FAD1018: Comprehensive Rapid-Fire Drill — Full Syllabus

Objective: Identify every weak spot across the entire FAD1018 chemistry syllabus.
Target: 2–3 min per problem. If you stall >4 minutes, skip and mark it.
Total problems: 84
Estimated time: 3–4 hours

Cheat Sheet (Memorize First)

Physical Chemistry

Topic Key Formula Notes
$K_c$ expression $\displaystyle K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ Pure solids/liquids omitted
$K_p$ relation $K_p = K_c(RT)^{\Delta n}$ For gas-phase reactions
Reaction quotient $Q$ same form as $K_c$ $Q<K\rightarrow$ forward; $Q>K\rightarrow$ reverse
Le Chatelier System counteracts disturbance Temp, pressure, concentration
Autoionisation of water $K_w = [H^+][OH^-] = 1.0\times10^{-14}$ at 25°C $K_w$ ↑ with temp
pH/pOH $pH=-\log[H^+]$, $pH+pOH=14$ At 25°C only
Degree of dissociation $\displaystyle \alpha = \sqrt{\frac{K_a}{c}}$ Valid only if $\alpha < 10%$
Ka/Kb relation $K_a \times K_b = K_w$ Conjugate pair
Henderson-Hasselbalch $\displaystyle pH = pK_a + \log\frac{[A^-]}{[HA]}$ Buffer solutions
Titration equivalence SA-SB: pH 7; WA-SB: pH>7; SA-WB: pH<7 Phenolphthalein / methyl red
$K_{sp}$ general $A_aB_b \rightleftharpoons aA^{m+}+bB^{n-}$, $K_{sp}=[A^{m+}]^a[B^{n-}]^b$
1:1 salt $K_{sp}=s^2$, $s=\sqrt{K_{sp}}$ e.g. AgCl
1:2 salt $K_{sp}=4s^3$, $s=\sqrt[3]{K_{sp}/4}$ e.g. CaF$_2$
Precipitation $Q > K_{sp} \rightarrow$ precipitate
Common ion effect $K_{sp}$ unchanged, solubility ↓
Raoult's Law $P_A = X_A P_A^\circ$ Ideal solutions
BP elevation $\Delta T_b = K_b m$ Colligative
FP depression $\Delta T_f = K_f m$ Colligative
Osmotic pressure $\Pi = MRT$ Colligative
Clausius-Clapeyron $\displaystyle \ln\frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$
Phase rule $F = C - P + 2$ Gibbs
Calorimetry $q = mc\Delta T$ Constant pressure → $\Delta H$
Hess's Law $\Delta H^\circ_{rxn} = \sum \Delta H^\circ_f(\text{prod}) - \sum \Delta H^\circ_f(\text{react})$ Path independent
Bond enthalpy $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$
Gibbs free energy $\Delta G = \Delta H - T\Delta S$ $\Delta G < 0$ = spontaneous
Standard cell potential $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
Nernst equation (25°C) $\displaystyle E_{cell} = E^\circ_{cell} - \frac{0.0592}{n}\log Q$
Gibbs & cell $\Delta G^\circ = -nFE^\circ_{cell}$ F = 96,485 C·mol$^{-1}$
Cell & equilibrium $\displaystyle E^\circ_{cell} = \frac{0.0592}{n}\log K$ At 25°C
Faraday's laws $m = \frac{QM}{nF} = \frac{ItM}{nF}$ $Q = It$
Selective discharge Cations: K$^+$ < Na$^+$ < ... < H$^+$ < Cu$^{2+}$ < Ag$^+$ Anions: F$^-$ < ... < OH$^-$
Rate law $\text{Rate} = k[A]^m[B]^n$ Orders from experiment only
Zero-order integrated $[A]_t = [A]_0 - kt$ $t_{1/2} = [A]_0/2k$ (↓)
First-order integrated $\ln[A]_t = \ln[A]_0 - kt$ $t_{1/2} = 0.693/k$ (constant)
Second-order integrated $1/[A]_t = 1/[A]_0 + kt$ $t_{1/2} = 1/k[A]_0$ (↑)
Arrhenius $\displaystyle k = Ae^{-E_a/RT}$ $\displaystyle \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
Michaelis-Menten $\displaystyle v = \frac{V_{max}[S]}{K_m + [S]}$ Low $K_m$ = high affinity

Organic Chemistry

Topic Key Fact Notes
Chirality 4 different groups on C Non-superimposable mirror image
Max stereoisomers $2^n$ $n$ = chiral centres
R/S priority Higher atomic number → higher priority 1→2→3 clockwise = R
Alcohol oxidation 1° → aldehyde → acid; 2° → ketone; 3° → no reaction PCC stops at aldehyde
Phenol acidity $pK_a \approx 10$ ($\sim 10^8\times$ more acidic than alcohols) Resonance-stabilised phenoxide
Carbonyl reactivity Aldehydes > Ketones (less steric + more electrophilic)
Aldehyde tests Tollens' → Ag mirror; Fehling's → brick-red Cu$_2$O Ketones: negative
2,4-DNP test Both aldehydes & ketones → orange/red ppt LEAK: Brady's reagent
Iodoform test CH$_3$CO- or CH$_3$CH(OH)- → yellow CHI$_3$ Methyl ketones + ethanol
Acidity order Carboxylic acid > Phenol > Water > Alcohol $pK_a$: 4–5 vs 10 vs 15.7 vs 16–18
Acyl derivative reactivity Acyl chloride > Anhydride > Ester $\sim$ Acid > Amide Leaving group ability
Fischer esterification RCOOH + R'OH $\rightleftharpoons$ RCOOR' + H$_2$O Acid-catalysed, reversible
Saponification RCOOR' + OH$^- \rightarrow$ RCOO$^-$ + R'OH Irreversible
Amine basicity Aliphatic 2° > 3° > 1° > NH$_3$ > Aromatic
HNO$_2$ test 1° aliphatic → N$_2$ gas; 1° aromatic → diazonium; 2° → yellow oil; 3° aromatic → solid
Amino acid Zwitterion at pH = pI Glycine only achiral
pI formula Neutral: $(pK_{a1}+pK_{a2})/2$
Polymer types Addition (C=C, no by-product) vs Condensation (functional groups, small molecule lost)
Natural rubber Polyisoprene; vulcanisation with S → cross-linked
Kolbe-Schmitt Phenol + NaOH/CO$_2$ → salicylic acid Industrial aspirin precursor
HVZ $\alpha$-H of carboxylic acid replaced by halogen (Br$_2$/PBr$_3$) Then NH$_3$ → $\alpha$-amino acid
Gabriel Phthalimide + KOH + RX → 1° amine Pure 1° amine, no over-alkylation
Hofmann rearrangement RCONH$_2$ + Br$_2$/KOH → RNH$_2$ + CO$_2$ Amide → amine, loss of carbonyl C
Claisen condensation 2 esters + base → $\beta$-keto ester + ROH Ester self-condensation
Cannizzaro 2 ArCHO + conc. base → alcohol + carboxylate No $\alpha$-H required

Part A: Chemical Equilibrium

Target: 2 min per problem. | Exam weight: ~7%

Set A1 — Kc/Kp & Le Chatelier (4 problems)

  1. Write the $K_c$ expression for: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$.
  2. For the reaction above, $K_c = 4.5$ at a given temperature. If $[\text{SO}_2] = 0.10$ M, $[\text{O}_2] = 0.10$ M, and $[\text{SO}_3] = 0.20$ M, calculate $Q$ and state which direction the reaction proceeds.
  3. $K_p = 0.82$ for $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ at 300 K. If $\Delta n = 1$, calculate $K_c$. (R = 0.0821 L·atm·mol$^{-1}$·K$^{-1}$)
  4. For the exothermic Haber process $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$, predict the shift when: (a) pressure is increased, (b) temperature is increased, (c) NH$_3$ is removed.

Score: ___/4

Set A2 — Degree of Dissociation & ICE (2 problems)

  1. A 0.10 M solution of weak acid HA has $K_a = 1.0 \times 10^{-5}$. Calculate $[H^+]$ and pH. Verify the $%\alpha < 10%$ assumption.
  2. $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$. Initially 1.0 mol N$_2$O$_4$ in a 5.0 L flask. At equilibrium, $[\text{NO}_2] = 0.060$ M. Calculate $K_c$.

Score: ___/2

Part B: Ionic Equilibria (pH, Buffers, Titrations)

Target: 2–3 min per problem. | Exam weight: ~13% (highest, every paper)

Set B1 — pH & Ka/Kb Calculations (4 problems)

  1. Calculate the pH of 0.01 M HCl (strong acid).
  2. A 0.50 M CH$_3$COOH solution has $K_a = 1.8 \times 10^{-5}$. Calculate the pH and $pK_a$.
  3. 0.20 M NH$_3$ has $K_b = 1.8 \times 10^{-5}$. Calculate $[OH^-]$ and pH.
  4. 0.10 M HNO$_2$ ($K_a = 5.0 \times 10^{-4}$). Solve for $[H^+]$ using a quadratic (assumption fails).

Score: ___/4

Set B2 — Salt Hydrolysis (2 problems)

  1. Classify each salt as acidic, basic, or neutral: (a) NH$_4$NO$_3$, (b) NaCH$_3$COO, (c) NaCl, (d) KF.
  2. 0.20 M NH$_4$Cl solution. $K_a$ for NH$_4^+$ is $5.6 \times 10^{-10}$. Calculate the pH.

Score: ___/2

Set B3 — Buffer Solutions (3 problems)

  1. A buffer contains 0.25 M CH$_3$COOH and 0.35 M CH$_3$COONa ($pK_a = 4.74$). Calculate the pH.
  2. To the buffer above, 0.010 mol of HCl is added to 1.0 L. Calculate the new pH.
  3. A basic buffer of NH$_3$ (0.30 M) and NH$_4$Cl (0.20 M) has $pK_b = 4.74$. Calculate pH. After adding 0.020 mol NaOH to 1.0 L, what is the new pH?

Score: ___/3

Set B4 — Acid-Base Titrations (3 problems)

  1. 50 mL of 0.10 M NaOH is titrated with 0.10 M HCl. Calculate pH at: (a) 0 mL, (b) 25 mL, (c) 50 mL (equivalence), (d) 60 mL.
  2. 100 mL of 0.10 M CH$_3$COOH ($K_a = 1.8 \times 10^{-5}$) is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Which indicator is suitable? (Phenolphthalein range 8.3–10.0, methyl red range 4.2–6.3)
  3. 40 mL of 0.10 M NH$_3$ ($K_b = 1.8 \times 10^{-5}$) is titrated with 0.10 M HCl. Calculate pH at equivalence. Which indicator?

Score: ___/3

Part C: Solubility Product

Target: 2 min per problem. | Exam weight: ~5% (declining but reappears)

Set C1 — Ksp & Molar Solubility (3 problems)

  1. The solubility of PbS is $8.36 \times 10^{-14}$ M. Calculate $K_{sp}$.
  2. $K_{sp}$ of Pb(IO$_3$)$_2$ is $2.56 \times 10^{-13}$. Calculate the molar solubility.
  3. $K_{sp}$ of Ag$_2$CrO$_4$ is $1.9 \times 10^{-12}$. Find $[Ag^+]$ and $[CrO_4^{2-}]$ in a saturated solution.

Score: ___/3

Set C2 — Precipitation & Common Ion (3 problems)

  1. $[Mg^{2+}] = 1.5 \times 10^{-6}$ M, $[OH^-] = 1.0 \times 10^{-4}$ M. $K_{sp}$ Mg(OH)$_2$ = $1.5 \times 10^{-14}$. Will a precipitate form?
  2. A solution contains 0.010 M Ag$^+$ and 0.020 M Pb$^{2+}$. Cl$^-$ is added slowly. $K_{sp}$ AgCl = $1.8 \times 10^{-10}$, PbCl$_2$ = $1.7 \times 10^{-5}$. Which precipitates first? At what [Cl$^-$]?
  3. Calculate the molar solubility of CaF$2$ ($K{sp} = 3.9 \times 10^{-11}$): (a) in pure water, (b) in 0.10 M NaF.

Score: ___/3

Part D: Phase Equilibria

Target: 2 min per problem. | Exam weight: ~8%

Set D1 — Colligative Properties (4 problems)

  1. 218 g of glucose (C$6$H${12}$O$_6$, M = 180 g/mol) is dissolved in 460 mL water. Vapour pressure of pure water = 31.82 mmHg. Calculate $\Delta P$ and the vapour pressure of the solution.
  2. 1.60 g naphthalene (C$_{10}$H$_8$, M = 128 g/mol) in 20.0 g benzene. $K_f = 5.12$ °C·m$^{-1}$. Freezing point of pure benzene = 5.50 °C. Find the freezing point of the solution.
  3. 651 g ethylene glycol (C$_2$H$_6$O$_2$, M = 62 g/mol) in 2505 g water. $K_f = 1.86$ °C·m$^{-1}$, $K_b = 0.512$ °C·m$^{-1}$. Find the freezing point and boiling point.
  4. 46.0 g/L glycerin solution at 25°C. R = 0.0821 L·atm·mol$^{-1}$·K$^{-1}$. Calculate the osmotic pressure.

Score: ___/4

Set D2 — Raoult's Law & Phase Diagrams (3 problems)

  1. At a given temperature, $P_A^\circ = 60$ kPa, $P_B^\circ = 30$ kPa. For a solution with $X_A = 0.3$, calculate the total vapour pressure (assuming ideal).
  2. A solution of CS$_2$ and acetone has a total pressure of 433 torr. Pure CS$_2$ has $P^\circ = 512$ torr. Is this positive or negative deviation? What type of azeotrope forms?
  3. Water has a triple point at 0.01°C, 0.006 atm and a critical point at 374°C, 218 atm. The solid-liquid equilibrium line has a negative slope. Explain what the negative slope means and give one element on the phase diagram.

Score: ___/3

Part E: Thermochemistry

Target: 2–3 min per problem. | Exam weight: ~9%

Set E1 — Calorimetry & Enthalpy Definitions (2 problems)

  1. 0.562 g of graphite is burned in a bomb calorimeter with heat capacity 12.05 kJ/°C. The temperature rises from 25.00°C to 26.89°C. Calculate $\Delta U$ and $\Delta H$ per mole. (M = 12.0 g/mol)
  2. Define: (a) standard enthalpy of formation, (b) standard enthalpy of combustion, (c) enthalpy of neutralisation, (d) lattice energy.

Score: ___/2

Set E2 — Hess's Law & Bond Enthalpies (3 problems)

  1. Given:
    • $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H^\circ = -393.5$ kJ
    • $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$, $\Delta H^\circ = -285.8$ kJ
    • $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$, $\Delta H^\circ = -1366.8$ kJ Calculate $\Delta H^\circ_f$ of ethanol.
  2. Given bond enthalpies (kJ/mol): C–C = 347, C=C = 614, C–H = 413, H–H = 436. Calculate $\Delta H$ for: $\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$.
  3. State whether each reaction is spontaneous or non-spontaneous: (a) $\Delta H = -$, $\Delta S = +$; (b) $\Delta H = -$, $\Delta S = -$ at low T; (c) $\Delta H = +$, $\Delta S = -$ at all T.

Score: ___/3

Part F: Kinetic Chemistry

Target: 2–3 min per problem. | Exam weight: ~10% (historically heavy)

Set F1 — Rate Laws & Order Determination (4 problems)

  1. For the reaction $\text{S}_2\text{O}_8^{2-} + 3\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_3^-$, the initial rate data shows: doubling [$\text{S}_2\text{O}_8^{2-}$] doubles the rate; doubling [I$^-$] doubles the rate. Write the rate law and overall order.
  2. A reaction: [A] doubling causes an 8× rate increase. What is the order with respect to A
  3. $\text{N}_2\text{O}5$ decomposes by first-order kinetics with $k = 5.7 \times 10^{-4}$ s$^{-1}$. Calculate $t{1/2}$ and the time for 90% decomposition.
  4. Given concentration-time data for a reaction, which plot would you use to test for 2nd order? What slope tells you $k$?

Score: ___/4

Set F2 — Half-life & Integrated Rate Laws (2 problems)

  1. A first-order reaction has $t_{1/2} = 40$ minutes. What is $k$? How much remains after 120 minutes if $[A]_0 = 1.0$ M?
  2. A second-order reaction (one reactant) starts with $[A]0 = 0.50$ M and $k = 0.040$ M$^{-1}$s$^{-1}$. Calculate $t{1/2}$ and $[A]_t$ after 100 s.

Score: ___/2

Set F3 — Arrhenius & Mechanisms (3 problems)

  1. The rate constant doubles when T increases from 300 K to 310 K. Calculate $E_a$. (R = 8.314 J·mol$^{-1}$·K$^{-1}$)
  2. A proposed mechanism has a fast equilibrium step followed by a slow step. Write the rate law for: $2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ if the slow step is $\text{N}_2\text{O}_2 + \text{H}_2 \xrightarrow{k_2} \text{N}_2\text{O} + \text{H}_2\text{O}$ and the fast equilibrium is $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$.
  3. Given the Michaelis-Menten parameters $V_{max} = 100$ $\mu$M/min and $K_m = 5$ mM for an enzyme, calculate $v$ when $[S] = 5$ mM and when $[S] = 20$ mM.

Score: ___/3

Part G: Electrochemistry

Target: 2–3 min per problem. | Exam weight: ~12% (trending up, 28% in 2024/25)

Set G1 — Cell Potential & Notation (4 problems)

  1. Given: Mg$^{2+}$/Mg $E^\circ = -2.37$ V, Ca$^{2+}$/Ca $E^\circ = -2.76$ V. Which is the anode? Which is the cathode? Calculate $E^\circ_{cell}$ and state if spontaneous.
  2. Write the cell notation for the voltaic cell: Zn(s) | Zn$^{2+}$(aq) || Cu$^{2+}$(aq) | Cu(s). Identify anode and cathode. Given $E^\circ_{Zn^{2+}/Zn} = -0.76$ V, $E^\circ_{Cu^{2+}/Cu} = +0.34$ V, find $E^\circ_{cell}$.
  3. In a galvanic cell, where does oxidation occur? What is the charge on the anode? What is the function of the salt bridge?
  4. Given $E^\circ_{cell} = 1.10$ V for the Zn-Cu cell, calculate $\Delta G^\circ$ (F = 96,485 C·mol$^{-1}$, n = 2). Is the reaction spontaneous?

Score: ___/4

Set G2 — Nernst Equation (3 problems)

  1. For the Zn-Cu cell above, if $[Zn^{2+}] = 0.010$ M and $[Cu^{2+}] = 1.0$ M, use the Nernst equation at 25°C to calculate $E_{cell}$. ($E^\circ_{cell} = 1.10$ V, n = 2)
  2. A concentration cell has two Zn/Zn$^{2+}$ half-cells with $[Zn^{2+}] = 0.10$ M and 1.0 M. Calculate $E_{cell}$. Which side is the anode?
  3. A cell has $E^\circ_{cell} = 0.46$ V with n = 2. Calculate the equilibrium constant $K$ at 25°C.

Score: ___/3

Set G3 — Electrolysis & Faraday's Laws (3 problems)

  1. How many grams of copper are deposited by a current of 2.00 A flowing for 30 minutes through a CuSO$_4$ solution? (Cu = 63.5 g/mol, F = 96,485 C·mol$^{-1}$)
  2. Predict the products of electrolysis of: (a) molten NaCl, (b) aqueous NaCl (with Pt electrodes). Explain the difference.
  3. Explain the overpotential effect in the electrolysis of aqueous NaCl. Why can Cl$2$ be produced at the anode despite $E^\circ{Cl_2/Cl^-} > E^\circ_{O_2/H_2O}$?

Score: ___/3

Part H: Stereochemistry

Target: 2 min per problem. | Exam weight: ~8%

Set H1 — Chirality & Identification (3 problems)

  1. How many chiral centres does 2-bromobutane have? Determine whether it is chiral or achiral. Mark the chiral carbon(s) with *.
  2. The maximum number of stereoisomers for a compound with 3 chiral centres is?
  3. What is a meso compound? Give an example. Why is it optically inactive despite having chiral centres?

Score: ___/3

Set H2 — R/S & Fischer Projections (3 problems)

  1. For the molecule with chiral centre C(b)(Cl)(Br)(F)(H), assign priorities and determine R/S configuration.
  2. Draw the Fischer projection of (2R,3R)-butane-2,3-diol. How does it differ from the meso form?
  3. The specific rotation of pure (R)-enantiomer is $+25^\circ$. A mixture has observed rotation $+10^\circ$. Calculate the enantiomeric excess and the % of each enantiomer.

Score: ___/3

Part I: Alcohol & Phenol

Target: 2 min per problem. | Exam weight: ~5%

Set I1 — Alcohol Reactions (3 problems)

  1. Classify each as 1°, 2°, or 3° alcohol: (a) ethanol, (b) propan-2-ol, (c) 2-methylpropan-2-ol.
  2. Predict the product of: (a) ethanol + K$_2$Cr$_2$O$_7$/H$^+$ (excess), (b) propan-2-ol + PCC, (c) 2-methylpropan-2-ol + K$_2$Cr$_2$O$_7$/H$^+$.
  3. Propan-1-ol is heated with conc. H$_2$SO$_4$ at: (a) 140°C, (b) 180°C. What are the two different products? Name the reaction type.

Score: ___/3

Set I2 — Phenol Acidity & Tests (2 problems)

  1. Why is phenol ($pK_a \approx 10$) much more acidic than cyclohexanol ($pK_a \approx 18$)? Use resonance structures to explain.
  2. Phenol gives a white precipitate with bromine water. Draw the product. What colour does phenol give with FeCl$_3$? Does phenol dissolve in NaOH? In NaHCO$_3$?

Score: ___/2

Part J: Carbonyl Compounds — LEAK INCLUDED

Target: 2–3 min per problem. | Exam weight: ~7%

Set J1 — Nucleophilic Addition (3 problems)

  1. Cyclohexanone reacts with HCN. Draw the product (cyanohydrin). What is the mechanism type?
  2. Rank in order of reactivity toward nucleophilic addition: formaldehyde, acetaldehyde, acetone. Explain.
  3. Propanal reacts with NaBH$_4$. Draw the product. Write the reagent combination that would instead give the alkane product (reduction to hydrocarbon).

Score: ___/3

Set J2 — Identification Tests & Leak (4 problems)

  1. LEAK: What is Brady's reagent? Draw its structure (2,4-dinitrophenylhydrazine). What colour change do you observe when it reacts with a carbonyl compound?
  2. A compound gives: (i) orange ppt with 2,4-DNP, (ii) silver mirror with Tollens' reagent, (iii) yellow ppt with I$_2$/NaOH. Identify the functional group(s) present.
  3. Distinguish between benzaldehyde and acetophenone using two chemical tests.
  4. Propanal and propanone are in separate unlabelled bottles. Describe a test to identify which is which.

Score: ___/4

Set J3 — Aldol & Cannizzaro (2 problems)

  1. Two molecules of acetaldehyde react in dilute NaOH. Draw the product and name the reaction. On heating, what further product forms?
  2. Benzaldehyde treated with concentrated NaOH gives benzyl alcohol and benzoic acid. Name the reaction. Why does acetaldehyde not undergo this reaction?

Score: ___/2

Part K: Carboxylic Acids & Derivatives

Target: 2–3 min per problem. | Exam weight: ~6%

Set K1 — Acidity & Preparation (2 problems)

  1. Rank in order of increasing $pK_a$: acetic acid, chloroacetic acid, trichloroacetic acid. Explain the trend.
  2. Complete: (a) CH$_3$CN + H$_2$O (H$^+$, heat) → ?, (b) CH$_3$CH$_2$OH + K$_2$Cr$_2$O$_7$/H$^+$ (excess) → ?, (c) CH$_3$MgBr + CO$_2$, then H$^+$ → ?

Score: ___/2

Set K2 — Derivative Interconversions (3 problems)

  1. Rank the following in order of reactivity toward nucleophilic acyl substitution: acetyl chloride, acetic anhydride, ethyl acetate, acetamide, acetic acid. Justify.
  2. Draw the product of Fischer esterification between propanoic acid and ethanol with H$^+$ catalyst.
  3. The ester above undergoes saponification with NaOH. Draw the products. Why is saponification irreversible while Fischer esterification is reversible?

Score: ___/3

Part L: Amines & Amino Acids

Target: 2–3 min per problem. | Exam weight: ~7%

Set L1 — Amine Reactions (3 problems)

  1. Classify each as 1°, 2°, or 3° amine: (a) CH$_3$CH$_2$NH$_2$, (b) (CH$_3$)$_2$NH, (c) (CH$_3$)$_3$N.
  2. Predict the observation when each amine is treated with NaNO$_2$/HCl at 0–5°C: (a) CH$_3$NH$_2$, (b) C$_6$H$_5$NH$_2$, (c) (CH$_3$)$_2$NH, (d) C$_6$H$_5$N(CH$_3$)$_2$.
  3. Aniline reacts with bromine water. Draw the product. Why does aniline not require a Lewis acid catalyst for electrophilic aromatic substitution?

Score: ___/3

Set L2 — Amino Acids (3 problems)

  1. Draw the structure of alanine at: (a) low pH (below pI), (b) at pI (zwitterion), (c) high pH (above pI).
  2. $pK_{a1}$ (COOH) of glycine = 2.34, $pK_{a2}$ (NH$_3^+$) = 9.60. Calculate the isoelectric point (pI).
  3. At pH = 7.0, will glycine migrate toward the cathode or anode in electrophoresis? What about at pH = 1.0?

Score: ___/3

Part M: Polymer Chemistry

Target: 2 min per problem. | Exam weight: ~4%

Set M1 — Polymer Types & Reactions (3 problems)

  1. Classify each polymerisation as addition or condensation: (a) ethene → polyethene, (b) hexane-1,6-diamine + adipic acid → nylon 6,6, (c) styrene → polystyrene.
  2. Draw the repeating unit of PVC (from chloroethene monomer). What type of polymerisation is this?
  3. What is vulcanisation? How does it change the properties of natural rubber?

Score: ___/3

Part N: Biochemistry (Carbohydrates, Lipids, Nucleic Acids)

Target: 2 min per problem. | Exam weight: ~4%

Set N1 — Carbohydrates (2 problems)

  1. D-glucose has 4 chiral centres. How many stereoisomers are possible? How many are D-sugars?
  2. What is the difference between a reducing sugar and a non-reducing sugar? Give one example of each.

Score: ___/2

Set N2 — Lipids & Nucleic Acids (2 problems)

  1. Explain why saturated fatty acids have higher melting points than unsaturated fatty acids of the same chain length.
  2. Name the four bases found in DNA. Which are purines and which are pyrimidines? How many H-bonds form between A–T and G–C pairs?

Score: ___/2

Final Scorecard

Part Topic Problems Raw Score
A Chemical Equilibrium 6 ___/6
B Ionic Equilibria 12 ___/12
C Solubility Product 6 ___/6
D Phase Equilibria 7 ___/7
E Thermochemistry 5 ___/5
F Kinetic Chemistry 9 ___/9
G Electrochemistry 10 ___/10
H Stereochemistry 6 ___/6
I Alcohol & Phenol 5 ___/5
J Carbonyl Compounds (incl. LEAK) 9 ___/9
K Carboxylic Acids & Derivatives 5 ___/5
L Amines & Amino Acids 6 ___/6
M Polymer Chemistry 3 ___/3
N Biochemistry 4 ___/4
TOTAL Full Syllabus 93 ___/93

Proficiency Benchmarks

  • 65/93 (70%) — Proficient on most topics. Review the parts you scored lowest on.
  • 79/93 (85%) — Solid across the board. Your weak spots are small.
  • 86/93 (93%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <2 h 20 min: Excellent mechanical fluency.
  • 2 h 20 min – 3 h 30 min: Good pace. Review patterns that slowed you down.
  • >3 h 30 min: You're stalling. Drill the specific parts you scored lowest on again tomorrow.

Part-by-Part Diagnosis

If you scored low in... Your weak spot is... Suggested action
Part A Chemical Equilibrium Review W1 lecture, $K_c$/$K_p$ conversions, Le Chatelier
Part B Ionic Equilibria Review W2-W3 lectures, Tutorial 3 — Highest weight topic, every paper
Part C Solubility Product Review W4 lecture, selective precipitation problems
Part D Phase Equilibria Review W5-W6 lecture, Tutorial 5, colligative properties
Part E Thermochemistry Review W14-W15 lectures, Hess's Law, Born-Haber cycle
Part F Kinetic Chemistry Review W16 lectures, Tutorials 1-2, half-life patterns
Part G Electrochemistry Review L1-L5 lectures, Tutorial 4, drill pack — Trending up
Part H Stereochemistry Review W6 lecture, Tutorial 6, Fischer projections
Part I Alcohol & Phenol Review W7 lecture, Tutorial 7, Lucas test, oxidation
Part J Carbonyl Compounds Review W8-W10 lecture, Tutorial 8, LEAK: 2,4-DNP
Part K Carboxylic Acids Review W11 lecture, Tutorial 9, derivative reactivity
Part L Amines & Amino Acids Review W12 lecture, Tutorial 10, HNO$_2$ test, pI
Part M Polymer Chemistry Review W14 lecture, addition vs condensation
Part N Biochemistry Review Tutorials 12-14, reducing sugars, base pairing

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 28 forgot colligative

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Chemical Equilibrium

  1. $K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$

  2. $Q = \frac{(0.20)^2}{(0.10)^2(0.10)} = 40$, $Q > K_c$ → reverse (toward reactants)

  3. $K_c = K_p(RT)^{-\Delta n} = 0.82 / (0.0821 \times 300)^1 = 0.82 / 24.63 = 0.0333$

  4. (a) forward (fewer gas moles), (b) reverse (exothermic), (c) forward

  5. $[H^+] = \sqrt{K_a c} = \sqrt{1.0\times10^{-5} \times 0.10} = \sqrt{1.0\times10^{-6}} = 1.0\times10^{-3}$ M. pH = 3.00. $%\alpha = \frac{1.0\times10^{-3}}{0.10} = 1.0% < 10%$ ✓

  6. ICE table: N$_2$O$_4$ initial = 1.0/5.0 = 0.20 M. Change: -x, +2x. $2x = 0.060$, $x = 0.030$. $[\text{N}_2\text{O}4]{eq} = 0.20 - 0.030 = 0.170$ M. $K_c = \frac{(0.060)^2}{0.170} = 0.0212$

Part B — Ionic Equilibria

  1. $[H^+] = 0.01$ M, pH = 2.00

  2. $[H^+] = \sqrt{1.8\times10^{-5} \times 0.50} = \sqrt{9.0\times10^{-6}} = 3.0\times10^{-3}$ M. pH = $-\log(3.0\times10^{-3}) = 2.52$. $pK_a = -\log(1.8\times10^{-5}) = 4.74$

  3. $[OH^-] = \sqrt{1.8\times10^{-5} \times 0.20} = \sqrt{3.6\times10^{-6}} = 1.90\times10^{-3}$ M. pOH = 2.72, pH = 14 - 2.72 = 11.28

  4. $x^2/(0.10 - x) = 5.0\times10^{-4}$. $x^2 = 5.0\times10^{-5} - 5.0\times10^{-4}x$. $x^2 + 5.0\times10^{-4}x - 5.0\times10^{-5} = 0$. $x = \frac{-5.0\times10^{-4} + \sqrt{25\times10^{-8} + 2.0\times10^{-4}}}{2} = 6.84\times10^{-3}$ M. pH = 2.16

  5. (a) acidic (SA-WB), (b) basic (SB-WA), (c) neutral (SA-SB), (d) basic (SB-WA)

  6. $[H^+] = \sqrt{K_a c} = \sqrt{5.6\times10^{-10} \times 0.20} = \sqrt{1.12\times10^{-10}} = 1.06\times10^{-5}$ M. pH = 4.98

  7. $pH = 4.74 + \log(0.35/0.25) = 4.74 + \log(1.4) = 4.74 + 0.146 = 4.89$

  8. After HCl: [HA] = 0.26 M, [A$^-$] = 0.34 M. pH = 4.74 + log(0.34/0.26) = 4.74 + 0.117 = 4.86

  9. $pOH = 4.74 + \log(0.20/0.30) = 4.74 + \log(0.667) = 4.74 - 0.176 = 4.56$. pH = 14 - 4.56 = 9.44. After NaOH: [B] = 0.32 M, [BH$^+$] = 0.18 M. $pOH = 4.74 + \log(0.18/0.32) = 4.74 - 0.250 = 4.49$. pH = 9.51

  10. (a) pH = 13.00, (b) 25 mL: pH = 12.52, (c) 50 mL: pH = 7.00, (d) 60 mL: $[H^+] = \frac{10 \times 0.10}{110} = 0.00909$, pH = 2.04

  11. At equivalence: $[CH_3COO^-] = \frac{100\times0.10}{200} = 0.050$ M. $K_b = 10^{-14}/1.8\times10^{-5} = 5.56\times10^{-10}$. $[OH^-] = \sqrt{5.56\times10^{-10} \times 0.050} = 5.27\times10^{-6}$ M. pOH = 5.28, pH = 8.72. Indicator: phenolphthalein (range 8.3–10.0)

  12. At equivalence: $[NH_4^+] = \frac{40\times0.10}{80} = 0.050$ M. $K_a = 10^{-14}/1.8\times10^{-5} = 5.56\times10^{-10}$. $[H^+] = \sqrt{5.56\times10^{-10} \times 0.050} = 5.27\times10^{-6}$ M. pH = 5.28. Indicator: methyl red (range 4.2–6.3)

Part C — Solubility Product

  1. $K_{sp} = s^2 = (8.36\times10^{-14})^2 = 6.99\times10^{-27}$

  2. $K_{sp} = 4s^3$, $s = \sqrt[3]{2.56\times10^{-13}/4} = \sqrt[3]{6.4\times10^{-14}} = 4.0\times10^{-5}$ M

  3. $K_{sp} = [Ag^+]^2[CrO_4^{2-}] = (2s)^2(s) = 4s^3$. $s = \sqrt[3]{1.9\times10^{-12}/4} = 7.8\times10^{-5}$ M. $[Ag^+] = 1.56\times10^{-4}$ M, $[CrO_4^{2-}] = 7.8\times10^{-5}$ M

  4. $Q = [Mg^{2+}][OH^-]^2 = (1.5\times10^{-6})(1.0\times10^{-4})^2 = 1.5\times10^{-14} = K_{sp}$. Saturated — no ppt (at the boundary)

  5. AgCl: $[Cl^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.8\times10^{-10}}{0.010} = 1.8\times10^{-8}$ M. PbCl$_2$: $[Cl^-] = \sqrt{\frac{1.7\times10^{-5}}{0.020}} = 0.029$ M. AgCl precipitates first at $[Cl^-] = 1.8\times10^{-8}$ M

  6. (a) $s = \sqrt[3]{3.9\times10^{-11}/4} = \sqrt[3]{9.75\times10^{-12}} = 2.14\times10^{-4}$ M. (b) In 0.10 M F$^-$: $K_{sp} = [Ca^{2+}][F^-]^2 = s(0.10+2s)^2 \approx s(0.10)^2$. $s = 3.9\times10^{-11}/0.01 = 3.9\times10^{-9}$ M

Part D — Phase Equilibria

  1. $n_{glucose} = 218/180 = 1.211$ mol. $n_{water} = 460/18 = 25.56$ mol. $X_{glucose} = 1.211/(1.211+25.56) = 0.0452$. $\Delta P = 0.0452 \times 31.82 = 1.44$ mmHg. $P_{soln} = 30.38$ mmHg

  2. $n_{naph} = 1.60/128 = 0.0125$ mol. $m = 0.0125/0.020 = 0.625$ m. $\Delta T_f = 5.12 \times 0.625 = 3.20$°C. FP = 5.50 - 3.20 = 2.30°C

  3. $n_{EG} = 651/62 = 10.5$ mol. $m = 10.5/2.505 = 4.19$ m. $\Delta T_f = 1.86 \times 4.19 = 7.79$°C. FP = -7.79°C. $\Delta T_b = 0.512 \times 4.19 = 2.15$°C. BP = 102.15°C

  4. $M = 92$ g/mol. $c = 46.0/92 = 0.50$ M. $\Pi = 0.50 \times 0.0821 \times 298 = 12.23$ atm

  5. $P_{total} = 0.3(60) + 0.7(30) = 18 + 21 = 39$ kPa

  6. $P_{ideal} = X_{CS_2}P^\circ_{CS_2} + X_{acetone}P^\circ_{acetone}$. Observed (433) < ideal → negative deviation → maximum boiling azeotrope

  7. Negative slope means ice is less dense than water — solid-liquid equilibrium line slopes left. This is why ice floats.

Part E — Thermochemistry

  1. $q = C\Delta T = 12.05 \times 1.89 = 22.77$ kJ (released). $n = 0.562/12.0 = 0.04683$ mol. $\Delta U = -22.77/0.04683 = -486.2$ kJ/mol. $\Delta H \approx \Delta U$ (solids, negligible $\Delta n_g$)

  2. (a) Enthalpy change when 1 mol of compound forms from elements in standard states. (b) Enthalpy change when 1 mol burns completely in O$_2$. (c) Enthalpy change when acid and base neutralise to form 1 mol water. (d) Enthalpy change when gaseous ions form 1 mol ionic solid.

  3. Target: $2\text{C} + 3\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}$. $\Delta H_f = 2(-393.5) + 3(-285.8) - (-1366.8) = -787 - 857.4 + 1366.8 = -277.6$ kJ/mol

  4. Bonds broken: C=C (614) + H-H (436) + 4 C-H (4×413) = 614+436+1652 = 2702 kJ. Bonds formed: C-C (347) + 6 C-H (6×413) = 347+2478 = 2825 kJ. $\Delta H = 2702 - 2825 = -123$ kJ

  5. (a) spontaneous at all T, (b) spontaneous at low T, (c) non-spontaneous at all T

Part F — Kinetic Chemistry

  1. Rate $= k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$, overall order = 2

  2. Third order with respect to A (since $2^3 = 8$)

  3. $t_{1/2} = 0.693 / 5.7\times10^{-4} = 1216$ s = 20.3 min. For 90%: $\ln(1.0/0.1) = kt$, $\ln(10) = 5.7\times10^{-4}t$, $t = 2.303/5.7\times10^{-4} = 4040$ s = 67.3 min

  4. Plot $1/[A]$ vs $t$. Linear with positive slope $= k$.

  5. $k = 0.693/40 = 0.0173$ min$^{-1}$. $\ln([A]_t/1.0) = -0.0173 \times 120 = -2.076$. $[A]_t = 1.0 \times e^{-2.076} = 0.125$ M (or $\frac{1}{8}$ — three half-lives)

  6. $t_{1/2} = 1/(0.040 \times 0.50) = 50$ s. $1/[A]{100} = 1/0.50 + 0.040 \times 100 = 2.0 + 4.0 = 6.0$. $[A]{100} = 0.167$ M

  7. $\ln(2) = (E_a/8.314)(1/300 - 1/310)$. $0.693 = (E_a/8.314)(0.003333 - 0.003226) = (E_a/8.314)(1.075\times10^{-4})$. $E_a = 0.693 \times 8.314 / 1.075\times10^{-4} = 53,600$ J/mol = 53.6 kJ/mol

  8. $K_{eq} = [\text{N}_2\text{O}_2]/[\text{NO}]^2$, so $[\text{N}_2\text{O}2] = K{eq}[\text{NO}]^2$. Rate = $k_2[\text{N}_2\text{O}_2][\text{H}2] = k_2 K{eq}[\text{NO}]^2[\text{H}_2] = k[\text{NO}]^2[\text{H}_2]$

  9. $v = V_{max}[S]/(K_m+[S])$. At $[S]=5$ mM: $v = 100(5)/(5+5) = 50$ $\mu$M/min. At $[S]=20$ mM: $v = 100(20)/(20+5) = 80$ $\mu$M/min

Part G — Electrochemistry

  1. Ca has more negative $E^\circ$ → oxidation → anode. Mg → cathode. $E^\circ_{cell} = (-2.37) - (-2.76) = +0.39$ V. Spontaneous ($E^\circ_{cell} > 0$).

  2. Zn anode (oxidation, -0.76 V), Cu cathode (reduction, +0.34 V). $E^\circ_{cell} = 0.34 - (-0.76) = 1.10$ V

  3. Oxidation at anode (negative in galvanic). Salt bridge: maintains charge balance by allowing ion migration.

  4. $\Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96,485 \times 1.10 = -212,267$ J = $-212.3$ kJ. Spontaneous ($\Delta G^\circ < 0$)

  5. $E = 1.10 - \frac{0.0592}{2}\log\frac{0.010}{1.0} = 1.10 - 0.0296\log(0.01) = 1.10 - 0.0296(-2) = 1.10 + 0.0592 = 1.159$ V

  6. $E = 0 - \frac{0.0592}{2}\log\frac{0.10}{1.0} = -0.0296\log(0.1) = -0.0296(-1) = 0.0296$ V. Anode = dilute side (0.10 M)

  7. $\log K = \frac{nE^\circ}{0.0592} = \frac{2 \times 0.46}{0.0592} = 15.54$. $K = 10^{15.54} = 3.47 \times 10^{15}$

  8. $Q = It = 2.00 \times 30 \times 60 = 3600$ C. $m = \frac{3600 \times 63.5}{2 \times 96485} = \frac{228600}{192970} = 1.185$ g

  9. (a) Molten NaCl: cathode → Na(l), anode → Cl$_2$(g). (b) Aqueous NaCl: cathode → H$_2$(g) (H$^+$ discharged before Na$^+$), anode → Cl$_2$(g) (Cl$^-$ discharged before OH$^-$ due to overpotential)

  10. Overpotential: O$_2$ evolution requires extra 0.4–0.6 V due to slow kinetics. This shifts the effective potential of OH$^-$ oxidation above that of Cl$^-$, so Cl$_2$ forms preferentially even though $E^\circ$ suggests OH$^-$ should oxidise first.

Part H — Stereochemistry

  1. 2-bromobutane has 1 chiral centre (C2). Chiral. Structure: CH$_3$-*CH(Br)-CH$_2$-CH$_3$

  2. $2^3 = 8$ stereoisomers

  3. Meso compound: chiral centres but internal plane of symmetry (achiral overall). Example: meso-butane-2,3-diol (same substituents on both chiral centres, opposite configuration). Optically inactive because one half cancels the rotation of the other.

  4. Priorities: Br(35) > Cl(17) > F(9) > H(1). If 1→2→3 clockwise → R; anticlockwise → S

  5. (2R,3R)-butane-2,3-diol: Fischer projection with OH on right at C2 and right at C3. Meso form: OH on right at C2, left at C3 (internal plane of symmetry)

  6. $ee = (10/25) \times 100% = 40%$. (R) = 70%, (S) = 30%

Part I — Alcohol & Phenol

  1. (a) 1°, (b) 2°, (c) 3°

  2. (a) CH$_3$COOH (acetic acid), (b) CH$_3$COCH$_3$ (propanone/acetone), (c) no reaction (3° alcohol resists oxidation)

  3. (a) 140°C: ether (dipropyl ether) — intermolecular dehydration. (b) 180°C: alkene (propene) — intramolecular dehydration (E1, Zaitsev product)

  4. Phenoxide ion is resonance-stabilised — the negative charge is delocalised into the aromatic ring. Cyclohexanol alkoxide has no such resonance. This stabilisation makes phenol more willing to lose H$^+$.

  5. 2,4,6-tribromophenol (white precipitate). With FeCl$_3$: light purple colour. Phenol dissolves in NaOH (forms phenoxide) but NOT in NaHCO$_3$ (weaker acid than H$_2$CO$_3$).

Part J — Carbonyl Compounds

  1. Cyanohydrin: (cyclohexanone)-C(OH)(CN). Mechanism: nucleophilic addition.

  2. HCHO > CH$_3$CHO > CH$_3$COCH$_3$. Formaldehyde has least steric hindrance and most electrophilic carbonyl C. Ketones have two alkyl groups (steric + electronic hindrance).

  3. Propan-1-ol (CH$_3$CH$_2$CH$_2$OH). For alkane reduction: Clemmensen reduction (Zn(Hg)/HCl) or Wolff-Kishner reduction (NH$_2$NH$_2$/KOH, heat).

  4. Brady's reagent = 2,4-dinitrophenylhydrazine (2,4-DNPH). Orange/red precipitate (hydrazone) forms with aldehydes and ketones.

  5. Compound has: (i) carbonyl group (C=O), (ii) aldehyde group (Tollens' positive), (iii) methyl carbonyl or methyl carbinol group (iodoform positive). Most likely: CH$_3$CHO (acetaldehyde) or CH$_3$COCH$_3$ cannot be (no Tollens').

  6. Tollens': benzaldehyde → silver mirror, acetophenone → no reaction. Iodoform: acetophenone (methyl ketone) → yellow CHI$_3$, benzaldehyde → no reaction.

  7. Tollens' test: propanal → silver mirror; propanone → no reaction. Alternatively Fehling's test: propanal → brick-red Cu$_2$O; propanone → no reaction.

  8. 3-hydroxybutanal ($\beta$-hydroxy aldehyde) — aldol addition. On heating: crotonaldehyde (but-2-enal) — aldol condensation (dehydration).

  9. Cannizzaro reaction. Benzaldehyde has no $\alpha$-hydrogen, so it undergoes disproportionation instead of aldol. Acetaldehyde has $\alpha$-H, so it does aldol condensation instead.

Part K — Carboxylic Acids & Derivatives

  1. $pK_a$: trichloroacetic (lowest) < chloroacetic < acetic (highest). More Cl atoms = stronger $-I$ effect = more stabilised carboxylate = stronger acid = lower $pK_a$.

  2. (a) CH$_3$COOH (acetic acid), (b) CH$_3$COOH (acetic acid — overoxidation of 1° alcohol), (c) CH$_3$COOH (from Grignard + CO$_2$)

  3. Acetyl chloride > acetic anhydride > ethyl acetate ∼ acetic acid > acetamide. Justification: leaving group ability (Cl$^-$ > RCOO$^-$ > RO$^-$ ∼ OH$^-$ > NH$_2^-$). Good leaving groups make the carbonyl more electrophilic.

  4. CH$_3$CH$_2$COOCH$_2$CH$_3$ (ethyl propanoate) + H$_2$O

  5. Products: CH$_3$CH$_2$COO$^-$Na$^+$ (sodium propanoate) + CH$_3$CH$_2$OH (ethanol). Saponification is irreversible because the carboxylate product is resonance-stabilised and does not react with alcohols. Fischer esterification is reversible because the products (ester + water) can react back.

Part L — Amines & Amino Acids

  1. (a) 1° (RNH$_2$), (b) 2° (R$_2$NH), (c) 3° (R$_3$N)

  2. (a) CH$_3$NH$_2$: N$_2$ gas bubbles (1° aliphatic). (b) C$_6$H$_5$NH$_2$: stable diazonium salt at 0–5°C (no bubbles). (c) (CH$_3$)$_2$NH: yellow oil (N-nitrosamine). (d) C$_6$H$_5$N(CH$_3$)$_2$: solid precipitate (p-nitroso product)

  3. 2,4,6-tribromoaniline (white precipitate). The NH$_2$ group is strongly activating (ortho/para-directing, resonance donor), so the ring is highly activated — no Lewis acid needed for bromination.

  4. (a) Low pH: NH$_3^+$-CH(CH$_3$)-COOH (fully protonated, + charge). (b) pI: NH$_3^+$-CH(CH$_3$)-COO$^-$ (zwitterion, neutral). (c) High pH: NH$_2$-CH(CH$_3$)-COO$^-$ (deprotonated, - charge).

  5. pI = (2.34 + 9.60)/2 = 5.97

  6. At pH 7.0 (pH > pI = 5.97): negative charge → migrates toward anode (+). At pH 1.0 (pH < pI): positive charge → migrates toward cathode (-).

Part M — Polymer Chemistry

  1. (a) addition, (b) condensation, (c) addition
  2. Repeating unit: $-$CH$_2$-CHCl$-$ (from CH$_2$=CHCl). Addition polymerisation.
  3. Vulcanisation: heating natural rubber with sulfur. Creates cross-links between polymer chains → harder, more elastic, heat-resistant, less sticky.

Part N — Biochemistry

  1. $2^4 = 16$ stereoisomers. $2^{4-1} = 8$ D-sugars (all D-sugars have OH on right at the bottom chiral centre in Fischer projection).
  2. Reducing sugar: has a free anomeric carbon (aldehyde or hemiacetal) that can reduce Tollens'/Fehling's. Example: glucose (reducing), sucrose (non-reducing — both anomeric carbons in glycosidic bond).
  3. Saturated fatty acids have straight chains that pack tightly → stronger intermolecular forces → higher MP. Unsaturated fatty acids have cis double bonds that introduce kinks → prevent tight packing → lower MP.
  4. Purines: Adenine (A), Guanine (G). Pyrimidines: Cytosine (C), Thymine (T). A–T: 2 H-bonds. G–C: 3 H-bonds.

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