ACS Organic Chemistry Practice Test 13

Correct: (B)

This is a Claisen–Schmidt crossed aldol: the acetone enolate attacks benzaldehyde, and the initial β-hydroxy ketone dehydrates under the basic, heated conditions to the conjugated enone.

Correct: (A)

Intramolecular Claisen (Dieckmann) condensation of this 1,6-diester closes a five-membered ring to give the cyclic β-keto ester.

Correct: (C)

Grignard reagents add twice to esters: the first equivalent gives acetophenone as an intermediate, and the second equivalent adds to the ketone to give the tertiary alcohol.

Correct: (A)

NaBH4 reduces the more electrophilic aldehyde in preference to the ketone, so only the aldehyde carbonyl is reduced under these controlled conditions, giving the hydroxy-ketone.

Correct: (B)

The Wittig reaction converts the ketone carbonyl into a C=C bond, replacing C=O with C=CH2 to give the 1,1-disubstituted alkene.

Correct: (B)

A secondary amine condenses with a ketone under dehydrating conditions to give an enamine, not an imine.

Correct: (B)

Acid-catalyzed reaction of an aldehyde with a 1,2-diol gives the cyclic five-membered acetal (a 1,3-dioxolane).

Correct: (B)

Reactivity follows leaving-group ability: Cl− > CH3COO− > CH3CH2O− > NH2−, the weaker the base, the better the leaving group.

Correct: (D)

The strongly electron-withdrawing CCl3 group greatly favors the gem-diol form; chloral is normally isolated and handled as chloral hydrate.

Correct: (A)

LiAlH4 is a powerful hydride donor that reduces nitriles completely to primary amines; stopping at the aldehyde requires DIBAL-H at low temperature.

Correct: (B)

The initial addition product, 3-hydroxybutanal, dehydrates under the acidic, heated conditions to give the conjugated α,β-unsaturated aldehyde.

Correct: (B)

Aldehydes are more electrophilic than ketones, so one equivalent of Grignard reagent attacks the aldehyde selectively.

Correct: (B)

Mono-α-methylation followed by ester hydrolysis and thermal decarboxylation gives CH3COCH2CH3.

Correct: (B)

The cyclohexanone enolate attacks acetone, and dehydration gives the more substituted, conjugated exocyclic alkene, 2-isopropylidenecyclohexanone.

Correct: (B)

The slow, rate-determining step of SN1 is unimolecular ionization of the alkyl halide, so the rate depends only on the substrate concentration.

Correct: (B)

SN2 rates decrease sharply with steric hindrance at the electrophilic carbon, so the unhindered methyl bromide reacts fastest.

Correct: (C)

The small base gives the most substituted (Zaitsev) alkene, 2-methyl-2-butene, which has no E/Z isomers.

Correct: (A)

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

Correct: (B)

Cyanide displaces bromide by a concerted backside SN2 attack, which inverts the configuration at the stereocenter from S to R.

Correct: (C)

A rate independent of nucleophile concentration is the kinetic signature of the unimolecular SN1 mechanism.

Correct: (C)

E2 rates increase with the stability of the alkene being formed; the tertiary bromide gives the most substituted alkene through the lowest-energy transition state.

Correct: (B)

Leaving-group ability increases down the periodic group as the anion becomes larger and more stable; iodide is the best leaving group listed.

Correct: (C)

Ionization gives a primary carbocation that undergoes a 1,2-methyl shift to the more stable tertiary carbocation, which is captured by water to give 2-methyl-2-butanol.

Correct: (A)

Tracing priorities OH(1) > Et(2) > Me(3) with OH on the right and H on the left (forward bond) gives an apparent clockwise (R) sequence that inverts to S; placing OH on the left gives the true (R) configuration.

Correct: (B)

Every stereocenter is inverted between the two structures, giving non-superimposable mirror images.

Correct: (B)

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

Correct: (C)

The anti staggered conformation places the two bulky methyl groups 180° apart, minimizing steric repulsion and torsional strain.

Correct: (A)

ee = 8.1/13.5 × 100% = 60%, so the major enantiomer is (100 + 60)/2 = 80% (S) and the minor is 20% (R).

Correct: (C)

The racemic 50:50 mixture has zero net rotation because the equal and opposite rotations of the enantiomers cancel exactly.

Correct: (C)

The molecule has two stereocenters with different substituents, so all four stereoisomers are distinct with no meso form, giving two pairs of enantiomers.

Correct: (B)

Stability increases with alkyl substitution on the double bond (hyperconjugation), and trans alkenes are more stable than their cis isomers.

Correct: (A)

Both fragments are identical three-carbon aldehydes, so the alkene must be symmetric: CH3CH2CH=CHCH2CH3.

Correct: (A)

Oxymercuration–demercuration adds OH to the more substituted alkene carbon (Markovnikov) without carbocation rearrangement, giving the tertiary alcohol.

Correct: (C)

Hydroboration–oxidation adds H and OH syn to the double bond; on the cyclohexene ring this syn addition places the OH trans to the methyl group.

Correct: (B)

The peroxide effect gives anti-Markovnikov radical addition: Br adds to the less substituted carbon via the more stable secondary radical, with no rearrangement.

Correct: (B)

Dissolving-metal reduction of an internal alkyne proceeds by anti addition through a vinyl radical anion, giving the trans alkene.

Correct: (B)

Secondary orbital interactions between the dienophile's carbonyl π systems and the diene favor the endo transition state, making it the kinetic product.

Correct: (B)

The methoxy group is a strong ortho/para activator, so electrophilic substitution occurs at the ortho and para positions.

Correct: (A)

Electron-donating groups activate the ring toward electrophiles (methoxy more than methyl), while the nitro group strongly deactivates.

Correct: (D)

The nitro group is a powerful deactivator, withdrawing electron density by both resonance and induction.

Correct: (D)

Two nitro groups ortho and para to the leaving group best stabilize the anionic Meisenheimer intermediate, greatly accelerating SNAr.

Correct: (C)

The primary propyl cation rearranges by hydride shift to the more stable isopropyl cation, so n-propylbenzene cannot be obtained directly.

Correct: (A)

Friedel–Crafts acylation with the cyclic anhydride opens the ring to give 4-oxo-4-phenylbutanoic acid, and Clemmensen reduction converts the ketone to CH2.

Correct: (A)

The 3H singlet at 2.2 ppm is an acetyl methyl group, and the quartet/triplet pattern is an ethyl group adjacent to the carbonyl.

Correct: (B)

Symmetry makes the two methyl carbons equivalent, the four ring CH carbons equivalent, and the two ipso carbons equivalent, giving three distinct environments.

Correct: (B)

Propylbenzene (M = 120) undergoes benzylic cleavage to give the stable tropylium ion C7H7+ at m/z 91.

Correct: (B)

The broad IR band and the far-downfield 1H singlet are diagnostic of a carboxylic acid O–H, and the 3H singlet is the acetyl methyl group.

Correct: (B)

Five aromatic protons indicate a monosubstituted benzene, and the quartet/triplet pair is an ethyl group attached to the ring.

Correct: (B)

The trans-1,2 isomer can place both methyl groups equatorial (one up, one down), avoiding all 1,3-diaxial interactions.

Correct: (C)

The nitrogen is nearly planar with its lone pair partially delocalized into the ring; it is best described as sp2 hybridized.

Correct: (B)

The cyclopentadienyl anion is planar, cyclic, fully conjugated, and has 6 π electrons (4n+2, n=1), satisfying Hückel's rule.

Correct: (B)

Cyclobutadiene is a planar conjugated ring with 4 π electrons (4n), making it antiaromatic and highly unstable.

Correct: (B)

Each additional electron-withdrawing chlorine further stabilizes the carboxylate conjugate base by induction, progressively lowering the pKa.

Correct: (C)

Aliphatic amines are far more basic than aromatic amines (whose lone pairs delocalize into the ring) or amides (whose lone pairs delocalize into the carbonyl).

Correct: (C)

The amide ion's conjugate acid (NH3, pKa ~35) is the weakest acid listed, making NH2− the strongest base.

Correct: (B)

The benzyl radical is resonance-stabilized, and among alkyl radicals stability increases with substitution (tertiary > secondary > primary).

Correct: (B)

Abstraction of the single tertiary hydrogen is fastest because the resulting tertiary radical is most stable.

Correct: (B)

Propagation is the repeated addition of the chain-carrying radical to fresh monomer molecules.

Correct: (B)

Head-to-tail addition polymerization of propene (CH2=CHCH3) gives a chain with a methyl substituent on every other carbon.

Correct: (C)

Nylon-6,6 forms by condensation of the six-carbon diamine with the six-carbon diacid, releasing water at each amide linkage.

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