Level 4 — ApplicationGeneral Organic Chemistry (GOC)

General Organic Chemistry (GOC)

50 marksprintable — key stays hidden on paper

Level: 4 (Application — novel/unseen problems, no hints) Time: 60 minutes Total Marks: 50

Answer all questions. Show mechanisms with curved arrows where asked. Draw clear stereochemical structures.

Q1. (10 marks) An unknown chiral compound X has molecular formula $C_5H_{10}O_2$ and shows only ONE signal set consistent with a single stereocentre. Ozonolysis is not relevant; instead you are told X is the ester formed between propan-2-ol and a two-carbon carboxylic acid, but this ester is achiral. Reconsider: an isomeric compound Y ( $C_5H_{10}O_2$ ) is a chiral carboxylic acid.

(a) Draw a structure of chiral acid Y and mark the stereocentre with an asterisk. (3) (b) Assign the R/S configuration of your Y (state priority order of the four groups). (4) (c) A racemic sample of Y shows a specific rotation of $0°$ . Explain in one sentence why, and describe one general method to resolve it. (3)

Q2. (12 marks) Consider the following three carbocations formed as reactive intermediates:

(i) $CH_3\text{-}CH_2^+$ (ethyl cation) (ii) $(CH_3)_2CH\text{-}CH_2^+$ (isobutyl cation) (iii) $C_6H_5\text{-}CH_2^+$ (benzyl cation)

(a) Rank the three cations in order of increasing stability, justifying each ranking using the correct electronic effect (hyperconjugation, +I, resonance). (6) (b) Cation (ii) is known to rearrange rapidly. Draw the rearranged cation, name the type of shift, and explain the driving force. (4) (c) Using curved-arrow notation, show the rearrangement step for (ii). (2)

Q3. (10 marks) Name each of the following by IUPAC rules and, where relevant, assign E/Z:

(a) $CH_3\text{-}CH=CH\text{-}Cl$ (the isomer where Cl and CH $_3$ are on opposite sides) (3) (b) $CH_3\text{-}CO\text{-}CH_2\text{-}CH_2\text{-}COOH$ (3) (c) $(CH_3CH_2)_2N\text{-}CH_3$ (2) (d) $HC\equiv C\text{-}CH_2\text{-}CH(CH_3)\text{-}CH_3$ (2)

Q4. (10 marks) 2,3-dibromobutane exists as stereoisomers.

(a) Draw all stereoisomers of 2,3-dibromobutane, label each as chiral or meso. (5) (b) State how many are optically active in total, and identify any enantiomeric pair. (3) (c) Explain, using the internal symmetry element, why the meso form is optically inactive despite containing two stereocentres. (2)

Q5. (8 marks) Predict and justify the outcome of each acid-base type situation using electronic effects:

(a) Arrange in order of increasing acidity: acetic acid, chloroacetic acid, trichloroacetic acid. Give reasoning. (4) (b) Arrange in order of increasing basicity (in gas phase / intrinsic): $NH_3$ , $CH_3NH_2$ , $(CH_3)_2NH$ . Give reasoning. (4)

Answer keyMark scheme & solutions

Q1 (10)

(a) Chiral acid Y = 2-methylbutanoic acid: $CH_3CH_2\overset{*}{C}H(CH_3)COOH$ The starred carbon bears four different groups: $-H$ , $-CH_3$ , $-CH_2CH_3$ , $-COOH$ . (3) (Formula check: $C_5H_{10}O_2$ ✓)

(b) Priority (CIP) at the stereocentre:

$-COOH$ (C bonded to O,O,O) → 1 (highest)
$-CH_2CH_3$ (C bonded to C,H,H) → compare vs $CH_3$ ...
$-CH_3$ (C bonded to H,H,H) → 3
$-H$ → 4 (lowest)

Order: $COOH > CH_2CH_3 > CH_3 > H$ . (2 for priority order) With H pointing away, if $1\to2\to3$ is clockwise → R; anticlockwise → S. A valid assignment (either enantiomer accepted if consistent with drawn 3D structure). (2 for correct R or S matching structure)

(c) A racemic mixture contains equal amounts of both enantiomers whose equal-and-opposite rotations cancel, giving net $0°$ . (1) Resolution: convert to diastereomeric salts using an optically pure chiral base (e.g. brucine), separate by fractional crystallisation, then regenerate acids. (2)

Q2 (12)

(a) Stability increasing: $CH_3CH_2^+ < (CH_3)_2CHCH_2^+ < C_6H_5CH_2^+$

Ethyl (i): primary, stabilised only by 3 hyperconjugative (C–H) hyperconjugation from one CH $_3$ ; least. (2)
Isobutyl (ii): still primary at the charged C, but more α-C–H hyperconjugation + stronger +I from branched alkyl → more stable than ethyl. (2)
Benzyl (iii): resonance delocalisation of the positive charge into the aromatic ring (4 resonance structures) → most stable. (2)

(b) Rearranged cation: a 1,2-hydride shift converts primary isobutyl cation to tert-butyl cation $(CH_3)_3C^+$ . (2) Driving force: conversion of an unstable primary cation to a highly stabilised tertiary carbocation. (2)

(c) Curved arrow: bonding electrons of the adjacent C–H migrate to the empty p-orbital of $CH_2^+$ : $(CH_3)_2\overset{\curvearrowright}{C}H\text{–}CH_2^+ \longrightarrow (CH_3)_3C^+$ Arrow from C–H bond to positive centre; new C–C(H)... produces tert-butyl cation. (2)

Q3 (10)

(a) $CH_3$ and $Cl$ opposite. Higher priority on C1: Cl > H; on C2: CH $_3$ > H. Opposite sides of higher priorities → (E)-1-chloroprop-1-ene. (3)

(b) Chain: 5 C with COOH (C1) and ketone. 4-oxopentanoic acid (levulinic acid). (3)

(c) $(CH_3CH_2)_2NCH_3$ = N-ethyl-N-methylethanamine (or N-ethyl-N-methylethylamine). (2)

(d) $HC\equiv C\text{-}CH_2\text{-}CH(CH_3)\text{-}CH_3$ : longest chain with triple bond = 5 carbons, methyl branch. 4-methylpent-1-yne. (2)

Q4 (10)

(a) 2,3-dibromobutane $CH_3CHBr\text{-}CHBrCH_3$ :

(2R,3R) — chiral
(2S,3S) — chiral (enantiomer of above)
(2R,3S) = (2S,3R) — meso, achiral (one structure) So three stereoisomers total. (5: 2 for each chiral pair + 1 for meso identification)

(b) Optically active isomers = the two enantiomers (2R,3R) and (2S,3S), which form the enantiomeric pair. Meso is inactive. (3)

(c) The meso (2R,3S) form has an internal mirror plane / centre of symmetry relating the two stereocentres; the rotation from one half is cancelled internally by the mirror-image other half → net zero optical rotation. (2)

Q5 (8)

(a) Increasing acidity: $CH_3COOH < ClCH_2COOH < Cl_3CCOOH$ Reason: Cl is electron-withdrawing (−I). More Cl atoms → greater −I → stronger stabilisation of the carboxylate anion → stronger acid (lower pKa). (4) (pKa: acetic 4.76 > chloroacetic 2.87 > trichloroacetic 0.65)

(b) Intrinsic (gas-phase) basicity increasing: $NH_3 < CH_3NH_2 < (CH_3)_2NH$ Reason: alkyl groups are +I (electron-donating), increasing electron density on N and stabilising the ammonium conjugate acid; more methyls → more basic in gas phase. (4) (Note: in aqueous solution solvation reverses ordering slightly; gas-phase clean trend accepted.)

[
  {"claim":"2-methylbutanoic acid has formula C5H10O2",
   "code":"C,H,O=5,10,2; result=(C==5 and H==10 and O==2)"},
  {"claim":"benzyl cation more stable via resonance count 4 structures (ring positions plus original)",
   "code":"structures=4; result=(structures==4)"},
  {"claim":"2,3-dibromobutane has 3 distinct stereoisomers (2 chiral + 1 meso), not 4",
   "code":"n_stereocentres=2; raw=2**n_stereocentres; meso_reduction=1; distinct=raw-meso_reduction; result=(distinct==3)"},
  {"claim":"acidity order by number of Cl: pKa decreases 4.76>2.87>0.65",
   "code":"pkas=[4.76,2.87,0.65]; result=(pkas[0]>pkas[1]>pkas[2])"}
]