ACS Organic Chemistry Practice Test 7

Correct: (C)

Two molecules of benzaldehyde (no alpha-H) condense with one acetone; each aldehyde carbonyl undergoes aldol addition then dehydration at one end of acetone, giving the bis-benzylidene product dibenzalacetone.

Correct: (C)

LDA deprotonates 3-pentanone to give the secondary enolate CH3CH(−)COCH2CH3, which attacks acetaldehyde to give CH3CH2COCH(CH3)CH(OH)CH3 (5-hydroxy-4-methylhexan-3-one); dehydration does not occur at −78 °C.

Correct: (B)

2,6-heptanedione closes to a six-membered ring via intramolecular aldol; five- and six-membered ring closures are strongly favored over four- or seven-membered ones.

Correct: (A)

Self-Claisen of ethyl acetate gives the beta-keto ester ethyl acetoacetate, CH3COCH2COOEt.

Correct: (A)

The six-carbon diester closes to a five-membered ring; the alternative seven-membered closure is disfavored.

Correct: (D)

Aldehydes are more electrophilic than esters, so 1 equivalent of PhMgBr attacks the aldehyde selectively.

Correct: (B)

The carboxylic acid proton consumes one equivalent as base; the ketone adds two equivalents to give a tertiary alkoxide, while the carboxylate forms a stable tetrahedral intermediate that collapses to a methyl ketone only upon aqueous workup.

Correct: (B)

LiAlH4 reduces esters to primary alcohols and does not reduce isolated C=C bonds; NaBH4 cannot reduce esters and H2/Pt would hydrogenate the alkene.

Correct: (C)

NaBH4 is mild enough to reduce the more reactive aldehyde preferentially; LiAlH4 reduces both and PCC is an oxidant.

Correct: (C)

The Wittig reagent converts the aldehyde carbonyl to the terminal alkene C6H5CH=CH2.

Correct: (D)

Secondary amines condense with ketones to form enamines: the C=C is between C1 and C2 of the ring with N attached to C1.

Correct: (B)

Ketones form cyclic ketals with diols under acid catalysis with water removal.

Correct: (C)

LiAlH4 reduces nitriles fully to primary amines, C6H5CH2NH2.

Correct: (A)

Acid protonates the carbonyl oxygen, methanol attacks, proton transfers give the hemiacetal, then protonation of OH allows water loss and second methanol attack.

Correct: (C)

Intramolecular Barbier-type reaction: the organomagnesium formed at C6 attacks the C2 ketone, giving a cyclic tertiary alcohol after workup.

Correct: (A)

SN1 rate follows carbocation stability: primary < secondary < benzylic (resonance) < tertiary.

Correct: (D)

Secondary alkyl halide with a strong nucleophile in polar aprotic solvent undergoes SN2: first order in each reactant.

Correct: (A)

Bulky bases favor the Hofmann product (least substituted alkene) by abstracting the most accessible proton.

Correct: (D)

Debromination requires anti-periplanar C–Br bonds; the trans isomer achieves this in its diaxial chair conformation, while the cis isomer cannot, so the trans isomer reacts far faster.

Correct: (B)

Protonated alcohol leaves as water giving a primary carbocation that undergoes methyl shift to the more stable tertiary carbocation, captured by Br−.

Correct: (B)

Pair the alkoxide of the hindered alcohol with the unhindered primary halide; option A undergoes E2 instead of SN2.

Correct: (A)

Under basic conditions nucleophiles attack the less substituted epoxide carbon: CH3CH(OH)CH2OCH3.

Correct: (D)

Under acidic conditions the nucleophile attacks the more substituted carbon (more partial positive character in the protonated epoxide), giving CH3CH(OCH3)CH2OH.

Correct: (C)

In a Fischer projection the horizontal bonds project forward, so H is forward; the 1→2→3 priority sequence (OH → CH2CH3 → CH3) traces clockwise (R), and with the lowest-priority group forward the true configuration is inverted to S.

Correct: (A)

They have the same connectivity but differ in configuration at one or both centers without being mirror images, so they are diastereomers.

Correct: (D)

Rotation about a single bond changes only the conformation, not the configuration at either stereocenter, so I and II are rotamers (conformational isomers) of one stereoisomer.

Correct: (C)

The enantiomeric excess is 8.1/13.5 = 60%, so the major enantiomer makes up (100 + 60)/2 = 80% of the mixture.

Correct: (B)

trans-1,2-Dimethylcyclohexane has no internal mirror plane and exists as a pair of enantiomers; each of the others possesses a plane of symmetry and is achiral.

Correct: (D)

On C2, Br outranks the ethyl carbon; on C3, the methyl carbon outranks H. The two higher-priority groups are on the same side, so the alkene is Z.

Correct: (A)

Priorities are OH (1) > vinyl (2) > CH3 (3) > H (4); with H on a dashed bond, the 1→2→3 sequence (OH → vinyl → CH3) runs clockwise, giving R.

Correct: (B)

Alkene stability increases with the degree of substitution through hyperconjugation: monosubstituted < disubstituted < trisubstituted < tetrasubstituted.

Correct: (A)

Cleaving (CH3)2C=CHCH3 gives acetone from the (CH3)2C= fragment and acetaldehyde from the =CHCH3 fragment; only 2-butene cleaves to give two molecules of acetaldehyde and nothing else.

Correct: (D)

Hydroboration–oxidation adds H and OH syn across the double bond with anti-Markovnikov regiochemistry (B, then OH, at the less substituted C2), placing the new OH trans to the C1 methyl group.

Correct: (A)

A cyclic bromonium ion forms and water attacks the more substituted carbon (C2), so OH ends up on C2 and Br on C3, giving 3-bromo-2-methyl-2-butanol.

Correct: (B)

Peroxides initiate radical addition: Br· adds to the terminal carbon to generate the more stable secondary carbon radical, and hydrogen abstraction places Br on C1 (anti-Markovnikov product).

Correct: (C)

The Diels–Alder reaction is a concerted suprafacial cycloaddition, and secondary orbital interactions favor the endo transition state, placing the anhydride substituents under the diene π system.

Correct: (D)

Lindlar hydrogenation delivers both hydrogens syn to the same face (cis alkene), whereas dissolving-metal reduction proceeds by anti addition through a radical anion, giving the trans alkene.

Correct: (B)

The –CO2H group is a deactivating meta director, so electrophilic substitution occurs predominantly at the meta position.

Correct: (D)

Electron-donating groups activate the ring toward electrophilic attack (anisole > toluene > benzene) while electron-withdrawing groups deactivate it (nitrobenzene slowest).

Correct: (C)

The primary propyl carbocation rearranges by hydride shift to the more stable isopropyl cation, so 1-chloropropane gives isopropylbenzene rather than n-propylbenzene.

Correct: (A)

Nitro groups ortho and para to the leaving group stabilize the anionic Meisenheimer intermediate by resonance, greatly accelerating SNAr; a meta nitro group cannot do this.

Correct: (B)

The electron-withdrawing p-nitro group stabilizes the carboxylate and increases acidity, while electron-donating p-methyl and p-methoxy groups destabilize it and decrease acidity.

Correct: (D)

Both methyl groups are ortho/para directors; attack at C4 is ortho to one methyl and para to the other and is less sterically hindered than attack at C2 between the two methyl groups.

Correct: (B)

The 3.7 ppm singlet (3H) is a methoxy group attached to a carbonyl, the 2.3 ppm quartet (2H) is a CH₂ alpha to the carbonyl, and the 1.1 ppm triplet (3H) is its neighboring methyl, matching CH₃CH₂COOCH₃.

Correct: (D)

Symmetry makes the two ring ipso carbons equivalent, the four ring ortho/meta carbons equivalent, and the two methyl carbons equivalent, giving 3 distinct carbon environments.

Correct: (C)

M⁺ = 74 fits C₃H₆O₂; α-cleavage of methyl acetate gives the acylium ion CH₃CO⁺ at m/z = 43 (base peak), and loss of a methyl radical gives m/z = 59.

Correct: (A)

The broad IR band plus C=O stretch and the 11.0 ppm singlet identify a carboxylic acid; the triplet at 2.4 ppm is the CH₂ alpha to COOH and the remaining signals are a straight four-carbon alkyl chain, matching CH₃CH₂CH₂CH₂COOH.

Correct: (D)

Five aromatic protons indicate a monosubstituted benzene ring; the quartet/triplet pair (2H/3H) is an ethyl group, matching C₆H₅CH₂CH₃.

Correct: (B)

The trans-1,2 isomer can adopt a diequatorial chair, avoiding all 1,3-diaxial interactions; the diaxial trans conformer and the cis axial-equatorial forms are higher in energy, and a diaxial cis-1,2 chair does not exist.

Correct: (A)

The two alkene carbons each have three sigma bonds and one pi bond (sp²), while the nitrile carbon has two sigma bonds and two pi bonds (sp).

Correct: (D)

Cyclobutadiene is a planar, cyclic, conjugated system with 4 π electrons (a 4n system), making it antiaromatic; the others are 6 π-electron (4n + 2) aromatic systems.

Correct: (C)

The π molecular orbitals of butadiene are ψ₁ (0 nodes), ψ₂ (1 node), ψ₃ (2 nodes), ψ₄ (3 nodes); the HOMO is ψ₂ with 1 node and the LUMO is ψ₃ with 2 nodes.

Correct: (D)

Carboxylic acids (pKₐ ~4.8) are more acidic than phenols; the nitro group stabilizes the phenoxide ion, making p-nitrophenol (pKₐ ~7.2) more acidic than phenol (pKₐ ~10), and alcohols (pKₐ ~16) are the least acidic.

Correct: (C)

Aliphatic amines are the most basic; delocalization of the nitrogen lone pair into the ring makes aniline less basic, the nitro group withdraws further electron density in p-nitroaniline, and amides are the least basic due to strong resonance delocalization into the carbonyl.

Correct: (A)

The strongest base has the conjugate acid with the highest pKₐ; acetylene (pKₐ = 25) gives the acetylide ion, a far stronger base than hydroxide, ammonia, or chloride.

Correct: (B)

Resonance delocalization makes the benzyl radical most stable; among alkyl radicals, stability increases with substitution (tertiary > secondary > primary) due to hyperconjugation.

Correct: (D)

Bromination is highly selective for the weakest C−H bond; abstraction of the single tertiary hydrogen at C-2 gives the most stable radical, yielding 2-bromo-2-methylbutane as the major product.

Correct: (B)

Addition to the unsubstituted carbon places the new radical on the benzylic position, where it is resonance-stabilized by the phenyl ring, so head-to-tail propagation dominates.

Correct: (C)

Head-to-tail addition polymerization of propene (CH₂=CHCH₃) gives a chain with a methyl substituent on every other carbon, matching the repeat unit −[CH₂−CH(CH₃)]ₙ−.

Correct: (A)

Kevlar is poly(p-phenylene terephthalamide), formed by condensation of the aromatic diacid chloride (terephthaloyl chloride) with the aromatic diamine (1,4-phenylenediamine).