ACS Organic Chemistry Practice Test 15

Correct: (A)

LDA at −78 °C irreversibly forms the kinetic (less substituted) enolate at C-1 of 2-butanone, which attacks benzaldehyde; the mild conditions give the aldol addition product with no dehydration.

Correct: (C)

Benzaldehyde has no α-hydrogens, so it acts only as the electrophile while acetone's enolate adds; heating drives dehydration to the conjugated enone (trans-benzalacetone).

Correct: (B)

Intramolecular aldol favors the five-membered ring (a terminal enolate attacks the other carbonyl), and heat dehydrates the β-hydroxy ketone to the conjugated cyclopentenone.

Correct: (D)

Ethoxide generates the ester enolate, which attacks a second molecule of ethyl acetate; collapse of the tetrahedral intermediate ejects ethoxide, giving ethyl acetoacetate after protonation.

Correct: (A)

This intramolecular Claisen (Dieckmann) condensation closes the favored five-membered ring, placing the ketone adjacent to the ester group.

Correct: (C)

The aldehyde is more electrophilic than the ketone (less steric hindrance, one fewer electron-donating alkyl group), so one equivalent of Grignard adds selectively to the aldehyde carbonyl.

Correct: (B)

The Grignard reagent adds to the ketone carbonyl to give a tertiary alkoxide, which is protonated on workup to the tertiary alcohol; no elimination or rearrangement occurs.

Correct: (D)

NaBH4 is a mild hydride donor that reduces aldehydes and ketones but leaves esters untouched under normal conditions, so only the ketone becomes a secondary alcohol.

Correct: (A)

LiAlH4 is a powerful hydride source that reduces both the ketone (to a secondary alcohol) and the ester (all the way to a primary alcohol), giving the diol after workup.

Correct: (B)

PCC oxidizes a primary alcohol to the aldehyde and stops there, because the anhydrous conditions prevent hydrate formation and further oxidation to the carboxylic acid.

Correct: (C)

The ylide carbon bonds to the carbonyl carbon via the oxaphosphetane intermediate, converting C=O into an exocyclic C=C bond (Ph3P=O is the byproduct).

Correct: (D)

A secondary amine condenses with the ketone but cannot form an imine (no N–H remains); loss of water from the α-position gives the enamine, driven by water removal.

Correct: (A)

The strongly electron-withdrawing CCl3 group destabilizes the carbonyl π-bond, shifting the hydration equilibrium far toward the gem-diol; chloral is isolated as chloral hydrate.

Correct: (C)

DIBAL-H delivers a single hydride to the nitrile, forming an imine intermediate that hydrolyzes on workup to the aldehyde; full reduction to the amine requires stronger conditions.

Correct: (B)

A tertiary substrate in a polar protic solvent with a weak nucleophile ionizes by SN1; backside SN2 attack is sterically blocked.

Correct: (A)

A primary alkyl halide with a good nucleophile in a polar aprotic solvent reacts by concerted SN2 with no rearrangement.

Correct: (B)

The bulky base abstracts the least hindered β-hydrogen, giving the Hofmann (least substituted) alkene.

Correct: (C)

A small, unhindered base follows Zaitsev's rule, giving the most substituted alkene, with the more stable trans isomer predominating.

Correct: (C)

The rate law is first order in each reactant, so the rate multiplies by 2 × 3 = 6.

Correct: (A)

E2 requires H and Br anti-periplanar; the staggered conformer placing the methyl groups anti is lowest in energy and leads to trans-2-butene.

Correct: (A)

Pairing the hindered alkoxide with an unhindered methyl halide favors SN2; tert-butyl bromide with base (B, D) undergoes E2 elimination instead.

Correct: (B)

A rate independent of nucleophile concentration and accelerated by ionizing solvents points to rate-determining carbocation formation (SN1).

Correct: (A)

E2 is a concerted single-step process that requires anti-periplanar alignment of the breaking C–H and C–X bonds.

Correct: (D)

Carbons 1 and 2 are alkene carbons (sp2); carbons 3 and 4 are alkyne carbons (sp).

Correct: (C)

Rotation about the C2–C3 single bond interconverts staggered conformers without changing the (2R,3R) configuration.

Correct: (A)

1-bromobutane (CH3CH2CH2CH2Br) has no stereocenter, so it is achiral; the others all have chiral stereoisomers.

Correct: (D)

The enantiomeric excess is 12.0/48.0 = 25%, so %R = (100 + 25)/2 = 62.5%.

Correct: (D)

The (2R,3R) stereoisomer is chiral with no improper symmetry; the meso form has an inversion center and is achiral.

Correct: (C)

The enantiomer inverts every stereocenter: (2R,3R) → (2S,3S). The (2R,3S) form is the meso diastereomer.

Correct: (B)

(2R,3R) and (2R,3S) differ at only one stereocenter, making them diastereomers; (A) are enantiomers and (C) are identical meso structures.

Correct: (C)

The acetaldehyde fragment comes from a =CHCH3 unit and cyclopentanone from the ring C= carbon, so the double bond is exocyclic with a methyl group: cyclopentane-C(H)=CH3.

Correct: (A)

Only (CH3)2C=C(CH3)2 cleaves into two identical (CH3)2C=O fragments; 2-methyl-2-butene would also give acetaldehyde.

Correct: (D)

Boron adds to the less substituted carbon (anti-Markovnikov) and syn addition places the new OH trans to the methyl group already on the ring.

Correct: (B)

The endo rule favors the transition state with the carbonyl π system tucked under the diene, placing CHO endo on the newly formed ring.

Correct: (A)

Retro-Diels–Alder cleavage of the two new σ bonds returns a methyl-substituted diene and unsubstituted maleic anhydride, matching only isoprene + maleic anhydride.

Correct: (C)

Radical addition puts Br on the less substituted carbon (anti-Markovnikov), and the tert-butyl group blocks any favorable hydride shift, so no rearrangement occurs.

Correct: (B)

A bromonium ion forms and water opens it by backside attack (anti), giving the trans halohydrin; water outcompetes bromide as the nucleophile in aqueous solution.

Correct: (C)

The isopropyl group is an ortho/para director, and its bulk sterically suppresses ortho attack, making the para product dominate.

Correct: (B)

The acetyl group withdraws electron density by resonance and is a deactivating meta director.

Correct: (D)

Strongly activating –OCH3 accelerates EAS most, weakly activating –CH3 accelerates modestly, and deactivating –NO2 severely retards it.

Correct: (A)

The deactivating acetyl group directs the incoming electrophile meta; bromination still proceeds since Br2/FeBr3 is a strong electrophile.

Correct: (C)

The chloride is activated toward nucleophilic aromatic substitution by the ortho and para nitro groups and is displaced by hydroxide via a Meisenheimer complex.

Correct: (A)

Electron-withdrawing groups stabilize the carboxylate conjugate base (nitro most strongly), while the electron-donating methyl group destabilizes it.

Correct: (D)

The methine proton couples to six equivalent methyl protons, so n + 1 = 7 (a septet).

Correct: (B)

The downfield 1H singlet is a carboxylic acid O–H and the 3H singlet is a methyl adjacent to a carbonyl, matching CH₃CO₂H.

Correct: (A)

Symmetry makes the two methyl carbons equivalent, the two ring C–CH₃ carbons equivalent, and the four ring CH carbons equivalent: 3 signals.

Correct: (C)

Alpha cleavage next to the carbonyl gives the resonance-stabilized acylium ion CH₃C≡O⁺ at m/z 43.

Correct: (B)

A ketone carbonyl (IR, no aldehyde proton or O–H) with a methyl-ketone singlet at δ 2.10 and a propyl chain matches CH₃COCH₂CH₂CH₃.

Correct: (B)

The nitrogen lone pair delocalizes into the carbonyl π system, giving the C–N bond partial double-bond character and a planar geometry.

Correct: (D)

An axial tert-butyl group is destabilized by ~5 kcal/mol of 1,3-diaxial interactions, making that conformer essentially unpopulated.

Correct: (A)

C₇H₇⁺ is cyclic, planar, and conjugated with 6 π electrons (4n + 2, n = 1).

Correct: (C)

Cyclobutadiene is planar and conjugated with 4 π electrons (4n), making it antiaromatic; cyclooctatetraene avoids this by adopting a nonplanar tub shape.

Correct: (A)

Fluorine's stronger −I effect stabilizes the carboxylate conjugate base more than chlorine does, and both exceed the unsubstituted acid.

Correct: (C)

The weaker conjugate acid (higher pKₐ) corresponds to the stronger base, so methylamine > pyridine in basicity.

Correct: (B)

NH₃ accepts the proton from acetic acid in the forward direction.

Correct: (A)

Benzyl (extensive resonance over the ring) > allyl (resonance) > tert-butyl (hyperconjugation) > ethyl (primary).

Correct: (D)

The bromine radical is highly selective and abstracts the weakest (tertiary) C–H bond, giving the tertiary bromide.

Correct: (B)

Propagation is repeated radical addition to monomer; (A) is termination, (C) is initiation, and (D) is chain transfer.

Correct: (B)

Opening the C=C bond of chloroethene (CH₂=CHCl) gives exactly the –CH₂–CHCl– repeating unit.

Correct: (C)

Nylon-6,6 forms from the 6-carbon diamine and the 6-carbon diacid with loss of water.