ACS Organic Chemistry Practice Test 11

Correct: (B)

The initial aldol adduct (diacetone alcohol) dehydrates on heating to give the conjugated α,β-unsaturated ketone, mesityl oxide.

Correct: (C)

Only ethyl acetate has α-hydrogens, so its enolate attacks the carbonyl of ethyl benzoate in a crossed Claisen condensation to give ethyl benzoylacetate.

Correct: (D)

Grignard reagents add twice to esters: the first addition gives acetophenone, which undergoes a second addition to furnish the tertiary alcohol after workup.

Correct: (D)

NaBH4 reduces ketones and aldehydes but is not strong enough to reduce carboxylic acids, so only the ketone is reduced.

Correct: (A)

The Wittig reaction replaces the carbonyl oxygen with the ylide carbon, giving (CH3)2C=CHCH3.

Correct: (C)

Intramolecular Claisen (Dieckmann) cyclization of the 1,6-diester gives the five-membered cyclic β-keto ester.

Correct: (B)

LiAlH4 reduces esters all the way to primary alcohols, whereas NaBH4 does not reduce esters at all.

Correct: (A)

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

Correct: (B)

The ketone functional group would be attacked by the organomagnesium species as it forms, so the Grignard reagent destroys itself.

Correct: (D)

Dilute aqueous acid hydrolyzes acetals back to the parent carbonyl compound plus the alcohol.

Correct: (D)

The strongly electron-withdrawing trichloromethyl group greatly stabilizes the gem-diol; chloral hydrate is a stable, isolable compound.

Correct: (B)

Crossed aldol condensation with acetone gives benzalacetone (PhCH=CHCOCH3), and catalytic hydrogenation selectively reduces the C=C bond.

Correct: (A)

Enamine formation, α-alkylation with methyl iodide, and acidic hydrolysis installs a single methyl group at the α position.

Correct: (C)

DIBAL-H at low temperature reduces esters only to the aldehyde oxidation state, stopping before the alcohol.

Correct: (B)

Cyanide displaces bromide by an SN2 mechanism, which proceeds with complete inversion of configuration at the stereocenter.

Correct: (A)

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

Correct: (C)

Following Zaitsev's rule, the most substituted (trisubstituted) alkene predominates in the E2 elimination.

Correct: (A)

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

Correct: (D)

SN1 reactivity parallels carbocation stability, which increases with alkyl substitution: methyl < primary < secondary < tertiary.

Correct: (B)

Debromination requires an anti-periplanar arrangement of the two C–Br bonds, and for the meso diastereomer this geometry leads to the trans alkene.

Correct: (C)

Ionization gives a secondary carbocation that undergoes a methyl shift to the more stable tertiary carbocation before capture by bromide.

Correct: (D)

A tertiary substrate, a polar protic solvent, and a weak nucleophile together strongly favor unimolecular ionization.

Correct: (A)

A strong base with heating favors E2 elimination over SN2 substitution for a secondary alkyl halide.

Correct: (B)

The 1→2→3 priority order (Br → CH2CH3 → CH3) appears clockwise, but H is on a horizontal (forward) bond, so the apparent R inverts to S.

Correct: (D)

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

Correct: (C)

They are stereoisomers with the same connectivity but are not non-superimposable mirror images.

Correct: (A)

The anti staggered conformation minimizes both torsional strain and steric repulsion between the methyl groups.

Correct: (C)

C3 of pentan-3-ol is bonded to two identical ethyl groups, so the molecule is achiral; the others each contain a stereocenter.

Correct: (A)

ee = (45/150) × 100% = 30%, and the major enantiomer makes up (100 + 30)/2 = 65% of the mixture.

Correct: (D)

The higher-priority ethyl groups are trans to each other across the double bond, which is designated E.

Correct: (B)

Markovnikov protonation gives a secondary carbocation, which undergoes a hydride shift to the more stable tertiary carbocation before bromide capture.

Correct: (C)

Rejoining the two carbonyl carbons gives (CH3)2C=CHCH3, the only alkene that cleaves into exactly those two fragments.

Correct: (A)

Boron adds syn to the less substituted carbon of the double bond, and oxidation replaces boron with OH with retention, placing OH trans to the methyl group (anti-Markovnikov).

Correct: (D)

Secondary orbital interactions between the diene and the carbonyl π system stabilize the endo transition state, making it the kinetic product.

Correct: (B)

Dissolving-metal reduction delivers the trans alkene via the more stable trans vinyl radical/anion intermediate, whereas Lindlar gives the cis alkene.

Correct: (A)

Stability increases with alkyl substitution on the double bond, and trans isomers are more stable than cis due to reduced steric strain.

Correct: (C)

Bromine adds anti through a cyclic bromonium ion, and anti addition to a cis alkene produces the meso diastereomer.

Correct: (D)

Halogens direct ortho/para by resonance donation but deactivate overall because their inductive electron withdrawal dominates.

Correct: (B)

Strong resonance donors activate most, alkyl groups activate weakly by hyperconjugation, and halogens are net deactivating.

Correct: (A)

The strongly deactivating nitro group makes the ring too electron-poor for Friedel–Crafts electrophiles.

Correct: (C)

The powerfully activating –NH2 group causes rapid polysubstitution at all available ortho and para positions, precipitating as a white solid.

Correct: (D)

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

Correct: (B)

All four unsubstituted ring positions in p-xylene are symmetry-equivalent, so only 2-nitro-1,4-dimethylbenzene forms.

Correct: (A)

The data fit CH3CH2C(=O)OCH3: the OCH3 singlet near 3.7 ppm, the CH2 quartet alpha to the carbonyl, and the terminal CH3 triplet.

Correct: (C)

Symmetry makes the ring carbons pairwise equivalent (C1/C4 and C2/C3/C5/C6), giving two ring signals plus one methyl signal.

Correct: (D)

Cleavage at the benzylic position gives the resonance-stabilized C7H7+ tropylium cation, a characteristic base peak for alkylbenzenes.

Correct: (B)

The 1715 cm−1 band indicates a ketone C=O, and a single 6H singlet requires two equivalent methyl groups, matching (CH3)2C=O.

Correct: (C)

By the n+1 rule, coupling to the six equivalent methyl protons splits the methine signal into seven lines.

Correct: (A)

The two alkene carbons are trigonal planar (sp2) and the nitrile carbon is linear (sp).

Correct: (D)

The tert-butyl group has an enormous A-value, so the axial conformer is vanishingly populated and the ring is locked with t-Bu equatorial.

Correct: (B)

The cyclopentadienyl anion is planar and cyclic with 6 π electrons, satisfying Hückel's 4n+2 rule.

Correct: (C)

Extending conjugation raises the HOMO and lowers the LUMO, so the triene with three conjugated double bonds absorbs at the longest wavelength.

Correct: (D)

Acidity follows pKa: HCl (~−7) < acetic acid (4.76) < phenol (~10) < ethanol (~16).

Correct: (A)

Basicity tracks conjugate-acid pKa (~11.3 > 5.2 > 4.6 > −0.5); amide resonance makes acetamide essentially nonbasic.

Correct: (B)

The strongest base has the weakest conjugate acid; acetylene (pKa ~25) is far less acidic than ethanol, water, or acetic acid.

Correct: (C)

Radical stability increases with alkyl substitution due to hyperconjugation delocalizing the unpaired electron.

Correct: (D)

Bromination is highly selective for the weaker tertiary C–H bond (abstraction selectivity ~1600:1), and radical intermediates do not rearrange.

Correct: (A)

Isobutylene forms a stable tertiary carbocation on protonation, and its electron-rich double bond is ideal for cationic propagation.

Correct: (B)

Head-to-tail polymerization of CH2=CHCl gives the PVC repeat unit –[CH2–CHCl]–n.

Correct: (D)

Condensation of the 6-carbon diamine with the 6-carbon diacid gives the nylon-6,6 polyamide.

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