ACS Organic Chemistry Practice Test 14

Correct: (B)

Acetophenone forms the enolate and attacks benzaldehyde in a Claisen–Schmidt crossed aldol; dehydration of the β-hydroxy ketone gives the conjugated enone chalcone.

Correct: (B)

The classic Claisen self-condensation of an ester with α-hydrogens gives the β-keto ester ethyl acetoacetate.

Correct: (A)

Intramolecular Dieckmann condensation of this 1,8-diester closes a seven-membered ring, the only geometrically accessible ring size here.

Correct: (A)

Grignard addition to an aldehyde gives a secondary alkoxide; a second equivalent cannot add, so protonation yields the secondary alcohol.

Correct: (A)

NaBH4 is too weak to reduce carboxylic acids, so it chemoselectively reduces the ketone to a secondary alcohol.

Correct: (B)

The stabilized Wittig ylide olefinates the ketone to give the α,β-unsaturated ester with the double bond exocyclic to the ring.

Correct: (B)

A ketone plus a secondary amine under dehydrating conditions gives the enamine, with the C=C bond in the ring.

Correct: (A)

Aldehydes are more reactive than ketones, steric hindrance and electron-donating groups slow addition, and phenyl conjugation stabilizes the benzaldehyde carbonyl.

Correct: (B)

Cyanide catalyzes the benzoin condensation, in which two molecules of benzaldehyde couple to give the α-hydroxy ketone.

Correct: (A)

LiAlH4 reduces the ketone to the corresponding secondary alcohol; it cannot reduce the C–O bond further to an alkane.

Correct: (B)

The enolate of one propanal molecule attacks the carbonyl of another, forming the β-hydroxy aldehyde with the hydroxyl group on C3 and the methyl branch on C2.

Correct: (B)

The electron-withdrawing nitro group makes the aldehyde carbon more electrophilic, accelerating nucleophilic attack by the acetone enolate.

Correct: (D)

Simple ester hydrolysis of ethyl acetoacetate gives acetoacetic acid, which spontaneously decarboxylates on heating to give butan-2-one directly.

Correct: (B)

LDA quantitatively forms the kinetic enolate, which adds to benzaldehyde; at −78 °C the aldol addition product does not dehydrate.

Correct: (A)

The rate-determining step is carbocation formation, so the tertiary chloride ionizes fastest.

Correct: (B)

Second-order kinetics: (2)(2) = 4-fold rate increase.

Correct: (C)

The small base gives the more substituted Zaitsev alkene, and the E isomer predominates due to lower steric strain in the transition state.

Correct: (A)

With a small base, the more substituted Zaitsev alkene is the major product of the E2 elimination.

Correct: (B)

SN2 proceeds with backside attack, inverting the configuration at the stereocenter.

Correct: (B)

Unimolecular loss of the leaving group to generate the carbocation is the slow step that determines the overall rate.

Correct: (B)

More polar protic solvents better solvate and stabilize the developing ions in the ionization transition state.

Correct: (D)

Bromomethane has no β-hydrogens, so elimination is impossible regardless of the base used.

Correct: (B)

SN2 is blocked by steric hindrance, so the reaction proceeds through a carbocation that rearranges by methyl shift; elimination then gives the tetrasubstituted alkene.

Correct: (A)

With H pointing away, priorities OH (1) > CH2CH3 (2) > CH3 (3) trace a clockwise 1→2→3 sequence, giving (R).

Correct: (B)

Inverting every stereocenter gives the non-superimposable mirror image, the definition of enantiomers.

Correct: (B)

The (2R,3S) isomer has an internal plane of symmetry, making it achiral despite two stereocenters.

Correct: (C)

Maximum torsional strain and methyl–methyl steric repulsion occur when the two methyl groups are directly eclipsed.

Correct: (B)

ee = (observed [α]/[α] of pure enantiomer) × 100% = (2.70/13.5) × 100% = 20%.

Correct: (B)

The meso diol has an internal plane of symmetry and is achiral, so it is optically inactive.

Correct: (B)

There are two enantiomers (2R,3R and 2S,3S) plus one meso form (2R,3S), for a total of three stereoisomers.

Correct: (D)

The tetrasubstituted alkene is the most stable due to maximum hyperconjugation.

Correct: (A)

Two identical four-carbon ketone fragments join at their carbonyl carbons to give CH3CH2C(CH3)=C(CH3)CH2CH3.

Correct: (B)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry, placing the OH trans to the methyl group.

Correct: (B)

Epoxidation followed by base-catalyzed ring opening gives anti addition of the two OH groups, producing the trans diol.

Correct: (C)

Protonation gives a secondary carbocation that undergoes methyl shift to the tertiary carbocation, which captures bromide.

Correct: (A)

Markovnikov addition occurs twice: the first HBr gives the vinyl bromide, and the second adds to the same carbon, giving the gem-dibromide.

Correct: (B)

Electron-donating groups raise the diene HOMO energy, accelerating the normal-demand Diels–Alder reaction most.

Correct: (B)

The –CO2H group is a deactivating meta director.

Correct: (C)

The methoxy group is a strong activator, donating electron density by resonance.

Correct: (B)

More nitro groups ortho/para to the leaving group better stabilize the Meisenheimer intermediate.

Correct: (B)

The nitro group stabilizes the carboxylate by resonance withdrawal; benzoic acid is stronger than acetic acid due to the sp2 carbon.

Correct: (A)

At low temperature the kinetic 1-sulfonic acid predominates; at high temperature the thermodynamic 2-isomer forms.

Correct: (B)

Nitration first installs the meta-directing nitro group, so chlorination occurs meta to it.

Correct: (B)

Two equivalent ethyl groups with no other signals indicate the symmetric ketone CH3CH2COCH2CH3.

Correct: (B)

Three distinct carbon environments: the three ipso carbons, the three ring CH carbons, and the three methyl carbons.

Correct: (B)

Alpha cleavage ejects an ethyl radical, leaving the oxonium ion CH3CH=OH+ at m/z 45.

Correct: (B)

The acetyl singlet plus the ethyl quartet/triplet and ketone IR band match CH3COCH2CH3.

Correct: (A)

The 9H singlet is a tert-butyl group attached to oxygen, and the 3H singlet is the OCH3 group.

Correct: (B)

The trans-1,3 isomer can place both methyl groups equatorial (both up or both down), avoiding 1,3-diaxial strain.

Correct: (B)

The cationic carbon is trigonal planar with its empty p orbital conjugated with the double bond.

Correct: (B)

The cyclopentadienyl anion is planar with 6 π electrons, satisfying Hückel's rule.

Correct: (B)

Planar cyclobutadiene has 4 π electrons (4n), making it antiaromatic.

Correct: (B)

The nitro group stabilizes the phenoxide anion by resonance, making p-nitrophenol the strongest acid listed.

Correct: (B)

Cyclohexylamine is an aliphatic amine (conjugate acid pKa ~10.6); aromatic amines and pyrrole are far less basic due to lone-pair delocalization.

Correct: (D)

Its conjugate acid (NH3, pKa 35) is the weakest acid, so NH2− is the strongest base.

Correct: (B)

Radical stability increases with alkyl substitution through hyperconjugation.

Correct: (B)

The bromine radical selectively abstracts the weakest (tertiary) C–H bond.

Correct: (A)

Propagation is the repeated addition of the growing radical to monomer.

Correct: (B)

Polyethylene is the addition polymer of ethylene (ethene).

Correct: (B)

The diacid plus diamine condense to nylon-6,6 with loss of water; the others undergo addition polymerization.