ACS Organic Chemistry Practice Test 3

Correct: (C)

This is a Claisen–Schmidt crossed aldol condensation; the initial beta-hydroxy ketone dehydrates under the basic, heated conditions to give the conjugated enone chalcone.

Correct: (A)

A Claisen condensation between two molecules of an ester bearing alpha hydrogens gives a beta-keto ester; two ethyl acetates self-condense to ethyl acetoacetate.

Correct: (D)

The Grignard reagent adds to the aldehyde carbonyl to give an alkoxide, which protonates to the secondary alcohol 1-phenylethanol.

Correct: (B)

NaBH4 reduces aldehydes much faster than ketones, so one equivalent selectively reduces the aldehyde to a primary alcohol while the ketone survives.

Correct: (C)

The Wittig reagent converts the ketone carbonyl into a C=C bond, replacing C=O with C=CH2 to give 2-methylpropene.

Correct: (A)

Intramolecular Claisen condensation of this six-carbon diester closes a five-membered ring, giving the cyclic beta-keto ester.

Correct: (D)

The strongly electron-withdrawing CCl3 group destabilizes the carbonyl and stabilizes the gem-diol, so chloral hydrate dominates at equilibrium.

Correct: (B)

A ketone plus a secondary amine forms an enamine (not an imine), giving the C=C–N product 1-(cyclohex-1-en-1-yl)pyrrolidine.

Correct: (A)

LiAlH4 is a strong hydride donor that reduces esters all the way to primary alcohols; the reaction does not stop at the aldehyde stage.

Correct: (C)

The propanal enolate attacks a second propanal carbonyl, forming the beta-hydroxy aldehyde 3-hydroxy-2-methylpentanal.

Correct: (D)

Grignard addition gives 1-methylcyclohexanol, dehydration gives 1-methylcyclohexene, and Br2 adds across the double bond to give the vicinal dibromide.

Correct: (B)

Grignard reagents add twice to esters: the first addition gives a ketone intermediate that reacts with the second equivalent to give the tertiary alcohol Ph2C(OH)CH3.

Correct: (A)

Deprotonation alpha to the ketone at C5 followed by attack on the aldehyde closes a favored five-membered ring, which dehydrates to the cyclopentenone.

Correct: (C)

A ketone plus a diol under acid with water removal gives the cyclic ketal (1,4-dioxaspiro[4.5]decane), a common carbonyl protecting group.

Correct: (A)

The SN1 rate-determining step is unimolecular ionization of the alkyl halide, so the rate law is first order in substrate only.

Correct: (B)

SN2 reactivity is governed by steric hindrance at the electrophilic carbon: methyl > primary > secondary > tertiary.

Correct: (D)

First-order dependence on substrate only plus a substitution product identifies an SN1 mechanism with rate-determining carbocation formation.

Correct: (C)

The bulky base favors abstraction of the least hindered beta-hydrogen (Hofmann orientation), giving 1-butene as the major product.

Correct: (A)

Loss of water gives a secondary carbocation that undergoes a methyl shift to the more stable tertiary carbocation, which deprotonates to the tetrasubstituted alkene.

Correct: (B)

Leaving-group ability increases down the halogen group as the anion becomes larger and more stable; iodide is the best leaving group listed.

Correct: (D)

Polar protic solvents stabilize the carbocation intermediate and the leaving group in the SN1 transition state, accelerating ionization.

Correct: (C)

Cyanide is a strong nucleophile in polar aprotic DMSO, so SN2 occurs with complete inversion of configuration at the stereocenter.

Correct: (A)

Stability follows tertiary > resonance-stabilized allyl > secondary > primary, reflecting hyperconjugation and resonance delocalization.

Correct: (B)

Priorities are OH (1) > COOH (2) > CH3 (3) > H (4); a counterclockwise 1→2→3 sequence with the lowest priority in back is S.

Correct: (C)

They are stereoisomers that are not mirror images (the 2R,3S isomer is meso), so they are diastereomers.

Correct: (D)

(2R,3S)-2,3-dichlorobutane has stereocenters but an internal plane of symmetry, making it achiral and meso.

Correct: (C)

The syn-eclipsed conformation maximizes torsional strain and methyl–methyl steric repulsion, making it the energy maximum.

Correct: (B)

(R)-2-butanol has a single stereocenter with no symmetry plane, so it is chiral and rotates plane-polarized light; the others are achiral.

Correct: (D)

Enantiomeric excess equals observed rotation divided by the rotation of the pure enantiomer: 6.8/13.6 × 100% = 50%.

Correct: (A)

With two stereocenters and no internal plane of symmetry, there are 2² = 4 stereoisomers, existing as two pairs of enantiomers.

Correct: (C)

Alkene stability increases with the degree of substitution through hyperconjugation, so the monosubstituted 1-butene is less stable than any of the disubstituted or trisubstituted isomers.

Correct: (A)

Cleaving CH2=C(CH3)CH2CH3 gives formaldehyde from the =CH2 end and 2-butanone, CH3COCH2CH3, from the trisubstituted end; the other isomers would give different carbonyl fragments.

Correct: (B)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry, placing OH on the less substituted carbon; the newly added H and OH end up cis, so the methyl and OH groups are trans to each other.

Correct: (D)

The peroxide effect gives anti-Markovnikov radical addition: Br· adds to the less substituted carbon to form the more stable tertiary radical, which then abstracts H to give 2-bromo-3-methylbutane.

Correct: (A)

A cyclic bromonium ion forms and water attacks the more substituted carbon from the back side, so OH ends up on C-2 and Br on C-1.

Correct: (C)

Secondary orbital interactions between the dienophile's carbonyl π systems and the diene favor the endo transition state, making the endo adduct the kinetic major product.

Correct: (B)

Dissolving-metal reduction of an alkyne proceeds by anti addition of two hydrogens via a vinyl radical anion intermediate, giving the trans alkene selectively.

Correct: (D)

All meta-directing groups are deactivating, and all activating groups are ortho/para directors, so no substituent is both meta-directing and activating.

Correct: (A)

The methoxy group is a strong activator that donates electron density by resonance, so anisole is nitrated far faster than toluene, benzene, or the deactivated nitrobenzene.

Correct: (C)

SNAr requires electron-withdrawing groups ortho or para to the leaving group to stabilize the anionic Meisenheimer intermediate, and two nitro groups activate far more than one or none.

Correct: (B)

The methyl group is an ortho/para director, and the para product predominates over the ortho product because of steric hindrance at the ortho positions.

Correct: (D)

The initially formed primary propyl carbocation rearranges by hydride shift to the more stable secondary isopropyl carbocation, which then alkylates benzene.

Correct: (C)

In strong acid the –NH2 group is protonated to –NH3+, which is a deactivating meta director, so nitration occurs mainly at the meta position.

Correct: (A)

In (CH3)2CHBr the single CH proton is split by six equivalent methyl protons into a septet, and the six methyl protons are split by the one CH proton into a doublet.

Correct: (C)

Symmetry makes the two methyl carbons equivalent, the two ipso ring carbons equivalent, and the four remaining ring carbons equivalent, giving three distinct carbon environments.

Correct: (D)

The ~3:1 ratio of M to M+2 matches the natural abundance of 35Cl and 37Cl, the classic isotope signature of a single chlorine atom.

Correct: (B)

The IR band indicates a ketone carbonyl, and the NMR pattern matches CH3COCH2CH3: an acetyl methyl singlet, a CH2 quartet, and a CH3 triplet.

Correct: (A)

The 5H aromatic signal plus two unsplit singlets fit C6H5CH2OCH3, and the absence of carbonyl and O–H bands rules out the alcohol and carbonyl alternatives.

Correct: (B)

The cyclopentadienyl anion is planar, cyclic, fully conjugated, and has 6 π electrons (4n+2, n=1), satisfying Hückel's rule.

Correct: (C)

The cis-1,3 relationship allows both methyl groups to be equatorial simultaneously after a ring flip, which is the lowest-energy chair.

Correct: (D)

The pyridine nitrogen is trigonal planar with its lone pair in an sp2 orbital, contributing one p electron to the aromatic sextet.

Correct: (A)

The cyclopentadienyl cation has 4 π electrons in a planar conjugated ring, making it antiaromatic rather than aromatic.

Correct: (C)

The carboxylate conjugate base is resonance-stabilized, making acetic acid (pKa ~4.7) far more acidic than phenol (~10) or the alcohols (~16–18).

Correct: (B)

Cyclohexylamine is an aliphatic amine whose conjugate acid has pKa ~10.6; the others are weakened by resonance delocalization of the nitrogen lone pair into an aromatic ring or carbonyl.

Correct: (D)

Basicity increases as the pKa of the conjugate acid increases, and NH3 (pKa ~35) is a far weaker acid than water (~15.7), acetic acid (~4.7), or HF (~3.2).

Correct: (C)

Radical stability increases with alkyl substitution through hyperconjugation, so the tertiary tert-butyl radical is the most stable of the choices.

Correct: (A)

Abstraction of the single tertiary hydrogen is fastest because the tertiary C–H bond is weakest and the resulting tertiary radical is most stable.

Correct: (B)

Propagation is the repeated addition of monomer to the radical chain end; coupling and disproportionation are termination steps.

Correct: (B)

The repeating unit –CH2–CH(C6H5)– is exactly what results from head-to-tail addition polymerization of styrene, CH2=CH(C6H5).

Correct: (C)

Nylon-6,6 forms by condensation of the six-carbon diamine (hexamethylenediamine) with the six-carbon diacid (adipic acid), releasing water.

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