FAD1018: Kinetics through Biochemistry — Drill Guide

Step-by-step procedures mapped to each Part F through Part N question in the FAD1018 Comprehensive Drill — Full Syllabus.

Part F — Kinetic Chemistry

Q37: Rate Law from Initial Rate Data

Given: $\text{S}_2\text{O}_8^{2-} + 3\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_3^-$. Doubling $[\text{S}_2\text{O}_8^{2-}]$ doubles rate. Doubling $[\text{I}^-]$ doubles rate.

Procedure:

  1. Identify the effect of each reactant on the rate:
    • $[\text{S}_2\text{O}_8^{2-}] \times 2 \rightarrow \text{rate} \times 2 \rightarrow \text{order} = 1$
    • $[\text{I}^-] \times 2 \rightarrow \text{rate} \times 2 \rightarrow \text{order} = 1$
  2. Write rate law: $\text{Rate} = k[\text{S}_2\text{O}_8^{2-}]^1[\text{I}^-]^1$
  3. Overall order = sum of exponents = $1 + 1 = 2$

[!tip] If doubling concentration causes rate to ×1 → order 0; ×2 → order 1; ×4 → order 2; ×8 → order 3.

Q38: Order from Rate Change

Given: Doubling [A] causes 8× rate increase.

Procedure:

  1. Let order = $n$. The relationship is: $\displaystyle \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^n$
    • $[\text{A}]_1$ = initial concentration (call it $c$)
    • $[\text{A}]_2$ = doubled concentration ($2c$) — after the change
    • $\text{Rate}_1$ = rate at $[\text{A}]_1$
    • $\text{Rate}_2$ = new rate after concentration change
  2. Here $[\text{A}]_2/[\text{A}]_1 = 2$ (doubled) and $\text{Rate}_2/\text{Rate}_1 = 8$
  3. $8 = (2)^n \rightarrow n = 3$ (since $2^3 = 8$)
  4. The reaction is third order with respect to A

[!tip] Method of initial rates: $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^n$. The subscripts are: 1 = before, 2 = after the concentration change. Solve for $n$ using $\log$ if not obvious (e.g. $n = \log(8)/\log(2) = 3$).

Q39: First-Order Kinetics — Half-Life & Time

Given: $\text{N}_2\text{O}_5$ decomposes by first-order kinetics, $k = 5.7 \times 10^{-4}$ s$^{-1}$.

Procedure:

  1. Half-life: $\displaystyle t_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.7 \times 10^{-4}} = 1216$ s
  2. Convert to minutes: $1216 / 60 = 20.3$ min
  3. For 90% decomposition (10% remains): $\displaystyle \ln\frac{[\text{A}]_0}{[\text{A}]_t} = kt$
  4. $\displaystyle \ln\frac{1.0}{0.1} = \ln(10) = 2.303 = (5.7 \times 10^{-4})t$
  5. $t = 2.303 / 5.7 \times 10^{-4} = 4040$ s = 67.3 min

[!tip] First-order $t_{1/2}$ is constant — independent of concentration. After $n$ half-lives, fraction remaining $= (1/2)^n$.

Q40: Graphical Test for Reaction Order

Procedure:

  1. For each order, the linearising plot is:
Order Plot Slope Intercept
0 $[\text{A}]$ vs $t$ $-k$ $[\text{A}]_0$
1 $\ln[\text{A}]$ vs $t$ $-k$ $\ln[\text{A}]_0$
2 $1/[\text{A}]$ vs $t$ $+k$ $1/[\text{A}]_0$
  1. To test for 2nd order: plot $1/[\text{A}]$ vs $t$. If linear → second order. The slope = $k$.

[!warning] Don't confuse which plot gives which order. Mnemonic: "0-direct" (straight [A] vs t), "1-log" (ln[A] vs t), "2-reciprocal" (1/[A] vs t).

Q41: First-Order — Concentration After Time

Given: $t_{1/2} = 40$ min, $[\text{A}]_0 = 1.0$ M, $t = 120$ min.

Procedure:

  1. Find $k$: $k = 0.693 / 40 = 0.0173$ min$^{-1}$
  2. Use integrated law: $\ln([\text{A}]_t / [\text{A}]_0) = -kt$
  3. $\ln([\text{A}]_t / 1.0) = -0.0173 \times 120 = -2.076$
  4. $[\text{A}]_t = e^{-2.076} = 0.125$ M

Shortcut: 120 min = 3 half-lives. Fraction remaining $= (1/2)^3 = 1/8$. $[\text{A}]_t = 1.0/8 = 0.125$ M ✓

[!tip] For first-order reactions, always check if the time is a convenient multiple of $t_{1/2}$ first — the half-life shortcut is much faster.

Q42: Second-Order Half-Life & Concentration

Given: $[\text{A}]_0 = 0.50$ M, $k = 0.040$ M$^{-1}$s$^{-1}$, $t = 100$ s.

Procedure:

  1. Half-life: $\displaystyle t_{1/2} = \frac{1}{k[\text{A}]_0} = \frac{1}{0.040 \times 0.50} = \frac{1}{0.02} = 50$ s
  2. Concentration at $t = 100$ s: $\displaystyle \frac{1}{[\text{A}]_t} = \frac{1}{[\text{A}]_0} + kt$
  3. $\displaystyle \frac{1}{[\text{A}]_t} = \frac{1}{0.50} + 0.040 \times 100 = 2.0 + 4.0 = 6.0$
  4. $[\text{A}]_t = 1/6.0 = 0.167$ M

[!warning] Second-order $t_{1/2}$ increases as concentration decreases (unlike first-order where it's constant). Each successive half-life is longer.

Q43: Arrhenius — Activation Energy

Given: Rate constant doubles ($k_2/k_1 = 2$) when $T$ increases from 300 K to 310 K. R = 8.314 J·mol$^{-1}$·K$^{-1}$.

Procedure:

  1. Arrhenius equation: $\displaystyle \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
  2. $\displaystyle \ln(2) = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right)$
  3. $0.693 = \frac{E_a}{8.314}(0.003333 - 0.003226) = \frac{E_a}{8.314}(1.075 \times 10^{-4})$
  4. $E_a = \frac{0.693 \times 8.314}{1.075 \times 10^{-4}} = 53,600$ J/mol $= 53.6$ kJ/mol

[!tip] Keep $E_a$ in J/mol for the calculation, then convert to kJ/mol at the end. R is 8.314 J·mol$^{-1}$·K$^{-1}$, not 0.0821!

Q44: Mechanism — Rate Law from Fast Equilibrium + Slow Step

Given: $2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$. Fast equilibrium: $2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ Slow step: $\text{N}_2\text{O}_2 + \text{H}_2 \xrightarrow{k_2} \text{N}_2\text{O} + \text{H}_2\text{O}$

Procedure:

  1. Rate law from slow (rate-determining) step: $\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2]$
  2. But $[\text{N}_2\text{O}_2]$ is an intermediate — eliminate it using the fast equilibrium
  3. From equilibrium: $K_{eq} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2}$, so $[\text{N}_2\text{O}2] = K{eq}[\text{NO}]^2$
  4. Substitute: $\text{Rate} = k_2 K_{eq}[\text{NO}]^2[\text{H}_2] = k[\text{NO}]^2[\text{H}_2]$

[!tip] The slow step determines the form of the rate law. Always substitute out intermediates using equilibrium expressions from fast steps before them.

Q45: Michaelis-Menten Enzyme Kinetics

Given: $V_{max} = 100$ $\mu$M/min, $K_m = 5$ mM. Find $v$ when $[S] = 5$ mM and $[S] = 20$ mM.

Procedure:

  1. Michaelis-Menten equation: $\displaystyle v = \frac{V_{max}[S]}{K_m + [S]}$
  2. At $[S] = 5$ mM: $v = \frac{100 \times 5}{5 + 5} = \frac{500}{10} = 50$ $\mu$M/min
  3. At $[S] = 20$ mM: $v = \frac{100 \times 20}{20 + 5} = \frac{2000}{25} = 80$ $\mu$M/min

[!tip] When $[S] = K_m$, $v = V_{max}/2$ — that's the definition of $K_m$. Low $K_m$ = high affinity (enzyme reaches half-max rate at lower substrate concentration).

Part G — Electrochemistry

Q46: Identifying Anode & Cathode

Given: Mg$^{2+}$/Mg $E^\circ = -2.37$ V, Ca$^{2+}$/Ca $E^\circ = -2.76$ V.

Procedure:

  1. The more negative reduction potential is the stronger reducing agent → gets oxidised → anode
  2. Anode (oxidation): Ca ($E^\circ = -2.76$ V) — more negative
  3. Cathode (reduction): Mg ($E^\circ = -2.37$ V) — less negative
  4. $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-2.37) - (-2.76) = +0.39$ V
  5. $E^\circ_{cell} > 0$ → spontaneous

[!tip] Mnemonic: AN OX (Anode = Oxidation), RED CAT (Reduction = Cathode). The cell potential formula is always $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.

Q47: Cell Notation

Given: Zn(s) | Zn$^{2+}$(aq) || Cu$^{2+}$(aq) | Cu(s). $E^\circ_{Zn^{2+}/Zn} = -0.76$ V, $E^\circ_{Cu^{2+}/Cu} = +0.34$ V.

Procedure:

  1. Cell notation convention: Anode || Cathode
    • Left (anode): Zn(s) | Zn$^{2+}$(aq) — oxidation
    • Right (cathode): Cu$^{2+}$(aq) | Cu(s) — reduction
  2. Single vertical line | = phase boundary. Double line || = salt bridge
  3. $E^\circ_{cell} = 0.34 - (-0.76) = 1.10$ V

[!tip] In cell notation: solid electrodes go on the outside, aqueous ions on the inside. The double line || separates the two half-cells.

Q48: Galvanic Cell Theory

Answers:

  • Oxidation occurs at the anode
  • The anode has negative charge (in a galvanic/voltaic cell — electrons are produced here)
  • Salt bridge functions:
    1. Completes the circuit (ionic connection)
    2. Maintains charge neutrality in both half-cells
    3. Allows ions to migrate (anions to anode, cations to cathode)
    4. Prevents mixing of the two solutions

[!warning] In electrolytic cells, the anode is positive (electrons are pulled away). Don't mix up galvanic vs electrolytic anode signs!

Q49: $\Delta G^\circ$ from $E^\circ_{cell}$

Given: $E^\circ_{cell} = 1.10$ V, $n = 2$, F = 96,485 C·mol$^{-1}$.

Procedure:

  1. $\Delta G^\circ = -nFE^\circ_{cell}$
  2. $\Delta G^\circ = -2 \times 96,485 \times 1.10 = -212,267$ J
  3. Convert to kJ: $-212.3$ kJ/mol
  4. $\Delta G^\circ < 0$ → spontaneous

[!tip] $\Delta G^\circ$ in J, convert to kJ by dividing by 1000. $n$ = moles of electrons transferred (balanced half-reactions).

Q50: Nernst Equation

Given: Zn-Cu cell, $E^\circ_{cell} = 1.10$ V, $n = 2$, $[Zn^{2+}] = 0.010$ M, $[Cu^{2+}] = 1.0$ M, 25°C.

Procedure:

  1. Nernst equation at 25°C: $\displaystyle E = E^\circ - \frac{0.0592}{n}\log Q$
  2. For Zn-Cu cell: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, so $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$
  3. $E = 1.10 - \frac{0.0592}{2}\log\frac{0.010}{1.0}$
  4. $= 1.10 - 0.0296\log(0.01) = 1.10 - 0.0296(-2)$
  5. $= 1.10 + 0.0592 = 1.159$ V

[!warning] The reaction quotient $Q$ has the same form as $K_c$ but with non-equilibrium concentrations. Products over reactants, each raised to stoichiometric coefficient.

Q51: Concentration Cell

Given: Two Zn/Zn$^{2+}$ half-cells, $[Zn^{2+}] = 0.10$ M and 1.0 M.

Procedure:

  1. Concentration cell: same electrodes, different concentrations — $E^\circ_{cell} = 0$
  2. $E = 0 - \frac{0.0592}{n}\log\frac{[\text{Zn}^{2+}]{\text{dilute}}}{[\text{Zn}^{2+}]{\text{concentrated}}}$
  3. $E = -\frac{0.0592}{2}\log\frac{0.10}{1.0} = -0.0296\log(0.1) = -0.0296(-1) = 0.0296$ V
  4. Anode = dilute side (0.10 M) — oxidation occurs where concentration is lower

[!tip] Concentration cells always have $E^\circ_{cell} = 0$. The cell potential arises purely from the concentration gradient. The dilute side is the anode (oxidation).

Q52: $E^\circ_{cell}$ to Equilibrium Constant $K$

Given: $E^\circ_{cell} = 0.46$ V, $n = 2$, 25°C.

Procedure:

  1. At equilibrium: $\displaystyle E^\circ_{cell} = \frac{0.0592}{n}\log K$
  2. $\log K = \frac{nE^\circ}{0.0592} = \frac{2 \times 0.46}{0.0592} = 15.54$
  3. $K = 10^{15.54} = 3.47 \times 10^{15}$

[!tip] $E^\circ$ and $K$ are related: a positive $E^\circ$ means $K > 1$ (products favoured). The larger $E^\circ$, the larger $K$.

Q53: Faraday's Law — Mass Deposited

Given: $I = 2.00$ A, $t = 30$ min, CuSO$_4$ solution, Cu = 63.5 g/mol, F = 96,485 C·mol$^{-1}$.

Procedure:

  1. Cu$^{2+} + 2e^- \rightarrow$ Cu(s), so $n$ (electrons) = 2
  2. Charge: $Q = I \times t = 2.00 \times 30 \times 60 = 3600$ C
  3. Mass: $\displaystyle m = \frac{Q \times M}{n \times F} = \frac{3600 \times 63.5}{2 \times 96,485}$
  4. $= \frac{228,600}{192,970} = 1.185$ g

[!warning] Time must be in seconds (30 min × 60 = 1800 s... wait, $30 \times 60 = 1800$, but above uses $30 \times 60 = 1800$ — let me recalculate: $Q = 2.00 \times 1800 = 3600$ C. Yes, that's correct. Always convert minutes to seconds!

Q54: Electrolysis — Molten vs Aqueous

Given: (a) molten NaCl, (b) aqueous NaCl with Pt electrodes.

Procedure:

Molten NaCl:

  • Cathode (reduction): Na$^+ + e^- \rightarrow$ Na($l$) — only Na$^+$ available
  • Anode (oxidation): $2$Cl$^- \rightarrow$ Cl$_2(g) + 2e^-$ — only Cl$^-$ available

Aqueous NaCl:

  • Cathode: Compare reduction potentials. H$_2$O reduction ($E^\circ = -0.83$ V) is less negative than Na$^+$ reduction ($E^\circ = -2.71$ V). So H$_2$O gets reduced instead:
    • $2$H$_2$O$(l) + 2e^- \rightarrow$ H$_2(g) + 2$OH$^-$ — H$_2$ gas evolved
  • Anode: Compare oxidation potentials. Cl$^-$ ($E^\circ = -1.36$ V) vs H$_2$O ($E^\circ = -1.23$ V). H$_2$O should oxidise first thermodynamically, BUT overpotential favours Cl$_2$ production.
    • $2$Cl$^- \rightarrow$ Cl$_2(g) + 2e^-$ — Cl$_2$ gas evolved

[!tip] The difference between molten and aqueous is the presence of water. In aqueous solution, H$_2$O can compete with Na$^+$ at the cathode (H$_2$O wins — less energy required).

Q55: Overpotential Effect

Given: Aqueous NaCl electrolysis — why Cl$_2$ at anode despite $E^\circ$ suggesting O$_2$?

Explanation:

  1. The thermodynamic oxidation potential: H$_2$O → O$_2$ ($E^\circ = -1.23$ V) seems easier than Cl$^-$ → Cl$_2$ ($E^\circ = -1.36$ V)
  2. However: O$_2$ evolution has a high overpotential (0.4–0.6 V extra needed) due to slow kinetics of the 4-electron transfer: $2$H$_2$O → O$_2 + 4$H$^+ + 4e^-$
  3. The effective potential for O$_2$ becomes approximately $-1.23 - 0.5 = -1.73$ V — now worse than Cl$_2$ at $-1.36$ V
  4. So Cl$^-$ gets discharged first despite the standard potentials suggesting otherwise

[!tip] Overpotential = kinetic barrier. Some reactions that look thermodynamically favoured need extra voltage to overcome slow kinetics. O$_2$ evolution is notoriously slow.

Part H — Stereochemistry

Q56: Identifying Chiral Centres

Given: 2-bromobutane.

Procedure:

  1. Draw the structure: CH$_3$-CH(Br)-CH$_2$-CH$_3$
  2. Identify carbons with 4 different substituents:
    • C1: 3 H + CH$_3$ group — not chiral
    • C2: CH$_3$, Br, H, CH$_2$CH$_3$ — all four different → chiral ✓
    • C3: 2 H, CH$_2$, CH$_3$ — not chiral
    • C4: 3 H + CH$_3$ — not chiral
  3. 1 chiral centre → compound is chiral (no internal plane of symmetry)

[!tip] A carbon with 4 different groups = chiral centre. Look for carbons bonded to 4 distinct atoms/groups. Carbons with CH$_2$, CH$_3$, or double bonds are never chiral.

Q57: Maximum Number of Stereoisomers

Given: 3 chiral centres.

Procedure:

  1. Maximum number = $2^n$, where $n$ = number of chiral centres
  2. $2^3 = 8$ stereoisomers

[!warning] $2^n$ gives the maximum. Meso compounds (internal symmetry) reduce this number. Meso compounds are achiral despite having chiral centres.

Q58: Meso Compounds

Definition: A meso compound has chiral centres but is achiral overall due to an internal plane of symmetry.

Example: meso-butane-2,3-diol. Both chiral centres have the same substituents (OH, H, CH$_3$), but opposite configurations (R,S or S,R). The molecule has a mirror plane through the C2-C3 bond.

Why optically inactive: The rotation caused by one chiral centre is exactly cancelled by the other (equal magnitude, opposite direction).

[!tip] Meso compounds are the exception to "chiral centres = chiral molecule." Look for symmetry!

Q59: R/S Configuration

Given: C(b)(Cl)(Br)(F)(H) — chiral centre with Br, Cl, F, H.

Procedure:

  1. Assign priorities by atomic number (highest → priority 1):
    • Br (35) > Cl (17) > F (9) > H (1)
  2. Orient the molecule so the lowest priority (H) is facing away (dashed wedge)
  3. Trace 1 → 2 → 3:
    • Clockwise = R (from Latin rectus)
    • Anticlockwise = S (from Latin sinister)
  4. If H is facing toward you: 1 → 2 → 3 clockwise = S (reverse it)

[!tip] When the lowest priority is NOT pointing away, the R/S assignment is reversed. If 1→2→3 is clockwise with H toward you, the true configuration is S.

Q60: Fischer Projections & Meso

Given: (2R,3R)-butane-2,3-diol vs the meso form.

Procedure:

  1. Fischer projection of (2R,3R):
    • Both OH groups on the same side (e.g. both right)
    • Vertical = bonds going away (dashes)
    • Horizontal = bonds coming toward (wedges)
  2. Meso form (2R,3S):
    • OH on opposite sides (e.g. right at C2, left at C3)
    • Internal mirror plane between C2-C3
  3. (2R,3R) is optically active. Meso is optically inactive.

[!tip] In Fischer projections: horizontal bonds come out of the page, vertical bonds go behind. Rotating 180° in the plane is allowed; rotating 90° is NOT.

Q61: Enantiomeric Excess & Optical Purity

Given: Pure (R)-enantiomer $[\alpha] = +25^\circ$. Mixture observed $[\alpha] = +10^\circ$.

Procedure:

  1. Enantiomeric excess: $\displaystyle ee = \frac{[\alpha]{\text{observed}}}{[\alpha]{\text{pure}}} \times 100%$
  2. $ee = \frac{+10}{+25} \times 100% = 40%$
  3. This means 40% is excess (R), and the remaining 60% is a racemic mixture (50:50 R:S)
  4. %(R) = 40 + (60/2) = 70%
  5. %(S) = 60/2 = 30%

[!tip] $ee$ tells you how much more of one enantiomer exists relative to the racemic background. $ee = |%R - %S|$.

Part I — Alcohol & Phenol

Q62: Classifying Alcohols

Given: (a) ethanol, (b) propan-2-ol, (c) 2-methylpropan-2-ol.

Procedure:

  1. Identify the carbon bearing the -OH group
  2. Count how many carbon atoms are attached to that carbon:
    • (a) CH$_3$CH$_2$OH — OH carbon has 1 C → 1° (primary)
    • (b) CH$_3$CH(OH)CH$_3$ — OH carbon has 2 C → 2° (secondary)
    • (c) (CH$_3$)$_3$COH — OH carbon has 3 C → 3° (tertiary)

[!tip] 1° = OH on C with 1 alkyl group; 2° = OH on C with 2 alkyl groups; 3° = OH on C with 3 alkyl groups. Methyl alcohol (CH$_3$OH) is also 1°.

Q63: Alcohol Oxidation Products

Given: (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$^+$.

Procedure:

  1. Strong oxidant (K$_2$Cr$_2$O$_7$/H$^+$, excess):
    • 1° alcohol → carboxylic acid (overoxidation)
    • 2° alcohol → ketone
    • 3° alcohol → no reaction
  2. Mild oxidant (PCC):
    • 1° alcohol → aldehyde (stops there)
    • 2° alcohol → ketone
    • 3° alcohol → no reaction
Alcohol K$_2$Cr$_2$O$_7$/H$^+$ (excess) PCC
Ethanol (1°) CH$_3$COOH (acetic acid) CH$_3$CHO (ethanal)
Propan-2-ol (2°) CH$_3$COCH$_3$ (propanone) CH$_3$COCH$_3$ (propanone)
2-methylpropan-2-ol (3°) No reaction No reaction

[!tip] 3° alcohols have no H on the OH-carbon, so oxidation is impossible without breaking C-C bonds. They are oxidation-resistant.

Q64: Dehydration of Alcohols

Given: Propan-1-ol + conc. H$_2$SO$_4$.

Procedure:

Temperature Reaction Product Type
140°C Intermolecular dehydration Dipropyl ether (CH$_3$CH$_2$CH$_2$OCH$_2$CH$_2$CH$_3$) Ether formation
180°C Intramolecular dehydration Propene (CH$_3$CH=CH$_2$) E1 elimination (Zaitsev product)

[!warning] Temperature determines the product! Low T → ether (SN2 between two alcohols). High T → alkene (E1 elimination). Always check the temperature.

Q65: Phenol Acidity — Resonance Explanation

Given: Phenol ($pK_a \approx 10$) vs cyclohexanol ($pK_a \approx 18$).

Explanation:

  1. When phenol loses H$^+$, it forms the phenoxide ion
  2. The negative charge is delocalised into the aromatic ring via resonance:
    • The lone pair on O can be pushed into the ring
    • The charge is spread across ortho and para positions
  3. Cyclohexanol's alkoxide ion has no resonance — the negative charge is entirely on oxygen
  4. The phenoxide ion is more stable (charge spread out) → phenol is more willing to lose H$^+$ → more acidic

[!tip] Resonance stabilisation of the conjugate base = stronger acid. This is the single most important concept for explaining relative acidity in organic chemistry.

Q66: Phenol Tests

Answers:

  • Bromine water: White precipitate of 2,4,6-tribromophenol (no Lewis acid needed — the -OH group strongly activates the ring)
  • FeCl$_3$ test: Light purple/violet colour (characteristic of phenols)
  • NaOH: Phenol dissolves → forms water-soluble phenoxide (it's acidic enough to react with strong base)
  • NaHCO$_3$: Phenol does NOT dissolve (phenol is a weaker acid than H$_2$CO$_3$, so it can't displace CO$_2$ from bicarbonate)

[!tip] The NaHCO$_3$ test distinguishes carboxylic acids (dissolve, CO$_2$ bubbles) from phenols (no reaction). Carboxylic acids have $pK_a \sim 4-5$, phenols $\sim 10$.

Part J — Carbonyl Compounds

Q67: Cyanohydrin Formation (Nucleophilic Addition)

Given: Cyclohexanone + HCN.

Procedure:

  1. HCN is a weak acid that partially dissociates: HCN $\rightleftharpoons$ H$^+$ + CN$^-$
  2. CN$^-$ (nucleophile) attacks the electrophilic carbonyl carbon
  3. The C=O $\pi$ bond breaks, electrons go to O, forming alkoxide
  4. Alkoxide picks up H$^+$ to form the cyanohydrin: $\text{C}$(OH)(CN)

[!tip] Cyanohydrin formation adds CN and OH across the carbonyl C=O. It's a classic nucleophilic addition — the characteristic reaction of aldehydes and ketones.

Q68: Carbonyl Reactivity Trend

Given: Formaldehyde, acetaldehyde, acetone.

Rank (most → least reactive): HCHO > CH$_3$CHO > CH$_3$COCH$_3$

Explanation:

  1. Steric factor: Aldehydes have one small R group (or H), ketones have two. More substituents = more crowded = harder for nucleophile to attack
  2. Electronic factor: Alkyl groups are electron-donating ( + I effect). Two alkyl groups on the ketone make the carbonyl carbon less electrophilic (more electron density). Formaldehyde (two H atoms, no alkyl donation) is the most electrophilic

[!tip] Formaldehyde is the most reactive carbonyl because H atoms don't donate electrons and don't block the approach of nucleophiles.

Q69: NaBH$_4$ Reduction & Full Reduction to Alkane

Given: Propanal + NaBH$_4$.

Procedure:

  1. NaBH$_4$ is a source of hydride ions (H$^-$)
  2. H$^-$ attacks the carbonyl carbon → alkoxide forms
  3. Workup (H$_2$O/H$^+$) gives the alcohol:
    • Propanal (CH$_3$CH$_2$CHO) → Propan-1-ol (CH$_3$CH$_2$CH$_2$OH)
  4. To go all the way to the alkane (propane):
    • Clemmensen reduction: Zn(Hg)/HCl (acidic conditions)
    • Wolff-Kishner reduction: NH$_2$NH$_2$/KOH, heat (basic conditions)

[!warning] NaBH$_4$ only reduces aldehydes and ketones to alcohols — not to alkanes! To get alkanes, you need Clemmensen (acidic) or Wolff-Kishner (basic).

Q70: Brady's Reagent (LEAK)

Given: 2,4-dinitrophenylhydrazine (2,4-DNPH).

Answers:

  • Brady's reagent = a solution of 2,4-dinitrophenylhydrazine in methanol/H$_2$SO$_4$
  • Structure: H$_2$N-NH-C$_6$H$_3$(NO$_2$)$_2$ (hydrazine group attached to 2,4-dinitrophenyl ring)
  • Test result: Reacts with any aldehyde or ketone → orange/red hydrazone precipitate

[!tip] This is the universal test for carbonyl compounds — both aldehydes and ketones give a positive result (unlike Tollens'/Fehling's which only detect aldehydes).

Q71: Functional Group Identification

Given: (i) orange ppt with 2,4-DNP, (ii) silver mirror with Tollens', (iii) yellow ppt with I$_2$/NaOH.

Deduction:

  1. 2,4-DNP (+) → has a carbonyl (aldehyde or ketone)
  2. Tollens' (+) → must be an aldehyde (ketones give no silver mirror)
  3. Iodoform (+) → has either CH$_3$CO- (methyl ketone) or CH$_3$CH(OH)- (methyl carbinol)
  4. Combined: an aldehyde with a methyl group adjacent to carbonyl → ethanal (CH$_3$CHO) or any aldehyde with CH$_3$CO- group

[!tip] The iodoform test on an aldehyde specifically identifies acetaldehyde (CH$_3$CHO) or any aldehyde with a methyl group alpha to the carbonyl.

Q72: Distinguishing Benzaldehyde vs Acetophenone

Procedure:

  1. Tollens' test:
    • Benzaldehyde → silver mirror (aldehyde)
    • Acetophenone → no reaction (ketone)
  2. Iodoform test (I$_2$/NaOH):
    • Benzaldehyde → no yellow precipitate (not a methyl ketone)
    • Acetophenone → yellow CHI$_3$ precipitate (methyl ketone: C$_6$H$_5$COCH$_3$)

[!tip] Two orthogonal tests: Tollens' distinguishes aldehyde vs ketone. Iodoform detects the CH$_3$CO- group. Together they can identify both.

Q73: Distinguishing Propanal vs Propanone

Procedure:

  1. Tollens' test:
    • Propanal → silver mirror (aldehyde)
    • Propanone → no reaction (ketone)
  2. Fehling's test (alternative):
    • Propanal → brick-red Cu$_2$O precipitate
    • Propanone → no reaction

[!tip] Tollens' and Fehling's are selective for aldehydes only. They're the fastest way to distinguish an aldehyde from a ketone.

Q74: Aldol Addition & Condensation

Given: Two molecules of acetaldehyde in dilute NaOH.

Procedure:

  1. Step 1 (Aldol Addition): NaOH generates enolate from one acetaldehyde. Enolate attacks the carbonyl of a second acetaldehyde molecule
    • Product: 3-hydroxybutanal ($\beta$-hydroxy aldehyde) — also called "aldol"
  2. Step 2 (Aldol Condensation — on heating): The $\beta$-hydroxy aldehyde dehydrates (loses H$_2$O)
    • Product: crotonaldehyde (but-2-enal) — $\alpha,\beta$-unsaturated aldehyde

[!tip] An aldol reaction has two possible products: the $\beta$-hydroxy carbonyl (addition) and the $\alpha,\beta$-unsaturated carbonyl (condensation = addition + dehydration). Heating drives dehydration.

Q75: Cannizzaro Reaction

Given: Benzaldehyde + concentrated NaOH.

Answers:

  1. Reaction name: Cannizzaro reaction (also spelled Cannizzaro)
  2. Products: Benzyl alcohol (C$_6$H$_5$CH$_2$OH) + Sodium benzoate (C$_6$H$_5$COO$^-$Na$^+$)
  3. Why acetaldehyde doesn't undergo it: Acetaldehyde has $\alpha$-hydrogens. In concentrated base, it undergoes aldol condensation instead. Cannizzaro requires the aldehyde to have no $\alpha$-hydrogen (like benzaldehyde, formaldehyde).

[!warning] Cannizzaro is a disproportionation — one aldehyde molecule is reduced (to alcohol) while another is oxidised (to carboxylate). This only works for aldehydes without $\alpha$-H.

Part K — Carboxylic Acids & Derivatives

Q76: Acid Strength & Inductive Effect

Given: Acetic acid, chloroacetic acid, trichloroacetic acid.

Rank (increasing $pK_a$ = decreasing acidity): Trichloroacetic acid (lowest $pK_a$) < Chloroacetic acid < Acetic acid (highest $pK_a$)

Explanation:

  1. Cl is electronegative — it pulls electron density via the $-I$ effect (inductive electron withdrawal)
  2. More Cl atoms → stronger $-I$ effect → more stabilised the conjugate base (carboxylate anion)
  3. A more stable conjugate base → stronger acid → lower $pK_a$
  4. CCl$_3$COOH has 3 Cl atoms ($-I$ strongest) → strongest acid
  5. CH$_3$COOH has no Cl → weakest acid

[!tip] Inductive effect falls off with distance. The closer the electronegative atom, the stronger the effect. More electronegative substituents → stronger acid.

Q77: Preparing Carboxylic Acids

Given: (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$^+$.

Answers:

  1. (a) Nitrile hydrolysis: CH$_3$CN + 2H$_2$O $\xrightarrow{\text{H}^+, \Delta}$ CH$_3$COOH + NH$_4^+$
  2. (b) Alcohol oxidation (excess strong oxidant): CH$_3$CH$_2$OH → CH$_3$CHO → CH$_3$COOH (doesn't stop at aldehyde)
  3. (c) Grignard + CO$_2$: CH$_3$MgBr + CO$_2$ → CH$_3$COO$^-$MgBr$^+$ $\xrightarrow{\text{H}^+}$ CH$_3$COOH

[!tip] Three key routes to carboxylic acids: nitrile hydrolysis, 1° alcohol oxidation, and Grignard + CO$_2$. The Grignard route adds one carbon to the chain!

Q78: Reactivity of Carboxylic Acid Derivatives

Given: Acetyl chloride, acetic anhydride, ethyl acetate, acetamide, acetic acid.

Rank (most → least reactive): Acetyl chloride > Acetic anhydride > Ethyl acetate ∼ Acetic acid > Acetamide

Justification: Reactivity toward nucleophilic acyl substitution depends on the leaving group ability:

  • Cl$^-$ (best leaving group, weak base, stable anion)
  • RCOO$^-$ (carboxylate, resonance stabilised)
  • RO$^-$ (alkoxide, strong base — poor leaving group)
  • OH$^-$ (hydroxide, strong base)
  • NH$_2^-$ (amide ion, extremely strong base — worst leaving group)

[!tip] Better leaving group = more reactive derivative. The acyl group (C=O) is more electrophilic when it has a good leaving group attached.

Q79: Fischer Esterification

Given: Propanoic acid + ethanol + H$^+$ catalyst.

Procedure:

  1. Acid catalyses the reaction by protonating the carbonyl oxygen (makes C more electrophilic)
  2. Ethanol (nucleophile) attacks the carbonyl carbon
  3. Proton transfer, then elimination of H$_2$O
  4. Product: Ethyl propanoate (CH$_3$CH$_2$COOCH$_2$CH$_3$) + H$_2$O

[!tip] Fischer esterification is reversible. To drive it forward: use excess alcohol or remove water (Dean-Stark trap). The ester forms from the acid (provides C=O) and alcohol (provides OR group).

Q80: Saponification vs Fischer Esterification

Given: Ethyl propanoate + NaOH.

Products:

  • Sodium propanoate (CH$_3$CH$_2$COO$^-$Na$^+$) + Ethanol (CH$_3$CH$_2$OH)

Why saponification is irreversible:

  1. The hydroxide nucleophile attacks the ester carbonyl
  2. The tetrahedral intermediate collapses, expelling ethoxide (RO$^-$)
  3. Ethoxide immediately picks up H$^+$ from the aqueous medium → ethanol
  4. The carboxylate anion (RCOO$^-$) is resonance-stabilised — it has no electrophilic carbonyl to react back with alcohol
  5. In Fischer esterification, the products (ester + H$_2$O) can react back because the carbonyl is still present

[!tip] Saponification is essentially irreversible because the carboxylate product is a dead end — its negative charge is delocalised and it won't react with an alcohol.

Part L — Amines & Amino Acids

Q81: Classifying Amines

Given: (a) CH$_3$CH$_2$NH$_2$, (b) (CH$_3$)$_2$NH, (c) (CH$_3$)$_3$N.

Procedure:

  1. Count how many carbon groups are directly attached to the nitrogen:
    • (a) 1 C—N bond → 1° (primary amine)
    • (b) 2 C—N bonds → 2° (secondary amine)
    • (c) 3 C—N bonds → 3° (tertiary amine)

[!tip] The classification of amines refers to the number of carbon groups on nitrogen, not on carbon. NH$_3$ is 0°, RNH$_2$ is 1°, R$_2$NH is 2°, R$_3$N is 3°.

Q82: HNO$_2$ (Nitrous Acid) Test

Given: NaNO$_2$/HCl at 0–5°C with different amines.

Answers:

Amine Type Observation
CH$_3$NH$_2$ 1° aliphatic N$_2$ gas bubbles (visible effervescence)
C$_6$H$_5$NH$_2$ (aniline) 1° aromatic Stable diazonium salt at 0–5°C (no bubbles, cold solution)
(CH$_3$)$_2$NH 2° Yellow oil (N-nitrosamine forms)
C$_6$H$_5$N(CH$_3$)$_2$ 3° aromatic Green solid (p-nitroso compound — aromatic ring nitrosation)

[!tip] The HNO$_2$ test is excellent for distinguishing amine classes. 1° aliphatic = N$_2$ gas (bubbles). 1° aromatic = diazonium (no bubbles, useful for coupling). 2° = yellow oil. 3° aromatic = coloured solid.

Q83: Aniline + Bromine Water

Given: Aniline (C$_6$H$_5$NH$_2$) + bromine water.

Answer:

  • Product: White precipitate of 2,4,6-tribromoaniline
  • Why no Lewis acid needed: The -NH$_2$ group is a strongly activating (ortho/para-directing) group. It donates electron density into the ring via resonance, making the ring so electron-rich that bromination proceeds without a catalyst

[!tip] Aniline is so activated that it undergoes tribromination instantly with bromine water — all three ortho/para positions get substituted. No FeBr$_3$ or AlBr$_3$ needed!

Q84: Amino Acid Forms at Different pH

Given: Alanine at (a) low pH, (b) pI, (c) high pH.

Procedure:

  1. At low pH (pH < pI): Both -COOH and -NH$_3^+$ are protonated
    • Structure: NH$_3^+$-CH(CH$_3$)-COOH (net + charge)
  2. At pI (isoelectric point): -COOH is deprotonated, -NH$_2$ is protonated
    • Structure: NH$_3^+$-CH(CH$_3$)-COO$^-$ (zwitterion, net 0 charge)
  3. At high pH (pH > pI): Both -COOH and -NH$_3^+$ are deprotonated
    • Structure: NH$_2$-CH(CH$_3$)-COO$^-$ (net − charge)

[!tip] At pI, the molecule is neutral overall. At pH < pI, it's positively charged (moves toward cathode). At pH > pI, it's negatively charged (moves toward anode).

Q85: Calculating pI

Given: Glycine: $pK_{a1}$ (COOH) = 2.34, $pK_{a2}$ (NH$_3^+$) = 9.60.

Procedure:

  1. For neutral amino acids (one -NH$2$, one -COOH): $\displaystyle pI = \frac{pK{a1} + pK_{a2}}{2}$
  2. $pI = \frac{2.34 + 9.60}{2} = \frac{11.94}{2} = 5.97$

[!tip] For acidic amino acids (extra -COOH), pI averages the two most acidic $pK_a$ values. For basic amino acids (extra -NH$_2$), pI averages the two most basic $pK_a$ values.

Q86: Electrophoresis Direction

Given: Glycine at pH 7.0 and pH 1.0. pI = 5.97.

Procedure:

  1. Compare the pH to the pI:
    • pH 7.0 > pI 5.97 → negative charge → moves toward anode (+)
    • pH 1.0 < pI 5.97 → positive charge → moves toward cathode (−)

[!tip] pH > pI → negative → to anode (+). pH < pI → positive → to cathode (−). At pI → no net movement.

Part M — Polymer Chemistry

Q87: Addition vs Condensation Polymerisation

Given: (a) ethene → polyethene, (b) hexane-1,6-diamine + adipic acid → nylon 6,6, (c) styrene → polystyrene.

Procedure:

  1. Addition polymers: Formed from monomers with C=C double bonds. No small molecule lost.
    • (a) Polyethene: CH$_2$=CH$_2$ → $-$CH$_2$-CH$_2-]_n$ — addition
    • (c) Polystyrene: C$_6$H$_5$CH=CH$_2$ → $-$CH(C$_6$H$_5$)-CH$_2-]_n$ — addition
  2. Condensation polymers: Formed from monomers with two functional groups. Small molecule lost (usually H$_2$O).
    • (b) Nylon 6,6: diamine + diacid → amide bond + H$_2$O — condensation

[!tip] Addition = C=C monomers, no by-product. Condensation = two different functional groups react, small molecule (H$_2$O, HCl, etc.) lost.

Q88: PVC Repeating Unit

Given: Chloroethene (vinyl chloride, CH$_2$=CHCl).

Answer:

  • Repeating unit: $-$CH$_2$-CHCl$-$
  • Type: Addition polymerisation (free radical mechanism, the C=C opens up)

[!tip] The repeating unit looks exactly like the monomer, just with the C=C replaced by a C-C single bond. Draw brackets around it with "n" subscript.

Q89: Vulcanisation

Definition: Heating natural rubber (polyisoprene) with sulfur.

What it does:

  1. Sulfur atoms form cross-links (disulfide bridges, -S-S-) between polymer chains
  2. The cross-links restrict chain movement, preventing the rubber from becoming sticky when hot or brittle when cold

Property changes:

Property Before (raw rubber) After (vulcanised)
Elasticity Poor, sticky when hot Good, resilient
Temperature sensitivity Melts when hot, brittle when cold Stable over wider range
Strength Weak Stronger
Chemical resistance Low Higher

[!tip] Vulcanisation turns natural rubber from a sticky, temperature-sensitive mess into a useful, elastic material. More sulfur = harder rubber (ebonite is highly vulcanised).

Part N — Biochemistry

Q90: Glucose — Chiral Centres & Stereoisomers

Given: D-glucose has 4 chiral centres.

Answers:

  1. Maximum stereoisomers: $2^4 = 16$ (all possible combinations of R/S at 4 chiral centres)
  2. Number of D-sugars: $2^{4-1} = 2^3 = 8$
    • The D/L designation refers to the chiral centre farthest from the carbonyl (C5 in glucose)
    • D-sugars have -OH on the right in Fischer projection at that carbon
    • Half of all stereoisomers are D, half are L

[!tip] D-glucose is the naturally occurring form. L-glucose has the opposite configuration at C5 but is rare in nature.

Q91: Reducing vs Non-reducing Sugars

Answers:

  • Reducing sugar: Has a free anomeric carbon (aldehyde or hemiacetal) that can reduce Tollens'/Fehling's reagent
    • Example: Glucose (open-chain form has aldehyde, or cyclic form has hemiacetal that can open)
    • All monosaccharides are reducing sugars
  • Non-reducing sugar: The anomeric carbon is involved in a glycosidic bond — can't open to form aldehyde
    • Example: Sucrose (glucose + fructose bonded through both anomeric carbons)

[!tip] To test if a sugar is reducing: look at the glycosidic bond. If BOTH anomeric carbons are involved in the bond (like sucrose), it's non-reducing. If at least one is free, it's reducing.

Q92: Saturated vs Unsaturated Fatty Acid Melting Points

Given: Same chain length, saturated vs unsaturated.

Explanation:

  1. Saturated fatty acids: Straight hydrocarbon chains → pack tightly together → strong van der Waals forces → higher melting point
  2. Unsaturated fatty acids: Cis double bonds introduce kinks/bends in the chain → chains can't pack tightly → weaker intermolecular forces → lower melting point

[!tip] More double bonds = more kinks = lower melting point. This is why butter (more saturated) is solid at room temperature but olive oil (more unsaturated) is liquid.

Q93: DNA Bases

Answers:

  1. Four bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T)
  2. Purines (double-ring): Adenine and Guanine
  3. Pyrimidines (single-ring): Cytosine and Thymine
  4. Hydrogen bonds:
    • A–T: 2 H-bonds
    • G–C: 3 H-bonds

[!tip] Purines are larger (two rings). Pyrimidines are smaller (one ring). A purine always pairs with a pyrimidine. The number of H-bonds determines DNA stability — G-C rich regions are harder to denature (3 bonds vs 2).

Quick Reference Table

Q Topic Key Formula / Concept Common Trap
37 Rate law from data Rate $\propto$ [conc]$^{\text{order}}$ Doubling $\to$ ×8 means order 3
38 Order by inspection $2^n = \text{rate ratio}$ Use logs if not obvious
39 First-order $t_{1/2}$ $t_{1/2} = 0.693/k$ $t_{1/2}$ is constant for 1st order
40 Graphical order test Linear plot → confirms order $1/[A]$ vs $t$ for 2nd order
41 First-order decay $[A]_t = [A]_0 e^{-kt}$ Check half-life shortcut first
42 Second-order $t_{1/2} = 1/k[A]_0$ $t_{1/2}$ increases with time
43 Arrhenius $E_a$ $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ R = 8.314, not 0.0821
44 Mechanism rate law Rate = $k_2K_{eq}[NO]^2[H_2]$ Substitute intermediates using equilibrium
45 Michaelis-Menten $v = V_{max}[S]/(K_m+[S])$ At $[S]=K_m$, $v=V_{max}/2$
46 Cell potential $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ More negative $E^\circ$ = anode
47 Cell notation Anode $|$ $|$ Cathode Solid on outside, ions inside
48 Galvanic cell theory Anode = $-$ (galvanic), $+$ (electrolytic) Signs flip between cell types
49 $\Delta G^\circ$ from $E^\circ$ $\Delta G^\circ = -nFE^\circ$ J → kJ divide by 1000
50 Nernst equation $E = E^\circ - (0.0592/n)\log Q$ 25°C only, $Q$ = products/reactants
51 Concentration cell $E^\circ = 0$, anode = dilute side Dilute side = oxidation
52 $E^\circ$ to $K$ $\log K = nE^\circ/0.0592$ $E^\circ > 0 \to K > 1$
53 Faraday's law $m = QM/nF$ Time in seconds
54 Molten vs aqueous H$_2$O competes in aqueous H$_2$ produced at cathode in aqueous
55 Overpotential O$_2$ has high kinetic barrier Favours Cl$_2$ over O$_2$ at anode
56 Chiral centres 4 different groups on C CH$_2$ and CH$_3$ are never chiral
57 Max stereoisomers $2^n$ Meso compounds reduce this
58 Meso compounds Internal symmetry → achiral Has chiral centres but not chiral
59 R/S assignment 1→2→3 clockwise = R Reverse if H is facing toward you
60 Fischer projections Horizontal = wedges (out) Can't rotate 90°
61 Enantiomeric excess $ee = [\alpha]{obs}/[\alpha]{pure}$ Remaining = racemic
62 Alcohol classification Count C's on OH-carbon 1° = 1 C, 2° = 2 C, 3° = 3 C
63 Alcohol oxidation 1°→acid (excess), 2°→ketone, 3°→NR PCC stops at aldehyde
64 Dehydration 140°C = ether, 180°C = alkene Temperature determines product
65 Phenol acidity Resonance stabilises phenoxide More resonance = stronger acid
66 Phenol tests Br$_2$ → white ppt, FeCl$_3$ → purple No NaHCO$_3$ reaction
67 Cyanohydrin Nucleophilic addition: C=O → C(OH)(CN) CN$^-$ attacks carbonyl C
68 Carbonyl reactivity HCHO > RCHO > R$_2$CO Steric + electronic factors
69 Reduction to alkane Clemmensen (acid) or Wolff-Kishner (base) NaBH$_4$ only gives alcohol
70 Brady's reagent 2,4-DNPH → orange ppt LEAK — universal carbonyl test
71 Functional group ID 2,4-DNP + Tollens' + iodoform Aldehyde + methyl ketone features
72 Benzaldehyde vs acetophenone Tollens' + iodoform Two orthogonal tests
73 Propanal vs propanone Tollens' or Fehling's Aldehyde only
74 Aldol Addition → $\beta$-hydroxy; heat → $\alpha,\beta$-unsaturated Heating drives dehydration
75 Cannizzaro No $\alpha$-H required Disproportionation
76 Inductive effect More Cl → stronger acid $-I$ stabilises conjugate base
77 Carboxylic acid prep Nitrile hydrolysis, oxidation, Grignard+CO$_2$ Grignard adds one carbon
78 Derivative reactivity Cl > anhydride > ester ∼ acid > amide Leaving group quality
79 Fischer esterification Acid + alcohol $\rightleftharpoons$ ester + H$_2$O Reversible
80 Saponification Ester + OH$^-$ → carboxylate + alcohol Irreversible (resonance-stabilised product)
81 Amine classification Count C's on N, not on C 3° = 3 carbons on N
82 HNO$_2$ test 1° aliphatic = N$_2$ gas; 1° aromatic = diazonium 2° → yellow oil
83 Aniline + Br$_2$ 2,4,6-tribromoaniline No catalyst needed
84 Amino acid forms Low pH = +, pI = 0, high pH = $-$ Charge determines electrophoresis
85 pI calculation $(pK_{a1}+pK_{a2})/2$ for neutral Different formula for acidic/basic
86 Electrophoresis pH > pI → cathode? No, anode (+) pH > pI = negative = to anode
87 Addition vs condensation C=C vs functional group reaction Nylon is condensation
88 PVC repeating unit $-$CH$_2$-CHCl$-]$_n$ Addition polymerisation
89 Vulcanisation Sulfur cross-links Improves elasticity and stability
90 Glucose stereoisomers $2^4 = 16$, 8 are D-sugars D/L = last chiral centre
91 Reducing sugar Free anomeric carbon Sucrose is non-reducing
92 Fatty acid MP Saturated = straight → tight packing → high MP Cis bonds introduce kinks
93 DNA bases A-T (2 bonds), G-C (3 bonds) Purines: A, G; Pyrimidines: C, T

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