ACS Organic Chemistry Practice Test 5

Correct: (B)

LDA forms the kinetic enolate of acetone, which undergoes 1,2-addition to the aldehyde carbonyl of benzaldehyde, giving the crossed aldol addition product.

Correct: (D)

Intramolecular aldol condensation of the 1,4-diketone closes a five-membered ring, and dehydration gives the conjugated cyclopentenone.

Correct: (A)

This is the Claisen condensation: the enolate of one ester attacks the carbonyl of another, giving ethyl acetoacetate.

Correct: (C)

Intramolecular Claisen (Dieckmann) condensation closes a five-membered ring, giving the cyclic beta-keto ester.

Correct: (B)

Grignard addition to an aldehyde gives a secondary alkoxide; a second equivalent cannot add further, so protonation yields 1-phenylethanol.

Correct: (A)

NaBH4 reduces the more reactive aldehyde selectively, leaving the ketone untouched under these mild conditions.

Correct: (D)

The Wittig reaction converts the ketone carbonyl into an exocyclic C=C bond, giving methylenecyclohexane.

Correct: (C)

A ketone plus a secondary amine under dehydrating conditions forms an enamine, not an imine.

Correct: (B)

The ketone is protected as a cyclic ketal (ethylene ketal) under acid-catalyzed dehydrating conditions.

Correct: (D)

Gilman reagents give 1,4-conjugate addition to enones, placing the methyl group at the beta carbon.

Correct: (A)

LiAlH4 reduces nitriles fully to primary amines, adding two hydrogens to the nitrile carbon.

Correct: (C)

The more reactive aldehyde must be masked as an acetal so the Grignard reagent attacks only the ketone; aqueous acid later restores the aldehyde.

Correct: (C)

Grignard reagents add twice to esters: the first addition gives a ketone intermediate, which reacts with the second equivalent to give a tertiary alcohol.

Correct: (B)

Alpha-methylation of acetoacetic ester, followed by ester hydrolysis and thermal decarboxylation, gives CH3COCH2CH3.

Correct: (C)

A first-order rate law independent of nucleophile concentration indicates unimolecular ionization to a carbocation (SN1).

Correct: (A)

The rate-determining step of an SN1 reaction is ionization of the substrate, which does not involve the nucleophile.

Correct: (D)

A bulky base abstracts the least hindered beta-hydrogen, giving the less substituted Hofmann alkene.

Correct: (B)

SN2 proceeds with backside attack, so the configuration at carbon inverts from R to S.

Correct: (A)

Ionization gives a planar, achiral carbocation that is attacked from either face, producing a racemic mixture.

Correct: (C)

The anti-periplanar arrangement allows the developing p orbitals to overlap as the pi bond forms.

Correct: (D)

Iodide is the best leaving group (weakest C–X bond, most stable anion), so ionization is fastest.

Correct: (B)

SN2 rates decrease with steric hindrance at the electrophilic carbon; methyl bromide is the least hindered.

Correct: (A)

Protonation of the alcohol and loss of water give a stable tert-butyl cation, which then loses a beta-proton (E1).

Correct: (D)

With H in the back, priorities are OH (1) > ethyl (2) > methyl (3); the 1→2→3 sequence (up→right→left) is clockwise, giving R.

Correct: (C)

They are stereoisomers that are not mirror images (one is chiral, the other is the meso form), so they are diastereomers.

Correct: (A)

The (2R,3S) stereoisomer of 2,3-dichlorobutane has an internal plane of symmetry and is achiral (meso).

Correct: (B)

The anti staggered conformation minimizes steric repulsion between the two methyl groups.

Correct: (D)

The enantiomeric excess is 20/80 = 25%, so the major enantiomer is (100 + 25)/2 = 62.5% of the mixture.

Correct: (C)

The meso diol has an internal plane of symmetry, making it achiral and optically inactive despite having stereocenters.

Correct: (B)

The meso (2R,3S) isomer with anti methyl groups places the bromines anti as well, giving a centrosymmetric, achiral conformer.

Correct: (A)

Alkene stability increases with substitution; the tetrasubstituted alkene is the most stable.

Correct: (C)

Ozonolysis cleaves the C=C bond: the (CH3)2C= carbon becomes acetone and the CHCH3 carbon becomes acetaldehyde.

Correct: (D)

Working backwards, the =CH2 fragment (formaldehyde) and the cyclohexanone fragment join at the carbonyl carbons to give the exocyclic alkene.

Correct: (B)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry; the OH ends up trans to the C1 methyl group.

Correct: (C)

Peroxides cause radical addition with anti-Markovnikov regiochemistry: Br adds to the terminal carbon and no carbocation rearrangement occurs.

Correct: (D)

Dissolving-metal reduction of an alkyne gives the trans alkene via anti addition of the two hydrogens.

Correct: (C)

The Diels–Alder reaction under kinetic control favors the endo adduct due to secondary orbital interactions (the endo rule).

Correct: (B)

The tert-butyl group is an ortho/para activator; the bulky group strongly favors para substitution for steric reasons.

Correct: (D)

The nitro group is strongly deactivating, so nitrobenzene is by far the least reactive toward electrophilic substitution.

Correct: (A)

Oxidizing the methyl group first gives benzoic acid, whose –CO2H group is meta-directing, so nitration then gives the meta isomer.

Correct: (C)

SNAr is accelerated by electron-withdrawing groups ortho/para to the leaving group, which stabilize the Meisenheimer complex.

Correct: (B)

The para nitro group stabilizes the carboxylate conjugate base by electron withdrawal, lowering the pKa.

Correct: (D)

Bromine is deactivating but ortho/para directing; para substitution dominates over ortho for steric reasons.

Correct: (A)

The 9H singlet indicates a tert-butyl group and the 1H singlet near 9.4 ppm is an aldehyde proton with no neighbors, matching pivaldehyde.

Correct: (C)

The triplet–quartet pattern is characteristic of an ethyl group, and the downfield quartet (4.1 ppm) shows the CH2 is attached to an electronegative atom.

Correct: (B)

Symmetry makes the two ipso (C–CH3) carbons equivalent, all four ring CH carbons equivalent, and the two methyl carbons equivalent: 3 distinct carbon environments.

Correct: (D)

Ethylbenzene (M = 106) undergoes benzylic cleavage to give the stable tropylium ion C7H7+ at m/z 91.

Correct: (A)

The carbonyl IR band without O–H indicates an ester, and the two uncoupled methyl singlets (acetyl CH3 and OCH3) match methyl acetate.

Correct: (C)

The very bulky tert-butyl group strongly prefers the equatorial position to avoid 1,3-diaxial interactions.

Correct: (B)

Resonance donation of the nitrogen lone pair into the carbonyl gives the C–N bond partial double-bond character, making nitrogen planar (sp2).

Correct: (D)

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

Correct: (A)

The central carbon forms two pi bonds (one to each terminal carbon) and no sigma bonds to hydrogen, requiring two sp hybrid orbitals.

Correct: (B)

The three fluorines strongly stabilize the carboxylate anion by induction, making trifluoroacetic acid far more acidic than acetic acid.

Correct: (D)

Aliphatic amines are much more basic than aromatic amines (lone pair delocalized into the ring) or amides (lone pair delocalized into the carbonyl).

Correct: (C)

The amide ion is the conjugate base of the extremely weak acid NH3 (pKa ≈ 38), making it a far stronger base than acetate, water, or chloride.

Correct: (A)

The triphenylmethyl radical is stabilized by resonance delocalization over three phenyl rings, far exceeding alkyl hyperconjugation effects.

Correct: (C)

The bromine radical selectively abstracts the weakest (tertiary) C–H bond, giving the tert-butyl bromide as the major monobromination product.

Correct: (D)

Propagation is the repeated addition of the growing carbon-centered radical to the pi bond of a new styrene monomer.

Correct: (B)

The polymer backbone is built by addition across the double bond of vinyl chloride, preserving the –CH2–CHCl– repeat unit.

Correct: (A)

Kevlar is the condensation polymer of a para-disubstituted diacid chloride and a para-disubstituted diamine, giving rigid para-linked amide chains.

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