ACS Organic Chemistry Practice Test 2

Correct: (B)

Crossed aldol addition occurs between the enolate of acetone and benzaldehyde, and heating dehydrates the β-hydroxy ketone to the conjugated enone.

Correct: (B)

The aldehyde is more electrophilic, so it must be masked as an acetal (which is inert to Grignard reagents) while the ketone reacts, then the acetal is hydrolyzed back.

Correct: (C)

LiAlH4 reduces the ester to a primary alcohol but does not reduce isolated alkenes; NaBH4 cannot reduce esters and H2/Pd would saturate the double bond.

Correct: (D)

The Wittig reaction converts the ketone carbonyl into a C=C bond bearing the ylide carbon, giving the ethylidene-substituted ring.

Correct: (C)

Intramolecular Claisen (Dieckmann) condensation of the 1,6-diester closes a five-membered ring to give the cyclic β-keto ester.

Correct: (A)

NaBH4 reduces aldehydes much faster than ketones, so the aldehyde is chemoselectively reduced while the ketone survives.

Correct: (C)

A ketone plus a secondary amine under dehydrating acidic conditions forms an enamine, with the double bond between the α-carbon and the former carbonyl carbon.

Correct: (D)

Grignard addition of a methyl group to the ketone carbonyl gives a tertiary alkoxide, which protonates to the tertiary alcohol.

Correct: (D)

An aldehyde plus a 1,2-diol under acid with water removal gives the five-membered cyclic acetal (a 1,3-dioxolane).

Correct: (A)

Reactivity follows leaving-group ability: chloride (weakest base) > carboxylate > alkoxide > amide (strongest base, poorest leaving group).

Correct: (C)

Hydration is favored by minimal steric hindrance and the absence of electron-donating groups that stabilize the carbonyl; formaldehyde hydrates most extensively.

Correct: (A)

The Grignard reagent carboxylates to pentanoic acid (one carbon added), which forms the acid chloride and then the tertiary amide with dimethylamine.

Correct: (A)

Benzaldehyde cannot form an enolate, so the acetaldehyde enolate adds to it (crossed aldol); dehydration gives the conjugated aldehyde cinnamaldehyde.

Correct: (D)

One equivalent of DIBAL-H at low temperature stops at the aldehyde stage; LiAlH4 reduces all the way to the primary alcohol.

Correct: (B)

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

Correct: (B)

SN2 rates decrease with steric hindrance at the α-carbon, so the unbranched primary bromide reacts fastest.

Correct: (D)

With a small base, Zaitsev's rule applies: the most substituted alkene predominates, and the E isomer is favored over Z for steric reasons.

Correct: (C)

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

Correct: (A)

Debromination requires the two C–Br bonds to be anti-periplanar; anti elimination from the meso dibromide stereospecifically gives the trans alkene.

Correct: (B)

Under acidic conditions the epoxide oxygen is protonated and the nucleophile attacks the more substituted carbon, placing the methoxy group at C-2.

Correct: (C)

Pairing the tert-butoxide with a primary alkyl halide allows clean SN2; the reverse pairing with a tertiary halide would give mainly elimination.

Correct: (D)

Loss of water gives a secondary carbocation that undergoes a methyl shift to the more stable tertiary carbocation before bromide capture.

Correct: (A)

The rate is independent of nucleophile concentration, consistent with unimolecular ionization of the tertiary bromide in the rate-determining step.

Correct: (A)

Priorities are OH(1) > CH2CH3(2) > CH3(3) > H(4); the 1→2→3 sequence (right → bottom → top) is clockwise (apparent R), but H lies on a horizontal (forward) bond, so the assignment inverts to S.

Correct: (C)

The (2R,3S) isomer of 2,3-butanediol has an internal plane of symmetry, making it achiral despite having two stereocenters.

Correct: (D)

In the Fischer convention, the L enantiomer of glyceraldehyde has the hydroxyl group on the left with the aldehyde carbon at the top.

Correct: (A)

ee = (24/80) × 100% = 30%, so the major enantiomer is (100 + 30)/2 = 65% (R).

Correct: (A)

They have the same connectivity but differ in configuration at only one of two stereocenters, so they are stereoisomers that are not mirror images.

Correct: (D)

Alkene stability increases with substitution; the tetrasubstituted alkene is more stable than any di- or trisubstituted isomer, and E is more stable than Z.

Correct: (C)

There are a pair of enantiomers (3R,4R and 3S,4S) plus one meso form (3R,4S), for a total of three stereoisomers.

Correct: (C)

Each doubly bonded carbon becomes a carbonyl retaining its substituents, so (CH3)2C= gives acetone and =CHCH3 gives acetaldehyde.

Correct: (A)

The OH group adds to the more substituted alkene carbon (Markovnikov) through a mercurinium ion, so no carbocation rearrangement occurs.

Correct: (B)

OsO4 adds two OH groups syn to the same face; syn addition to a trans alkene gives the meso diastereomer, while anti addition via an epoxide gives the racemic pair.

Correct: (D)

The kinetic endo product is favored by secondary orbital overlap between the diene π system and the anhydride carbonyls in the transition state.

Correct: (A)

Markovnikov hydration places the OH on the more substituted carbon, giving an enol that tautomerizes to the methyl ketone, acetone.

Correct: (C)

More substituted double bonds are lower in energy, so the trisubstituted alkene (I) is most stable, and the endocyclic disubstituted alkene (III) is slightly more stable than the exocyclic methylene (II).

Correct: (B)

Oxidative cleavage opens the ring, converting the methyl-bearing carbon to a methyl ketone and the other alkene carbon to a carboxylic acid, giving CH3CO(CH2)4COOH.

Correct: (D)

The activating methyl group outcompetes the deactivating acetyl group, so nitration occurs ortho to the methyl group (positions 2 and 6, which are equivalent).

Correct: (A)

The nitro group is a strongly deactivating meta director, so further electrophilic substitution occurs meta to it.

Correct: (B)

Sulfonation is reversible, so at high temperature the thermodynamic 2-sulfonic acid predominates over the kinetic 1-isomer formed at low temperature.

Correct: (C)

Diazotization gives benzenediazonium chloride, which the Sandmeyer reaction with CuBr converts to bromobenzene with loss of nitrogen gas.

Correct: (C)

Nitro groups stabilize the phenoxide by electron withdrawal, with the para isomer additionally stabilized by resonance, while the methyl group destabilizes the anion.

Correct: (B)

SNAr requires nitro groups ortho or para to the leaving group to stabilize the anionic Meisenheimer intermediate; two nitro groups activate far more than one.

Correct: (A)

Symmetry makes the two methyl groups equivalent and all four aromatic protons equivalent, giving just two signals.

Correct: (C)

A plane of symmetry through the carbonyl carbon and C4 makes C2/C6 and C3/C5 equivalent, giving four distinct carbon environments.

Correct: (D)

Alpha cleavage next to the carbonyl gives the resonance-stabilized acylium ion CH3CO+ at m/z 43.

Correct: (A)

The carbonyl stretch near 1740 cm⁻¹ plus a C–O stretch near 1200 cm⁻¹, with no aldehydic C–H stretches or broad O–H band, indicates an ester.

Correct: (B)

The CH2 group is coupled to three equivalent methyl protons, giving a quartet by the n + 1 rule.

Correct: (C)

The central carbon uses two sp orbitals for the sigma framework and two perpendicular p orbitals for the π bonds, so the terminal CH2 groups lie in perpendicular planes.

Correct: (C)

In the cis-1,4 isomer one substituent must be axial, so the very bulky tert-butyl group takes the equatorial position and forces the methyl group axial.

Correct: (A)

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

Correct: (C)

The central carbon is sp hybridized, giving a linear C=C=C arrangement.

Correct: (B)

The amide ion's conjugate acid (NH3, pKa ≈ 35) is the weakest acid listed, making NH2⁻ the strongest base.

Correct: (D)

The electron-withdrawing chlorine stabilizes the carboxylate by induction, making chloroacetic acid stronger than acetic acid, phenol, and ethanol.

Correct: (A)

At a pH below the carboxyl pKa, both the amino group (as NH3⁺) and the carboxyl group (as neutral COOH) are protonated.

Correct: (C)

The vinyl radical places the unpaired electron in an sp2 orbital with high s-character and gains no resonance stabilization, making it less stable than even primary alkyl radicals.

Correct: (B)

NBS maintains a low, steady concentration of Br2, favoring allylic hydrogen abstraction and substitution at the allylic position.

Correct: (D)

A propagation step consumes one radical and generates a new one, continuing the chain; (A) is initiation and (B) and (C) are termination steps.

Correct: (A)

Each addition occurs head-to-tail so the growing radical stays on the more stable carbon adjacent to chlorine, giving the regular –CH2–CHCl– repeat unit.

Correct: (B)

The diacid and diamine link by amide formation with loss of water at each step, the hallmark of condensation polymerization, while the alkenes undergo addition polymerization.