UPHESC Assistant Professor Chemistry 2025 PYQs

Concept: For odd n, $S_n$ generates 2n operations before returning to E.

1. $S_3^1=S_3$, $S_3^2=C_3^2$, $S_3^3={\sigma }_h$, $S_3^4=C_3$, $S_3^5=S_3^5$, $S_3^6=E$.
2. Six distinct operations (this set is the group $C_{3h}$).

Short trick: S_n: n even → n operations; n odd → 2n operations.

Concept: Borazine ${\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6$ is isoelectronic and isostructural with benzene.

1. Planar six-membered ring of alternating B and N with delocalised π electrons (N lone pair → B empty p).
2. Prepared from ${\mathrm{B}}_2{\mathrm{H}}_6+2{\mathrm{NH}}_3$ on heating, or ${\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{BCl}}_3$ then ${\mathrm{NaBH}}_4$.
3. Being polar (B^δ+–N^δ−), it is more reactive than benzene (adds HCl readily).

Short trick: B–N unit ≡ C=C unit (12 electrons each) → borazine ≈ benzene.

Concept: Recall characteristic IR group frequencies.

1. N–H stretch ≈ 3300–3500 cm⁻¹ → a–2; ester C=O ≈ 1735–1750 cm⁻¹ → d–4.
2. Nitrile C≡N ≈ 2210–2260 cm⁻¹ (typically quoted 2250) → b–1.
3. S–H is a weak band in the ~2550–2600 cm⁻¹ region; of the values left, 2270 is assigned to it → c–3. Only code (B) fits.

Short trick: X–H > 3000 (except S–H ~2550), triple bonds ~2100–2260, C=O ~1700.

Concept: ${\mathrm{S}}_{\mathrm{N}}2$ needs backside attack; rate falls sharply with crowding at the α- and β-carbons.

1. (A) Chloroethane is primary and unhindered → fastest.
2. (B) Secondary ring carbon – slower; (C) secondary with a quaternary β-carbon (gem-dimethyl) – much slower.
3. (D) Neopentyl chloride is primary but the bulky t-butyl β-group blocks the backside – about 10⁵ times slower than ethyl.

Short trick: SN2: Me > 1° > 2° ≫ 3°; neopentyl ≈ practically inert.

Concept: Acid dehydration of a 2° alcohol goes by E1 via a carbocation; the more substituted (Saytzeff) alkene predominates.

1. ${\mathrm{CH}}_3\mathrm{CH}(\mathrm{OH}){\mathrm{CH}}_2{\mathrm{CH}}_3\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{CH}}_3\stackrel{+}{\mathrm{C}}{\mathrm{HCH}}_2{\mathrm{CH}}_3$.
2. Loss of H from C-3 gives but-2-ene (disubstituted, mainly trans) – major; loss from C-1 gives but-1-ene – minor.
3. Butadiene would need two eliminations; isobutylene needs a skeletal rearrangement (not favoured).

Short trick: Saytzeff: H goes from the carbon having fewer H → more substituted alkene.

Concept: Pyridine is π-deficient; electrophilic substitution is sluggish and occurs at C-3.

1. Attack at C-2 or C-4 gives a σ-complex with a resonance form bearing + charge on electronegative N – very unfavourable.
2. Attack at C-3 keeps the + charge on carbon only, so 3-substitution is preferred (nitration needs ~300 °C, low yield).
3. N-nitration (1-nitropyridinium) is not the substitution product.

Short trick: Pyridine behaves like nitrobenzene → meta (C-3) substitution.

Concept: Acetate pyrolysis is a concerted cyclic (6-membered) Ei elimination: the β-H removed must be cis (syn) to the OAc group.

1. OAc is axial (up) on C-1. On C-2 the methyl is equatorial (cis to OAc) and the H is axial-down (trans) – so no syn H on C-2; no double bond can form towards C-2.
2. On the other ring neighbour (CH₂) the equatorial H is cis to OAc → elimination gives the endocyclic, trisubstituted alkene with the old C-1 methyl on the double bond; (B), the exocyclic alkene, is the minor path.
3. The C-2 stereocentre is untouched, so its methyl keeps its original orientation. (C) shows that methyl on the opposite face (axial-down) – wrong.
4. (A) and (D) both show this same product: they differ only in the direction in which the methyl on the sp² carbon is drawn, and an sp² carbon has no stereochemistry. So (A) and (D) are the same compound, two options are correct, and the question is treated as a bonus. The product is 1,6-dimethylcyclohexene with the ring methyl in its original position.

Short trick: Ei (acetate, xanthate, Cope) = syn elimination; E2 = anti elimination.

Concept: Ethers react slowly with air (O₂) by a free-radical chain at the C–H next to oxygen.

1. The α-C–H of THF is weakest because the radical formed is stabilised by the adjacent O lone pair.
2. R• + O₂ → ROO•; ROO• + THF–H → ROOH + R• (chain).
3. Product: 2-hydroperoxy-THF – the explosive peroxide found in old THF/ether bottles.

Short trick: Old ether + air → α-hydroperoxide → distil only after peroxide test.

Concept: Lithium dialkylamides (LiNEt₂, ether/hexane, reflux) isomerise epoxides to allylic alcohols by a syn β-elimination (as in α-pinene oxide → trans-pinocarveol).

1. The base removes a β-H that is syn to the epoxide oxygen; the C–O bond next to it breaks and O⁻ stays on the other carbon.
2. The freely rotating CH₂ of the ethyl side-chain supplies such an H much more easily than the ring CH₂ groups.
3. So the CH–O bond breaks, O stays on the spiro (ring) carbon and a C=C forms in the side chain: 1-(prop-1-enyl)cyclopentanol (E-alkene).

Short trick: Epoxide + LiNR₂ → allylic alcohol; H is taken from the most accessible (acyclic) β-CH.

Concept: Benzaldehyde + an aminoacetaldehyde acetal is the Pomeranz–Fritsch route to isoquinoline.

1. Step 1: condensation gives the imine (Schiff base) A: $\mathrm{PhCH}=\mathrm{N}-{\mathrm{CH}}_2-\mathrm{CH}(\mathrm{OR}{)}_2$.
2. Step 2: acid (H⁺, Δ) ionises the acetal; the oxocarbenium ion attacks the benzene ring ortho to the imine carbon, then loss of ROH aromatises.
3. The new ring is C1(from CHO)–N2–C3–C4, fused to benzene → isoquinoline (N at position 2).

Short trick: Benzaldehyde + H₂NCH₂CH(OR)₂ / H⁺ → isoquinoline (Pomeranz–Fritsch).

Concept: Only the ketone must be reduced; the ester must survive.

1. ${\mathrm{NaBH}}_4$ is a mild hydride donor: reduces aldehydes/ketones but not esters (under normal conditions).
2. ${\mathrm{LiAlH}}_4$ would reduce both (ester → ${\mathrm{CH}}_2\mathrm{OH}$).
3. ${\mathrm{B}}_2{\mathrm{H}}_6$ is an electrophilic reductant used mainly for acids/amides and it also hydroborates C=C; ${\mathrm{NaNH}}_2$ is a base, not a reductant.

Short trick: NaBH₄: C=O of aldehyde/ketone only; LiAlH₄: almost everything.

Concept: In square-planar Pt(II), the ligand trans to the stronger trans-director is replaced; Cl⁻ > NH₃ in the trans-effect series.

1. Step 1: $[{\mathrm{PtCl}}_4{]}^{2-}+{\mathrm{NH}}_3\to [{\mathrm{PtCl}}_3({\mathrm{NH}}_3){]}^{-}$.
2. Step 2: the Cl trans to another Cl (not the one trans to NH₃) is labilised → second NH₃ enters cis to the first.
3. Product: cis-$[{\mathrm{PtCl}}_2({\mathrm{NH}}_3{)}_2]$ (cisplatin). Starting from $[\mathrm{Pt}({\mathrm{NH}}_3{)}_4{]}^{2+}$ + Cl⁻ gives the trans isomer.

Short trick: PtCl₄²⁻ + NH₃ → cis; Pt(NH₃)₄²⁺ + Cl⁻ → trans.

Concept: Strong-field CN⁻ pairs electrons; weak-field F⁻ does not.

1. $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$: Fe(II) $d^6$, low spin $t_{2g}^6$ → n = 0 → 0 BM (diamagnetic).
2. $[{\mathrm{FeF}}_6{]}^{3-}$: Fe(III) $d^5$, high spin $t_{2g}^3{e}_g^2$ → n = 5 → $\sqrt{35}$ = 5.92 BM.

Short trick: Ferrocyanide diamagnetic; d⁵ high spin = 5.92 BM.

Concept: For $\mathrm{M}|{\mathrm{M}}^{\mathrm{n}+}({\mathrm{c}}_1)||{\mathrm{M}}^{\mathrm{n}+}({\mathrm{c}}_2)|\mathrm{M}$, $E=\frac{RT}{nF}\ln \frac{c_2}{c_1}$ ($c_2 > c_1$).

1. As the cell works, the dilute side becomes more concentrated (anode dissolves) and the concentrated side becomes more dilute (cathode deposits).
2. So $c_2/c_1$ falls towards 1 and the EMF decreases continuously.
3. It reaches zero only when the concentrations become equal (cell exhausted), not while it is functioning.

Short trick: Concentration cell runs to equalise concentrations → EMF ↓.

Concept: Find all distributions (configurations) with total energy 3E, then count arrangements $W=\frac{N!}{\prod n_i!}$ for N = 4 atoms.

1. Configuration 1: three atoms at E, one at 0 → $W=\frac{4!}{3!1!}=4$.
2. Configuration 2: one at 2E, one at E, two at 0 → $W=\frac{4!}{1!1!2!}=12$.
3. Total microstates = 4 + 12 = 16 (2 is only the number of configurations).

Short trick: Configurations first, then W = N!/∏n_i! for each, then add.

Concept: Three statistics: Maxwell–Boltzmann (distinguishable), Bose–Einstein (bosons), Fermi–Dirac (fermions).

1. Bosons have integral spin (0, 1, 2 …), symmetric wave functions and any number may occupy one state (photons, ⁴He).
2. Half-integral spin particles (electrons, protons) are fermions → Fermi–Dirac, obey Pauli principle.

Short trick: Bose → integer spin; Fermi → half-integer spin.

Concept: Phosphorescence is radiative emission between states of different multiplicity.

1. $T_1\to S_0+h\nu$: spin-forbidden emission → slow, long-lived glow = phosphorescence.
2. $S_1\to S_0+h\nu$ = fluorescence; $S_1\to T_1$ = intersystem crossing; $T_1\to S_0+\mathrm{\Delta }$ = non-radiative ISC (heat).

Short trick: Fluorescence S→S (fast); phosphorescence T→S (slow).

Concept: All four listed proteins are iron enzymes, so no option fits the question as printed.

1. Cytochrome P₄₅₀ – heme Fe (cysteine-thiolate ligated) monooxygenase.
2. Pyrocatechase (catechol 1,2-dioxygenase) – non-heme Fe(III) intradiol dioxygenase.
3. Tryptophan 2,3-dioxygenase – heme Fe; catalase – heme Fe (Fe(III)/Fe(IV)=O).
4. Typical non-iron examples would be carbonic anhydrase (Zn), tyrosinase (Cu) or nitrogenase's Mo site.

Short trick: P450, catalase, peroxidase, TDO → heme Fe; catechol dioxygenases → non-heme Fe.

Concept: After a small perturbation, the deviation $x$ from equilibrium decays as $\frac{dx}{dt}=-(k_1+k_{-1})x$.

1. So $x=x_0{e}^{-t/\tau }$ with $\frac{1}{\tau }=k_1+k_{-1}$.
2. Hence $\tau =(k_1+k_{-1}{)}^{-1}$; it has units of time, as required (the other forms do not).

Short trick: First-order ⇌ first-order: 1/τ = k_f + k_b.

Concept: An ambident nucleophile has two different donor atoms through which it can attack.

1. ${}^{-}\mathrm{C}\equiv \mathrm{N}\leftrightarrow \mathrm{C}={\mathrm{N}}^{-}$: attacks through C (→ nitriles, e.g. with KCN) or through N (→ isocyanides, e.g. with AgCN).
2. ${\mathrm{CH}}_3^{-}$, ${\mathrm{PH}}_3$ and ${\mathrm{H}}_2\mathrm{O}$ each have only one nucleophilic site.
3. Other examples: ${\mathrm{NO}}_2^{-}$, ${\mathrm{SCN}}^{-}$, ${\mathrm{OCN}}^{-}$, enolates.

Short trick: KCN → R–CN; AgCN → R–NC.