ACS Organic Chemistry Practice Test 4

Correct: (B)

The initial aldol addition gives 3-hydroxybutanal, but heating causes dehydration of the β-hydroxy aldehyde to the conjugated α,β-unsaturated aldehyde (crotonaldehyde).

Correct: (A)

This is the Claisen condensation: enolate formation at one ester followed by nucleophilic acyl substitution at a second ester gives the β-keto ester ethyl acetoacetate.

Correct: (C)

Intramolecular Claisen (Dieckmann) condensation of this six-carbon diester closes a five-membered ring, giving the cyclic β-keto ester.

Correct: (D)

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

Correct: (B)

NaBH4 performs 1,2-reduction of the aldehyde carbonyl chemoselectively, leaving the conjugated C=C double bond untouched.

Correct: (A)

LiAlH4 reduces carboxylic acids all the way to primary alcohols; NaBH4 would be too weak, and the aldehyde is not isolable under these conditions.

Correct: (C)

The Wittig reaction replaces the carbonyl oxygen with the =CH2 group of the unstabilized ylide, giving the terminal alkene styrene.

Correct: (A)

Stabilized ylides give E-alkenes predominantly because the reaction proceeds through the more stable trans oxaphosphetane-like transition state.

Correct: (D)

Secondary amines add to ketones and, with loss of water, give enamines; imines form only from primary amines.

Correct: (B)

Hydrate formation is favored by strong electron-withdrawing groups and the absence of electron-donating alkyl groups; chloral has the largest hydration equilibrium constant of the group.

Correct: (C)

NaBH4 is a mild hydride donor that reduces aldehydes and ketones but is too weak to reduce esters, making it useful for chemoselective reductions.

Correct: (A)

The Grignard reagent adds once to the ketone carbonyl, and protonation of the resulting alkoxide gives the tertiary alcohol 2-phenyl-2-butanol.

Correct: (B)

The stabilized enolate undergoes C-alkylation preferentially under these conditions, and with one equivalent of base and alkyl halide the mono-alkylated product dominates.

Correct: (D)

Diacetone alcohol is the self-aldol addition product of acetone; its symmetry (two identical methyl groups on the hydroxyl-bearing carbon) reveals acetone as the only carbonyl component.

Correct: (A)

SN2 reactivity decreases with increasing steric hindrance at the α-carbon: methyl > primary > secondary > tertiary.

Correct: (C)

The SN1 rate is determined by carbocation stability, and the tertiary chloride forms the most stable tertiary carbocation in the rate-determining step.

Correct: (A)

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

Correct: (B)

Under E1 conditions the reaction proceeds through the most stable tertiary carbocation, and loss of a proton gives the most substituted (Saytzeff) alkene.

Correct: (C)

First-order kinetics plus racemization indicate a planar carbocation intermediate formed in the rate-determining ionization step, the hallmark of SN1.

Correct: (B)

Cyanide substitution in the polar aprotic solvent DMSO proceeds by SN2, which inverts the configuration at the stereocenter in a single backside attack.

Correct: (D)

E2 elimination requires at least one β-hydrogen to remove along with the leaving group, and methyl bromide has no β-carbon at all.

Correct: (A)

In protic solvents, smaller anions are more strongly solvated by hydrogen bonding, so the largest, least solvated iodide is the strongest nucleophile.

Correct: (C)

Ionization gives a secondary carbocation that undergoes a 1,2-methyl shift to the more stable tertiary carbocation before bromide capture, giving the rearranged bromide.

Correct: (B)

Assigning priorities (OH > ethyl > methyl > H) with H on a horizontal (forward) bond gives a clockwise 1→2→3 sequence that must be reversed, yielding S.

Correct: (C)

The (2R,3S) isomer is the meso form with an internal plane of symmetry, so it is a diastereomer—not the mirror image—of the chiral (2R,3R) isomer.

Correct: (A)

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

Correct: (B)

The meso diastereomer contains an internal plane of symmetry, so its two halves cancel and it does not rotate plane-polarized light.

Correct: (D)

The ee is 6.75/13.5 × 100% = 50%, and the negative sign of the observed rotation matches the pure (S) enantiomer, so (S) is in excess.

Correct: (C)

The two stereocenters are in different environments (no internal symmetry possible), giving 2² = 4 stereoisomers as two pairs of enantiomers.

Correct: (A)

trans-1,2-Dimethylcyclopropane lacks any plane or center of symmetry and exists as a pair of enantiomers; the cis isomer is meso and the others have symmetry planes through C1.

Correct: (A)

Alkene stability increases with substitution (hyperconjugation): monosubstituted < disubstituted < trisubstituted < tetrasubstituted, and trans isomers are more stable than cis.

Correct: (C)

Rejoining the two carbonyl fragments at their C=O carbons gives (CH3)2C=CHCH3, i.e., 2-methyl-2-butene; the other alkenes would give formaldehyde or a second equivalent of propanone.

Correct: (B)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry, placing OH on the less substituted terminal carbon to give 3-methyl-1-butanol.

Correct: (D)

The peroxide effect reverses the regiochemistry of HBr addition via a radical chain mechanism, so Br adds to the less substituted carbon, giving anti-Markovnikov 1-bromobutane.

Correct: (A)

Water opens the cyclic bromonium ion at the more substituted carbon with anti stereochemistry, giving the trans bromohydrin rather than the dibromide.

Correct: (B)

The para rule places the diene methyl group para to the anhydride carbonyls, giving the 4-methyl adduct; the dienophile's cis geometry is retained in the adduct.

Correct: (C)

Dissolving-metal reduction of alkynes proceeds by sequential electron/proton transfers that give the more stable trans (E) alkene, unlike Lindlar reduction which gives the cis isomer.

Correct: (A)

The nitro group is a deactivating meta director, so further nitration occurs meta to it; the other substituents are all ortho/para directors.

Correct: (D)

Electron-donating groups accelerate EAS and withdrawing groups retard it, so the strongly deactivated CF3 compound is slowest and the strongly activated anisole is fastest.

Correct: (B)

A nitro group ortho or para to the leaving group stabilizes the Meisenheimer intermediate by resonance; the meta isomer cannot delocalize the negative charge onto the nitro group.

Correct: (C)

Nitration first installs the meta-directing nitro group, so subsequent bromination occurs meta to it; bromination first would give ortho/para bromonitrobenzene instead.

Correct: (A)

The activating methyl group outcompetes the deactivating nitro group as director, so bromination occurs ortho to the methyl group at the less hindered C-2 position (meta to NO2).

Correct: (D)

Sulfonation gives benzenesulfonic acid, and alkali fusion of the sulfonate displaces the sulfo group with hydroxide to give phenol after acidic workup.

Correct: (B)

The 2.1 ppm singlet is a methyl ketone (CH3C=O) and the quartet/triplet pair is an ethyl group deshielded by the carbonyl, matching CH3COCH2CH3; butanal would show an aldehyde proton near 9–10 ppm.

Correct: (C)

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

Correct: (A)

Alpha cleavage next to the carbonyl expels an ethyl radical to give the resonance-stabilized acylium ion CH3C≡O+ at m/z 43.

Correct: (D)

The 1715 cm−1 band is a ketone C=O stretch, and a single 6H singlet requires two equivalent methyl groups adjacent to the carbonyl, which is acetone.

Correct: (B)

Five aromatic protons indicate a monosubstituted ring, and the quartet/triplet pair is an ethyl group attached to the ring; the xylenes would show only four aromatic protons.

Correct: (C)

The cyclopentadienyl anion is cyclic, planar, conjugated, and has 6 π electrons (Hückel 4n+2); cyclobutadiene and the Cp cation have 4 π electrons and are antiaromatic.

Correct: (A)

In cis-1,3-disubstituted cyclohexane both substituents point to the same face, so one chair places both equatorial, avoiding the 1,3-diaxial strain of the flipped chair.

Correct: (D)

The pyridine nitrogen is part of the planar aromatic ring with its lone pair in an sp2 orbital in the ring plane, contributing one p electron to the 6 π system.

Correct: (B)

The tropylium cation has 6 π electrons in a planar conjugated ring (4n+2, n=1); the CHT anion and Cp cation have 8 and 4 π electrons respectively and are antiaromatic if planar.

Correct: (C)

Ethanol (pKa ~16) is least acidic, and each electron-withdrawing chlorine inductively stabilizes the carboxylate conjugate base, making trichloroacetic acid the strongest.

Correct: (A)

Aliphatic amines are more basic than aromatic amines because the nitrogen lone pair in aniline and p-nitroaniline is delocalized into the ring, and pyrrole's lone pair is part of the aromatic sextet.

Correct: (D)

Base strength increases as the conjugate acid gets weaker; NH3 (pKa ~36) is a far weaker acid than H2O (~15.7), acetic acid (~4.8), or HCl (~−7).

Correct: (B)

Stability rises with hyperconjugation (methyl < tert-butyl) and resonance delocalization, with the benzyl radical more stabilized than allyl due to delocalization over the aromatic ring.

Correct: (C)

Abstraction of the single tertiary hydrogen is fastest because the tertiary radical intermediate is most stable, giving 2-chloro-2-methylbutane as the major product.

Correct: (A)

The phenyl radical adds to the terminal CH2 of styrene to generate the more stable benzylic radical, which then propagates the chain by adding to further styrene molecules.

Correct: (D)

Removing the single bond between repeat units reveals the alkene CH2=CH–CN, which is acrylonitrile.

Correct: (B)

Cleaving each ester bond gives a diol fragment (HO–CH2CH2–OH, ethylene glycol) and a diacid fragment (HOOC–C6H4–COOH, terephthalic acid).