ACS Organic Chemistry Practice Test 9

Correct: (C)

Benzaldehyde has no alpha hydrogens, so only acetophenone forms the enolate, which adds to benzaldehyde; under basic, heated conditions the aldol product dehydrates to the conjugated enone (chalcone).

Correct: (B)

Intramolecular aldol condensation favors the 5-membered ring: one enolate attacks the other carbonyl, then dehydration gives the conjugated cyclopentenone.

Correct: (A)

Dieckmann condensation of a 1,6-diester closes to the 5-membered beta-keto ester; the 6-membered closure would require attack across the wrong distance.

Correct: (D)

Classic Claisen condensation: enolate of ethyl acetate attacks a second ester molecule, expelling ethoxide to give the beta-keto ester.

Correct: (B)

Aldehydes are more electrophilic than ketones, so one equivalent of Grignard adds chemoselectively to the aldehyde, giving the secondary alcohol; the acetyl group survives.

Correct: (A)

NaBH4 is a mild hydride donor that reduces aldehydes and ketones but not esters (LiAlH4 would be needed for the ester).

Correct: (C)

The Wittig reagent converts the carbonyl carbon directly into an exocyclic C=C bond, giving methylenecyclohexane with triphenylphosphine oxide as byproduct.

Correct: (D)

Secondary amines condense with ketones under acid with water removal to give the enamine; the nitrogen adds and eliminates water, not reduced to the amine.

Correct: (B)

Aldehydes plus diols under acid give cyclic acetals; acetals are protecting groups precisely because they resist base and nucleophiles but hydrolyze in aqueous acid.

Correct: (C)

LiAlH4 reduces nitriles fully to primary amines; partial reduction to the aldehyde requires DIBAL-H at low temperature instead.

Correct: (A)

Reactivity follows leaving-group ability and resonance donation to the carbonyl: chloride leaves best and donates least, while the amide nitrogen donates most, making it least reactive.

Correct: (D)

One equivalent of DIBAL-H at low temperature reduces esters only to the aldehyde stage; the tetrahedral intermediate is stable until workup, unlike LiAlH4 which gives the primary alcohol.

Correct: (B)

Luche reduction: CeCl3 promotes 1,2-addition of hydride to the carbonyl, giving the allylic alcohol while leaving the C=C bond intact.

Correct: (C)

Cannizzaro reaction: aldehydes lacking alpha hydrogens undergo base-induced disproportionation into the alcohol and the carboxylate.

Correct: (A)

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

Correct: (D)

Polar aprotic solvents like DMSO solvate cations but leave the nucleophilic anion unsolvated and highly reactive; protic solvents hydrogen-bond to CN- and slow SN2.

Correct: (B)

SN2 proceeds with backside attack and complete inversion of configuration at the electrophilic carbon, so (R) substrate gives the (S) nitrile.

Correct: (C)

The bulky base abstracts the least hindered proton from a C1 methyl group, giving the Hofmann (less substituted) alkene 2-methyl-1-butene, CH2=C(CH3)CH2CH3; the Zaitsev product dominates only with small bases.

Correct: (D)

Loss of water gives a secondary carbocation, which undergoes a 1,2-methyl shift to the more stable tertiary carbocation; deprotonation then gives the most substituted (Zaitsev) alkene.

Correct: (C)

Tosylate is the best leaving group (very stable conjugate acid), and among halides leaving-group ability increases down the group as the anion becomes larger and more stable: I- > Br- > Cl- >> F-.

Correct: (B)

E2 with a small base gives the Zaitsev product; the more substituted alkene is stabilized by hyperconjugation, and the trans isomer dominates for steric reasons.

Correct: (D)

SN2 at the neopentyl carbon is extremely hindered, so ionization occurs with a 1,2-methyl shift to give the tertiary carbocation, which is trapped by water to give 2-methyl-2-butanol.

Correct: (A)

Tertiary substrate with a weak nucleophile/base in a polar protic solvent ionizes to a carbocation, which partitions between substitution (SN1) and elimination (E1).

Correct: (A)

Assigning priorities OH(1) > CH2CH3(2) > CH3(3) > H(4): with H on the hashed wedge (pointing away), the 1→2→3 sequence (OH → ethyl → methyl) traces clockwise, giving R.

Correct: (D)

The two structures differ in configuration at every stereocenter and are non-superimposable mirror images, the definition of enantiomers.

Correct: (A)

(2R,3S)-butane-2,3-diol has stereocenters but an internal plane of symmetry, making it achiral (meso) and optically inactive.

Correct: (C)

A 180-degree dihedral angle between the two methyl groups is the staggered anti conformation; 60 degrees is the staggered gauche conformation.

Correct: (B)

ee = (20/50) x 100% = 40%; the major enantiomer is (50% + 20%) = 70% R and the minor is 30% S.

Correct: (C)

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

Correct: (D)

Stability increases with alkyl substitution via hyperconjugation: trisubstituted > disubstituted, and trans disubstituted > cis (less steric strain) > monosubstituted.

Correct: (A)

The double bond cleaves so that (CH3)2C= becomes acetone and =CHCH2CH3 becomes propanal.

Correct: (B)

Hydroboration adds H and OH syn and anti-Markovnikov (B to the less substituted carbon), placing the new OH at C2; the syn addition to the ring gives the trans relationship between the C1 methyl and the C2 OH.

Correct: (C)

The cycloaddition is stereospecific, so the cis relationship of the two carbonyl groups in maleic anhydride is retained in the cyclohexene adduct.

Correct: (D)

The peroxide effect reverses the regiochemistry of HBr addition, so Br adds via a radical to the less substituted (terminal) carbon, giving the anti-Markovnikov bromide.

Correct: (A)

Water acts as the nucleophile and attacks the more substituted carbon of the bromonium ion, so Br ends up on the less substituted carbon and OH on the more substituted one.

Correct: (B)

Dissolving-metal reduction of an alkyne gives the trans (E) alkene, whereas Lindlar's catalyst gives the cis (Z) alkene.

Correct: (C)

Alkene stability increases with the number of alkyl substituents on the double bond (hyperconjugation), so the tetrasubstituted alkene is most stable and ethylene least.

Correct: (A)

The acetyl group is a deactivating meta director, so electrophilic substitution occurs meta to it.

Correct: (B)

Electron-donating groups activate the ring (anisole > toluene > benzene) while the strongly deactivating nitro group makes nitrobenzene the slowest.

Correct: (C)

The ring is activated toward nucleophilic aromatic substitution by two ortho/para nitro groups, so hydroxide displaces fluoride via the addition–elimination (SNAr) mechanism.

Correct: (D)

Strongly deactivated rings such as nitrobenzene do not undergo Friedel–Crafts alkylation or acylation because the ring is too electron-poor to attack the electrophile.

Correct: (A)

The methyl group is an ortho/para director, and the para product dominates because the bulky sulfonating electrophile is hindered at the ortho positions.

Correct: (B)

The strongly activating OH group makes phenol so reactive that all three available ortho and para positions are brominated without a Lewis acid catalyst.

Correct: (A)

The triplet/quartet pattern is a classic ethyl group attached to an aromatic ring (deshielded benzylic quartet at 2.6 ppm), and the 5H aromatic multiplet indicates a monosubstituted benzene.

Correct: (D)

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

Correct: (D)

The molecular weight is 72, and alpha-cleavage next to the carbonyl gives the resonance-stabilized acylium ion CH3CO+ at m/z 43, the most abundant fragment.

Correct: (B)

The IR band indicates a carbonyl, the 9.8 ppm triplet is a characteristic aldehyde proton split by the adjacent CH2, and the remaining signals fit a CH3CH2– group.

Correct: (A)

The 6H doublet plus 1H septet is the signature splitting pattern of an isopropyl group, and the 5H aromatic signal shows it is attached to a monosubstituted benzene ring.

Correct: (B)

An axial methyl group suffers 1,3-diaxial steric interactions with the axial hydrogens on C3 and C5, so the equatorial conformer is more stable by about 1.7 kcal/mol.

Correct: (C)

Resonance delocalization of the nitrogen lone pair into the carbonyl pi system gives the C–N bond partial double-bond character, making nitrogen sp2 hybridized and planar.

Correct: (A)

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

Correct: (D)

Planar cyclobutadiene is cyclic, fully conjugated, and has 4 pi electrons (4n with n = 1), making it antiaromatic and highly unstable.

Correct: (B)

The electron-withdrawing nitro group stabilizes the carboxylate conjugate base through resonance and induction, lowering the pKa relative to the other substituted benzoic acids.

Correct: (C)

In water, solvation effects make the basicity order secondary > primary > tertiary amine, and aniline is far weakest because its lone pair is delocalized into the ring.

Correct: (A)

Basicity increases as the conjugate acid gets weaker; NH3 (pKa ≈ 35) is a far weaker acid than water, acetic acid, or carbonic acid, so NH2− is the strongest base listed.

Correct: (B)

Radical stability follows the extent of delocalization: the triphenylmethyl radical is stabilized by three phenyl rings, followed by benzyl (one phenyl), allyl (one pi bond), and the hyperconjugation-only tert-butyl radical.

Correct: (C)

Under radical conditions, bromination occurs selectively at the weak benzylic C–H bond via a resonance-stabilized benzyl radical, not on the aromatic ring.

Correct: (D)

Propagation is the repeated addition of a growing carbon radical to another monomer alkene, extending the chain while regenerating a new radical at the chain end.

Correct: (A)

Polypropylene is an addition polymer formed by the polymerization of propene monomers across their carbon–carbon double bonds.

Correct: (B)

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

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