Free practice final
Organic Chemistry 2 practice final (free, with answers)
An Organic Chemistry 2 final usually covers conjugated dienes and the Diels-Alder reaction, aromaticity and electrophilic aromatic substitution, aldehydes, ketones, carboxylic acid derivatives and enolates, amines, and IR and NMR spectroscopy. These 15 questions mix mechanism reasoning with predicting products and reading spectra. Work each one on paper before you open the answer, and redraw the mechanism whenever an explanation surprises you.
Topics
- Conjugated Systems
- Aromaticity
- Electrophilic Aromatic Substitution
- Aldehydes and Ketones
- Carboxylic Acid Derivatives
- Enolate Chemistry
- Amines
- Spectroscopy
15 questions
Try each one before you open its answer.
1. Buta-1,3-diene reacts with one equivalent of HBr at 40 °C, a temperature at which the products can interconvert. What is the major product?
- A 3-Bromobut-1-ene
- B 4-Bromobut-1-ene
- C 1-Bromobut-2-ene
- D 1,4-Dibromobut-2-ene
Show answer
Answer: C. 1-Bromobut-2-ene
Protonation at C1 gives an allylic cation whose positive charge is shared by C2 and C4. Bromide at C2 gives the 1,2-adduct, which forms faster and dominates at low temperature, but at 40 °C the products equilibrate and the more stable 1,4-adduct, with its more substituted internal C=C, wins.
2. Buta-1,3-diene undergoes a Diels-Alder reaction with dimethyl fumarate, in which the two ester groups are trans across the C=C. What is the product?
- A cis-Dimethyl cyclohex-4-ene-1,2-dicarboxylate, a meso compound
- B trans-Dimethyl cyclohex-4-ene-1,2-dicarboxylate, as a racemic mixture
- C A roughly equal mixture of the cis and trans diesters
- D trans-Dimethyl cyclohex-4-ene-1,2-dicarboxylate, as a single enantiomer
Show answer
Answer: B. trans-Dimethyl cyclohex-4-ene-1,2-dicarboxylate, as a racemic mixture
The Diels-Alder reaction is a concerted, suprafacial cycloaddition, so the dienophile's geometry is kept: ester groups that were trans on the alkene end up trans on the new ring. The trans product is chiral, and the diene adds to either face of the achiral dienophile equally, so both enantiomers form in equal amounts.
3. Which species is aromatic?
- A Cyclobutadiene
- B Cyclopentadienyl cation
- C Cyclooctatetraene
- D Cyclopentadienyl anion
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Answer: D. Cyclopentadienyl anion
The cyclopentadienyl anion is cyclic, planar and fully conjugated with 6 π electrons, which fits Hückel's 4n + 2 rule with n = 1. The cation and cyclobutadiene each have 4 π electrons and are antiaromatic, and cyclooctatetraene (8 π electrons) puckers into a nonplanar tub shape, so it is nonaromatic.
4. Benzene is nitrated with HNO₃ and H₂SO₄, and the product is then brominated with Br₂ and FeBr₃. What is the major final product?
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Answer: 1-Bromo-3-nitrobenzene (m-bromonitrobenzene)
The nitro group is a strong deactivator and a meta director, so bromination of nitrobenzene puts the bromine meta to it. Doing the two steps in the reverse order would give mostly the ortho and para isomers, because bromine directs ortho and para.
5. Which substituent on a benzene ring makes the ring less reactive toward electrophilic aromatic substitution, yet directs the incoming electrophile to the ortho and para positions?
- A –Cl
- B –OCH₃
- C –NO₂
- D –CH₃
Show answer
Answer: A. –Cl
Halogens withdraw electron density by induction, which slows the reaction, but their lone pairs can donate by resonance to stabilize the carbocation formed by ortho or para attack. –OCH₃ and –CH₃ activate the ring, and –NO₂ deactivates it and directs meta.
6. What is the major organic product when cyclohexanone reacts with the Wittig reagent Ph₃P=CH₂?
- A 1-Methylcyclohexan-1-ol
- B 1-Methylcyclohexene
- C Methylenecyclohexane
- D Cycloheptanone
Show answer
Answer: C. Methylenecyclohexane
The ylide carbon adds to the carbonyl carbon to form an oxaphosphetane, which breaks apart into an alkene and triphenylphosphine oxide. The new C=C appears exactly where the C=O was, so the product is the exocyclic alkene, not the more substituted 1-methylcyclohexene.
7. Cyclohexanone is heated with ethylene glycol (HOCH₂CH₂OH) and a catalytic amount of p-toluenesulfonic acid while the water that forms is removed. What type of compound is the product?
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Answer: A cyclic acetal, also called a ketal (1,4-dioxaspiro[4.5]decane)
Acid catalyzes addition of one OH group to give a hemiacetal, then loss of water and ring closure by the second OH group. Removing water pulls the equilibrium toward the acetal, which is stable to bases and nucleophiles, so this reaction is used to protect a ketone that can later be recovered with aqueous acid.
8. Which ranking lists carboxylic acid derivatives from most to least reactive toward nucleophilic acyl substitution?
- A Acid chloride > ester > anhydride > amide
- B Amide > ester > anhydride > acid chloride
- C Anhydride > acid chloride > amide > ester
- D Acid chloride > anhydride > ester > amide
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Answer: D. Acid chloride > anhydride > ester > amide
Reactivity tracks how good the leaving group is (Cl⁻ > carboxylate > alkoxide > amide ion) and how strongly the attached atom donates electrons to the carbonyl by resonance, which is greatest for the nitrogen of an amide. That is why a more reactive derivative is easily converted into a less reactive one, such as an acid chloride into an ester or amide, but the reverse is uphill.
9. Methyl benzoate is treated with excess CH₃MgBr, followed by aqueous acid workup. What is the organic product?
Show answer
Answer: 2-Phenylpropan-2-ol
The first equivalent of Grignard reagent adds and expels methoxide, giving acetophenone. That ketone is more reactive than the starting ester, so it quickly adds a second methyl group, and protonation gives a tertiary alcohol whose carbinol carbon carries a phenyl and two methyl groups.
10. 2-Methylcyclohexanone is treated with LDA in THF at −78 °C, and then with one equivalent of CH₃I. What is the major product?
- A 2,2-Dimethylcyclohexanone
- B 2,6-Dimethylcyclohexanone
- C 1,2-Dimethylcyclohexan-1-ol
- D A roughly 1:1 mixture of 2,2- and 2,6-dimethylcyclohexanone
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Answer: B. 2,6-Dimethylcyclohexanone
LDA is a strong, bulky base that removes a proton quickly and irreversibly at low temperature, so it takes one from the less hindered α-carbon (C6) and forms the kinetic enolate. Methylation there gives 2,6-dimethylcyclohexanone; conditions that let the enolates equilibrate would favor the more substituted enolate and 2,2-dimethylcyclohexanone instead.
11. Ethyl acetate is treated with sodium ethoxide in ethanol, and the mixture is then acidified. What is the product of this Claisen condensation?
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Answer: Ethyl acetoacetate (ethyl 3-oxobutanoate)
Ethoxide removes an α-hydrogen from one ester, the enolate adds to the carbonyl of a second ester, and the tetrahedral intermediate expels ethoxide. The β-keto ester product (pKa about 11) is much more acidic than ethanol, so ethoxide deprotonates it, which drives the reaction forward; acid workup then gives the neutral product.
12. Which compound is the strongest base?
- A Cyclohexylamine
- B Aniline
- C 4-Nitroaniline
- D Acetamide
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Answer: A. Cyclohexylamine
Cyclohexylamine's lone pair sits on an sp³ nitrogen and is not delocalized, so its conjugate acid has a pKa near 10.6. In aniline the lone pair is partly delocalized into the ring (anilinium pKa about 4.6), a para nitro group pulls it away even more (about 1.0), and in acetamide it is delocalized onto the carbonyl oxygen, so acetamide is barely basic.
13. Which set of reagents converts cyclohexanone into N-methylcyclohexanamine by reductive amination?
- A Ammonia, with NaBH₃CN and mild acid
- B Dimethylamine, with NaBH₃CN and mild acid
- C Methanol, with an acid catalyst
- D Methylamine, with NaBH₃CN and mild acid
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Answer: D. Methylamine, with NaBH₃CN and mild acid
The amine and ketone condense to an imine (or iminium ion), which NaBH₃CN reduces much faster than it reduces the ketone itself. The nitrogen keeps the groups of the amine you start with, so methylamine gives the N-methyl product, ammonia gives cyclohexylamine, and dimethylamine gives N,N-dimethylcyclohexanamine.
14. A compound with formula C₃H₆O has a strong IR band near 1715 cm⁻¹, no broad band between 3200 and 3600 cm⁻¹, and no weak C–H bands near 2720 and 2820 cm⁻¹. Which compound is it?
- A Propanone (acetone)
- B Propanal
- C Prop-2-en-1-ol (allyl alcohol)
- D Methyloxirane (propylene oxide)
Show answer
Answer: A. Propanone (acetone)
A strong band near 1715 cm⁻¹ is the C=O stretch of a simple ketone. Propanal would also show the aldehyde C–H stretches near 2720 and 2820 cm⁻¹, allyl alcohol would show a broad O–H band and no C=O, and methyloxirane has no carbonyl at all.
15. A compound with formula C₄H₈O₂ has a strong IR band at 1740 cm⁻¹. Its ¹H NMR spectrum shows a triplet at δ 1.26 (3H), a singlet at δ 2.05 (3H) and a quartet at δ 4.12 (2H). Name the compound.
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Answer: Ethyl acetate (ethyl ethanoate)
The 3H singlet near 2.0 ppm is a CH₃ attached to a carbonyl with no neighboring hydrogens, and the quartet coupled to the triplet is an ethyl group. Its CH₂ at 4.1 ppm must be bonded to oxygen, giving CH₃COOCH₂CH₃; in the isomer methyl propanoate, the CH₂ would appear near 2.3 ppm and an OCH₃ singlet near 3.7 ppm.