ACS Organic Chemistry Practice Test 10

Correct: (D)

Directed cross-aldol: acetophenone enolate attacks p-tolualdehyde (no alpha-H, cannot self-condense), and warming dehydrates the aldol to the conjugated enone chalcone.

Correct: (A)

Secondary amines form enamines with ketones under dehydrating conditions; the C=O oxygen is lost as water and the double bond forms at the alpha carbon.

Correct: (D)

Reactivity follows leaving-group ability: Cl- > carboxylate > alkoxide > amide, so acid chloride > anhydride > ester > amide.

Correct: (A)

Both reagents reduce ketones to secondary alcohols; NaBH4 reduces ketones readily (only carboxylic derivatives require LiAlH4).

Correct: (D)

The ketone alpha-CH2 (C5) enolizes and attacks the aldehyde carbonyl in a favored 5-exo-trig cyclization; dehydration toward conjugation gives 2-methylcyclopent-2-en-1-one.

Correct: (D)

Non-stabilized ylides give predominantly Z-alkenes via the cis-oxaphosphetane; propanal + ethylidene ylide gives pent-2-ene, Z favored.

Correct: (B)

DIBAL-H reduces nitriles to imines at low temperature, which hydrolyze to aldehydes; full reduction to amine requires LiAlH4 or H2/catalyst.

Correct: (A)

Ketones with 1,2-diols under acid with water removal form cyclic ketals; ethylene glycol gives the 1,3-dioxolane derivative.

Correct: (C)

Dieckmann condensation: enolate at C2 attacks the C6 ester carbonyl, closing a 5-membered ring to give ethyl 2-oxocyclopentanecarboxylate.

Correct: (B)

Grignard reagents are chemoselective for the less hindered, more electrophilic aldehyde; benzaldehyde reacts much faster than acetophenone.

Correct: (D)

Grignard addition to the ketone carbonyl gives the tertiary alcohol: (CH3)2CHCOCH3 + CH3- → (CH3)2CHC(OH)(CH3)2, i.e., 2,3-dimethylbutan-2-ol.

Correct: (C)

Hydrate stability increases with electrophilicity and decreases with steric bulk: formaldehyde (no alkyl donors, unhindered) hydrates extensively, acetone barely.

Correct: (B)

The ambident enolate of a beta-keto ester undergoes C-alkylation preferentially with alkyl halides (C is the softer nucleophilic site in SN2).

Correct: (B)

Acid chlorides react with 2 equiv Grignard: first gives the ketone, which reacts again to the tertiary alcohol, 1-cyclohexyl-1-methylethanol.

Correct: (A)

Tertiary halide solvolysis is SN1: rate-determining ionization, first order in substrate only; solvent concentration change has negligible rate effect (though polarity matters).

Correct: (C)

Secondary halide + strong nucleophile in polar aprotic solvent = SN2 with complete inversion; (S) substrate gives (R) product.

Correct: (B)

E1 via tertiary carbocation; Zaitsev product (most substituted alkene) dominates: (CH3)2C=CHCH3.

Correct: (B)

E1 with hydride shift: secondary carbocation rearranges to the more stable tertiary carbocation (2,3-dimethylbutan-2-yl cation), giving the tetrasubstituted alkene.

Correct: (B)

SN1 rate follows carbocation stability: benzylic/allylic > tertiary > secondary > primary. The benzyl cation is resonance-stabilized, making it more stable than the tert-butyl cation; benzyl bromide also benefits from the superior bromide leaving group.

Correct: (A)

Bulky base with primary halide gives E2 (Hofmann orientation irrelevant — only one beta carbon set); 2-methylprop-1-ene is the sole alkene.

Correct: (C)

SN1 proceeds through a planar carbocation; solvent capture occurs from either face, giving racemic 2-butanol (which cannot be meso).

Correct: (C)

Leaving-group ability increases down the periodic table: I- > Br- > Cl- > F- >> OH-; weaker bases are better leaving groups, and larger ions stabilize the negative charge.

Correct: (C)

Small ethoxide favors the more substituted Zaitsev alkene; bulky t-BuOK abstracts the less hindered beta-H, giving the Hofmann alkene but-1-ene.

Correct: (A)

Priorities: OH (1) > CH2CH3 (2) > CH3 (3) > H (4). H is on the hashed wedge (pointing away), and tracing 1→2→3 goes clockwise, so the configuration is R.

Correct: (B)

Diastereomers are stereoisomers that are not mirror images: (2R,3R) vs (2R,3S) differ at one center only. A is enantiomers; C is the meso compound twice (identical).

Correct: (C)

Anti and gauche butane are rotamers (conformational isomers) of the same achiral molecule, interconverted by C-C bond rotation.

Correct: (A)

ee = 6.75/13.5 × 100% = 50%. %R = (50 + 100)/2 = 75%, %S = 25%.

Correct: (D)

(2R,3S)-2,3-dichlorobutane is meso (has inversion center); cis-1,2-dichlorocyclopropane has a plane of symmetry; 2,2-dichlorobutane has no stereocenter; (1R,2R)-1,2-dichlorocyclopropane is chiral with no symmetry plane.

Correct: (A)

Stability increases with substitution: monosubstituted (ethene, 0 alkyl groups on C=C) < disubstituted trans < trisubstituted (2-methylpropene) < tetrasubstituted (2,3-dimethylbut-2-ene).

Correct: (B)

In (E)-1-bromo-1-propene the bulky Br and CH3 are trans, minimizing steric clash; alkenes do not rotate about the pi bond, so E/Z are distinct configurational isomers.

Correct: (B)

Cleaving the C=C bond must regenerate ethanal and propanone, which only fits (CH3)2C=CHCH3, i.e. 2-methyl-2-butene.

Correct: (C)

Hydroboration-oxidation adds H and OH syn to the less substituted carbon (anti-Markovnikov); since H and OH add cis, the OH ends up trans to the C1 methyl group.

Correct: (A)

The Diels–Alder reaction is concerted and suprafacial, and the endo rule favors the –CHO group tucked under the diene π-system via secondary orbital interactions.

Correct: (D)

Retrosynthetic cleavage of the new σ-bonds gives butadiene and the acetyl-bearing dienophile; acrolein would give an aldehyde, not a ketone, and cyclopentadiene would give a norbornene skeleton.

Correct: (A)

Ionic addition follows Markovnikov's rule (most stable carbocation), while the peroxide-initiated radical addition gives anti-Markovnikov product via the more stable carbon radical.

Correct: (C)

Water opens the cyclic bromonium ion anti to the C–Br bond, giving the trans bromohydrin rather than the dibromide.

Correct: (D)

Stability increases with alkyl substitution (hyperconjugation), and trans-disubstituted alkenes are more stable than cis isomers due to reduced steric strain.

Correct: (A)

The acetamido group is a strongly activating ortho/para director; the para product dominates because ortho attack is sterically hindered.

Correct: (C)

Activating groups (–OCH3 > –CH3) speed EAS by stabilizing the arenium ion, while the strongly deactivating –NO2 group slows it dramatically.

Correct: (D)

The –CO2CH3 group is a deactivating meta director, so electrophilic substitution occurs meta to the ester.

Correct: (A)

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

Correct: (B)

Friedel–Crafts reactions fail on strongly deactivated rings; the nitro group makes nitrobenzene too electron-poor for electrophilic attack.

Correct: (D)

The primary butyl cation rearranges by hydride shift to the more stable secondary carbocation before attacking the ring.

Correct: (A)

The data fit CH3COCH2CH3 exactly: an acetyl singlet, plus an ethyl group (triplet/quartet); an aldehyde would show a signal near δ 9–10.

Correct: (A)

The 6H doublet plus 1H septet is the classic isopropyl pattern, and the 3H singlet is an acetate methyl: CH3COOCH(CH3)2.

Correct: (B)

Symmetry makes the two methyl carbons equivalent, the four ring CH carbons equivalent, and the two quaternary ring carbons equivalent: 3 unique environments.

Correct: (B)

Alpha cleavage next to the carbonyl ejects an ethyl radical, giving the resonance-stabilized acylium ion CH3CH2C≡O+ at m/z 57.

Correct: (C)

The broad IR O–H stretch plus C=O and the exchangeable δ 11.6 proton identify a carboxylic acid; the ethyl pattern gives CH3CH2COOH.

Correct: (D)

The trans-1,2 isomer can place both methyls equatorial (one up, one down), avoiding all 1,3-diaxial interactions.

Correct: (C)

Each terminal carbon has three σ-bonds (sp2) while the central carbon has two σ-bonds in a cumulated diene (sp).

Correct: (B)

The cyclopentadienyl anion is planar, cyclic, and conjugated with 6 π electrons, satisfying Hückel's 4n + 2 rule; the others are antiaromatic or nonplanar.

Correct: (C)

The bulky tert-butyl group locks the ring with itself equatorial; for the cis-1,4 relationship this forces the methyl group axial.

Correct: (A)

Electron-withdrawing p-NO2 stabilizes the carboxylate (strongest acid), while electron-donating p-OCH3 and p-CH3 destabilize it (weakest acids).

Correct: (C)

Alkyl groups donate electron density (dimethylamine strongest in water), while aniline's lone pair is delocalized into the ring, making it the weakest base.

Correct: (D)

The strongest base has the weakest conjugate acid; NH3 (pKa ~38) is far less acidic than CH3OH, H2O, or CH3COOH.

Correct: (B)

Radical stability increases with hyperconjugation and resonance: vinylic (least) < primary < tertiary < benzylic (most, resonance-delocalized).

Correct: (A)

Bromination is highly selective for the weakest C–H bond; abstraction of the single tertiary hydrogen gives the most stable radical and thus tert-butyl bromide.

Correct: (B)

Propagation is the repeated addition of the chain-carrying radical to monomer; (A) is initiation and (C) is termination.

Correct: (D)

Cleaving the repeating unit at the single bonds of the backbone regenerates the alkene monomer CH2=CH–CN (Orlon/polyacrylonitrile).

Correct: (A)

Kevlar is an aromatic polyamide formed from the diacid chloride and the para-diamine; (B) gives nylon-6,6 and (D) gives Dacron polyester.