ACS Organic Chemistry Practice Test 12

Correct: (B)

The first equivalent adds to give acetophenone after collapse of the tetrahedral intermediate, and the second equivalent adds to the ketone to give a tertiary alkoxide, which protonates to 2-phenyl-2-propanol.

Correct: (C)

Benzaldehyde has no alpha-hydrogens so it acts only as the electrophile; the acetaldehyde enolate adds to it, and dehydration gives the conjugated aldehyde cinnamaldehyde (Claisen-Schmidt reaction).

Correct: (A)

Intramolecular Claisen (Dieckmann) condensation of this 1,7-diester favors 6-membered ring closure; without heating, the beta-keto ester does not decarboxylate.

Correct: (B)

NaBH4 is a mild hydride donor that reduces aldehydes and ketones but is unreactive toward esters, whereas LiAlH4 and Grignard reagents reduce both functional groups.

Correct: (B)

Stabilized ylides do olefinate ketones (more slowly than aldehydes), giving the alpha,beta-unsaturated ester with an exocyclic double bond.

Correct: (A)

Two electron-donating alkyl groups plus steric hindrance make acetone least reactive; aldehydes are more reactive than ketones, and phenyl resonance deactivates benzaldehyde relative to formaldehyde.

Correct: (D)

Enamine formation activates the alpha-carbon toward SN2 alkylation (Stork enamine reaction), and acidic hydrolysis regenerates the ketone bearing one new alpha-methyl group.

Correct: (C)

The enolate of ethyl acetate attacks the carbonyl carbon of ethyl benzoate, which has no alpha-hydrogens and thus cannot self-condense, displacing ethoxide.

Correct: (B)

CeCl3 coordinates the carbonyl oxygen, hardening the carbonyl carbon so hydride adds 1,2 to give the allylic alcohol rather than the conjugate-addition product.

Correct: (A)

One equivalent of Grignard reagent adds to the nitrile carbon, and acidic hydrolysis of the resulting imine salt furnishes the ketone.

Correct: (C)

Dehydration occurs preferentially toward the aldehyde to give the more substituted, conjugated alpha,beta-unsaturated aldehyde (Zaitsev-type regioselectivity reinforced by conjugation).

Correct: (D)

Two bulky tert-butyl groups both donate electron density (stabilizing the carbonyl) and sterically crowd the tetrahedral gem-diol, strongly disfavoring hydration.

Correct: (A)

Aqueous acid hydrolyzes the enamine (the reverse of its formation), regenerating the parent ketone and the secondary amine.

Correct: (B)

These are Wolff-Kishner conditions, which fully reduce the ketone carbonyl to a methylene group, giving the ethyl-substituted arene.

Correct: (C)

The rate-determining step is unimolecular ionization, so the rate follows carbocation stability: tertiary > secondary > primary.

Correct: (D)

The bulky base abstracts the least hindered primary beta-hydrogen, giving the less substituted Hofmann alkene rather than the Zaitsev product.

Correct: (A)

Azide displaces iodide by a concerted SN2 backside attack, which proceeds with complete inversion of configuration from S to R.

Correct: (B)

Ionization gives a secondary carbocation that undergoes a methyl shift to the more stable tertiary carbocation; Zaitsev deprotonation then gives the tetrasubstituted alkene.

Correct: (D)

The slow step of this SN1 reaction is unimolecular ionization of the alkyl halide, so the rate depends only on the substrate concentration.

Correct: (C)

A more polar protic solvent better stabilizes the developing charges in the ionization transition state, accelerating the rate-determining step.

Correct: (A)

Zinc debromination is a stereospecific anti elimination; the (2R,3R) diastereomer can only align the C-Br bonds anti with the methyl and ethyl groups trans, giving trans-2-pentene.

Correct: (B)

SN2 is fastest at an unhindered primary carbon with the best leaving group; the neopentyl iodide is severely sterically hindered despite being primary.

Correct: (D)

A small base gives the more substituted Zaitsev alkene (2-pentene), and the trans isomer predominates because the anti transition state placing the alkyl groups opposite is lowest in energy.

Correct: (B)

In a Fischer projection horizontal bonds come forward; tracing priorities 1->2->3 (OH->Et->Me) goes counterclockwise, but with H projecting forward the sense inverts to (R).

Correct: (C)

(2R,3S)-2,3-butanediol possesses an internal plane of symmetry, making it achiral; the pentanediol analog lacks such symmetry because its two ends differ.

Correct: (A)

The anti staggered conformation minimizes both torsional strain and steric repulsion by placing the bulky methyl groups as far apart as possible.

Correct: (D)

Enantiomeric excess = (6.8/13.6) x 100% = 50%, and the major enantiomer constitutes (100 + 50)/2 = 75% of the mixture.

Correct: (C)

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

Correct: (A)

Only 3-chloro-1-butene (CH2=CH-CHCl-CH3) has a carbon bonded to four different groups; each of the others possesses a plane of symmetry.

Correct: (B)

The (2R,3R)/(2S,3S) pair are chiral enantiomers and optically active, whereas the meso (2R,3S) form has an internal plane of symmetry and is achiral despite having two stereocenters.

Correct: (C)

Hydroboration–oxidation adds H and OH syn with anti-Markovnikov regiochemistry, placing OH on the less substituted terminal carbon; no carbocation rearrangement occurs.

Correct: (B)

Oxidative ozonolysis cleaves each alkyne carbon to a carboxylic acid: the phenyl-bearing carbon gives benzoic acid and the methyl-bearing carbon gives acetic acid.

Correct: (A)

With a 2-substituted diene the "para rule" places the dienophile substituent para to the diene substituent, giving the 1,4-disubstituted cyclohexene.

Correct: (C)

More alkyl substituents stabilize the double bond by hyperconjugation, and for 1,2-disubstituted alkenes the trans isomer is more stable than the cis.

Correct: (B)

Lindlar's catalyst delivers both hydrogens syn to the same face of the alkyne, giving the cis (Z) alkene; dissolving-metal reduction would give the trans isomer.

Correct: (D)

Protonation gives a secondary carbocation that undergoes a hydride shift to the more stable tertiary carbocation before capture by chloride.

Correct: (D)

Markovnikov hydration forms the vinyl cation at the more substituted alkyne carbon (C3), so the carbonyl ends up at C3 rather than at the methyl-ketone position.

Correct: (C)

The methyl group is an electron-donating activator that increases the ring's nucleophilicity toward electrophiles; nitro and cyano groups strongly deactivate.

Correct: (A)

The methyl group is ortho/para-directing and the nitro group is meta-directing; both direct to C2/C6, which are equivalent by symmetry.

Correct: (D)

OH is a strong activator, CH3 a weak activator, and NO2 a strong deactivator of the aromatic ring.

Correct: (A)

The ortho and para nitro groups stabilize the Meisenheimer intermediate, allowing SNAr displacement of fluoride by the amine nucleophile.

Correct: (B)

The acetyl group strongly deactivates the ring, and strongly deactivated arenes do not react under Friedel–Crafts conditions.

Correct: (D)

The initially formed primary carbocation rearranges by hydride shift to the more stable secondary carbocation before electrophilic attack on the ring.

Correct: (A)

A monosubstituted benzene plus an ethyl group whose CH2 is deshielded to δ 4.0 by oxygen matches Ph–O–CH2CH3.

Correct: (B)

Equal-intensity M and M+2 peaks are the signature of one bromine atom (⁷⁹Br/⁸¹Br natural abundance).

Correct: (B)

Threefold symmetry makes the three methyl carbons equivalent, the three ring CH carbons equivalent, and the three quaternary ring carbons equivalent: three distinct environments.

Correct: (D)

The IR band indicates a ketone; the acetyl singlet plus the isopropyl septet/doublet pattern matches CH3COCH(CH3)2.

Correct: (C)

The 3H singlet is an OCH3 group, and the septet/doublet pair shows an isopropyl group attached to oxygen.

Correct: (A)

The central allene carbon participates in two cumulative double bonds, requiring two unhybridized p orbitals — sp hybridization.

Correct: (C)

For cis-1,2-disubstitution every chair conformation has one axial and one equatorial substituent, and the two chairs are degenerate.

Correct: (B)

Cyclobutadiene is planar with 4n (n = 1) π electrons; cyclooctatetraene is nonplanar (nonaromatic), while benzene and the cyclopentadienyl anion are aromatic.

Correct: (D)

Extending conjugation narrows the HOMO–LUMO gap, shifting the π→π* absorption to longer wavelength.

Correct: (C)

Lower pKa means a stronger acid: carboxylic acids (pKa ~5) are stronger than phenols (~10), which are stronger than alcohols (~16).

Correct: (C)

The aliphatic amine's lone pair is not delocalized into a ring; anilines are weakened by resonance, and the nitro group weakens p-nitroaniline further.

Correct: (A)

Equilibrium favors the weaker acid: acetic acid (pKa 4.8) is far stronger than water (pKa 15.7), so hydroxide deprotonates it quantitatively.

Correct: (D)

Resonance stabilization (benzyl > allyl) outweighs hyperconjugation, and among alkyl radicals more substitution means greater stability.

Correct: (A)

Bromine is highly selective for the weakest C–H bond; 2,3-dimethylbutane has two equivalent tertiary hydrogens, and abstraction gives the tertiary bromide.

Correct: (C)

Radical addition polymerization of vinyl chloride (CH2=CHCl) gives the repeat unit –[CH2–CHCl]n–.

Correct: (A)

Cleaving the polymer backbone at the repeat unit gives the alkene CH2=C(CH3)2, i.e., 2-methylpropene (isobutylene).

Correct: (A)

PET is the condensation polymer of benzene-1,4-dicarboxylic acid (terephthalic acid) and ethane-1,2-diol (ethylene glycol); option B gives nylon-6,6.