ACS Organic Chemistry Practice Test 8

Correct: (C)

Benzaldehyde has no α-hydrogens so it acts only as the electrophile; the acetone enolate adds once and dehydrates to the conjugated enone, and excess acetone suppresses the double-addition product (A).

Correct: (B)

Intramolecular aldol forms a five-membered β-hydroxy ketone, which dehydrates toward the conjugated α,β-unsaturated ketone with the methyl at C3.

Correct: (D)

Dieckmann cyclization of an 8-carbon diester closes a seven-membered ring (n−1 rule), giving 2-(ethoxycarbonyl)cycloheptanone.

Correct: (A)

The ester undergoes two successive additions: first to the ketone (acetophenone), then to the tertiary alkoxide, giving PhC(OH)(CH3)2 after workup.

Correct: (B)

The aldehyde carbonyl is far more electrophilic than the ester, so one equivalent of Grignard adds chemoselectively to the aldehyde, giving the secondary alcohol while the ester survives.

Correct: (C)

NaBH4 reduces the ketone but is too mild to reduce the ester under these conditions, so only the keto group becomes a secondary alcohol.

Correct: (A)

The Wittig reagent transfers its alkylidene fragment (CHCH3) to the carbonyl carbon, giving the exocyclic alkene C6H10=C(H)CH3.

Correct: (D)

Under thermodynamic control the more substituted (trisubstituted) C=C bond is favored; pyrrolidine is a secondary amine, so an enamine (not an imine) forms.

Correct: (B)

Aldehydes are favored over ketones, and the electron-withdrawing nitro group makes the carbonyl carbon more electrophilic, shifting the addition equilibrium furthest to the right.

Correct: (C)

In the Baeyer–Villiger oxidation the benzyl group migrates in preference to the primary ethyl group, inserting oxygen between the carbonyl carbon and the benzyl carbon.

Correct: (A)

Pentan-2-one is a methyl ketone, so it undergoes the haloform reaction: the CH3CO– fragment becomes CHI3 and the remaining propyl fragment becomes butanoate.

Correct: (D)

LDA forms the kinetic enolate quantitatively, which adds to benzaldehyde; acid and heat dehydrate the aldol adduct to the conjugated benzylidene product.

Correct: (B)

The aldehyde is protected as an acetal so the Grignard adds only to the ketone; aqueous acid then deprotects the aldehyde and protonates the alkoxide.

Correct: (C)

The methyl-ketone enolate attacks the aldehyde intramolecularly to give a six-membered β-hydroxy ketone, which dehydrates to the conjugated cyclohexenone.

Correct: (B)

Loss of water gives a secondary carbocation; a methyl shift produces the more stable tertiary carbocation, which is captured by chloride.

Correct: (A)

SN2 rates decrease with steric hindrance at the α-carbon; neopentyl bromide, though primary, is the slowest because its tert-butyl group blocks backside attack.

Correct: (D)

Alcoholic KOH with heat promotes E2 elimination to the alkene, whereas aqueous conditions favor SN1 substitution to tert-butanol.

Correct: (C)

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

Correct: (A)

Tosylate is an excellent leaving group for SN2; cyanide displaces it with inversion of configuration, so the (R) substrate gives the (S) nitrile.

Correct: (B)

Polar aprotic solvents like DMSO solvate cations but leave the anionic nucleophile unsolvated and highly reactive, accelerating SN2.

Correct: (C)

The primary cyclobutylmethyl cation undergoes ring expansion (alkyl shift) to the more stable secondary cyclopentyl cation, which captures bromide.

Correct: (D)

E2 transition states have alkene character, so substrates leading to more substituted alkenes react faster: tertiary > secondary > primary, with (d) giving the most substituted alkene.

Correct: (A)

SN1 rates increase with solvent ionizing power; water, the most polar protic solvent listed, best stabilizes the carbocation-like transition state.

Correct: (C)

BrCH2–CH(OH)–CH(OH)–CH2Br with (2R,3S) configuration has a plane of symmetry through the midpoint of the C2–C3 bond, making it achiral despite two stereocenters; the other three lack any internal symmetry element.

Correct: (B)

Priorities are OH (1), ethyl (2), methyl (3), H (4); the 1→2→3 path (right → bottom → top) traces clockwise, but H sits on a horizontal (forward) bond in the Fischer projection, so the apparent R inverts to S.

Correct: (C)

The (1R,2R) isomer is trans-1,2-dimethylcyclohexane and (1R,2S) is the cis isomer; they differ in relative stereochemistry, are not mirror images, and have different physical properties.

Correct: (D)

c = 0.60 g/10.0 mL = 6 g per 100 mL, so [α]obs = (100 × −0.864)/(1.0 × 6) = −14.4°; ee = |−14.4|/|−24| × 100% = 60%.

Correct: (A)

(E)-4-methyl-2-hexene has a stereocenter at C4 and the E geometry minimizes steric strain versus the Z isomer; (B) and (C) are achiral, and (D) is impossibly strained.

Correct: (B)

The chiral amine converts the two enantiomers into diastereomeric ammonium salts with different solubilities, allowing separation by crystallization; acid workup returns the resolved enantiomer.

Correct: (D)

The E isomer places the bulky Br and CH3 groups trans, minimizing steric strain; CH3–CH=CHBr has no stereocenter and a plane of symmetry in both geometries, so neither is chiral.

Correct: (B)

Oxymercuration proceeds through a bridged mercurinium ion (not a free carbocation), so no rearrangement occurs, giving the unrearranged Markovnikov alcohol 3,3-dimethyl-2-butanol; hydroboration adds anti-Markovnikov with no rearrangement, giving 3,3-dimethyl-1-butanol.

Correct: (A)

(E)-2-butene undergoes syn epoxidation to give the trans-epoxide (racemic 2R,3R/2S,3S); acid-catalyzed opening occurs by anti (backside) attack, inverting one center to give meso-2,3-butanediol.

Correct: (D)

Hg2+-catalyzed hydration follows Markovnikov's rule via the more stable vinyl cation, giving acetone; hydroboration places boron on the terminal carbon, and the resulting enol tautomerizes to the anti-Markovnikov aldehyde, propanal.

Correct: (A)

The electron-donating methyl group on the diene directs cycloaddition so the substituents end up para to each other (ortho/para rule), giving 4-methyl-3-cyclohexene-1-carboxaldehyde as the major regioisomer.

Correct: (D)

OsO4 dihydroxylation is a syn addition; syn addition to the (Z)-alkene delivers both OH groups to the same face, producing the chiral racemic (3R,4R)/(3S,4S) pair rather than the meso diastereomer.

Correct: (C)

The Simmons–Smith reagent delivers a CH2 unit that adds syn to the exocyclic double bond, forming a cyclopropane ring sharing the former alkene carbon: spiro[2.5]octane.

Correct: (A)

NBS effects allylic bromination at one of the twelve equivalent methyl hydrogens; of the two resonance contributors of the allyl radical, Br capture at the primary carbon gives the more substituted (tetrasubstituted) alkene, which predominates.

Correct: (A)

Sn/HCl reduces the nitro group to aniline, diazotization forms the benzenediazonium salt, and CuCN effects Sandmeyer-type replacement of N2 by CN to give benzonitrile.

Correct: (D)

Sulfonation installs a meta-directing SO3H group, nitration occurs meta to it, and superheated steam removes the sulfonic acid group — a classic indirect route to a meta-substituted product.

Correct: (B)

Friedel–Crafts acylation with acetyl chloride/AlCl3 installs the –COCH3 group directly in one step; (A) gives toluene (alkylation), not the ketone.

Correct: (A)

Both substituents reinforce the same outcome: positions ortho to CH3 are activated by the methyl group and are meta to NO2 (allowed), while positions ortho to NO2 are strongly deactivated.

Correct: (D)

With no strongly electron-withdrawing ortho/para group, the aryl halide undergoes base-induced elimination to a benzyne intermediate, then nucleophilic addition of NH2− and protonation gives aniline.

Correct: (A)

Cyclohexanol's alkoxide lacks resonance stabilization, making it least acidic; among the phenols, the electron-donating methyl group destabilizes the phenoxide while the electron-withdrawing nitro group stabilizes it.

Correct: (B)

The 2H doublet at δ 4.52 is an O–CH₂ group coupled to a single vinylic proton, the 3H doublet at δ 1.70 is a methyl coupled to a vinylic proton, and the 3H singlet at δ 2.04 is an acetate methyl; only CH₃CH=CHCH₂OCOCH₃ fits all these features.

Correct: (A)

Para symmetry makes the two tert-butyl groups equivalent and the two pairs of ring CH carbons equivalent, giving just four distinct carbons: ipso ring C, ring CH, quaternary (CH₃)₃C, and methyl.

Correct: (A)

The McLafferty rearrangement requires transfer of a γ-hydrogen to the carbonyl oxygen through a six-membered transition state; here the γ-carbon is quaternary, so only α-cleavage fragments (CH₃CO⁺ at m/z 43 and (CH₃)₃C⁺ at m/z 57) are seen.

Correct: (D)

The absence of a broad O–H stretch rules out the carboxylic acids, and the 1645 cm⁻¹ band requires a C=C bond, ruling out the diketone; only methyl methacrylate contains both an ester carbonyl and an alkene.

Correct: (C)

The triplet aldehyde proton is coupled to a neighboring CH₂, and the 3H triplet methyl with no doublet methyl signals indicates a straight chain; branched isomers would show doublet methyls and a differently split aldehyde proton, and the ketone would lack the δ 9.62 signal.

Correct: (D)

The tert-butyl group experiences severe 1,3-diaxial interactions when axial, giving it an A-value of roughly 5 kcal/mol — far larger than methyl, ethyl, or isopropyl — so it effectively locks the ring with itself equatorial.

Correct: (B)

In a cis-1,3-disubstituted cyclohexane one substituent must be axial and the other equatorial; the larger isopropyl group occupies the equatorial position to minimize 1,3-diaxial strain.

Correct: (A)

The central allenic carbon is sp hybridized with two mutually perpendicular p orbitals, so the two π systems (and the planes of the terminal groups) are orthogonal.

Correct: (C)

Pyrrole's nitrogen lone pair occupies a p orbital and counts toward the Hückel 6π system; N-protonation ties up that pair and breaks aromaticity, whereas pyridine's lone pair sits in an in-plane sp² orbital outside the sextet, making pyridine the stronger base.

Correct: (B)

The p-nitro group stabilizes the phenoxide by resonance electron withdrawal, while electron-donating groups destabilize it; p-methoxy donates more strongly by resonance than p-methyl does by hyperconjugation, making it the least acidic.

Correct: (D)

The p-methoxy group donates electron density into the ring by resonance, enriching the nitrogen and increasing basicity, whereas the p-nitro group strongly withdraws electron density and further delocalizes the nitrogen lone pair, sharply decreasing basicity.

Correct: (A)

Bulky LDA at –78 °C deprotonates the less hindered α-carbon irreversibly (kinetic control); the more substituted, lower-energy enolate is the thermodynamic product and forms only under equilibrating conditions.

Correct: (B)

Resonance delocalization stabilizes the allyl radical most, followed by hyperconjugation in the tert-butyl radical, then the primary ethyl radical; the vinyl radical is least stable because the unpaired electron resides in an sp² orbital with high s-character.

Correct: (A)

Abstraction of the C-4 allylic hydrogen gives an allylic radical delocalized over C-2 and C-4; bromine capture at the more substituted terminus (C-2) affords the more substituted, trisubstituted alkene as the major product.

Correct: (D)

The bromine radical adds to the less substituted carbon so that the resulting carbon radical is secondary rather than primary; this radical then abstracts H from HBr to give the anti-Markovnikov product and regenerate Br·.

Correct: (C)

A diol plus a diacid undergo esterification with loss of water at each linkage, a step-growth condensation process that yields a polyester (poly(hexamethylene adipate)).

Correct: (B)

Styrene polymerizes by radical chain growth with no small-molecule byproduct, whereas the diacid plus diamine combine with loss of water in a step-growth condensation to give nylon-6,6.