All 120 questions of the 2020 Combined Geo-Scientist Prelims Paper II (chemistry), each with the answer and a short explanation in the hint-sheet style: the key fact first, numericals worked with units. Tap a number in the key to jump to the question; every question also has its own page.
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Consider the following statements regarding radial and total angular function of H-atom:
The radial function R(r) depends on the quantum number \(n\) and \(l\)
The total angular function depends only on the direction and is independent of the distance from the nucleus (r)
Which of the statements given above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Answer: (c) Both 1 and 2
ψ = R(r)·Y(θ,φ). 1 ✔ The radial part R(r) depends on n and l (e.g. R₂₀ ≠ R₂₁). 2 ✔ The angular part Y(θ,φ) depends only on direction (l, m), not on r. Both correct → (c). Remember: radial nodes = n − l − 1; angular nodes = l.
Which one of the following is the correct electronic configuration of gold (ground state)?
(a) [Xe] 6s² 4f¹⁴ 5d⁹
(b) [Xe] 6s¹ 4f¹⁴ 5d¹⁰
(c) [Xe] 6s² 4f¹³ 5d¹⁰
(d) [Xe] 6s² 4f¹⁴ 5d¹⁰
Answer: (b) [Xe] 6s¹ 4f¹⁴ 5d¹⁰
Au (Z = 79) is like Cu and Ag: a full d¹⁰ is more stable, so one 6s electron moves to 5d. Ground state: [Xe] 4f¹⁴ 5d¹⁰ 6s¹ → (b). Relativistic contraction of 6s in Au supports this.
Which one of the following is the correct order of increasing acidic nature of oxides?
(a) SiO₂ < Al₂O₃ <SO₃ < Cl₂O₇
(b) SiO₂ < SO₃ <Al₂O₃ < Cl₂O₇
(c) Al₂O₃ <SiO₂ < SO₃ < Cl₂O₇
(d) Cl₂O₇ < SiO₂ < SO₃ < Al₂O₃
Answer: (c) Al₂O₃ <SiO₂ < SO₃ < Cl₂O₇
Across period 3, oxides go from basic → amphoteric → acidic as the oxidation state and electronegativity of the element rise: Al₂O₃ (amphoteric) < SiO₂ (weakly acidic) < SO₃ < Cl₂O₇ (strongest, gives HClO₄) → (c).
Which one of the following statements regarding actinoid metals is NOT correct?
(a) The actinoid metals are usually silvery in appearance
(b) Ionization enthalpies of the early actinoids are higher than that of the early lanthanoids
(c) The actinoids are highly reactive metals
(d) The actinoid metals in general show +3 oxidation state
Answer: (b) Ionization enthalpies of the early actinoids are higher than that of the early lanthanoids
(a) ✔ silvery metals. (c) ✔ highly reactive, esp. finely divided. (d) ✔ +3 is common. (b) ✘ 5f electrons are less shielded by inner core and less penetrating than 4f, so early actinoids have lower ionisation enthalpies than early lanthanoids → (b) is NOT correct.
Consider the following statements in respect of alkali metal halides:
The low solubility of LiF in water is due to its high lattice enthalpy
The low solubility of CsI in water is due to smaller hydration enthalpy of its two ions
Which of the statements given above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Answer: (c) Both 1 and 2
Solubility depends on lattice enthalpy vs hydration enthalpy. 1 ✔ LiF: tiny Li⁺ and F⁻ → very high lattice enthalpy → poorly soluble. 2 ✔ CsI: both ions large → low hydration enthalpy → poorly soluble. Both → (c).
Which one of the following is the correct order of thermal stabilities of metal carbonates?
(a) BaCO₃ > CaCO₃ > MgCO₃ > BeCO₃
(b) BaCO₃ > MgCO₃ > CaCO₃ > BeCO₃
(c) CaCO₃ > BaCO₃ > MgCO₃ > BeCO₃
(d) BaCO₃ > BeCO₃ > MgCO₃ > CaCO₃
Answer: (a) BaCO₃ > CaCO₃ > MgCO₃ > BeCO₃
Larger cation → less polarisation of CO₃²⁻ → more stable carbonate. Down group 2 stability rises: BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃ > BeCO₃ → (a). BeCO₃ is so unstable it is kept under CO₂.
Match List-I with List-II and select the correct answer using the code given below the lists:
List-I (Molecule/ion)
List-II (Shape)
A. H₂O
1. Linear
B. BF₃
2. Angular
C. I₃⁻
3. Trigonal planar
D. SF₄
4. See-saw
Code:
(a) A-4, B-1, C-3, D-2
(b) A-2, B-3, C-1, D-4
(c) A-2, B-1, C-3, D-4
(d) A-4, B-3, C-1, D-2
Answer: (b) A-2, B-3, C-1, D-4
A. H₂O: 2 bp + 2 lp → angular (2). B. BF₃: 3 bp → trigonal planar (3). C. I₃⁻: 2 bp + 3 lp → linear (1). D. SF₄: 4 bp + 1 lp → see-saw (4). Code 2-3-1-4 → (b).
How many resonating structures are possible for NO₃⁻ ion?
(a) 4
(b) 5
(c) 6
(d) 3
Answer: (c) 6
NO₃⁻ (24 valence e⁻): N forms one N=O and two N–O⁻. The double bond can sit on any of the 3 O atoms → 3 equivalent major contributors, each N–O bond order 4/3. UPSC's key counts 6, i.e. it also counts the minor charge-separated forms → (c).
Sodium nitroprusside when added to an alkaline solution of sulphide ion gives red coloured solution. The red colouration is due to the formation of:
(a) Na₄[Fe(CN)₅(NOS)]
(b) Na₂[Fe(CN)₅(NO)]
(c) Na₄[FeCl₅(NOS)]
(d) [Fe(H₂O)₅(NO)]SO₄
Answer: (a) Na₄[Fe(CN)₅(NOS)]
Sodium nitroprusside Na₂[Fe(CN)₅NO] + S²⁻ (alkaline) → Na₄[Fe(CN)₅NOS], a violet/purple-red complex. S²⁻ adds to the NO ligand → (a). Remember: same test is used for S in Lassaigne's extract.
Which one of the following complexes will NOT show geometrical isomerism?
(a) [Cr(NH₃)₄Cl₂]Cl
(b) [Co(en)₂Cl₂]Cl
(c) [Co(NH₃)₅NO₂]Cl₂
(d) Pt(NH₃)₂Cl₂
Answer: (c) [Co(NH₃)₅NO₂]Cl₂
MA₄B₂ ([Cr(NH₃)₄Cl₂]⁺), M(AA)₂B₂ ([Co(en)₂Cl₂]⁺) and square planar MA₂B₂ (Pt(NH₃)₂Cl₂) all show cis/trans. [Co(NH₃)₅NO₂]²⁺ is MA₅B: only one arrangement → no geometrical isomers → (c). (It shows linkage isomerism.)
Which one of the following complexes exhibits optical isomerism?
(a) cis-[PtCl₂(NH₃)₂]
(b) trans-[PtCl₂(NH₃)₂]
(c) trans-[Co(en)₂Cl₂]⁺
(d) [Co(EDTA)]⁻
Answer: (d) [Co(EDTA)]⁻
Square planar Pt complexes have a mirror plane → optically inactive. trans-[Co(en)₂Cl₂]⁺ has a plane of symmetry. [Co(EDTA)]⁻: hexadentate ligand wraps the metal in a chiral arrangement (like M(AA)₃) → optically active → (d).
A solution is obtained by mixing 10 ml of 0.2 N KOH solution with 30 ml of 0.1 N CH₃COOH. The pH of the solution would be: (Given: pKa of CH₃COOH = 4.75; log 2 = 0.30)
What will be the hydrogen ion concentration of a solution obtained by mixing 500 ml of 0.20 M acetic acid and 500 ml of 0.30 M sodium acetate? (Dissociation constant of acetic acid = 1.75 × 10⁻⁵)
(a) 2.75 × 10⁻⁵ mol.dm⁻³
(b) 1.17 × 10⁻⁵ mol.dm⁻³
(c) 3.50 × 10⁻⁵ mol.dm⁻³
(d) 1.75 × 10⁻³ mol.dm⁻³
Answer: (b) 1.17 × 10⁻⁵ mol.dm⁻³
After mixing (1 L total): [HA] = 0.10 M, [A⁻] = 0.15 M. [H⁺] = Ka·[HA]/[A⁻] = 1.75 × 10⁻⁵ × 0.10/0.15 = 1.17 × 10⁻⁵ mol dm⁻³ → (b).
Acidity of [M(H₂O)₆]ⁿ⁺ rises with charge density of Mⁿ⁺: Fe²⁺ is weakest. On simple charge/size (Al³⁺ 53.5 pm < Fe³⁺ 64.5 pm) the key ranks Fe²⁺ < Fe³⁺ < Al³⁺ → (b). Note: measured pKa values (Fe³⁺ ≈ 2.2, Al³⁺ ≈ 5.0) actually make Fe³⁺ the stronger acid, i.e. order (a).
The correct order of basicity of the above hydroxides will be:
(a) Sr(OH)₂ < Be(OH)₂ < Ba(OH)₂ < Mg(OH)₂
(b) Be(OH)₂ < Mg(OH)₂ < Ba(OH)₂ < Sr(OH)₂
(c) Mg(OH)₂ < Ba(OH)₂ < Sr(OH)₂ < Be(OH)₂
(d) Be(OH)₂ < Mg(OH)₂ < Sr(OH)₂ < Ba(OH)₂
Answer: (d) Be(OH)₂ < Mg(OH)₂ < Sr(OH)₂ < Ba(OH)₂
Down group 2, ionisation energy falls and M–OH bond becomes more ionic → basicity rises: Be(OH)₂ (amphoteric) < Mg(OH)₂ < Sr(OH)₂ < Ba(OH)₂ → (d). Solubility also rises down the group.
Which one of the following is the correct order of Lewis Acidity in Boron halides?
(a) BF₃ < BI₃ < BBr₃ < BCl₃
(b) BI₃ < BF₃ < BCl₃ < BBr₃
(c) BF₃ < BCl₃ < BBr₃ < BI₃
(d) BBr₃ < BI₃ < BF₃ < BCl₃
Answer: (c) BF₃ < BCl₃ < BBr₃ < BI₃
Expected from electronegativity: BF₃ strongest, but actual order is BF₃ < BCl₃ < BBr₃ < BI₃ → (c). Reason: strong pπ–pπ back-bonding from F to empty B 2p is lost when BF₃ becomes pyramidal on accepting a lone pair; overlap weakens with larger halogens.
What will be the value of mole fraction of the solute in 1.00 molal aqueous solution?
(a) 1.7700
(b) 1.7770
(c) 0.0180
(d) 0.0344
Answer: (c) 0.0180
1 molal = 1 mol solute in 1000 g water (1000/18 = 55.5 mol). x(solute) = 1/(1 + 55.5) = 0.0177 ≈ 0.018 → (c). Trick: mole fraction must be < 1, so (a) and (b) are out at once.
Which one of the following is a primary standard substance?
(a) KMnO₄
(b) K₂Cr₂O₇
(c) Na₂S₂O₃
(d) NaOH
Answer: (b) K₂Cr₂O₇
A primary standard must be pure, stable, non-hygroscopic, of known formula and high molar mass. K₂Cr₂O₇ fits → (b). KMnO₄ (decomposes, MnO₂ impurity), Na₂S₂O₃ (efflorescent, bacterial decay), NaOH (hygroscopic, absorbs CO₂) are secondary standards.
The equivalence point of titration of oxalate solution (acidic medium) with KMnO₄ solution will be:
(a) Pink
(b) Colourless
(c) Orange
(d) Light green
Answer: (b) Colourless
At the equivalence point all MnO₄⁻ added has just been reduced to Mn²⁺ (nearly colourless) and all oxalate is used up → solution is colourless → (b). The end point is one drop later, when excess MnO₄⁻ gives a permanent light pink. Trap: equivalence point ≠ end point.
In which of the following titrations, an external indicator is NOT necessary?
(a) Titration of oxalate solution with KMnO₄ solution
(b) Titration of HCl with NaOH solution
(c) Titration of Hard water with EDTA
(d) Titration of Na₂CO₃ with HCl solution
Answer: (a) Titration of oxalate solution with KMnO₄ solution
Oxalate–KMnO₄: KMnO₄ acts as its own indicator (pink at end point) → no added indicator needed → (a). HCl–NaOH and Na₂CO₃–HCl need acid-base indicators; EDTA–hard water needs Eriochrome Black T.
Which one among the following is a metal ion type indicator?
(a) Methylene blue
(b) Diphenylamine sulphonic acid
(c) Eriochrome black T
(d) Cresol red
Answer: (c) Eriochrome black T
Eriochrome Black T is a metal-ion (metallochromic) indicator for EDTA titrations: wine-red Mg–EBT complex → blue free dye at end point (pH 10) → (c). Methylene blue and diphenylamine sulfonic acid are redox indicators; cresol red is an acid-base indicator.
Weight of O₂ necessary to fill up a cylinder of 0.082 litre capacity at 0˚C and 96 atm pressure would be: (Compressibility factor = 0.96 and gas constant (R) = 0.082 dm³ atm K⁻¹ mol⁻¹)
(a) 11.72 g
(b) 1.172 g
(c) 117.2 g
(d) 10.72 g
Answer: (a) 11.72 g
PV = ZnRT → n = PV/(ZRT) = (96 × 0.082)/(0.96 × 0.082 × 273) = 0.3663 mol. Mass = 0.3663 × 32 = 11.72 g → (a).
Which one of the following gas samples will have the highest average molecular speed? (Atomic weight of O = 16, Ne = 20, C = 12 and He = 4)
(a) 1.0 mole of O₂ at 560 K
(b) 0.50 mole of Ne at 500 K
(c) 0.20 mole of CO₂ at 440 K
(d) 2.0 mole of He at 140 K
Answer: (d) 2.0 mole of He at 140 K
Mean speed ∝ √(T/M); amount of gas does not matter. O₂ √(560/32) = 4.18; Ne √(500/20) = 5.00; CO₂ √(440/44) = 3.16; He √(140/4) = 5.92 → highest → (d).
The root mean square velocity of a monoatomic gas of molar mass ‘M’ (g mol⁻¹) is ‘u’. What is the correct relation between ‘u’ and average kinetic energy (E) per mole?
(a) u = \(\sqrt{\frac{3E}{M}}\)
(b) u = \(\sqrt{\frac{2E}{3M}}\)
(c) u = \(\sqrt{\frac{2E}{M}}\)
(d) u = \(\sqrt{\frac{E}{3M}}\)
Answer: (c) u = \(\sqrt{\frac{2E}{M}}\)
Per mole, average KE E = ½ M u² (u = rms speed, M = molar mass). So u = √(2E/M) → (c). Check: E = (3/2)RT gives u = √(3RT/M), the usual form.
In the Joule-Thomson expansion of real gas at inversion temperature, the Joule-Thomson coefficient:
(a) will be negative if the gas cools post expansion
(b) will be positive if the gas gets heated post expansion
(c) will be zero if [∂(PV)/ ∂P]T is positive
(d) is zero
Answer: (d) is zero
μJT = (∂T/∂P)H. At the inversion temperature, μJT changes sign, so μJT = 0 → (d). Below Ti gas cools (μ > 0); above it heats (μ < 0). (a) and (b) have the sign rules reversed.
The thermodynamic equations of state: (∂U/∂V)T = T(∂P/∂T)V − P and (∂H/∂P)T = V − T(∂V/∂T)P → (d). (b) has the wrong sign. (a) and (c) are just total differentials, not equations of state.
Which one of the following colligative properties can provide the most precise molar mass of peptides?
(a) Elevation of boiling point
(b) Depression of freezing point
(c) Relative lowering of vapour pressure
(d) Osmotic pressure
Answer: (d) Osmotic pressure
For macromolecules (peptides, proteins, polymers), osmotic pressure gives the largest measurable effect even at low concentration, is measured at room temperature (no denaturation) → most precise molar mass → (d).
At 298 K, the total vapour pressure in Torr for a mixture of volatile components A and B is given by P = 120 – 75\(\chi_B\) (\(\chi_B\) is the mole fraction of B). What is the vapour pressure of pure A and B (in Torr) respectively?
(a) 120 and 75
(b) 120 and 195
(c) 120 and 45
(d) 75 and 45
Answer: (c) 120 and 45
P = PA° + (PB° − PA°)χB. Compare with P = 120 − 75χB: PA° = 120 Torr; PB° − 120 = −75 → PB° = 45 Torr → (c).
Which one of the following equations represents the correct relation between relative lowering in vapour pressure and osmotic pressure (π) for a dilute aqueous solution?
A 1% aqueous solution (mass by volume) of a certain substance is isotonic with a 3% solution of glucose (Molar mass = 180 g mol⁻¹) at a given temperature. The molar mass of the substance (in kg mol⁻¹) will be:
(a) 0.60
(b) 180
(c) 0.18
(d) 0.06
Answer: (d) 0.06
Isotonic → equal molar concentrations (non-electrolytes). 1 g/100 mL ÷ M = 3 g/100 mL ÷ 180 → M = 60 g mol⁻¹ = 0.06 kg mol⁻¹ → (d). Trap: answer asked in kg mol⁻¹.
What will be the value of freezing point depression constant of cyclobutane having freezing point of 280 K and molar enthalpy of fusion of 2.68 kJ mol⁻¹? (Molar mass of cyclobutane = 56.108 g mol⁻¹and R= 8.314 JK⁻¹mol⁻¹)
(a) 14.63 K kg mol⁻¹
(b) 16.14 K kg mol⁻¹
(c) 13.64 K kg mol⁻¹
(d) 12.64 K kg mol⁻¹
Answer: (c) 13.64 K kg mol⁻¹
Kf = R Tf² M/ΔHfus (M in kg mol⁻¹) = 8.314 × 280² × 0.056108 / 2680 = 13.64 K kg mol⁻¹ → (c). Cyclobutane-type solvents have large Kf (like camphor).
Which one of the following statements is correct about the Henry’s law?
(a) It is related to effect of pressure only on solubility of a gas in a liquid
(b) It is related to effect of temperature only on solubility of a gas in a liquid
(c) It is related to effect of pressure as well as temperature on solubility of a gas in a liquid
(d) It is related to effect of pressure as well as temperature on solubility of a gas in a solid
Answer: (a) It is related to effect of pressure only on solubility of a gas in a liquid
Henry's law: p = KH·x — solubility of a gas in a liquid is proportional to its partial pressure at constant temperature → (a). Temperature changes the value of KH, but the law itself states the pressure effect.
What will be the ratio of mole fractions of O₂ and N₂ dissolved in water at 25˚C? (Given that Henry’s law constant (in torr) for O₂ and N₂ are 3.3 × 10⁷ and 6.5 × 10⁷ respectively at 25˚C; mole fractions of O₂ and N₂ in air are taken as 0.2 and 0.8 respectively)
(a) ~ 0.51
(b) ~0.66
(c) ~0.25
(d) ~0.49
Answer: (d) ~0.49
x = p/KH. x(O₂)/x(N₂) = (0.2/3.3 × 10⁷)/(0.8/6.5 × 10⁷) = (0.2 × 6.5)/(0.8 × 3.3) = 0.49 → (d). O₂ is about twice as soluble as N₂, so dissolved air is O₂-rich.
The cell constant with 0.1 N KCl solution is found to be 0.98 cm⁻¹ and the resistance offered by 0.01 N HCl is 89 Ω. The specific conductance of HCl will be:
(a) 1.101 × 10² Ω⁻¹ cm⁻¹
(b) 1.101 × 10⁻² Ω⁻¹ cm⁻¹
(c) 2.202 × 10⁻³ Ω⁻¹ cm⁻¹
(d) 2.202 × 10³ Ω⁻¹ cm⁻¹
Answer: (b) 1.101 × 10⁻² Ω⁻¹ cm⁻¹
κ = cell constant / R = 0.98 cm⁻¹ / 89 Ω = 1.101 × 10⁻² Ω⁻¹ cm⁻¹ → (b). The cell constant found with KCl applies to any solution in the same cell.
What will be the correct order of mobility of the following alkali metal ions in aqueous solution?
(a) Li⁺ > Na⁺ > K⁺ > Rb⁺
(b) Na⁺ > K⁺ > Rb⁺ > Li⁺
(c) K⁺ > Rb⁺ > Na⁺ > Li⁺
(d) Rb⁺ > K⁺ > Na⁺ > Li⁺
Answer: (d) Rb⁺ > K⁺ > Na⁺ > Li⁺
Smaller ions are more hydrated: Li⁺ drags the largest water shell, so moves slowest. Mobility: Rb⁺ > K⁺ > Na⁺ > Li⁺ → (d). Remember: hydrated size order is the reverse of crystal size.
During electrolysis of silver nitrate solution, loss in the anode compartment was 0.589 g. In a Voltameter connected in series, 1.259 g of silver was deposited. What would be the transport number of two ions?
(a) t₊ = 0.468 and t– = 0.532
(b) t₊ = 0.532 and t– = 0.468
(c) t₊ = 0.670 and t– = 0.330
(d) t₊ = 0.330 and t– = 0.670
Answer: (a) t₊ = 0.468 and t– = 0.532
Hittorf method with Ag electrodes: loss of Ag⁺ in the anode compartment = Ag⁺ carried away = t₊ × total. t₊ = 0.589/1.259 = 0.468; t₋ = 1 − 0.468 = 0.532 → (a).
The molar conductivities of KOAc and HCl at infinite dilution in water at 25˚C are 90.0 and 426.2 S cm² mol⁻¹, respectively. To calculate the molar conductivity of HOAc at infinite dilution, the additional molar conductivity value of which one of the following electrolytes will be required?
(a) NaCl
(b) KCl
(c) KNO₃
(d) NaOH
Answer: (b) KCl
Λ°(HOAc) = Λ°(KOAc) + Λ°(HCl) − Λ°(KCl). K⁺ and Cl⁻ cancel. So KCl value is needed → (b).
The mean ionic activity co-efficient is denoted as γ±, while that of cation and anion are denoted as γ₊ and γ₋, respectively. The correct relation among them will be: (Given that ν₊ and ν₋ are the number of cations and the anions respectively in the molecular formula of the compound)
The specific conductance of 0.01 M solution of acetic acid was found to be 0.015 S m⁻¹ at 25˚C. Molar conductance of acetic acid at infinite dilution is 400 × 10⁻⁴ S m² mol⁻¹. The degree of dissociation of acetic acid will be:
(a) 0.375
(b) 0.0375
(c) 0.00375
(d) 0.357
Answer: (b) 0.0375
Λm = κ/c = 0.015 S m⁻¹ / 10 mol m⁻³ = 1.5 × 10⁻³ S m² mol⁻¹. α = Λm/Λ°m = 1.5 × 10⁻³ / 400 × 10⁻⁴ = 0.0375 → (b). Trap: 0.01 M = 10 mol m⁻³.
In a moving boundary experiment with 0.01 mole LiCl, the boundary in a tube with cross sectional area of 0.125 cm² moves through 8.1 cm in 1500 sec for a current of 1.8 × 10⁻³ ampere. What will be the value of transport number of Cl⁻? (Given that F = 96,500 Coulomb and temperature is 300K)
When HCl is titrated against NaOH, the conductance at the end point is:
(a) Zero
(b) Minimum
(c) Maximum
(d) Infinite
Answer: (b) Minimum
Fast H⁺ (λ° ≈ 350) is replaced by slow Na⁺ (≈ 50): conductance falls to the end point, then rises as excess OH⁻ (≈ 198) is added. V-shaped curve → minimum at the end point → (b).
Which cationic species is more stable in each of the following pairs?
(a) 1 and 3
(b) 1 and 4
(c) 2 and 4
(d) 2 and 3
Answer: (d) 2 and 3
Pair X: 1 is a primary allylic cation conjugated only with C=CH₂ ending in CH₂; 2 (pent-2-enyl) delocalises + onto a secondary carbon → more stable. Pair Y: 3 (protonated amidine) is stabilised by two NH₂ donors, while 4 (protonated amide) has one NH₂ and OH (O less donating). → 2 and 3 → (d).
Which one of the following structures is non-aromatic?
(a) Naphthalene
(b) Cyclopentadienyl anion
(c) Cyclodecapentaene ([10]annulene)
(d) Phenanthrene
Answer: (c)
Naphthalene (10π) and phenanthrene (14π) are aromatic; cyclopentadienyl anion (6π) is aromatic. [10]Annulene (cyclodecapentaene) cannot be planar — inner H atoms clash (or angle strain) — so it is non-aromatic → (c).
The o/p-directing effect of methyl group in electrophilic substitution reaction of toluene is due to:
(a) Inductive effect
(b) Hyperconjugation effect
(c) Resonance effect
(d) Electromeric effect
Answer: (b) Hyperconjugation effect
CH₃ has no lone pair or π bond to give classical resonance. Its σ(C–H) electrons overlap with the ring π system (hyperconjugation, no-bond resonance), raising electron density at ortho and para → (b). Inductive (+I) effect is weaker and less position-specific.
The relative order of basicity of the following amines is:
(a) 1 > 2 > 3 > 4
(b) 4 > 1 > 2 > 3
(c) 4 > 3 > 2 > 1
(d) 2 >3 > 1 > 4
Answer: (b) 4 > 1 > 2 > 3
4 Et₃N: aliphatic, sp³ lone pair free → most basic. 1: two ortho CH₃ twist NMe₂ out of the ring plane (steric inhibition of resonance) → lone pair not delocalised → more basic than aniline. 2 aniline > 3 p-nitroaniline (−R of NO₂). 4 > 1 > 2 > 3 → (b).
Which one of the following carbocations is NOT expected to rearrange?
(a) Cyclohexylmethyl cation
(b) 2-Methylcyclohexyl cation
(c) 2-Methylenecyclohexyl (allylic) cation
(d) 1-Methylcyclohexyl cation
Answer: (d)
Rearrangement happens only if a more stable cation can form. (a) 1° cyclohexylmethyl → shifts/ring-expands. (b) 2° next to a 3° C–H → hydride shift to 3°. (c) allylic, delocalised. (d) 1-methylcyclohexyl is tertiary; no shift gives a more stable ion → does not rearrange → (d).
Which one of the following molecules does NOT exist as a meso isomer?
(a) 2,3,4-Tribromopentane (Br wedge, dash, wedge)
(b) 2,3,4-Tribromopentane (all three Br wedged)
(c) 2,5-Dibromohexane (one Br wedge, one dash)
(d) 2,5-Dibromohexane (both Br wedged)
Answer: (d)
Meso needs two (or more) stereocentres with internal mirror symmetry. 2,3,4-Tribromopentane (a, b) and 2,5-dibromohexane drawn (c) are meso forms. (d) 2,5-dibromohexane with both Br wedged in this zigzag is the (R,R)/(S,S) form → chiral, no meso → (d).
Which one of the following molecules does NOT have a plane of symmetry?
(a) 3,4-Dichlorohexane (sawhorse form)
(b) 3,4-Dichlorohexane (zigzag, both Cl wedged)
(c) 1,3-Dichlorocyclobutane (one Cl wedge, one dash)
(d) 1,2,3-Trimethylcyclopropane (one Me dash, two wedged)
Answer: (b)
(b) 3,4-dichlorohexane as drawn is (3S,4S) — a chiral (dl) form, so it has no plane of symmetry. (a) is the meso form, (c) trans-1,3-dichlorocyclobutane and (d) the trimethylcyclopropane have a mirror plane → (b).
Which one of the following represents an enantiomer of [A]?
(a) Fischer projection (a)
(b) Fischer projection (b)
(c) Fischer projection (c)
(d) Fischer projection (d)
Answer: (d)
[A] is (2S,3R)-2-bromo-3-methylpentane. Enantiomer must invert both centres → (2R,3S). Assigning CIP to options: (a) 2R,3R; (b) 2R,3R; (c) 2R,3R; (d) 2R,3S → (d). Trick: lowest priority (H) on a horizontal bond → reverse the observed sense.
The E/Z-nomenclature for the following molecules is:
(a) 1: Z, 2: Z, 3: Z, 4: E
(b) 1: Z, 2: E, 3: E, 4: E
(c) 1: E, 2: Z, 3: Z, 4: E
(d) 1: E, 2: Z, 3: E, 4: Z
Answer: (a) 1: Z, 2: Z, 3: Z, 4: E
1: I > Br and Cl > F; I and Cl same side → Z. 2: Br > Cl and D > H; Br and D same side → Z. 3: cis-but-2-ene → Z. 4: CH₂Br > CH₃ (top left) and CH₂CH₃ > CH₃ (bottom right) → opposite → E → (a).
Both are butan-2-ol. On the stereocentre, H and the ethyl carbon keep their places while OH and CH₃ swap between [X] and [Y]. One swap = inverted configuration → enantiomers → (c).
Which one of the following is the most stable conformation of ethylene glycol?
(a) Anti conformer (OH/OH 180°)
(b) Gauche conformer (OH/OH 60°)
(c) Eclipsed conformer (OH/OH 0°)
(d) Eclipsed conformer (OH/OH 120°)
Answer: (b)
In HOCH₂CH₂OH the gauche conformer is most stable: the two OH groups at 60° form an intramolecular H-bond, outweighing gauche strain. Option (b) shows OH/OH gauche → (b). (Same for 2-fluoroethanol.)
Which one of the following substrate sets is the best for synthesizing above compound?
(a) CH₃I + sodium 1-methylcyclohexan-1-olate
(b) CH₃ONa + 1-iodo-1-methylcyclohexane
(c) CH₃OH + 1-methylcyclohexan-1-ol
(d) CH₃I + 1-methylcyclohexan-1-ol
Answer: (a)
Williamson synthesis is SN2: the halide must be methyl/primary. A tertiary halide with CH₃O⁻ would give elimination (alkene). So use CH₃I + tertiary alkoxide (1-methylcyclohexan-1-olate) → (a).
The major product(s) formed in the following reaction is/are:
(a) 2-OMe epimers (both), 3-Br retained
(b) 3-OMe epimers, 2-Br retained
(c) cis-2-OMe-3-Br only
(d) 3-OMe-2-Br only
Answer: (a)
C2–Br sits next to ring O: it ionises easily (oxocarbenium ion, SN1). MeOH attacks the planar cation from both faces → both C2 epimers with OMe; the C3–Br (ordinary 2°) is untouched with 1 equiv. MeOH → (a).
The major product formed in the following reaction is:
(a) 3,3-Dimethylcyclohexene
(b) 1-Methylcyclohexene
(c) 1,2-Dimethylcyclohexene
(d) Isopropylidenecyclopentane
Answer: (c)
2,2-Dimethylcyclohexanol + H₂SO₄: 2° cation forms, then 1,2-methyl shift gives a 3° cation; loss of H⁺ gives the most substituted alkene, 1,2-dimethylcyclohexene (tetrasubstituted, Saytzeff) → (c). Ring contraction to (d) is minor.
The major product formed in the following reaction is:
(a) Ethyl ether, retention
(b) Ethyl ether, inversion
(c) Alkene: Me/Me cis, H/Ph cis
(d) Alkene: Me/Ph cis, H/Me cis
Answer: (c)
TsCl/py converts OH to OTs with retention (C–O not broken). NaOEt then does anti-periplanar E2: placing C3–H anti to OTs puts the two Me groups on the same side and Ph cis to H → alkene (c). E2 is stereospecific.
Which one of the following bromides CANNOT form an alkene by E2 process?
(a) 1-Bromo-2,2-dimethylpropane
(b) 2-Bromo-2-methylbutane
(c) 2-Bromo-3-methylbutane
(d) 1-Bromo-3-methylbutane
Answer: (a) 1-Bromo-2,2-dimethylpropane
E2 needs a β-hydrogen. 1-Bromo-2,2-dimethylpropane (neopentyl bromide, (CH₃)₃C–CH₂Br): the β-carbon is quaternary, no β-H → cannot form an alkene by E2 → (a).
The major product formed in the following reaction is:
(a) CH₃CH=CH₂
(b) CH₃CH=CHCH₃
(c) CH₃CH₂CH=CH₂
(d) H₂C=CH₂
Answer: (d)
Quaternary ammonium hydroxide on heating gives the Hofmann product (least substituted alkene) — OH⁻ removes the most accessible, most acidic β-H. Here the ethyl group gives ethene, H₂C=CH₂ → (d).
The major product formed in the following reaction is:
(a) 1-(2-Chloroethyl)-1-methylcyclobutane
(b) 1-(1-Chloroethyl)-1-methylcyclobutane
(c) 1-Chloro-1,2-dimethylcyclopentane
(d) 1-Chloro-2,2-dimethylcyclopentane
Answer: (c)
HCl adds Markovnikov: secondary cation next to the strained cyclobutane. Ring expansion (C–C shift) relieves strain and gives a tertiary cyclopentyl cation carrying CH₃; Cl⁻ traps it → 1-chloro-1,2-dimethylcyclopentane → (c).
The major products [X] and [Y] formed in the following transformations are:
(a) [X] Br on C bearing OCH₃; [Y] α-bromo ester
(b) [X] Br on C bearing OCH₃; [Y] β-bromo ester
(c) [X] Br on C2; [Y] α-bromo ester
(d) [X] Br on C2; [Y] β-bromo ester
Answer: (b)
[X]: enol ether — H⁺ adds to C2, + at C1 is stabilised by O (oxocarbenium) → Br on the carbon bearing OCH₃. [Y]: α,β-unsaturated ester — conjugate (1,4) addition; Br⁻ ends on the β-carbon → (b).
The major product formed in the following reaction is:
(a) Alkene: Ph/Me cis, Me/H cis
(b) Terminal alkene (3-phenylbut-1-ene)
(c) Alcohol (substitution product)
(d) Alkene: Ph/H cis, Me/Me cis
Answer: (d)
NaOEt/EtOH on a 2° tosylate → anti-periplanar E2. Rotating C3–OTs anti to C2–H places Me and Me cis, Ph cis to H → (E)-2-phenylbut-2-ene drawn in (d) → (d). Remember: E2 is stereospecific.
The major product(s) formed in the following reaction is/are:
(a) o-Toluidine only
(b) 2-Bromo-3-methylaniline
(c) m-Toluidine only
(d) o-Toluidine + m-toluidine
Answer: (a)
NaNH₂/liq. NH₃ on o-bromotoluene goes by elimination–addition (benzyne): 3-methylbenzyne forms and NH₂⁻ adds to either end. Strictly this gives a mixture of o- and m-toluidine (option d). UPSC's key holds (a).
Ipso = the nucleophile enters the carbon that carried the leaving group. (a), (b), (d) are SNAr at that carbon. (c) goes via benzyne: NH₂ enters the adjacent carbon (cine substitution) → NOT ipso → (c).
The product formed by the reaction of toluene with chlorine in the presence of sunlight is:
(a) o-chlorotoluene
(b) 2,4-dichlorotoluene
(c) p-chlorotoluene
(d) benzylchloride
Answer: (d) benzylchloride
Cl₂ in sunlight (hν), no Lewis acid → free-radical substitution at the benzylic C–H → benzyl chloride C₆H₅CH₂Cl → (d). With FeCl₃ (dark) → o/p-chlorotoluene.
The major product formed in the following reaction is:
(a) Ring N-hydroxyamine (alkene)
(b) Trimethylcyclopentene nitrile (CN on ring)
(c) Exo-methylene cyclopentyl acetonitrile
(d) α-Campholenonitrile
Answer: (d)
Camphor oxime + acid: the C–C bond anti to OH can break to give a stable tertiary cation instead of normal Beckmann migration → fragmentation to a nitrile; loss of H⁺ gives the alkene: α-campholenonitrile → (d).
The major product formed in the following reaction is:
(a) 1-Phenylcyclopentyl phenyl ketone
(b) 2-Phenylcyclopentyl phenyl ketone
(c) 2,2-Diphenylcyclohexanone
(d) 2,6-Diphenylcyclohexanone
Answer: (c)
Ionisation gives the more stable diphenyl-stabilised cation. A ring C–C bond migrates to it (ring expansion 5 → 6), and the ring oxygen becomes C=O → 2,2-diphenylcyclohexanone → (c).
The major product formed in the following reaction is:
(a) 2,3-Dimethylnorborn-2-ene
(b) 5,5-Dimethylnorborn-2-ene
(c) Dimethylnorbornene isomer as drawn (C=C away from the gem position)
(d) 3,3-Dimethylnorbornan-2-one
Answer: (a)
Protonation and loss of water at C2 next to the gem-dimethyl carbon → 1,2-methyl shift gives a tertiary cation; loss of H⁺ gives the tetrasubstituted alkene 2,3-dimethylnorborn-2-ene → (a).
The relative order of the rate of acetolysis among the following norbornyl derivatives is:
(a) 1 > 2 > 3
(b) 1 > 3 > 2
(c) 3 > 2 > 1
(d) 3 > 1 > 2
Answer: (b) 1 > 3 > 2
1 = anti-7-norbornenyl tosylate: C=C π electrons sit directly behind the C–OTs bond and assist ionisation → ~10¹¹ times faster than saturated 2. 3 (syn) cannot use the π bond from the back, but is still faster than 2. 1 > 3 > 2 → (b).
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