.ΔU = q + w
Where ΔU is the change in internal energy, q is heat added to the system, and w is work done on the system
.4. (b) (i) Derive Kirchhoff's equation. 6
Kirchhoff's equation describes how the enthalpy of a reaction changes with temperature
.For a reaction: ΔH = H(products) - H(reactants)
.Differentiating with respect to Temperature (T) at constant Pressure (P):
d(ΔH)/dT = dH(prod)/dT - dH(react)/dT
.Since (dH/dT)p = Cp (heat capacity at constant pressure):
d(ΔH)/dT = ΔCp
Integrating from T1 to T2:
ΔH2 - ΔH1 = ΔCp(T2 - T1)
UNIT-III
5. (a) Distinguish between ideal and non-ideal solutions. 2
| Ideal Solution | Non-Ideal Solution |
|---|---|
| Obeys Raoult's law at all concentrations. | Does not obey Raoult's law. |
| ΔH_mixing = 0. | ΔH_mixing ≠ 0. |
| ΔV_mixing = 0. | ΔV_mixing ≠ 0. |
5. (c) Define Gibbs' phase rule. Explain terms. 2
F = C - P + 2
- F (Degrees of Freedom): Number of independent variables (P, T, concentration) that must be specified to define the system .
- C (Components): Minimum number of chemically independent constituents .
- P (Phases): Homogeneous, physically distinct part of the system .
6. (b) (i) Draw the phase diagram of water system. 6
The water system is a one-component system with three possible phases: ice, water, and vapor
.- Triple Point: Point where all three phases coexist in equilibrium (F=0) .
- Curves: Represent two phases in equilibrium (F=1) e.g., Vaporization curve .
- Areas: Represent a single phase (F=2) e.g., Liquid region .
UNIT-IV
7. (b) Propene treated with HBr (Peroxide effect). 2
- Absence of Peroxide (Markownikoff's Rule): H attaches to the carbon with more hydrogens. Product: 2-Bromopropane.
- Presence of Peroxide (Anti-Markownikoff's Rule): Br attaches to the carbon with more hydrogens. Product: 1-Bromopropane.
CH3-CH=CH2 + HBr → CH3-CHBr-CH3 (No Peroxide)
CH3-CH=CH2 + HBr (Peroxide) → CH3-CH2-CH2Br
8. (a) (ii) Friedel-Crafts Alkylation and Acylation. 6
Alkylation: Introduction of an alkyl group into the benzene ring using an alkyl halide and AlCl3
.C6H6 + CH3Cl (AlCl3) → C6H5-CH3 + HCl
Acylation: Introduction of an acyl group (RCO-) using an acid chloride and AlCl3
.C6H6 + CH3COCl (AlCl3) → C6H5-COCH3 + HCl
UNIT-V
9. (a) Why does CH3X favour SN2 reaction? 2
Methyl halides (CH3X) favour SN2 because they have the least steric hindrance
. The nucleophile can easily attack the central carbon atom from the backside. SN1 is not favored because the methyl carbocation is highly unstable.10. (a) (i) Discuss SN1 mechanism. 6
SN1 (Unimolecular Nucleophilic Substitution) is a two-step process
:- Formation of Carbocation: The leaving group departs (slowest step) .
- Nucleophilic Attack: The nucleophile attacks the flat carbocation from either side .
Racemic Mixture: Since the carbocation intermediate is planar, the nucleophile can attack from the front or back with equal probability, leading to an equal mixture of two enantiomers (racemization)
.