Organic Chemistry — Diagnostic Tests

Unit Tests

UT-1: $S_N1$ vs $S_N2$ Mechanism Prediction

Question: For each substrate, predict whether it will react faster via $S_N1$ or $S_N2$ with aqueous NaOH, and explain why: (a) $(\text{CH}_3)_3\text{CBr}$, (b) $\text{CH}_3\text{CH}_2\text{Br}$, (c) $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ (benzyl bromide).

Solution:

(a) $(\text{CH}_3)_3\text{CBr}$ (tertiary alkyl halide): $S_N1$ is favoured. The tertiary carbocation intermediate $(\text{CH}_3)_3\text{C}^+$ is stabilised by hyperconjugation from nine C--H bonds on the three methyl groups. Steric hindrance around the tertiary carbon makes the backside attack required for $S_N2$ extremely difficult.

(b) $\text{CH}_3\text{CH}_2\text{Br}$ (primary alkyl halide): $S_N2$ is favoured. The primary carbon has minimal steric hindrance, allowing the nucleophile ($\text{OH}^-$) easy backside access. A primary carbocation would be too unstable to form (no hyperconjugation stabilisation), making $S_N1$ unviable.

(c) $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ (benzyl bromide): Both mechanisms are fast, but $S_N1$ is particularly favoured. The benzyl carbocation $\text{C}_6\text{H}_5\text{CH}_2^+$ is exceptionally stabilised by resonance delocalisation of the positive charge into the benzene ring (the positive charge is delocalised over the ortho and para positions). Despite being a primary halide, the stability of the carbocation makes $S_N1$ competitive with $S_N2$.

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UT-2: E1 vs E2 Elimination and Zaitsev's Rule

Question: When 2-bromo-3-methylbutane is treated with a strong base (NaOEt in ethanol), two elimination products are possible. Draw both products, identify the major product using Zaitsev's rule, and explain whether E1 or E2 is the dominant mechanism.

Solution:

Substrate: $\text{CH}_3-\text{CH}(\text{Br})-\text{CH}(\text{CH}_3)_2$ (2-bromo-3-methylbutane).

E2 elimination removes H from a beta-carbon and Br from the alpha-carbon, forming a C=C double bond.

Possible products:

• H removed from $\text{C}_1$ ($\text{CH}_3$ group): 2-methylbut-2-ene (trisubstituted)
• H removed from the other beta-carbon (the $-\text{CH}(\text{CH}_3)_2$ group): 3-methylbut-1-ene (monosubstituted)

Major product: 2-methylbut-2-ene (Zaitsev's rule -- the more substituted alkene is more stable due to hyperconjugation and alkyl group electron-donating effects).

Mechanism: E2 is dominant because (1) NaOEt is a strong base, (2) EtOH is a polar protic solvent that favours E2 over $S_N2$ for secondary halides, and (3) E2 has a lower activation energy than E1 for secondary substrates. E1 would require carbocation formation and is disfavoured with a strong base present.

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UT-3: Functional Group Identification and Classification

Question: A compound $\text{C}_4\text{H}_8\text{O}$ has the following properties: (i) it reacts with 2,4-DNPH to form an orange precipitate, (ii) it gives a positive iodoform test, (iii) it does not react with Fehling's solution. Identify the compound, draw its structure, and explain each observation.

Solution:

(i) Positive 2,4-DNPH test: the compound contains a carbonyl group (C=O) -- it is either an aldehyde or a ketone.

(ii) Positive iodoform test: the compound has a $\text{CH}_3\text{CO}-$ group (methyl ketone) or a $\text{CH}_3\text{CH(OH)}-$ group that can be oxidised to a methyl ketone.

(iii) Negative Fehling's test: the compound is NOT an aldehyde (or not an aliphatic aldehyde). Since (ii) shows it is a methyl ketone, this is consistent.

The compound is butan-2-one ($\text{CH}_3\text{COCH}_2\text{CH}_3$).

Structure: $\text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_3$

Molecular formula check: $\text{C}_4\text{H}_8\text{O}$. Correct.

The iodoform reaction: $\text{CH}_3\text{COCH}_2\text{CH}_3 + 3\text{I}_2 + 4\text{NaOH} \to \text{CHI}_3 \downarrow + \text{CH}_3\text{CH}_2\text{COONa} + 3\text{NaI} + 3\text{H}_2\text{O}$. The yellow precipitate of iodoform ($\text{CHI}_3$) confirms the methyl ketone group.

Integration Tests

IT-1: Organic Synthesis and Redox (with Redox)

Question: Outline a two-step synthesis of benzoic acid from benzene using appropriate reagents and conditions. Write balanced equations for each step, identify oxidation number changes at the carbon atoms involved, and explain why direct oxidation of the alkyl side chain to a carboxylic acid is possible.

Solution:

Step 1: Friedel-Crafts alkylation (or acylation followed by reduction, but alkylation is simpler here):

$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$

This forms toluene (methylbenzene).

Step 2: Oxidation of the alkyl side chain:

$\text{C}_6\text{H}_5\text{CH}_3 + 2\text{KMnO}_4 \to \text{C}_6\text{H}_5\text{COOH} + 2\text{MnO}_2 + 2\text{KOH} + \text{H}_2\text{O}$

(Or using acidified $\text{K}_2\text{Cr}_2\text{O}_7$ as the oxidising agent.)

Oxidation number changes:

In toluene: the methyl carbon has oxidation state approximately $-3$ (as in $\text{CH}_3-$). In benzoic acid: the carboxyl carbon has oxidation state $+3$ (two bonds to O at $-2$ each, one bond to C at $0$, one bond to O at $0$ using average: $0 + 0 + (-2) + (-2) + 3 = -1$; actually for $-\text{COOH}$: the carbon is bonded to one C ($0$), one $=\text{O}$ ($-2 \times 2 = -4$) and one $-\text{O}^-$ ($-1 \times 1 = -1$); sum $= -5$ but with the C--C bond and corrections: $+3$). The change is from $-3$ to $+3$, a 6-electron oxidation.

Direct oxidation works because the benzylic position (carbon adjacent to the benzene ring) is activated. The benzene ring stabilises the intermediate benzylic radicals and cations through resonance, allowing the oxidising agent to sequentially remove hydrogen atoms from the methyl group, ultimately converting it to $-\text{COOH}$.

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IT-2: Reaction Mechanism and Kinetics (with Kinetics)

Question: The hydrolysis of 2-bromo-2-methylpropane follows first-order kinetics with respect to the alkyl halide but zero-order with respect to hydroxide. Explain this observation using the $S_N1$ mechanism. How would the rate change if the solvent was changed from water to a more polar solvent?

Solution:

The $S_N1$ mechanism has two steps:

Step 1 (rate-determining, slow): $(\text{CH}_3)_3\text{CBr} \to (\text{CH}_3)_3\text{C}^+ + \text{Br}^-$

Step 2 (fast): $(\text{CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \to (\text{CH}_3)_3\text{COH}_2^+$

Step 3 (fast): $(\text{CH}_3)_3\text{COH}_2^+ + \text{H}_2\text{O} \to (\text{CH}_3)_3\text{COH} + \text{H}_3\text{O}^+$

The rate is determined solely by step 1: $\text{rate} = k[(\text{CH}_3)_3\text{CBr}]$.

This is first-order in the alkyl halide and zero-order in $\text{OH}^-$ because the nucleophile ($\text{H}_2\text{O}$) only participates in the fast second step. The rate-determining step involves only the alkyl halide -- it is a unimolecular process.

Changing to a more polar solvent increases the rate. A more polar solvent stabilises the charged transition state (separated ions) and the carbocation intermediate more than it stabilises the neutral reactant. This lowers the activation energy for step 1, increasing the rate constant $k$. The Hammond postulate explains this: the transition state for an endothermic step resembles the products, so greater polar solvation of the ionic products means greater solvation of the transition state.

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IT-3: Stereochemistry and Nucleophilic Attack (with Chemical Bonding)

Question: $(R)$-2-bromobutane undergoes $S_N2$ reaction with NaCN. Draw the transition state, determine the configuration of the product, and explain the stereochemical outcome using orbital hybridisation.

Solution:

Substrate: $(R)$-2-bromobutane, $\text{CH}_3-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3$.

In the $S_N2$ transition state, the nucleophile ($\text{CN}^-$) approaches from the back side (opposite the leaving group $\text{Br}^-$) along the axis of the $sp^3$ orbital containing the C--Br bond. The carbon is partially bonded to both $\text{CN}$ and $\text{Br}$, with the three other substituents in a trigonal planar arrangement:

$\text{CN}^\delta-\cdots\text{C}^\delta+\cdots\text{Br}^\delta-$

with $\text{CH}_3$, $\text{H}$, and $\text{CH}_2\text{CH}_3$ in a plane.

As the $\text{CN}^-$ bonds and $\text{Br}^-$ leaves, the three substituents flip like an umbrella inverting. This produces $(S)$-2-methylbutanenitrile.

The stereochemical outcome is Walden inversion: the $S_N2$ mechanism at a chiral centre always inverts the configuration. The reaction goes through a single transition state (not an intermediate), so there is no opportunity for the configuration to racemise.

Using hybridisation: the central carbon's $sp^3$ orbital involved in the C--Br bond rehybridises towards $sp^2$ in the transition state (trigonal planar geometry), then back to $sp^3$ in the product with the nucleophile occupying the position vacated by the leaving group.