ACS Organic Chemistry Practice Test 19

Correct: (B)

Benzaldehyde has no alpha-hydrogens, so only acetophenone forms the enolate; the aldol addition product dehydrates under the basic, heated conditions to give the conjugated enone chalcone.

Correct: (D)

LDA forms the kinetic enolate of cyclohexanone, which adds to benzaldehyde; at -78 °C no dehydration occurs, so the beta-hydroxy ketone is isolated.

Correct: (A)

Intramolecular aldol addition forms a five-membered ring, and dehydration under the heated basic conditions gives 3-methylcyclopent-2-en-1-one.

Correct: (C)

This is the Claisen condensation: the ester enolate attacks a second molecule of ester, displacing ethoxide to give ethyl acetoacetate.

Correct: (B)

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

Correct: (D)

The ketone carbonyl is far more electrophilic than the aryl bromide toward nucleophilic addition, and Grignard reagents do not undergo halogen-metal exchange with aryl bromides under these conditions, so addition occurs only at the ketone to give the tertiary alcohol.

Correct: (A)

Grignard addition to an ester gives a ketone intermediate that is more reactive than the starting ester, so a second equivalent adds and the reaction cannot be stopped at acetophenone; the product is 2-phenylpropan-2-ol.

Correct: (C)

NaBH4 reduces aldehydes and ketones but not esters, so only the aldehyde is reduced to the primary alcohol while the methyl ester survives.

Correct: (D)

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

Correct: (B)

A ketone plus a secondary amine with acid catalysis and removal of water gives an enamine, not an imine (imines form from primary amines).

Correct: (A)

Under acid catalysis with water removal, the aldehyde and the diol form the cyclic acetal 2-phenyl-1,3-dioxolane.

Correct: (C)

Carboxylic acid derivative reactivity toward nucleophilic acyl substitution follows: acid chloride > anhydride > ester > amide.

Correct: (B)

DIBAL-H reduces nitriles to imines at low temperature, which hydrolyze to aldehydes; over-reduction to the primary amine requires LiAlH4.

Correct: (D)

One ketone is masked as the acetal protecting group, so the Grignard reagent adds only to the free ketone; acidic workup then unmasks the protected ketone, giving the hydroxy ketone.

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

Correct: (B)

DMSO is polar aprotic and N3- is a strong nucleophile but weak base, favoring SN2; backside attack inverts the (R) center to (S).

Correct: (D)

The tertiary bromide ionizes (SN1/E1 conditions); deprotonation of the carbocation gives the more substituted Zaitsev alkene as the major product.

Correct: (C)

Bulky KOtBu acts as a strong hindered base in E2, abstracting the less hindered beta-hydrogen to give the Hofmann (less substituted) alkene.

Correct: (A)

Ionization is the rate-determining step of SN1, and polar protic solvents stabilize the developing charges, accelerating it.

Correct: (B)

SN1 proceeds through a planar tertiary carbocation that is attacked from either face, giving racemic product.

Correct: (D)

Ag+ promotes ionization to a primary carbocation, which undergoes hydride shift to the more stable tertiary cation before capture by ethanol.

Correct: (A)

The SN1 rate law is rate = k[RX]; only the substrate appears in the rate-determining ionization step.

Correct: (C)

E2 is concerted and bimolecular, so both substrate and base appear in the rate law.

Correct: (B)

SN2 is fastest at the least hindered primary carbon; ethyl bromide has minimal steric hindrance to backside attack.

### Stereochemistry (Q24-Q30)

Correct: (A)

At C2 the priorities are -OH (1) > -COOH (2) > C3 (3) > H (4); with H on a horizontal bond (pointing forward), the 1-2-3 order appears counterclockwise (S) and is reversed to R.

Correct: (C)

The two structures are non-superimposable mirror images; (A) and (D) are diastereomer pairs, and (B) compares two different compounds.

Correct: (B)

(2R,3S)-2,3-butanediol has an internal plane of symmetry, making it achiral; the others lack any improper symmetry element.

Correct: (D)

The anti staggered conformer minimizes steric repulsion between the two bulky methyl groups; gauche lies higher and eclipsed forms higher still.

Correct: (B)

The like-on-like eclipsed geometry has a mirror plane perpendicular to the C-C bond, so it is achiral; only the meso (2R,3S) stereoisomer can adopt it.

Correct: (C)

ee = (6.75/13.5) x 100% = 50%, and the major enantiomer is (100 + 50)/2 = 75% (R).

Correct: (A)

Only a non-racemic chiral sample is optically active; (B) and (D) are achiral meso compounds and (C) is a racemate with zero net rotation.

### Alkenes/Alkynes (Q31-Q37)

Correct: (D)

Tetrasubstituted > trisubstituted > disubstituted (trans) > monosubstituted; more alkyl groups on the double bond give greater hyperconjugative stabilization.

Correct: (C)

Rejoining the two carbonyl fragments (CH3CH2CHO and CH3CHO) at the cleaved double bond gives CH3CH2CH=CHCH3.

Correct: (B)

Cleaving the C1=C2 bond of the C8 ring (both carbons bearing methyl groups) opens the ring to the C8 diketone CH3CO(CH2)4COCH3.

Correct: (A)

OH adds to the less substituted carbon (anti-Markovnikov) via a concerted syn addition with no carbocation intermediate, so no methyl shift occurs.

Correct: (C)

The peroxide effect gives anti-Markovnikov addition: Br radical adds to the less substituted carbon (C2), generating the more stable tertiary radical at C1, which then abstracts H.

Correct: (A)

The ortho rule places the electron-withdrawing CHO on the carbon adjacent to the new double bond (C4 relative to the double bond at C1), with the endo orientation favored.

Correct: (A)

Lindlar's catalyst effects syn addition of one equivalent of H2, stopping at the cis (Z) alkene stage.

### Aromatic Chemistry (Q38-Q43)

Correct: (A)

The tert-butyl group is an ortho/para director, but its bulk blocks the ortho positions, so electrophilic attack occurs almost exclusively para.

Correct: (B)

The nitro group is a strongly deactivating meta director, so bromine enters meta to NO2; the ring is too deactivated for a second bromination.

Correct: (C)

The -NH2 group is a powerful activating ortho/para director, so with excess Br2 all three available ortho and para positions are brominated.

Correct: (D)

Activating CH3 accelerates EAS, halogens deactivate by induction despite ortho/para direction, and NO2 is the strongest deactivator.

Correct: (A)

Nitro groups ortho and para to the leaving group stabilize the anionic Meisenheimer complex by resonance, enormously accelerating SNAr.

Correct: (B)

The electron-withdrawing NO2 stabilizes the carboxylate anion while electron-donating CH3 and OCH3 destabilize it; pKa values are 3.44, 4.20, 4.37, and 4.47 respectively.

### Spectroscopy (Q44-Q48)

Correct: (C)

Only two singlets are consistent with the high symmetry of p-xylene; ethylbenzene would show a quartet/triplet ethyl pattern, and o-/m-xylene show split aromatic multiplets.

Correct: (D)

The central CH2 couples to five neighboring protons (3 on CH3 plus 2 on CH2Br), so n + 1 = 6 lines.

Correct: (A)

The broad IR band plus D2O-exchangeable proton indicate an -OH, and the 2H singlet at 4.6 ppm is a benzylic CH2 attached to oxygen.

Correct: (B)

M+. 106 fits C8H10; base peak 91 is the benzyl/tropylium ion from benzylic cleavage, and only ethylbenzene gives six 13C signals (o-xylene 4, p-xylene 3, m-xylene 5).

Correct: (C)

The symmetric AA'BB' spin system of a para-disubstituted ring gives two 2H doublets.

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

Correct: (B)

The central carbon forms two pi bonds and no lone pairs (two electron domains, sp), while each terminal carbon has three domains (sp2).

Correct: (D)

The carbonyl carbon is trigonal planar (sp2) and the hydroxyl oxygen has two bonds plus two lone pairs in a bent arrangement (sp3).

Correct: (C)

Ring flip interconverts the diaxial and diequatorial chairs for cis-1,3; the diequatorial chair avoids severe 1,3-diaxial strain and is the global minimum.

Correct: (A)

The cyclopentadienyl anion is cyclic, planar, fully conjugated, with 6 pi electrons (Huckel 4n+2, n=1); cyclobutadiene is antiaromatic and the others are nonaromatic.

### Acid-Base (Q53-Q55)

Correct: (D)

Lower pKa means stronger acid, so acidity increases as pKa decreases: 25 > 16 > 10 > 4.8.

Correct: (A)

Piperidine is an aliphatic secondary amine; the others have lone pairs delocalized into a ring or carbonyl, reducing basicity.

Correct: (C)

NH2- is the conjugate base of the weakest acid listed (NH3, pKa ~35 vs. H2O 15.7), and weaker conjugate acids correspond to stronger bases.

### Radicals (Q56-Q58)

Correct: (B)

Radical stability follows 3° > 2° > 1° > methyl due to hyperconjugation and inductive donation from alkyl groups.

Correct: (C)

Bromination is highly selective for the weakest C-H bond: the tertiary hydrogen is abstracted far faster than primary hydrogens, so the tertiary bromide dominates.

Correct: (D)

Propagation repeatedly adds monomer to the chain-end radical; (A) is initiation, (B) is termination by coupling, and (C) is chain transfer.

### Polymers (Q59-Q60)

Correct: (D)

A diamine plus a diacid undergo condensation polymerization to a polyamide with loss of water.

Correct: (B)

Cleaving the repeating unit at the new C-C bonds of addition polymerization gives the alkene monomer vinylbenzene (styrene).

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