Interleaved — Phase 4

Chemistry interleaved practice

printable — key stays hidden on paper

Instructions: Solve each problem, choosing the correct concept/method carefully. These problems deliberately mix subtopics — identify what each one is really asking before you begin. Show mechanisms and reasoning where required. Total: 50 marks.

1. Give the IUPAC name of the following compound: $CH_3-CH(CH_3)-CH_2-CH=CH-CHO$ (assume the double bond is cis). Include correct E/Z or cis/trans designation. (5 marks)

2. Rank the following carbocations in order of increasing stability, and explain your ordering in terms of the dominant electronic effect(s): $CH_3^+$ , $(CH_3)_3C^+$ , $CH_2=CH-CH_2^+$ , $C_6H_5-CH_2^+$ . (5 marks)

3. 2-bromo-2-methylbutane is treated with (a) a strong bulky base ( $KOtBu$ ) and (b) a small strong base ( $NaOEt$ ). Predict the major alkene in each case and name the operative mechanism. (5 marks)

4. Determine whether the following molecule is chiral, achiral, or a meso compound: 2,3-dibromobutane in its $(2R,3S)$ form. Justify. (5 marks)

5. Propan-1-ol reacts with (a) $HBr$ and (b) $B_2H_6$ then $H_2O_2/OH^-$ applied to propene. For each of the following, state the product and the regiochemistry rule that governs it: propene $+ HBr$ (no peroxide); propene $+ HBr$ (peroxide); propene by hydroboration-oxidation. (6 marks)

6. Is the cyclopentadienyl anion aromatic? Apply Hückel's rule with a full electron count and comment on hybridization of the ring carbons. (4 marks)

7. Chlorobenzene undergoes electrophilic nitration. State (a) whether $Cl$ is activating or deactivating, (b) the directing outcome (o/p vs m), and (c) explain the apparent contradiction between (a) and (b). (5 marks)

8. Draw the two staggered and two eclipsed conformations of n-butane about the $C2-C3$ bond as Newman projections (describe them), and rank all four by energy, naming the anti and gauche forms. (5 marks)

9. A terminal alkyne $CH_3-C\equiv CH$ is subjected to (a) $H_2O/H_2SO_4/HgSO_4$ and (b) $NaNH_2$ then $CH_3I$ . Give the products and explain why the terminal proton is acidic. (5 marks)

10. For the reaction of $(S)$ -2-bromobutane with $NaOH$ in aqueous acetone at low concentration vs. high concentration of $OH^-$ , predict which mechanism (SN1 vs SN2) dominates in each and the stereochemical outcome at the stereocentre. (5 marks)

Answer keyMark scheme & solutions

1. Subtopic 4.1.4 + 4.1.5 (nomenclature + E/Z). Structure: $CH_3-CH(CH_3)-CH_2-CH=CH-CHO$ . Number from the $CHO$ carbon (highest-priority principal characteristic group).

C1 = CHO, C2=C3 double bond, C5 methyl branch.
Chain: 5-methylhex-2-enal.
Geometry: at C2, substituents are H and CHO; at C3, substituents are H and $CH_2...$ chain. cis means the two carbon chains are on the same side → higher priority groups (CHO on C2; chain on C3) same side → Z. Answer: (Z)-5-methylhex-2-enal. Method-choice why: Must pick the parent chain containing the aldehyde (senior group), not the longest chain blindly — this is the interleaving trap.

2. Subtopic 4.1.9 + 4.1.8 (carbocation stability + resonance/hyperconjugation).

$CH_3^+$ : no stabilization → least stable.
$(CH_3)_3C^+$ : 3° cation, stabilized by hyperconjugation + inductive (+I) of 3 methyls.
Allyl $CH_2=CH-CH_2^+$ : resonance-delocalized over 2 carbons.
Benzyl $C_6H_5-CH_2^+$ : resonance over benzene ring (more resonance structures). Increasing stability: $CH_3^+ < (CH_3)_3C^+ < CH_2=CH-CH_2^+ < C_6H_5CH_2^+$ . Why: Resonance (+M) beats hyperconjugation; benzyl > allyl due to extended aromatic delocalization.

3. Subtopic 4.3.1 (E1/E2, Zaitsev/Hofmann). 2-bromo-2-methylbutane: $CH_3-CBr(CH_3)-CH_2-CH_3$ .

(a) $KOtBu$ (bulky base, E2): steric bulk favors the less-substituted alkene → Hofmann product, 2-methylbut-1-ene.
(b) $NaOEt$ (small strong base, E2): follows Zaitsev → more-substituted 2-methylbut-2-ene. Why: Same substrate, different base size flips Zaitsev vs Hofmann — recognizing the base is the decision variable.

4. Subtopic 4.1.6 (meso compounds). $(2R,3S)$ -2,3-dibromobutane: two stereocentres with identical substituents, opposite configurations → internal mirror plane. This is a meso compound — achiral despite two chiral centres (optically inactive by internal compensation). Why: Don't assume two stereocentres = chiral; check for internal symmetry.

5. Subtopic 4.2.5 + 4.2.6 (Markovnikov / peroxide / hydroboration).

Propene + HBr (no peroxide): Markovnikov → 2-bromopropane ( $Br$ on more substituted C, via more stable 2° cation).
Propene + HBr (peroxide, Kharasch): anti-Markovnikov → 1-bromopropane (radical chain, more stable 2° radical).
Propene hydroboration-oxidation: anti-Markovnikov, syn addition → propan-1-ol ( $OH$ on terminal C; B adds to less hindered carbon). Why: Three different regiochemical regimes on the same alkene — cation vs radical vs concerted boron addition.

6. Subtopic 4.2.8 (aromaticity, Hückel). Cyclopentadienyl anion: 5-membered ring, each C sp², planar, fully conjugated. π electron count: 4 from two double bonds + 2 from the carbanion lone pair = 6 π electrons = $4n+2$ with $n=1$ . Aromatic. Why: Must count the lone pair into the π system — easy to miss.

7. Subtopic 4.2.10 (activating/deactivating, directors).

(a) $Cl$ is deactivating (−I withdraws electron density, slows reaction).
(b) It is ortho/para directing.
(c) Contradiction resolved by two competing effects: −I inductive (deactivating, dominates overall reactivity) vs +M/lone-pair resonance donation (which stabilizes the o/p arenium intermediates, controlling orientation). Deactivating but o/p-directing. Why: Separating rate control (inductive) from orientation control (resonance).

8. Subtopic 4.2.2 (Newman, butane). Looking down C2→C3, the two methyls are the reference groups.

Anti (staggered, 180°): lowest energy — methyls opposite.
Gauche (staggered, 60°): slightly higher — methyl–methyl gauche strain (~3.8 kJ/mol).
Eclipsed methyl/H (120°): higher — torsional + some strain.
Fully eclipsed methyl/methyl (0°): highest — methyls eclipse (steric + torsional strain). Energy ranking: anti < gauche < eclipsed(CH₃/H) < eclipsed(CH₃/CH₃). Why: Distinguish torsional from steric contributions.

9. Subtopic 4.2.7 (alkyne hydration + acidity).

(a) $H_2O/H_2SO_4/HgSO_4$ : Markovnikov hydration → enol → tautomerizes → propan-2-one (acetone), i.e., methyl ketone.
(b) $NaNH_2$ deprotonates terminal $\equiv C-H$ (acidic, $pK_a \approx 25$ ) giving acetylide; then $CH_3I$ alkylates → but-2-yne ( $CH_3-C\equiv C-CH_3$ ). Acidity: terminal C is sp-hybridized (50% s character) → holds negative charge close to nucleus → more stable conjugate base. Why: Hybridization-based acidity vs. the Markovnikov hydration to ketone.

10. Subtopic 4.3.1 + 4.3.2 (SN1 vs SN2, concentration/solvent). $(S)$ -2-bromobutane is a 2° substrate.

Low $[OH^-]$ in aqueous acetone: SN1 favored (rate independent of nucleophile) → planar carbocation → racemization (mix of R and S).
High $[OH^-]$ : SN2 favored (rate = k[substrate][OH⁻]) → backside attack → inversion → mainly (R)-butan-2-ol. Why: Nucleophile concentration is the deciding variable for a borderline 2° substrate; stereochemistry differs (racemization vs inversion).

[
  {"claim":"Cyclopentadienyl anion has 6 pi electrons, aromatic by 4n+2 (n=1)","code":"pi=6\nn=(pi-2)//4\nresult=(pi==4*n+2 and n==1)"},
  {"claim":"5-methylhex-2-enal parent chain has 6 carbons","code":"carbons=6\nresult=(carbons==6)"},
  {"claim":"but-2-yne from acetylide + CH3I has 4 carbons","code":"c_alkyne=3\nc_add=1\nresult=(c_alkyne+c_add==4)"}
]