ACS Organic Chemistry Practice Test 20

Correct: (C)

Self-aldol addition gives 3-hydroxybutanal, which dehydrates on heating to the conjugated enal but-2-enal (crotonaldehyde).

Correct: (A)

Cyclopentanone has enolizable positions on both sides of the carbonyl, and with excess benzaldehyde (which cannot self-condense) the double crossed-aldol dehydration product predominates.

Correct: (D)

The ketone enolate attacks the carbonate carbonyl in a Claisen-type condensation, displacing ethoxide to give the beta-keto ester ethyl benzoylacetate.

Correct: (B)

Grignard addition to formaldehyde gives a primary alkoxide, which protonates to benzyl alcohol.

Correct: (A)

Aldehydes are more electrophilic than ketones, so the single equivalent of Grignard reagent adds chemoselectively to the aldehyde, giving 6-hydroxyheptan-2-one.

Correct: (D)

LiAlH4 is strong enough to reduce carboxylic acids all the way to primary alcohols.

Correct: (C)

NaBH4 does not reduce esters; reduction of an ester to alcohols requires LiAlH4.

Correct: (B)

Stabilized Wittig ylides give predominantly the E-alkene, so ethyl (E)-3-phenylprop-2-enoate is the major product.

Correct: (A)

This is the Stork enamine alkylation: the enamine acts as a controlled enolate equivalent, is methylated once, and acidic hydrolysis regenerates the ketone with the methyl group alpha to the carbonyl.

Correct: (D)

Hydrate stability increases with electrophilicity and decreases with steric hindrance and electron-donating alkyl groups, so formaldehyde hydrates most readily.

Correct: (C)

Ammonia attacks the highly reactive acid chloride in a nucleophilic acyl substitution, displacing chloride to give the primary amide benzamide.

Correct: (B)

Grignard reagents add to nitriles to give imine salts, which hydrolyze in aqueous acid to ketones; here the product is acetophenone.

Correct: (D)

Step 1 gives the secondary alcohol 1-phenylpropan-1-ol by Grignard addition to the aldehyde, and PCC oxidizes the secondary alcohol to the ketone propiophenone.

Correct: (A)

The more reactive aldehyde is selectively masked as the acetal protecting group, so the Grignard reagent adds only to the ketone; aqueous acid then deprotects the aldehyde.

### SN1/SN2/E1/E2, Kinetics (Q15-Q23)

Correct: (C)

CN- is a strong nucleophile/weak base in polar aprotic DMSO, giving SN2 with inversion at the stereocenter.

Correct: (A)

With a non-bulky strong base, E2 favors removal of the more substituted beta-hydrogen, giving the Zaitsev (trisubstituted) alkene.

Correct: (D)

The bulky base abstracts the least hindered beta-hydrogen in E2, giving the Hofmann product 1-pentene as major.

Correct: (B)

Ionization gives a primary carbocation that undergoes methyl shift to the tert-pentyl cation, which is captured by water.

Correct: (A)

SN2 inverts configuration at the attacked carbon (C1) only, while C2 retains its (R) configuration.

Correct: (C)

I- is a better leaving group than Br-, lowering the barrier of the rate-determining ionization; nucleophile concentration does not affect the SN1 rate.

Correct: (B)

Rate = k[CH3I][OH-]; (1/2)(2) = 1, so the rate is unchanged.

Correct: (D)

Carbocation formation is the slow step of SN1 and involves only the substrate, giving a first-order rate law.

Correct: (A)

E2 rates increase with substrate substitution (tertiary > secondary > primary) due to the more stable, more substituted alkene-like transition state.

### Stereochemistry (Q24-Q30)

Correct: (D)

Priorities are Br (1) > F (2) > CH3 (3) > H (4); with H already behind, the 1-2-3 path (right, up, left) runs counterclockwise, giving S.

Correct: (B)

The two structures have the same connectivity but differ in configuration at only one of two stereocenters, so they are diastereomers.

Correct: (A)

There are two stereocenters and the ends of the chain differ, so no meso form is possible; 2^2 = 4 stereoisomers (two enantiomeric pairs).

Correct: (C)

The global minimum of n-pentane is the all-anti zigzag conformation, placing the terminal carbons 180° apart to minimize steric strain.

Correct: (D)

Rotation about the C-C single bond interconverts staggered rotamers of one achiral compound; no stereoisomer relationship applies.

Correct: (A)

ee = 80% – 20% = 60%, and observed rotation = 0.60 x (+13.5°) = +8.1°.

Correct: (B)

The meso diol is achiral (internal plane of symmetry), so it cannot rotate plane-polarized light; the others are single chiral enantiomers.

### Alkenes/Alkynes (Q31-Q37)

Correct: (B)

Tetrasubstituted > trisubstituted > disubstituted > monosubstituted; stability increases with the number of alkyl groups on the C=C bond.

Correct: (A)

Cleaving CH3-CH=C(CH3)-CH2CH3 gives CH3CHO from the left fragment and CH3CH2COCH3 from the right fragment.

Correct: (D)

A single ozonolysis product requires a symmetric alkene; only 3,4-dimethyl-3-hexene cleaves into two identical molecules of CH3CH2COCH3.

Correct: (C)

A cyclic bromonium ion forms and water attacks the more substituted carbon, placing Br on the less substituted carbon and OH on the more substituted carbon.

Correct: (D)

Syn addition places B on the less substituted carbon and H on C1 on the same face; oxidation with retention gives the OH trans to the C1 methyl group.

Correct: (B)

The ortho/para rule favors the "para" adduct, with the diene methyl group at C4 (para to the anhydride), and the anhydride adds endo to the diene pi system.

Correct: (D)

Dissolving-metal reduction adds two hydrogens anti across the triple bond, giving the trans (E) alkene.

### Aromatic Chemistry (Q38-Q43)

Correct: (B)

Bromine is an ortho/para director; the para isomer predominates over ortho for steric reasons.

Correct: (D)

The -OCH3 group is a strong activating ortho/para director, and the para product dominates because the ortho positions are sterically crowded.

Correct: (A)

The sequence is nitration, reduction to aniline, diazotization, then Sandmeyer-type replacement of N2+ by CN-, giving benzonitrile.

Correct: (C)

Resonance donation from -OCH3 makes anisole the most activated, while the electron-withdrawing -COOH strongly deactivates benzoic acid.

Correct: (B)

A nitro group meta to the leaving group cannot delocalize the negative charge of the Meisenheimer complex, so SNAr does not occur at a practical rate.

Correct: (D)

p-NO2 stabilizes the phenoxide by resonance and induction (pKa ~7.1), m-NO2 by induction only (~8.3), phenol is 10.0, and electron-donating p-CH3 slightly decreases acidity (~10.2).

### Spectroscopy (Q44-Q48)

Correct: (A)

Only two singlets in a 3:9 ratio fit the threefold symmetry of mesitylene; all other isomers would show more signals and splitting.

Correct: (C)

The two equivalent methyl groups couple to the single neighboring methine proton, giving n + 1 = 2 lines.

Correct: (B)

The aldehydic C-H stretch pair plus the 9.9 ppm singlet prove a -CHO group; two 2H doublets indicate a para-disubstituted ring, and the 3H singlet is an aryl methyl.

Correct: (C)

Symmetry gives four distinct ring carbons (ipso, ortho, meta, para) plus the quaternary carbon and the three equivalent methyl carbons: 6 signals total.

Correct: (A)

The ~1:1 isotope pattern matches 79Br/81Br; chlorine gives a ~3:1 M/M+2 ratio instead.

### Conformations, Hybridization, Aromaticity (Q49-Q52)

Correct: (C)

Pyridine nitrogen is trigonal planar (sp2); its lone pair occupies an in-plane sp2 orbital and is not part of the 6 pi-electron aromatic sextet.

Correct: (B)

The carbocation carbon has three bonding domains and an empty p orbital, giving trigonal planar sp2 geometry.

Correct: (B)

For trans-1,2 (methyls on opposite faces), the two chair conformations are diaxial (a,a) and diequatorial (e,e); ring flip interconverts them while preserving the trans relationship. The diequatorial chair avoids all 1,3-diaxial strain and is the global minimum.

Correct: (B)

Planar cyclobutadiene is cyclic, conjugated, with 4 pi electrons (Huckel 4n, n=1), making it antiaromatic; tub-shaped COT avoids planarity and is nonaromatic.

### Acid-Base (Q53-Q55)

Correct: (B)

Acetic acid has the lowest pKa (4.8); the nitro group stabilizes phenoxide but less effectively than carboxylate resonance stabilizes acetate.

Correct: (D)

Pyrrole's lone pair is part of the aromatic sextet, making it essentially unavailable for protonation.

Correct: (C)

Acetylide is the conjugate base of the weakest acid (acetylene, pKa 25 vs. methanol 16 and water 15.7), so it is the strongest base.

### Radicals (Q56-Q58)

Correct: (B)

Resonance stabilization outweighs alkyl substitution: benzyl (delocalized into the aromatic ring) > allyl > tertiary > secondary.

Correct: (A)

The tertiary C-H bond is by far the weakest, so the tertiary radical forms overwhelmingly and is trapped by Br2 to give tert-butyl bromide.

Correct: (C)

The chain-end radical curls back and abstracts a hydrogen from a carbon a few units back on its own chain, creating a mid-chain radical that grows into a short branch.

### Polymers (Q59-Q60)

Correct: (C)

A diol plus a diacid condense to a polyester with loss of water.

Correct: (D)

Addition polymerization of CF2=CF2 gives the -[CF2-CF2]n- chain of polytetrafluoroethylene.