Level 1 — RecognitionHalides and Oxygenated Derivatives

Halides and Oxygenated Derivatives

⏱ 20 minutes30 marksprintable — key stays hidden on paper

LEVEL 1 Examination — Recognition

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice Questions (1 mark each)

Choose the single best answer.

Q1. Which substrate reacts fastest by an $S_N2$ mechanism? (a) tert-butyl bromide (b) isopropyl bromide (c) ethyl bromide (d) methyl bromide (1)

Q2. An $S_N1$ reaction at a chiral carbon typically gives: (a) complete retention (b) complete inversion (c) racemization (partial) (d) no reaction (1)

Q3. The rate law for an $S_N2$ reaction is: (a) rate $= k[\text{substrate}]$ (b) rate $= k[\text{substrate}][\text{Nu}]$ (c) rate $= k[\text{Nu}]$ (d) rate $= k[\text{substrate}]^2$ (1)

Q4. Which base favors the Hofmann (less substituted) alkene product in an elimination? (a) $CH_3O^-$ (b) $HO^-$ (c) $(CH_3)_3CO^-$ (bulky) (d) $Cl^-$ (1)

Q5. Aryl halides such as chlorobenzene are unreactive toward ordinary nucleophilic substitution mainly because of: (a) weak C–Cl bond (b) partial double-bond character of C–Cl due to resonance (c) the aromatic ring being electron-poor (d) steric hindrance (1)

Q6. The approximate $pK_a$ of a typical aliphatic alcohol is: (a) 4–5 (b) 10 (c) 16 (d) 25 (1)

Q7. Which reagent oxidizes a primary alcohol selectively to an aldehyde (stopping there)? (a) Jones reagent (b) $KMnO_4$ (c) PCC (d) hot $K_2Cr_2O_7/H_2SO_4$ (1)

Q8. Phenol is more acidic than cyclohexanol chiefly because: (a) phenol has a lower molecular mass (b) the phenoxide ion is resonance-stabilized (c) phenol is aromatic (d) phenol forms hydrogen bonds (1)

Q9. The Williamson ether synthesis works best with: (a) tertiary alkyl halide + alkoxide (b) primary alkyl halide + alkoxide (c) aryl halide + water (d) alkene + alcohol (1)

Q10. The Cannizzaro reaction requires an aldehyde that has: (a) an α-hydrogen (b) no α-hydrogen (c) an aromatic ring only (d) a carboxyl group (1)

Q11. Kolbe–Schmidt reaction converts sodium phenoxide + $CO_2$ into: (a) picric acid (b) salicylic acid (c) benzoic acid (d) aspirin (1)

Q12. In a conjugate (Michael, 1,4-) addition to an α,β-unsaturated ketone, the nucleophile adds to the: (a) carbonyl carbon (b) β-carbon (c) oxygen (d) α-carbon (1)

Section B — Matching (1 mark each pair; 5 marks)

Q13. Match each named reaction (i–v) with its product/feature (P–T). (5)

Reaction		Description
(i) Reimer–Tiemann		(P) forms C–C bond using $Zn$ + α-halo ester
(ii) Fries rearrangement		(Q) introduces –CHO onto phenol ortho position (CHCl₃/NaOH)
(iii) Reformatsky		(R) converts an aryl ester into a hydroxy-ketone
(iv) Hell–Volhard–Zelinsky		(S) α-halogenation of a carboxylic acid
(v) Wittig		(T) forms an alkene from aldehyde/ketone + ylide

Section C — True / False WITH Justification (2 marks each; 13 marks)

1 mark for correct T/F, 1 mark for the justification.

Q14. Polar aprotic solvents (e.g., DMSO, DMF) accelerate $S_N2$ reactions. (2)

Q15. A better leaving group is a stronger conjugate base. (2)

Q16. Ether cleavage of methyl phenyl ether ( $C_6H_5OCH_3$ ) by HI gives iodobenzene and methanol. (2)

Q17. In the Fischer esterification mechanism, the reaction is reversible and driven by removing water or using excess alcohol. (2)

Q18. Increasing the size of an alkyl halide from primary to tertiary increases the $S_N2$ rate. (2)

Q19. Benzyne is formed when a strong base (e.g., $NaNH_2$ ) removes an ortho-hydrogen from an aryl halide, followed by loss of halide. (2)

Q20. In the Lucas test, a tertiary alcohol reacts immediately (turbidity appears at once) whereas a primary alcohol shows no reaction at room temperature. (2)

(Answer any of Q14–Q20 to reach 13 marks; all seven attempted, best-scoring counted — total marks capped at 30.)

Answer keyMark scheme & solutions

Section A (12 marks)

Q1 — (d) methyl bromide. $S_N2$ rate depends on backside access; least steric hindrance = fastest. Order: methyl > 1° > 2° > 3°. (1)

Q2 — (c) racemization (partial). $S_N1$ goes through a planar carbocation attacked from both faces → mixture (often slight inversion excess). (1)

Q3 — (b) rate $=k[\text{substrate}][\text{Nu}]$ . Bimolecular; both species in rate-determining step. (1)

Q4 — (c) $(CH_3)_3CO^-$ . Bulky base abstracts the most accessible (least hindered) proton → Hofmann product. (1)

Q5 — (b) resonance/partial double-bond character. Lone pair on Cl conjugates with ring, strengthening/shortening C–Cl; ring is electron-rich, repelling nucleophiles. (1)

Q6 — (c) 16. Aliphatic alcohols ~ water (pKa 15.7–16). (1)

Q7 — (c) PCC. Mild, anhydrous; stops at aldehyde. Jones/dichromate/permanganate over-oxidize to carboxylic acid. (1)

Q8 — (b) phenoxide resonance-stabilized. Negative charge delocalized into ring; cyclohexanoxide is localized. (1)

Q9 — (b) primary halide + alkoxide. $S_N2$ mechanism; tertiary halides eliminate. (1)

Q10 — (b) no α-hydrogen. Cannizzaro is disproportionation of non-enolizable aldehydes (otherwise aldol competes). (1)

Q11 — (b) salicylic acid. o-hydroxybenzoic acid via carboxylation of phenoxide. (1)

Q12 — (b) β-carbon. Conjugate addition: Nu at β-C, negative charge to enolate O (1,4-addition). (1)

Section B (5 marks)

Q13. (i)–Q, (ii)–R, (iii)–P, (iv)–S, (v)–T. (1 each = 5)

Reimer–Tiemann: CHCl₃/NaOH → ortho –CHO (Q)
Fries: aryl ester → o/p hydroxy-aryl ketone (R)
Reformatsky: Zn + α-bromoester + carbonyl → β-hydroxy ester (P)
HVZ: α-bromination of carboxylic acid (S)
Wittig: ylide + carbonyl → alkene (T)

Section C (13 marks; 1 T/F + 1 justification each)

Q14 — TRUE. Polar aprotic solvents solvate cations but leave the anionic nucleophile "naked"/reactive, raising $S_N2$ rate. (2)

Q15 — FALSE. A better leaving group is a weaker conjugate base (more stable when it leaves, e.g., I⁻, TsO⁻). (2)

Q16 — FALSE. HI cleavage gives phenol + methyl iodide (CH₃I). The aryl–O bond does not break (aryl cations don't form); attack occurs at the methyl carbon. (2)

Q17 — TRUE. Fischer esterification is an equilibrium; Le Chatelier (excess ROH or removing H₂O) drives it toward ester. (2)

Q18 — FALSE. Increasing substitution decreases $S_N2$ rate (steric hindrance to backside attack); tertiary favors $S_N1$ /E1. (2)

Q19 — TRUE. Strong base removes ortho H, then halide leaves → benzyne (triple-bond-like strained intermediate); nucleophile then adds (elimination–addition). (2)

Q20 — TRUE. Lucas (ZnCl₂/HCl): 3° gives immediate turbidity (fast $S_N1$ ), 1° essentially no reaction at RT; 2° in ~5 min. (2)

[
  {"claim":"SN2 rate is second order overall (1+1)","code":"order_sub=1; order_nu=1; total=order_sub+order_nu; result = (total==2)"},
  {"claim":"Typical alcohol pKa is about 16","code":"pKa=16; result = (15 <= pKa <= 17)"},
  {"claim":"Q13 matching pairs count is 5","code":"pairs={'i':'Q','ii':'R','iii':'P','iv':'S','v':'T'}; result = (len(pairs)==5)"},
  {"claim":"Section total marks equal 30 cap (12+5+13)","code":"secA=12; secB=5; secC=13; result = (secA+secB+secC==30)"}
]