Instructions:
- Attempt all questions.
- Show all your work clearly for calculations and mechanisms.
- Use clear, precise, and systematic language for explanations.
- No notes or textbooks are allowed.
- A periodic table, relevant spectroscopic correlation tables, and a list of common constants are provided at the end of the exam.
- Time allowed: 2 hours 30 minutes
Total Marks: 120
Section A: Multiple Choice & Short Answer (40 Marks)
Choose the best answer or provide a concise explanation.
Chirality & Stereochemistry (4 Marks)Which of the following statements about stereoisomers is false?a) Enantiomers are stereoisomers that are non-superimposable mirror images.b) Diastereomers are stereoisomers that are not mirror images of each other.c) A meso compound contains chiral centers but is achiral overall due to an internal plane of symmetry.d) All compounds with multiple chiral centers are necessarily optically active.
Answer: d) All compounds with multiple chiral centers are necessarily optically active.
❌ False because meso compounds have chiral centers but are not optically active due to internal symmetry.
Spectroscopy – 1H NMR (4 Marks)A compound with molecular formula C_6H_12O_2 shows a 1H NMR spectrum with a triplet at delta 1.2 (3H), a quartet at delta 4.1 (2H), and a singlet at delta 2.1 (3H), and a singlet at delta 3.4 (4H). What is a likely functional group present based on this data?a) Ketone and Etherb) Ester and Etherc) Aldehyde and Alcohold) Carboxylic Acid and Alkene
Answer: b) Ester and Ether
Triplet at δ 1.2 (3H) + quartet at δ 4.1 (2H) = ethyl group near electronegative atom
Singlet at δ 2.1 (3H) = methyl on carbonyl (acetate)
Singlet at δ 3.4 (4H) = two equivalent –CH₂– in an ether
Suggests an ethyl acetate moiety and an ether — consistent with b).
Reaction Mechanisms – SN1/SN2/E1/E2 (4 Marks)Consider the reaction of 2-bromo-2-methylpropane with sodium ethoxide (NaOCH_2CH_3) in ethanol. Which of the following is the major product and the dominant mechanism?a) 2-ethoxy-2-methylpropane, S_N1b) 2-ethoxy-2-methylpropane, S_N2c) 2-methylpropene, E1d) 2-methylpropene, E2
Answer: d) 2-methylpropene, E2
2-bromo-2-methylpropane is tertiary, favors elimination with strong base like ethoxide.
E2 is favored here due to strong base and polar protic solvent.
Thermodynamics & Equilibrium (4 Marks)For a reaction where $\\Delta H^\\circ \> 0$ and $\\Delta S^\\circ \< 0$:a) The reaction will be spontaneous at high temperatures.b) The reaction will be spontaneous at low temperatures.c) The reaction will never be spontaneous under standard conditions.d) The reaction will always be spontaneous under standard conditions.
Answer: c) The reaction will never be spontaneous under standard conditions.
ΔH° > 0 (endothermic), ΔS° < 0 → ΔG = ΔH – TΔS → always positive, never spontaneous.
Kinetics – Rate Laws (4 Marks)A reaction A+BrightarrowC is found to be first order with respect to A and second order with respect to B. If the concentration of A is doubled and the concentration of B is halved, the rate of the reaction will:a) Doubleb) Halvec) Remain unchangedd) Decrease by a factor of 4
Answer: b) Halve
Rate law: rate = k[A][B]^2
[A] doubles → ×2
[B] halves → ×(½)² = ×¼
Total change = 2 × ¼ = ½
IR Spectroscopy (4 Marks)A strong, broad IR absorption centered around 3300textcm−1 and another strong absorption at 1710textcm−1 in a compound suggest the presence of what two functional groups?a) Alcohol and Amineb) Carboxylic Acidc) Alcohol and Ketone/Aldehyded) Ether and Ester
Answer: b) Carboxylic Acid
Broad stretch at ~3300 cm⁻¹ = O–H (carboxylic acid)
Strong C=O stretch at ~1710 cm⁻¹ = C=O
Typical of carboxylic acid functional group.
Acid-Base Chemistry (4 Marks)Explain why pyrrole is significantly less basic than pyridine, even though both contain nitrogen atoms in a heterocyclic ring. (Concise explanation, 2-3 sentences)
Answer: Pyrrole is less basic than pyridine because the lone pair on pyrrole’s nitrogen is part of the aromatic π system, contributing to resonance. In pyridine, the nitrogen lone pair is not involved in aromaticity and is available for protonation, making it more basic.
Orbital Hybridization (4 Marks)Determine the hybridization of the oxygen atom in dimethyl ether (CH_3OCH_3) and the carbon atoms of the double bond in ethene (C_2H_4).
Answer:
- Oxygen in CH₃OCH₃: sp³ hybridized (2 sigma bonds, 2 lone pairs)
- Carbons in C=C of ethene: each sp² hybridized (1 sigma, 1 pi bond, 2 sigma bonds)
Resonance Structures (4 Marks)Draw all significant resonance structures for the carbonate ion (CO_32−). Indicate formal charges.
Answer:
Draw all three resonance forms where one double bond rotates among the three oxygens. Each structure has:
- One C=O double bond
- Two C–O single bonds
- Formal charges:
- Double-bonded O: 0
- Single-bonded O: –1
- Carbon: 0
All resonance forms are equivalent.
Conformational Analysis (4 Marks)Draw the most stable and least stable Newman projections for 1,2-dichloroethane looking down the C1-C2 bond. Label them.
Answer:
Draw Newman projections looking down the C1–C2 bond:
- Most stable: Anti conformation — Cl atoms 180° apart
- Least stable: Eclipsed conformation — Cl atoms eclipsed with each other and Hs
Label the views accordingly (anti = staggered; eclipsed = highest energy).
Section B: Problem Solving & Mechanisms (80 Marks)
Organic Synthesis & Retrosynthesis (25 Marks)a) Propose a plausible synthesis for the following target molecule starting from benzene and any other necessary reagents with 3 carbons or less. Show all intermediate steps and reagents. (15 Marks)(Target Molecule: 1-phenylpropan-1-ol)b) Explain why a direct S_N2 reaction of CH_3CH_2Br with OH− might not be the most efficient route to CH_3CH_2OH if competing E2 elimination is a concern, and suggest a better alternative using a milder nucleophile/base. (10 Marks)
a) Synthesize 1-phenylpropan-1-ol from benzene and reagents ≤ 3 carbon atoms
Target molecule: 1-phenylpropan-1-ol
Strategy:
We need to construct a 3-carbon side chain on a benzene ring ending in a primary alcohol. This suggests:
Friedel-Crafts acylation to introduce a 3-carbon chain (as a ketone)
Reduction of the ketone to an alcohol
(Optional) Alternatives might involve Grignard addition, but we’ll go with a more efficient and linear approach here.
Step-by-Step Synthesis:
Step 1: Friedel-Crafts Acylation
React benzene with propanoyl chloride (CH₃CH₂COCl) and AlCl₃
→ Forms: 1-phenylpropan-1-one (Ph–COCH₂CH₃)
Step 2: Reduction of Ketone to Alcohol
Use NaBH₄ or LiAlH₄ to reduce the ketone to a secondary alcohol
→ Forms: 1-phenylpropan-1-ol
Reaction Scheme:Benzene │ └─(1) CH₃CH₂COCl, AlCl₃ → 1-phenylpropan-1-one │ └─(2) NaBH₄ or LiAlH₄ → 1-phenylpropan-1-ol ✅
Alternative Synthesis via Grignard (if allowed):
Step 1: Brominate propane → CH₃CH₂CH₂Br
Step 2: Form Grignard → CH₃CH₂CH₂MgBr
Step 3: Add to benzaldehyde → PhCH=O → addition of Grignard → 1-phenylpropan-1-ol
However, propyl bromide is >3 carbons (C₃ chain + Br), so not acceptable under current reagent constraints.
b) Why SN2 with ethyl bromide + OH⁻ may not be ideal, and a better alternative
Issue with SN2:
Ethyl bromide + OH⁻:
CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
Competing E2 reaction can occur under basic conditions
E2 competes especially with secondary/strong base + heat
Solution:
Use a weaker base/nucleophile, such as water or alkoxide in alcohol solvent.
Best alternative: Use NaOAc (sodium acetate) followed by acidic hydrolysis:
Step 1: CH₃CH₂Br + NaOAc → CH₃CH₂OAc
Step 2: Hydrolyze ester with acid or base → CH₃CH₂OH
This avoids elimination and gives cleaner substitution via acetate displacement, then hydrolysis.
Spectroscopy – Structure Elucidation (25 Marks)A compound, X, has the molecular formula C_9H_10O_2.The following spectroscopic data were obtained:
- IR Spectrum: Strong absorption at 1735textcm−1, strong absorption at 1210textcm−1, and several weak to moderate absorptions in the 3000−3100textcm−1 range. No significant absorption above 3000textcm−1 other than the ones noted.
- 1H NMR Spectrum (delta ppm):
- 7.3 (5H, multiplet)
- 4.3 (2H, triplet)
- 2.6 (2H, triplet)
- 2.1 (3H, singlet)
- $^{13}$C NMR Spectrum ($\\delta$ ppm):
- 172.0 (s)
- 136.5 (s)
- 129.5 (d)
- 128.5 (d)
- 127.0 (d)
- 65.0 (t)
- 34.0 (t)
- 20.5 (q)
Given:
- Molecular formula: C₉H₁₀O₂
- IR spectrum:
- 1735 cm⁻¹ (strong): typical ester or saturated ketone C=O
- 1210 cm⁻¹: C–O stretch (supportive of ester)
- 3000–3100 cm⁻¹: aromatic C–H
- No broad O–H (no alcohol or acid)
1H NMR:
| δ (ppm) | Integration | Multiplicity | Interpretation |
|---|---|---|---|
| 7.3 | 5H | multiplet | Mono-substituted benzene ring (C₆H₅–) |
| 4.3 | 2H | triplet | –CH₂–O (typical for ethyl ester) |
| 2.6 | 2H | triplet | –CH₂–C=O |
| 2.1 | 3H | singlet | Methyl group near carbonyl, likely acetyl |
13C NMR:
| δ (ppm) | Assignment |
|---|---|
| 172.0 | C=O (ester) |
| 136.5 | Aromatic C–C (quaternary) |
| 129.5, 128.5, 127.0 | Aromatic CH (×3) |
| 65.0 | CH₂–O (alkoxy) |
| 34.0 | CH₂–C=O |
| 20.5 | Methyl (CH₃–C=O) |
Step-by-step Interpretation
I. Unsaturation Calculation
C₉H₁₀O₂:
- Degree of unsaturation =
= (2C + 2 – H)/2 = (18 + 2 – 10)/2 = 5
→ One benzene ring = 4
→ One carbonyl = 1
✅ Fits perfectly.
II. Functional Group Identification
- Strong C=O at 1735 cm⁻¹ = ester
- 1210 cm⁻¹ = C–O stretch (ester support)
- No O–H → eliminates acids and alcohols
III. Proposed Structure: Ethyl 3-phenylpropanoate
Structure:
O
||
Ph–CH₂–CH₂–C–OCH₂CH₃
Name: Ethyl 3-phenylpropanoate
IV. Assigning NMR Peaks
| Peak (δ) | Integration | Assignment | Explanation |
|---|---|---|---|
| 7.3 ppm | 5H | Aromatic (C₆H₅) | Mono-substituted phenyl ring |
| 4.3 ppm | 2H | CH₂–O (from ethyl ester) | Triplet, OCH₂–CH₃ |
| 2.6 ppm | 2H | CH₂ next to aromatic (Ph–CH₂–) | Triplet |
| 2.1 ppm | 3H | Actually this is slightly off! (see below) | |
| → BUT: 2.1 ppm is unexpected for ethyl ester. It suggests a methyl near carbonyl. So: |
🔄 Revised Hypothesis: Methyl 3-phenylpropanoate
Structure:
O
||
Ph–CH₂–CH₂–C–OCH₃
- C₉H₁₀O₂ = 9C (6 ring, 3 chain) + 10H
- All spectral features match better
- The 2.1 ppm singlet = methyl ester (OCH₃)
- CH₂ at 4.3 ppm = CH₂ near oxygen
- CH₂ at 2.6 ppm = CH₂ next to carbonyl
- 7.3 ppm = aromatic (5H)
✅ Final Structure: Methyl 3-phenylpropanoate
IUPAC Name: Methyl 3-phenylpropanoate
Structure:
Ph–CH₂–CH₂–C(=O)–OCH₃
Final NMR Table:
| δ (ppm) | #H | Multiplicity | Assignment |
|---|---|---|---|
| 7.3 | 5 | multiplet | Aromatic (Ph–) |
| 4.3 | 2 | triplet | CH₂–OCH₃ |
| 2.6 | 2 | triplet | CH₂–CO |
| 2.1 | 3 | singlet | OCH₃ |
13C NMR Assignments:
| δ (ppm) | Type | Assignment |
|---|---|---|
| 172.0 | s | C=O (ester) |
| 136.5 | s | Aromatic C (quaternary) |
| 129.5, 128.5, 127.0 | d | Aromatic CH |
| 65.0 | t | CH₂–O |
| 34.0 | t | CH₂–CO |
| 20.5 | q | OCH₃ |
Kinetics & Thermodynamics (15 Marks)Consider the gas-phase decomposition of dinitrogen pentoxide:2N2O5(g)→4NO2(g)+O2(g)The reaction is first-order with respect to N_2O_5. At 300 K, the rate constant is k=3.38times10−5texts−1, and at 350 K, k=1.63times10−3texts−1.a) Calculate the activation energy (E_a) for this reaction. (8 Marks)b) If the initial concentration of N_2O_5 is 0.500textM at 300 K, what will be its concentration after 1 hour? (7 Marks)
Given:
Reaction: 2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)
- First-order reaction
- k1=3.38×10−5 s−1k_1 = 3.38 \times 10^{-5} \,\text{s}^{-1} at T₁ = 300 K
- k2=1.63×10−3 s−1k_2 = 1.63 \times 10^{-3} \,\text{s}^{-1} at T₂ = 350 K
- Initial [N₂O₅] = 0.500 M, time = 1 hour = 3600 s
Part (a): Calculate activation energy EaE_a
Using Arrhenius equation: ln(k2k1)=−EaR(1T2−1T1)\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left( \frac{1}{T_2} – \frac{1}{T_1} \right)
Plug in values:
- k2/k1=1.63×10−33.38×10−5≈48.22k_2 / k_1 = \frac{1.63 \times 10^{-3}}{3.38 \times 10^{-5}} \approx 48.22
- ln(48.22)≈3.875\ln(48.22) \approx 3.875
- 1T2−1T1=1350−1300≈0.002857−0.003333=−0.000476 K−1\frac{1}{T_2} – \frac{1}{T_1} = \frac{1}{350} – \frac{1}{300} \approx 0.002857 – 0.003333 = -0.000476 \,\text{K}^{-1}
Now solve: 3.875=−Ea8.314×(−0.000476)3.875 = -\frac{E_a}{8.314} \times (-0.000476) Ea=3.8750.000476×8.314≈68,000 J/mol=68.0 kJ/molE_a = \frac{3.875}{0.000476} \times 8.314 \approx 68,000 \,\text{J/mol} = \boxed{68.0 \,\text{kJ/mol}}
✅ Answer: Ea=68.0 kJ/molE_a = 68.0 \, \text{kJ/mol}
Part (b): Find [N₂O₅] after 1 hour at 300 K
First-order integrated rate law: [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}
- [A]0=0.500 M[A]_0 = 0.500 \, \text{M}
- k=3.38×10−5 s−1k = 3.38 \times 10^{-5} \, \text{s}^{-1}
- t=3600 st = 3600 \, \text{s}
[A]t=0.500×e−3.38×10−5×3600[A]_t = 0.500 \times e^{-3.38 \times 10^{-5} \times 3600} =0.500×e−0.1217≈0.500×0.8855=0.443 M= 0.500 \times e^{-0.1217} \approx 0.500 \times 0.8855 = \boxed{0.443 \, \text{M}}
✅ Answer: [N₂O₅] = 0.443 M after 1 hour
✅ Summary of Answers:
a) Ea=68.0 kJ/molE_a = 68.0 \, \text{kJ/mol}
b) [N2O5]=0.443 M[N_2O_5] = 0.443 \, \text{M}
Reaction Mechanisms – Advanced (15 Marks)Propose a detailed, step-by-step mechanism for the acid-catalyzed formation of an acetal from an aldehyde and an alcohol. Show all intermediates and electron-pushing arrows.(Reactants: Ethanal (CH_3CHO) and Methanol (CH_3OH))(Product: 1,1-dimethoxyethane (CH_3CH(OCH_3)_2))
Overview of Mechanism:
This reaction is acid-catalyzed and proceeds in four major stages:
- Protonation of the carbonyl oxygen
- Nucleophilic attack by methanol
- Formation of hemiacetal → proton transfer
- Elimination of water → second alcohol addition → acetal
Step-by-Step Mechanism with Arrow-Pushing:
Step 1: Protonation of Carbonyl
CH₃CHO+H+→CH3C(OH+)H\text{CH₃CHO} + H⁺ → CH₃C(OH⁺)H
Carbonyl oxygen gets protonated, increasing electrophilicity.
Step 2: Nucleophilic Attack by CH₃OH
CH3OH+CH3C(OH+)H→CH3CH(OH)(OCH3+)CH₃OH + CH₃C(OH⁺)H → CH₃CH(OH)(OCH₃⁺)
Methanol attacks carbonyl carbon → forms protonated hemiacetal intermediate.
Step 3: Deprotonation
CH3CH(OH)(OCH3+)→CH3CH(OH)(OCH3)+H+CH₃CH(OH)(OCH₃⁺) → CH₃CH(OH)(OCH₃) + H⁺
Loss of a proton gives neutral hemiacetal.
Step 4: Protonation of –OH group
CH3CH(OH)(OCH3)+H+→CH3CH(OH2+)(OCH3)CH₃CH(OH)(OCH₃) + H⁺ → CH₃CH(OH₂⁺)(OCH₃)
Makes –OH into a good leaving group.
Step 5: Loss of Water
CH3CH(OH2+)(OCH3)→CH3C+(OCH3)+H2OCH₃CH(OH₂⁺)(OCH₃) → CH₃C⁺(OCH₃) + H₂O
Forms a resonance-stabilized carbocation (oxonium ion).
Step 6: Second CH₃OH Attack
CH3OH+CH3C+(OCH3)→CH3CH(OCH3)(OCH3+)CH₃OH + CH₃C⁺(OCH₃) → CH₃CH(OCH₃)(OCH₃⁺)
Second molecule of methanol attacks → protonated acetal.
Step 7: Final Deprotonation
CH3CH(OCH3)(OCH3+)→CH3CH(OCH3)2+H+CH₃CH(OCH₃)(OCH₃⁺) → CH₃CH(OCH₃)₂ + H⁺
Yields final product: 1,1-dimethoxyethane
Key Features to Include in Diagram:
- Curved arrows for all nucleophilic/electrophilic steps
- Indicate protonation and deprotonation steps clearly
- Label intermediates like hemiacetal, oxonium ion, etc.
- Emphasize regeneration of the acid catalyst
Final Product:
CH3CH(OCH3)2(1,1-dimethoxyethane)\boxed{CH₃CH(OCH₃)₂} \quad \text{(1,1-dimethoxyethane)}
Constants and Data:
- Gas Constant (R): 8.314textJ/(molcdottextK) or 0.008314textkJ/(molcdottextK)
- Faraday Constant (F): 96485textC/mol
- Avogadro’s Number (N_A): 6.022times1023textmol−1
- Standard Temperature and Pressure (STP): 273.15 K and 1 bar
- ln(k_2/k_1)=−E_a/R(1/T_2−1/T_1)
- For first-order reactions: ln[A]_t−ln[A]_0=−kt or [A]_t=[A]_0e−kt
(A detailed periodic table and a table of common IR stretches, 1H NMR chemical shifts, and $^{13}$C NMR chemical shifts would be provided here.)
Answer Key (For Instructor Use Only)
Section A: Multiple Choice & Short Answer
- d) All compounds with multiple chiral centers are necessarily optically active. (False, meso compounds are an exception).
- b) Ester and Ether (or more specifically, an ester where one part is an ethyl group and the other is an ethoxy group attached to the carbonyl). Let’s re-evaluate the NMR for specificity:
- Triplet delta 1.2 (3H): CH_3
- Quartet delta 4.1 (2H): CH_2These two together are an ethyl group (CH_3CH_2−). The delta 4.1 for CH_2 is highly deshielded, characteristic of being next to an oxygen in an ester (RCOOCH_2R′).
- Singlet delta 2.1 (3H): CH_3 next to an electron-withdrawing group, typically CH_3C(=O)−.
- Singlet delta 3.4 (4H): This signal is tricky. If it’s a singlet, it means no neighbors. If it’s 4H, it could be two equivalent CH_2 groups without neighbors, or a CH_2CH_2 system with no neighbors, or two equivalent methyl ethers, etc.Let’s reconsider C_6H_12O_2.If it’s an ester CH_3COOCH_2CH_2CH_2CH_3 (butyl acetate) – no, this wouldn’t fit.What if it’s Methyl 3-ethoxypropanoate? CH_3CH_2OCH_2CH_2COOCH_3 (Too many carbons).Re-evaluation of NMR and MC options:The delta 1.2 (3H, triplet) and delta 4.1 (2H, quartet) strongly indicate an ethyl ester (−textCOOCH∗2CH_3).The delta 2.1 (3H, singlet) suggests a methyl ketone or acetate (CH_3C(=O)−R).The delta 3.4 (4H, singlet) is the key. 4H and a singlet indicates two equivalent, uncoupled CH_2 groups. This is highly suggestive of a symmetrical ether.However, the formula is C_6H∗12O_2.Let’s try to piece it together. If we have CH_3CO_2CH_2CH_3 (ethyl acetate, C_4H_8O_2).We need two more carbons and four more hydrogens.If we have an ester part like CH_3COOCH_2CH_3, it uses C_4H_8O_2. We need C_2H_4.The delta 3.4 (4H, singlet) is unusual. It hints at (CH_2)_2 not coupled. This often means symmetrically placed CH_2 units.Consider Methyl 2-(2-methoxyethyl)acrylate (No, C_7 not C_6).Ethyl 2-methoxyacetate (CH_3OCH_2COOCH_2CH_3)
- CH_3 (ether): singlet, ~delta 3.3 (3H)
- OCH_2: singlet, ~delta 3.5 (2H)
- OCH_2CH_3: quartet, ~delta 4.2 (2H)
- OCH_2CH_3: triplet, ~delta 1.3 (3H)This is C_5H_10O_3. Still doesn’t fit.
- a) Ketone and Ether: Maybe CH_3−CO−CH_2−O−CH_2−CH_3. No, the CH_2 next to CO would be a triplet, not 4H singlet.
- b) Ester and Ether:Let’s try Ethyl 3-methoxypropanoate: CH_3OCH_2CH_2COOCH_2CH_3
- CH_3 (ethyl ester): delta 1.2 (3H, triplet) – Matches
- CH_2 (ethyl ester): delta 4.1 (2H, quartet) – Matches
- CH_3O: delta 3.4 (3H, singlet) – Only 3H, not 4H.
- CH_2 next to OCH_3: delta ~3.6 (2H, triplet)
- CH_2 next to CO: delta ~2.5 (2H, triplet)This still doesn’t explain the 4H singlet.
- CH_3 (ethyl ether): delta 1.2 (3H, t)
- CH_2 (ethyl ether): delta 3.4 (2H, q)
- CH_2 (next to ester carbonyl and ether O): delta 4.1 (2H, s) – No, this would be a singlet.
- CH_2 (ethyl ester): delta 4.1 (2H, q)
- CH_3 (ethyl ester): delta 1.2 (3H, t)This molecule is C_6H_12O_3. Still wrong.
- delta 1.2 (3H, triplet) impliesCH_3CH_2 (part of ethyl group)
- delta 4.1 (2H, quartet) impliesCH_3CH_2 (part of ethyl group) – This fits an ester’s OCH_2
- delta 2.1 (3H, singlet) impliesCH_3 (adjacent to C=O or similar)
- delta 3.4 (4H, singlet) implies two equivalent CH_2 groups that are very deshielded, and no protons on adjacent carbons. This is the problematic one.
- CH_3 (ester ethyl): 1.2 (3H, t)
- CH_2 (ester ethyl): 4.1 (2H, q)
- CH_3 (butyrate): 0.9 (3H, t)
- CH_2 (alpha to carbonyl): 2.3 (2H, t)
- CH_2 (beta to carbonyl): 1.6 (2H, sextet)Doesn’t match.
- d) 2-methylpropene, E2. (Tertiary halide, strong bulky base, favors E2).
- c) The reaction will never be spontaneous under standard conditions. (DeltaG=DeltaH−TDeltaS. If DeltaH0 and $\\Delta S \< 0$, then −TDeltaS0. So DeltaG will always be positive).
- b) Halve. Rate = k[A]1[B]2. New rate = k[2A][0.5B]2=k[2A][0.25B2]=0.5k[A][B2]. So the rate is halved.
- c) Alcohol and Ketone/Aldehyde (broad 3300textcm−1 for O-H, sharp 1710textcm−1 for C=O). Note: Carboxylic acid also has both, but the O-H is often broader and the C=O is slightly lower, plus it has unique NMR/C13 signals. The question specifies two functional groups.
- Pyrrole vs. Pyridine Basicity: Pyrrole is significantly less basic than pyridine because the lone pair of electrons on the nitrogen atom in pyrrole is part of the aromatic pi-system (contributing 2 electrons to the 6$\pi$ aromatic system). If this lone pair is protonated, the aromaticity is destroyed, which is energetically unfavorable. In contrast, the lone pair on the nitrogen in pyridine is in an sp2 orbital and is not part of the aromatic system, so it is readily available for protonation without losing aromaticity.
- Dimethyl ether (CH_3OCH_3): Oxygen is sp3 hybridized (two lone pairs and two sigma bonds). Ethene (C_2H_4): Each carbon of the double bond is sp2 hybridized.
- Carbonate Ion (CO_32−):
O=C(-O^-)-O^- <--> O^--C(=O)-O^- <--> O^--C(-O^-)=O (Each oxygen with single bond has -1 formal charge. The carbon has 0. Sum: -2)Draw three resonance structures, where the double bond formally moves between each oxygen. - 1,2-Dichloroethane Newman Projections:
- Most Stable (Anti-periplanar): Cl-C-C-Cl dihedral angle is 180°.
Cl / H---C---H | H---C---H \ Cl(Imagine looking down the C-C bond. Cl is opposite Cl.) - Least Stable (Eclipsed – Cl-Cl): Cl-C-C-Cl dihedral angle is 0°.
Cl / H---C---H | (eclipsed) H---C---H \ Cl(Imagine looking down the C-C bond. Cl is directly behind Cl.)
- Most Stable (Anti-periplanar): Cl-C-C-Cl dihedral angle is 180°.
Section B: Problem Solving & Mechanisms
- Organic Synthesis & Retrosynthesisa) Target: 1-phenylpropan-1-ol. This is a secondary alcohol.
- Retrosynthesis: A secondary alcohol can be formed from a ketone via Grignard reaction or reduction. If from a ketone, it’s a phenyl ketone.
- 1-phenylpropan-1-ol xleftarrowreduction 1-phenylpropan-1-one (propiophenone)
- 1-phenylpropan-1-ol xleftarrowGrignard benzaldehyde + ethylmagnesium bromide (CH_3CH_2MgBr) OR acetophenone + phenylmagnesium bromide (less direct for “from benzene”).
- Let’s choose the ketone route (propiophenone) from benzene.Step 1: Acylation of Benzene (Friedel-Crafts Acylation)To get a ketone with an ethyl group, we need a propionyl group.
Benzene + Propanoyl chloride (CH3CH2COCl) --(AlCl3)--> Propiophenone (C6H5COCH2CH3) + HCl- Reagents: Benzene, Propanoyl chloride (CH_3CH_2COCl, derived from propanoic acid, 3 carbons), AlCl_3 (Lewis acid catalyst).Step 2: Reduction of KetoneReduction of a ketone to a secondary alcohol.
Propiophenone (C6H5COCH2CH3) --(NaBH4 or LiAlH4, then H2O)--> 1-phenylpropan-1-ol (C6H5CH(OH)CH2CH3)- Reagents: NaBH_4 (sodium borohydride) in CH_3OH or EtOH, followed by aqueous workup (H_2O/acid). (LiAlH_4 is stronger but also works).
CH3CH2Br + CH3COOK --(polar aprotic solvent, e.g., DMSO/DMF)--> CH3COOCH2CH3 (ethyl acetate) + KBr CH3COOCH2CH3 + NaOH (aq), heat --> CH3CH2OH + CH3COONaThis two-step process ensures a clean S_N2 displacement in the first step (acetate is a good nucleophile, weak base), followed by hydrolysis to the alcohol. - Retrosynthesis: A secondary alcohol can be formed from a ketone via Grignard reaction or reduction. If from a ketone, it’s a phenyl ketone.
- Spectroscopy – Structure ElucidationMolecular Formula: C_9H_10O_2Degrees of Unsaturation (DoU): (2C+2+N−H−X)/2=(2times9+2−10)/2=(18+2−10)/2=10/2=5.
- DoU of 5 strongly suggests a benzene ring (4 DoU) plus one more (likely a carbonyl or another ring).
- Strong absorption at 1735textcm−1: Characteristic of a carbonyl (C=O). The high frequency (above 1700) is typical for an ester.
- Strong absorption at 1210textcm−1: C−O stretch, consistent with an ester.
- Weak/moderate absorptions 3000−3100textcm−1: Aromatic C−H stretches.
- No significant absorption above 3000textcm−1 other than noted: No O−H (alcohol/acid) or N−H.
- 7.3 (5H, multiplet): Classic for a monosubstituted benzene ring (C_6H_5−). This accounts for 6 carbons and 5 hydrogens.
- 4.3 (2H, triplet): CH_2 group. Triplet means 2 neighbors (CH_2CH_2). Deshielded (3.0 ppm) strongly suggests it’s next to an oxygen, specifically an ester oxygen (−textCOOCH_2− or −OCH_2textCO).
- 2.6 (2H, triplet): CH_2 group. Triplet means 2 neighbors (CH_2CH_2). Chemical shift suggests it’s next to a carbonyl (−textCH_2textCO−).
- 2.1 (3H, singlet): CH_3 group. Singlet means no neighbors. Chemical shift suggests it’s next to a carbonyl (CH_3textCO−).
- 172.0 (s): Carbonyl carbon of an ester. (Quaternary C, no attached protons, so it’s a singlet).
- 136.5 (s): Quaternary carbon of the aromatic ring (the carbon directly attached to the substituent).
- 129.5 (d), 128.5 (d), 127.0 (d): Three different types of protonated aromatic carbons (para, ortho, meta or similar pattern for a monosubstituted benzene ring).
- 65.0 (t): CH_2 carbon (triplet implies 2 attached protons) highly deshielded, next to oxygen (O−CH_2).
- 34.0 (t): CH_2 carbon (triplet implies 2 attached protons) relatively deshielded, consistent with being next to a carbonyl or another electron-withdrawing group.
- 20.5 (q): CH_3 carbon (quartet implies 3 attached protons), consistent with an aliphatic methyl group, particularly if next to a carbonyl.
- We have a C_6H_5− (benzene).
- We have an ester (COO).
- We have a CH_3− (singlet in NMR at 2.1, quartet in C13 at 20.5).
- We have two CH_2− groups that are triplets in NMR (meaning they have 2 neighbors each), one at delta 4.3 (next to O), one at delta 2.6 (next to C=O).
- Total carbons: 6 (benzene) + 1 (carbonyl) + 1 (CH_3) + 2 (CH_2 groups) = 10 carbons. Our formula is C_9H_10O_2. This means my interpretation of CH_3 and CH_2 groups needs refinement or one of the CH_2 groups is part of the aromatic substituent.
- Benzene ring accounts for 6 carbons, 5 hydrogens. So remaining are C_3H_5O_2.
- Ester C=O (172.0textppm) is one carbon. So C_2H_5O.
- CH_3 at 2.1textppm and 20.5textppm. So one CH_3 group. Remaining C_1H_2O.
- A CH_2 at 4.3textppm (65.0textppm C13). This fits CH_2−O−.
- A CH_2 at 2.6textppm (34.0textppm C13). This fits CH_2−C(=O)−.
- Phenyl (C_6H_5−)
- Ester C=O
- CH_3
- O−CH_2−
- −CH_2−C=O
- Phenyl C_6H_5− attached to the oxygen of the ester (phenyl ester): textR−COO−Phenyl.
- This would give: textR−C(=O)−O−C_6H_5.
- Then the CH_3 (2.1 ppm) is CH_3C(=O)−. So R is CH_3.
- So far: CH_3C(=O)O−textPhenyl. This is Phenyl acetate (C_8H_8O_2). We have C_9H_10O_2. Not this.
- Phenyl attached to the CH_2 chain.Phenyl is attached to the chain. The CH_2 at delta 2.6 (triplet) would be next to another CH_2. The CH_2 at delta 4.3 (triplet) is next to another CH_2 AND oxygen.So, a sequence like textPhenyl−CH_2−CH_2−.Then an ester: C=O.And a methyl group: CH_3.Consider Phenylpropanoic acid and its derivatives. No, textR−CH_2−textCH_2−textPhenyl.If it’s an ester of phenylpropanoic acid.Phenylpropanoic acid is C_6H_5−CH_2−CH_2−COOH.If the compound is an ester of phenylpropanoic acid with methanol:Methyl 3-phenylpropanoate: C_6H_5−CH_2−CH_2−COOCH_3
- Formula: C_9H_10O_2. Matches!
- DoU: 5. Matches!
- C_6H_5: delta 7.3 (5H, multiplet) – Matches!
- COOCH_3: delta 3.6-3.8 (3H, singlet) – The given is delta 2.1 (3H, singlet). This points to an acetyl methyl CH_3CO−, not a methoxy methyl. So, it’s not Methyl 3-phenylpropanoate.
- Formula: C_2H_3O_2 (acetate) + C_8H_9 (phenylethyl). C_2+C_8=C_10. No, too many carbons.
- C_9H_10O_2. We have phenyl (C_6H_5), then C_3H_5O_2 remaining.
- Aromatic C_6H_5: 6C, 5H. (from delta 7.3, 5H and 13C signals 136.5, 129.5, 128.5, 127.0). This uses C_6H_5. Remaining: C_3H_5O_2.
- Ester C=O: 1C. (from delta 172.0). Remaining: C_2H_5O.
- CH_3: 1C, 3H. (from delta 2.1, 3H singlet; 20.5textppm). Remaining: C_1H_2O.
- CH_2 (A): 1C, 2H. (from delta 4.3, 2H triplet; 65.0textppm). Remaining: H_0O. This CH_2 is clearly OCH_2.
- CH_2 (B): 1C, 2H. (from delta 2.6, 2H triplet; 34.0textppm). Remaining: No atoms left.
- C_6H_5−
- Ester C=O
- CH_3− (an acetyl methyl, since singlet at 2.1 ppm)
- −O−CH_2−
- −CH_2− (the one at 2.6 ppm, next to C=O)
- C_2H_3O_2 (acetyl group CH_3CO_2−)
- C_6H_5 (phenyl)
- CH_2CH_2 (ethyl bridge: C_2H_4)Total C: 2+6+2=10. Total H: 3+5+4=12. Total O: 2.This formula is C_10H_12O_2. Still not C_9H_10O_2!
- Formula: C_6H_5 (6C, 5H) + CH (1C, 1H) + CH_3 (1C, 3H) + COOCH_3 (2C, 3H) = C_10H_12O_2. No.
- Phenyl ring C_6H_5− (6C, 5H)
- C=O (1C)
- OCH_2− (1C, 2H) (from delta 4.3)
- −CH_2− (1C, 2H) (from delta 2.6)
- −CH_3 (1C, 3H) (from delta 2.1)
- Molecular Formula: C_10H_12O_2.
- IR: C=O ester 1735textcm−1, C−O 1210textcm−1, aromatic 3000−3100textcm−1. Fits!
- 1H NMR:
- C_6H_5: delta 7.3 (5H, multiplet). Fits!
- CH_2 (next to O): delta 4.3 (2H, triplet, coupled to adjacent CH_2). Fits!
- CH_2 (next to C_6H_5): delta 2.6 (2H, triplet, coupled to adjacent CH_2). Fits!
- CH_3 (acetyl): delta 2.1 (3H, singlet). Fits!
- $^{13}$C NMR:
- C=O: delta 172.0 (s). Fits!
- Aromatic carbons: delta 136.5 (s, C_ipso), delta 129.5 (d), delta 128.5 (d), delta 127.0 (d). Fits!
- O−CH_2: delta 65.0 (t). Fits!
- CH_2−Ar: delta 34.0 (t). Fits!
- CH_3 (acetyl): delta 20.5 (q). Fits!
- 1735textcm−1: C=O stretch, highly characteristic of an ester.
- 1210textcm−1: C−O stretch, consistent with an ester.
- 3000−3100textcm−1: Aromatic C−H stretching vibrations.
- Kinetics & Thermodynamicsa) Activation Energy (E_a): Use the Arrhenius equation (or its integrated form):ln(k_2/k_1)=−E_a/R(1/T_2−1/T_1)Given:k_1=3.38times10−5texts−1 at T_1=300textKk_2=1.63times10−3texts−1 at T_2=350textKR=8.314textJ/(molcdottextK)
$ln((1.63 \times 10^{-3}) / (3.38 \times 10^{-5})) = -E_a / 8.314 \text{ J/(mol} \cdot \text{K)} (1/350 \text{ K} - 1/300 \text{ K})$ $ln(48.22485) = -E_a / 8.314 \text{ J/(mol} \cdot \text{K)} (0.002857 - 0.003333) \text{ K}^{-1}$ $3.876 = -E_a / 8.314 \text{ J/(mol} \cdot \text{K)} (-0.000476) \text{ K}^{-1}$ $3.876 = E_a / 8.314 \times 0.000476$ $E_a = (3.876 \times 8.314) / 0.000476$ $E_a = 32.22 / 0.000476$ $E_a = 67689 \text{ J/mol}$ $E_a = 67.69 \text{ kJ/mol}$b) Concentration after 1 hour (at 300 K):Reaction is first-order. Use [A]_t=[A]_0e−kt.Given:[N_2O_5]_0=0.500textMk=3.38times10−5texts−1 (at 300 K)t=1texthour=3600texts$[N_2O_5]_t = 0.500 \text{ M} \times e^{-(3.38 \times 10^{-5} \text{ s}^{-1} \times 3600 \text{ s})}$ $[N_2O_5]_t = 0.500 \text{ M} \times e^{-0.12168}$ $[N_2O_5]_t = 0.500 \text{ M} \times 0.8853$ $[N_2O_5]_t = 0.44265 \text{ M}$ Therefore, $[N_2O_5]$ after 1 hour is approximately $0.443 \text{ M}$. - Reaction Mechanisms – Acid-Catalyzed Acetal FormationReactants: Ethanal (CH_3CHO) and Methanol (CH_3OH)Product: 1,1-dimethoxyethane (CH_3CH(OCH_3)_2)Catalyst: Acid (H+)Mechanism:
- Step 1: Protonation of the Carbonyl Oxygen. The carbonyl oxygen of ethanal is protonated by the acid catalyst. This activates the carbonyl carbon towards nucleophilic attack.
CH3-CH=O + H+ <=> CH3-CH=O+H (resonance stabilized oxonium ion)(Draw arrows: O lone pair attacks H+, forming O−H bond. Or show H_3O+ acting as acid) - Step 2: Nucleophilic Attack by Alcohol. Methanol, a nucleophile, attacks the activated (electrophilic) carbonyl carbon. This forms a hemiacetal intermediate, which is protonated at the oxygen derived from the carbonyl.
CH3-CH=O+H + CH3OH --> CH3-CH(O+H)-OCH3 (hemiacetal oxonium ion)(Draw arrows: lone pair on CH_3OH oxygen attacks C, pi electrons from C=O move to O−H). - Step 3: Deprotonation of the Hemiacetal (Hemi-acetal formation). The protonated oxygen of the hemiacetal intermediate loses a proton to regenerate the acid catalyst (or to another methanol molecule), forming the neutral hemiacetal.
CH3-CH(O+H)-OCH3 + CH3OH <=> CH3-CH(OH)-OCH3 + CH3O+H2(Draw arrows: CH_3OH abstracts proton from protonated oxygen). - Step 4: Protonation of the Hemiacetal Hydroxyl Group. The hydroxyl group of the hemiacetal is protonated by the acid catalyst. This turns it into a good leaving group (water).
CH3-CH(OH)-OCH3 + H+ <=> CH3-CH(O+H2)-OCH3(Draw arrows: O lone pair attacks H+). - Step 5: Loss of Water to Form an Oxocarbenium Ion. Water departs as a leaving group, forming a resonance-stabilized oxocarbenium ion (or an equivalent carbocation stabilized by the adjacent oxygen lone pair). This is typically the rate-determining step.
CH3-CH(O+H2)-OCH3 --> CH3-CH=O+CH3 + H2O (resonance stabilized)(Draw arrows: O−H_2+ bond breaks, H_2O leaves, lone pair on other oxygen moves to form double bond, stabilizing positive charge on carbon). - Step 6: Nucleophilic Attack by Second Alcohol Molecule. A second molecule of methanol attacks the electrophilic carbon of the oxocarbenium ion.
CH3-CH=O+CH3 + CH3OH --> CH3-CH(O+H)-OCH3(Draw arrows: lone pair on CH_3OH oxygen attacks carbon, forming C-O bond). - Step 7: Deprotonation to Form the Acetal. The protonated acetal product loses a proton to regenerate the acid catalyst, yielding the final acetal product.
CH3-CH(O+H)-OCH3 + CH3OH <=> CH3-CH(OCH3)2 + CH3O+H2(Draw arrows: CH_3OH abstracts proton from protonated oxygen).
- Step 1: Protonation of the Carbonyl Oxygen. The carbonyl oxygen of ethanal is protonated by the acid catalyst. This activates the carbonyl carbon towards nucleophilic attack.
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