ACS Organic Chemistry Practice Test 16

Correct: (B)

The first equivalent of Grignard converts the ester to the ketone (acetophenone), but with excess reagent the ketone is attacked a second time, giving the tertiary alcohol after workup.

Correct: (D)

The first addition collapses to benzaldehyde, which is attacked by the second equivalent of PhMgBr, giving the secondary alcohol Ph2CHOH after protonation.

Correct: (A)

The stabilized enolate is an ambident nucleophile, but carbon alkylation dominates with alkyl halides, giving the C-alkylated β-keto ester.

Correct: (C)

Monoalkylation gives CH(Et)(CO2Et)2, hydrolysis gives the substituted malonic acid, and heating a β-diacid drives decarboxylation to butanoic acid.

Correct: (D)

Gilman reagents add 1,4 (conjugate addition) to enones, delivering the methyl group to the β-carbon; protonation of the resulting enolate gives 3-methylcyclohexanone.

Correct: (B)

The more reactive aldehyde is selectively masked as a cyclic acetal, so the Grignard attacks only the ketone; aqueous acid unmasks the aldehyde and protonates the alkoxide.

Correct: (A)

Acetal protection of the aldehyde prevents its intramolecular aldol, letting LDA form the ketone enolate cleanly; it adds to benzaldehyde, and hot acid dehydrates the aldol while restoring the aldehyde.

Correct: (D)

Reactivity follows leaving-group ability and inversely the resonance donation into the carbonyl: chloride is the best leaving group and the amide nitrogen donates most strongly.

Correct: (C)

Organolithiums open epoxides by SN2 attack at the less hindered carbon, so n-butyl adds to the CH2 terminus and the alkoxide forms on C-2, giving heptan-2-ol.

Correct: (B)

LDA at low temperature irreversibly removes the less hindered C-6 proton (kinetic enolate), so methylation occurs at C-6 rather than at the more substituted C-2 position.

Correct: (D)

In protic, equilibrating conditions the more substituted (thermodynamic) enolate at C-2 predominates, so alkylation occurs there to give 2,2-dimethylcyclohexanone.

Correct: (A)

Acid protonates the carbonyl oxygen, making the carbonyl carbon electrophilic enough for water (a weak nucleophile) to attack.

Correct: (C)

The ketone condenses with methylamine to the imine, which NaBH3CN reduces selectively (it reduces iminium ions far faster than ketones), giving the secondary amine.

Correct: (A)

The Gilman reagent performs one nucleophilic acyl substitution, displacing chloride; unlike Grignard reagents it does not add again to the resulting ketone, so acetone is isolated.

Correct: (D)

The primary carbocation rearranges by a methyl shift to the more stable tertiary 2-methyl-2-butyl cation, which is captured by water.

Correct: (C)

Loss of Br− gives a secondary carbocation that undergoes a methyl shift to a tertiary carbocation, captured by methanol at the tertiary carbon.

Correct: (B)

SN2 proceeds by backside attack, so the single stereocenter undergoes complete inversion of configuration.

Correct: (B)

E1 of the tertiary substrate follows Zaitsev's rule strongly; the bulky base in E2 shifts the mixture toward the less substituted Hofmann alkene.

Correct: (C)

SN2 rates decrease with steric hindrance at the α-carbon, and the allyl system is fastest because π-overlap stabilizes the SN2 transition state.

Correct: (B)

Two maxima with an intervening intermediate indicates a two-step mechanism, and the higher first barrier means carbocation formation is rate-determining.

Correct: (D)

Tosylate is an excellent leaving group, and a bulky strong base drives E2 elimination to the alkene.

Correct: (D)

SN1 rates follow carbocation stability: tertiary > secondary > primary > methyl.

Correct: (A)

E2 requires an axial leaving group; the cis isomer reaches the reactive Cl-axial/Me-equatorial chair readily, while the trans isomer must adopt a high-energy diaxial chair.

Correct: (B)

Priorities: OH (1) > isopropyl (2) > ethyl (3). The 1→3→2 order is clockwise, so 1→2→3 is counterclockwise = S.

Correct: (D)

Stability increases with alkyl substitution on the double bond, and trans disubstituted alkenes are more stable than cis.

Correct: (C)

The anti conformation maximizes the separation of the two bulky methyl groups, minimizing steric strain.

Correct: (D)

With identical substituents at C2 and C3, the (2R,3S) configuration has an internal plane of symmetry and is meso (achiral).

Correct: (B)

Lindlar hydrogenation adds H2 syn to the alkyne, giving the cis (Z) alkene.

Correct: (A)

Enantiomers rotate plane-polarized light by equal magnitudes in opposite directions.

Correct: (C)

P is the chiral (2S,3S) stereoisomer and its non-superimposable mirror image Q is (2R,3R); the pair are enantiomers.

Correct: (D)

Dissolving-metal reduction of an alkyne proceeds through a trans vinyl radical/anion sequence, giving the trans-alkene (opposite of Lindlar).

Correct: (B)

Hydroboration adds boron to the less substituted carbon (anti-Markovnikov); oxidation gives the enol PhCH=CHOH, which tautomerizes to the aldehyde.

Correct: (A)

Cleaving both double bonds of CH2=C(CH3)–C(CH3)=CH2 releases the two terminal CH2 groups as formaldehyde and the central C2 unit as CH3COCOCH3.

Correct: (C)

The ortho rule pairs the electron-donating methoxy group ortho to the electron-withdrawing aldehyde, and the endo rule orients the aldehyde under the diene.

Correct: (D)

OH adds to the more substituted carbon (Markovnikov) via a mercurinium ion, so no carbocation forms and the skeleton does not rearrange.

Correct: (A)

OsO4 delivers both oxygens syn to the same face of the cis-alkene, generating the achiral meso diol (the trans-alkene would give the racemic pair).

Correct: (B)

Electron-donating groups raise the diene HOMO and accelerate the normal-demand Diels–Alder, while the strongly withdrawing nitro group lowers it and retards reaction.

Correct: (A)

Nitro is reduced to the amine, diazotized, and the diazonium group is replaced by Br in the Sandmeyer reaction.

Correct: (B)

Acetylation converts –NH2 into the less activating acetamido group so only para monobromination occurs; basic hydrolysis then regenerates the free amine.

Correct: (C)

The aliphatic amine's lone pair is fully available, aniline's is delocalized into the ring, and the p-nitro group withdraws it even further.

Correct: (D)

The methyl group (ortho/para director) and the nitro group (meta director) both direct the incoming nitro to C2/C6, ortho to the methyl group.

Correct: (A)

Birch reduction adds two hydrogens 1,4 across the ring, giving the non-conjugated 1,4-cyclohexadiene rather than the conjugated isomer.

Correct: (B)

Any alkyl side chain bearing a benzylic hydrogen is oxidatively cleaved all the way to benzoic acid under vigorous conditions.

Correct: (C)

The δ 9.8 triplet is an aldehyde proton coupled to a CH₂, and the remaining ethyl pattern (CH₂ multiplet, CH₃ triplet) gives CH₃CH₂CHO.

Correct: (D)

A ~1:1 M/M+2 pattern indicates one bromine; 136 − 79 = 57, which corresponds to C₄H₉.

Correct: (A)

Ketone carbonyl carbons appear near δ 205–220, while esters, acids, and amides appear near δ 165–180.

Correct: (B)

Both show the C=O stretch, but only aldehydes show the Fermi doublet from the aldehydic C–H stretch.

Correct: (D)

The ethoxy quartet/triplet plus an acetyl singlet match CH₃CO₂CH₂CH₃, and alpha cleavage gives CH₃CO⁺ at m/z 43.

Correct: (C)

The central carbon forms two double bonds using two mutually perpendicular p orbitals, leaving sp hybridization.

Correct: (B)

The cis isomer can place both methyl groups equatorial in one chair, avoiding all 1,3-diaxial interactions.

Correct: (D)

Pyrrole's lone pair in a p orbital completes the sextet (making it weakly basic); pyridine's lone pair is not part of the sextet and is readily protonated.

Correct: (A)

Extending conjugation raises the HOMO and lowers the LUMO, so the gap narrows: hexatriene < butadiene < ethylene.

Correct: (D)

The electron-withdrawing nitro group stabilizes the phenoxide ion, while the electron-donating methyl group destabilizes it.

Correct: (A)

Ethylamine's lone pair is localized; aniline's is conjugated with the ring; acetamide's is strongly delocalized into the carbonyl, making it the weakest base.

Correct: (C)

Its conjugate acid (acetylene, pKₐ ~25) is the weakest acid of the set, so acetylide is the strongest base.

Correct: (C)

The allyl radical is resonance-stabilized, while the vinyl radical's unpaired electron sits in an sp² orbital with no delocalization.

Correct: (A)

Br• adds to give the more substituted (more stable) carbon radical, which then abstracts H from HBr.

Correct: (D)

The benzyl radical delocalizes the unpaired electron over the aromatic ring, exceeding allylic resonance and hyperconjugation effects.

Correct: (A)

1,4-addition places the double bond in the backbone between C2 and C3 of each monomer unit.

Correct: (D)

Kevlar is poly(p-phenylene terephthalamide), from the para diacid and para diamine.