General Organic Chemistry (GOC)

Question 1

(a) Proof of $2^n$ (4 marks) Each independent stereocentre has exactly 2 configurations ($R$ or $S$). Configurations are chosen independently, so by the multiplication principle the number of configurational arrangements is $2\times2\times\cdots = 2^n$. (2) Every stereoisomer has a unique mirror image obtained by inverting all centres ($R\leftrightarrow S$ at each). Since no internal symmetry, mirror image $\ne$ original, so they pair up: $2^n/2 = 2^{n-1}$ enantiomeric pairs. (2)

(b) Tartaric acid enumeration (6 marks) Two stereocentres (C2, C3), formula naive count $2^2 = 4$. (1) Configurations: $(R,R)$, $(S,S)$, $(R,S)$, $(S,R)$.

• $(R,R)$ and $(S,S)$ are the enantiomeric pair $(+)$ and $(-)$ tartaric acid — both optically active. (2)
• $(R,S)$ and $(S,R)$ are identical (superimposable) because the molecule has an internal mirror plane between C2–C3 (same two halves): this is the single meso compound, optically inactive (internally compensated). (2) Total distinct stereoisomers $= 3$ (a $\pm$ pair + one meso); 2 are optically active, 1 is inactive. (1)

(c) Code (7 marks) Closed forms:

• No symmetry: $N = 2^n$.
• Symmetric, $n$ even: $N = 2^{n-1} + 2^{n/2 - 1}$ (the added term counts meso forms; pairs $(config,\text{mirror})$ that coincide).
• Symmetric, $n$ odd: central carbon lies on the mirror plane and is not a stereocentre in the symmetric case; $N = 2^{n-1} + 2^{(n-1)/2}$.

def count_stereoisomers(n, symmetric=False):
    if not symmetric:
        return 2**n
    if n % 2 == 0:              # even: no central carbon on plane
        return 2**(n-1) + 2**(n//2 - 1)
    else:                       # odd: central carbon on mirror plane
        return 2**(n-1) + 2**((n-1)//2)

Marks: correct non-symmetric (1), even meso formula (2), odd formula (2), working code (1), explanation of odd subtlety — the central carbon sits on the symmetry axis so its two "halves" are mirror-related, giving extra meso forms (1). Check: $n=2$ symmetric $\to 2^1 + 2^0 = 3$ ✓ (tartaric acid).

(d) Specific rotation (3 marks) $[\alpha] = \dfrac{\alpha_{obs}}{l\cdot c}$. Only the $(+)$ form rotates (meso inactive). Let mass fraction of active form $= f$, so its concentration $= 0.20f$ g/mL. $1.20 = 12.0 \times 1 \times (0.20 f) \Rightarrow 1.20 = 2.4 f \Rightarrow f = 0.50.$ Mass fraction $(+)$-form $= 0.50$ (50%). (3)

Question 2

(a) Stability ordering (6 marks) Increasing stability: $\text{CH}_3^+ < (\text{CH}_3)_2\text{CH}^+ < \text{CH}_2=CH\text{-}CH_2^+ \approx (\text{CH}_3)_3\text{C}^+ < \text{C}_6\text{H}_5\text{CH}_2^+$ Reasoning: $\text{CH}_3^+$ has no stabilisation ($n_\beta=0$). Isopropyl ($n_\beta=6$) > methyl via hyperconjugation + $+I$. tert-Butyl ($n_\beta=9$) most stabilised of the simple alkyls. Allyl is resonance-stabilised (+M, 2 equivalent structures) — comparable to/greater than tert-butyl. Benzyl has resonance delocalisation over the aromatic ring (multiple resonance structures) → most stable. (6) [accept allyl vs t-Bu either order with justification].

(b) Model fitting (8 marks) With $n_{res}=0$ for both, $\Delta E = a\,n_\beta$. Ethyl: $36 = 3a$; t-butyl: $68 = 9a$. (2) Least-tension: use both. From ethyl $a=12$; from t-Bu $a = 68/9 = 7.56$. Best linear fit through both (least squares, intercept 0): $a = \frac{\sum n_\beta E}{\sum n_\beta^2} = \frac{3(36)+9(68)}{9+81} = \frac{108+612}{90} = \frac{720}{90} = 8.0.$ So $a = 8.0$ kcal/mol per $\beta$-C–H. (3) Predict isopropyl ($n_\beta=6$): $\Delta E = 8.0\times6 = 48$ kcal/mol. (2) Experiment $= 54$; error $= 54-48 = 6$ kcal/mol ($\approx 11\%$ underestimate). The purely additive $n_\beta$ model ignores that hyperconjugation per bond is non-linear and neglects $+I$ saturation; hence deviation. (1)

(c) Neopentyl rearrangement (6 marks) $(\text{CH}_3)_3\text{C-CH}_2^+$ (primary) → a methyl group migrates with its bonding pair from the quaternary carbon to the adjacent cationic carbon (1,2-methyl shift): $(\text{CH}_3)_3\text{C-}\overset{+}{\text{CH}}_2 \xrightarrow{\text{1,2-CH}_3 \text{ shift}} (\text{CH}_3)_2\overset{+}{\text{C}}\text{-CH}_2\text{CH}_3$ producing the tertiary tert-pentyl cation. (3) Classification: 1,2-methyl shift. (1) Driving force: primary → tertiary carbocation, large gain in stability (hyperconjugation + $+I$). (1) The methyl migrates with its electron pair because the reaction is heterolytic (curved arrow = 2 electrons); the σ C–CH₃ bonding pair moves to form the new bond, leaving the positive charge on the now-tertiary carbon. (Homolysis would give radicals, not observed here.) (1)

Question 3

(a) C₄H₈ isomers (8 marks) DoU $= (2\cdot4+2-8)/2 = 1$ → one ring OR one double bond. (1) Alkenes (chain/position/geometrical):

1. but-1-ene $\text{CH}_2=CH\text{-}CH_2\text{-}CH_3$
2. cis-but-2-ene ($Z$)
3. trans-but-2-ene ($E$)
4. 2-methylprop-1-ene (isobutylene) Cyclic:
5. cyclobutane
6. methylcyclopropane (5, ½ each accept) Relationships: but-1-ene vs but-2-ene = position isomers; but-2-ene vs 2-methylpropene = chain; cis/trans-but-2-ene = geometrical (cis–trans); alkenes vs cycloalkanes = functional/ring–chain isomers. (2)

(b) DoU derivation (6 marks) Saturated acyclic $\text{C}_nH_{2n+2}$ is the reference (max H). Each degree of unsaturation (ring or $\pi$-bond) removes 2 H. (2) For $\text{C}_nH_{2n}$: missing H $= (2n+2)-2n = 2$, so DoU $= 2/2 = 1$. (1) General: $N$ (trivalent, like C-analog but 3 bonds) adds capacity for +1 H each → $+N$; halogen $X$ (monovalent, like H) → $-X$; O (divalent) inserts into a bond without changing H count → does not appear. (2) $\text{DoU} = \frac{2C+2+N-H-X}{2}.$ (1)

(c) Code + application (6 marks)

def dou(C, H, N=0, X=0):
    return (2*C + 2 + N - H - X) // 2

(i) Benzene $\text{C}_6H_6$: $(12+2-6)/2 = 4$ → 3 $\pi$-bonds + 1 ring (aromatic). (2) (ii) Pyridine $\text{C}_5H_5N$: $(10+2+1-5)/2 = 4$ → ring + 3 $\pi$ (aromatic N-heterocycle). (2) (iii) Chloroacetic acid $\text{C}_2H_3\text{ClO}_2$: $(4+2-3-1)/2 = 1$ → one $\pi$-bond (the C=O). (2)

[
  {"claim":"Tartaric acid symmetric n=2 gives 3 stereoisomers",
   "code":"def cs(n,sym):\n    if not sym: return 2**n\n    return 2**(n-1)+ (2**(n//2-1) if n%2==0 else 2**((n-1)//2))\nresult = (cs(2,True)==3) and (cs(3,True)==6) and (cs(4,False)==16)"},
  {"claim":"Specific rotation mass fraction of active form is 0.50",
   "code":"f = symbols('f'); sol = solve(Eq(1.20, 12.0*1*(0.20*f)), f); result = abs(sol[0]-0.5) < 1e-9"},
  {"claim":"Least-squares slope a=8.0 and isopropyl prediction 48 kcal/mol",
   "code":"a = (3*36+9*68)/(9+81); result = (a==8.0) and (a*6==48.0)"},
  {"claim":"DoU values: benzene 4, pyridine 4, chloroacetic acid 1",
   "code":"def dou(C,H,N=0,X=0): return (2*C+2+N-H-X)//2\nresult = dou(6,6)==4 and dou(5,5,1)==4 and dou(2,3,0,1)==1"}
]