ACS Organic Chemistry Practice Test 1

Correct: (C)

Crossed aldol addition gives the β-hydroxy ketone, which dehydrates under the basic, warmed conditions to the conjugated enone (E)-4-phenylbut-3-en-2-one.

Correct: (A)

Claisen condensation of two molecules of ethyl acetate gives the β-keto ester ethyl acetoacetate after protonation of the stabilized enolate.

Correct: (D)

Intramolecular Dieckmann condensation of the adipate ester closes a five-membered ring, giving ethyl 2-oxocyclopentanecarboxylate.

Correct: (B)

Grignard addition of PhMgBr to the aldehyde carbonyl gives, after protonation, the secondary alcohol Ph2CHOH.

Correct: (A)

NaBH4 reduces the more electrophilic aldehyde far faster than the ketone, so only the aldehyde is reduced under these controlled conditions.

Correct: (C)

LiAlH4 is a powerful hydride donor that reduces esters all the way to primary alcohols; it does not stop at the aldehyde stage.

Correct: (B)

The Wittig reagent converts the ketone carbonyl into an exocyclic C=C bond, giving methylenecyclohexane and triphenylphosphine oxide.

Correct: (A)

A secondary amine condenses with a ketone to give an enamine: the iminium intermediate loses an α-proton to form the C=C bond adjacent to nitrogen.

Correct: (B)

Acid-catalyzed reaction with two equivalents of alcohol converts the aldehyde fully to its diethyl acetal with loss of water.

Correct: (C)

Alkyl groups donate electron density and add steric crowding, destabilizing the gem-diol; formaldehyde, with no alkyl groups, hydrates most readily.

Correct: (B)

Reactivity follows leaving-group ability: Cl⁻ > CH3CO2⁻ > EtO⁻ > NH2⁻, the weaker the conjugate base, the better it leaves.

Correct: (A)

Grignard addition gives the secondary alcohol 1-phenylethanol, and PCC oxidizes a secondary alcohol to the ketone acetophenone.

Correct: (D)

NaBH4 is too mild to reduce esters, so only the ketone is reduced, leaving the ester functional group intact.

Correct: (C)

Retrosynthetic disconnection of the C3–C4 bond gives the enolate of acetone plus acetone, so dimerization of acetone via aldol addition furnishes the target β-hydroxy ketone.

Correct: (B)

A tertiary halide with a strong base undergoes E2 elimination; SN2 is impossible at a tertiary carbon and substitution is outcompeted by the strong base.

Correct: (D)

For rate = k[CH3I][CN⁻], the rate changes by (2)(3) = 6 when both concentrations are scaled.

Correct: (C)

The SN1 rate-determining step is ionization of the alkyl halide alone, so the rate law is first order in substrate and zero order in the nucleophilic solvent.

Correct: (A)

The bulky base abstracts the least sterically hindered β-hydrogen, giving the less substituted Hofmann product as the major alkene.

Correct: (D)

SN2 proceeds by backside attack, which inverts the configuration at the electrophilic carbon, converting (R) substrate to (S) product.

Correct: (B)

Bromine adds anti via a cyclic bromonium ion, giving the trans dibromide as a pair of enantiomers.

Correct: (D)

Polar aprotic solvents like DMSO solvate cations but leave the anionic nucleophile poorly solvated and highly reactive.

Correct: (A)

In protic solvents, small anions are heavily hydrogen-bonded and deactivated, so the large, polarizable iodide is the most nucleophilic.

Correct: (C)

The small base gives the more substituted Zaitsev alkene, and the trans isomer predominates because it avoids the steric strain of the cis isomer.

Correct: (B)

Priorities are Br > CO2H > CH3 > H; the 1→2→3 sequence is counterclockwise (S), but H lies on a horizontal (forward) bond, so the assignment reverses to R.

Correct: (D)

Every stereocenter is inverted, giving non-superimposable mirror images, the definition of enantiomers.

Correct: (A)

The cis-1,2-dichloro isomer has an internal plane of symmetry through C3, making it achiral despite having two stereocenters.

Correct: (C)

Rotation about the C2–C3 single bond interconverts conformers without breaking any bonds, so the configuration and the compound are unchanged.

Correct: (B)

Enantiomeric excess = (observed rotation / rotation of pure enantiomer) × 100% = (18.0/60.0) × 100% = 30%.

Correct: (D)

The racemate is optically inactive because the equal and opposite rotations cancel, and the meso compound is achiral with an internal plane of symmetry.

Correct: (A)

Greater alkyl substitution stabilizes the double bond through hyperconjugation, and trans alkenes avoid the steric strain present in cis alkenes.

Correct: (D)

The single 4-carbon cleavage product means two identical halves joined at the double bond, giving CH₃CH₂C(CH₃)=C(CH₃)CH₂CH₃, which is 3,4-dimethylhex-3-ene.

Correct: (B)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry; syn addition to the two faces of symmetric cis-3-hexene creates the single stereocenter at C3 in both configurations equally.

Correct: (C)

Markovnikov protonation of 2-methyl-1-butene at the terminal carbon gives the tertiary 2-methyl-2-butyl cation, which captures bromide to form 2-bromo-2-methylbutane.

Correct: (D)

Water opens the cyclic bromonium ion anti at the more substituted carbon, giving the trans bromohydrin with the OH on the more substituted carbon.

Correct: (A)

Isoprene (2-methyl-1,3-butadiene) plus methyl acrylate places the diene methyl para to the ester (ortho/para rule), giving the 4-methylcyclohex-3-ene-1-carboxylate skeleton.

Correct: (B)

Dissolving-metal reduction of an internal alkyne proceeds through a trans vinyl radical anion, giving the E (trans) alkene; Lindlar's catalyst would instead give the Z alkene.

Correct: (C)

Stability increases with alkyl substitution via hyperconjugation: tetrasubstituted (I) > trisubstituted (II) > disubstituted trans (III) > monosubstituted (IV).

Correct: (D)

The tert-butyl group is an ortho/para activator, but its bulk blocks ortho attack, so para substitution dominates.

Correct: (A)

Methoxy is a stronger activator than methyl, and nitro strongly deactivates the ring, so the order is anisole > toluene > benzene > nitrobenzene.

Correct: (C)

This is nucleophilic aromatic substitution: hydroxide displaces the chloride activated by the two ortho/para nitro groups, giving 2,4-dinitrophenol after workup.

Correct: (B)

Nitration gives nitrobenzene, reduction gives aniline, and the strongly activating amino group directs excess bromine to all three ortho/para positions.

Correct: (D)

Acetylation first installs the meta-directing acetyl group, so subsequent bromination occurs meta to it, giving m-bromoacetophenone.

Correct: (C)

SNAr requires electron-withdrawing groups ortho or para to the leaving group to stabilize the Meisenheimer complex, and two nitro groups activate far more than one.

Correct: (A)

The singlet methyl next to a carbonyl plus the ethyl quartet/triplet pattern matches CH₃C(=O)CH₂CH₃; an aldehyde proton (~9–10 ppm) is absent.

Correct: (B)

1,4-Dimethylbenzene has D2h symmetry: the two methyl carbons are equivalent, the two ipso (C–CH3) ring carbons are equivalent, and all four ring CH carbons are equivalent, giving 3 distinct signals.

Correct: (B)

Alpha cleavage ejects an ethyl radical, leaving the resonance-stabilized oxonium ion (CH₃)₂C=OH⁺ at m/z 59 as the base peak.

Correct: (C)

The ketone IR band plus the high symmetry of CH₃C(=O)CH₂CH₂C(=O)CH₃ accounts for the equivalent methyls (6H singlet) and equivalent methylenes (4H singlet).

Correct: (A)

The ester carbonyl plus a deshielded –OCH₂– quartet at 4.1 ppm and an acetyl methyl singlet at 2.0 ppm match CH₃C(=O)OCH₂CH₃.

Correct: (D)

The methyl-bearing carbon is part of the C=C double bond, so it has three sigma bonds and one pi bond, giving sp² hybridization.

Correct: (B)

In cis-1,3-disubstituted cyclohexane one chair conformation places both substituents equatorial (the ring flip would make both axial), and the diequatorial chair is strongly favored.

Correct: (B)

The cyclopentadienyl anion has 6 π electrons (4n+2, n=1) in a planar conjugated ring and is aromatic; the cation (4 π) would be antiaromatic.

Correct: (C)

Butadiene's π MOs ψ₁ through ψ₄ have 0, 1, 2, and 3 nodes respectively, and the HOMO is ψ₂, which has one node between C2 and C3.

Correct: (D)

The electron-withdrawing nitro group stabilizes the carboxylate (strongest acid), while electron-donating groups destabilize it, with resonance-donating methoxy weakening the acid more than methyl.

Correct: (A)

Aliphatic amines exceed ammonia in basicity, aniline's lone pair is delocalized into the ring (weaker), and the p-nitro group withdraws further electron density, making p-nitroaniline weakest.

Correct: (B)

Ammonia's lone pair accepts a proton to form NH₄⁺; methane cannot accept a proton, BF₃ is a Lewis acid, and concentrated sulfuric acid acts as a proton donor.

Correct: (C)

The benzyl radical is resonance-stabilized (most stable), followed by tertiary > secondary > primary alkyl radicals via hyperconjugation.

Correct: (D)

The bromine radical is highly selective for the weakest C–H bond, so abstraction of the single tertiary hydrogen gives the tertiary radical and thus 2-bromo-2-methylbutane.

Correct: (A)

Propagation is the repeated addition of the chain-carrying benzylic radical to the double bond of a fresh monomer, extending the chain.

Correct: (C)

PVC is the chain-growth addition polymer of vinyl chloride, CH₂=CHCl.

Correct: (B)

Nylon-6,6 forms by condensation of the diacid HOOC(CH₂)₄COOH (adipic acid) with the diamine H₂N(CH₂)₆NH₂ (hexamethylenediamine), releasing water.

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