ACS Organic Chemistry Practice Test 18
60 original exam-style multiple-choice questions with answers & explanations — covering every major topic tested on the ACS Organic Chemistry exam.
Note: These are original practice questions written in the ACS exam style — not reproduced exam questions. Use them to test your readiness for the real ACS exam.
📱 How to use: Try each question first, then click “Show Answer” to check. Covers: carbonyl chemistry, substitution & elimination, stereochemistry, alkenes & alkynes, aromatics, spectroscopy and more.
🧪 Carbonyl Chemistry
Q1What is the major product of the self-aldol ADDITION of propanal (CH3CH2CHO) in dilute aqueous NaOH at room temperature (no heating)?
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Correct: (B)
The enolate formed at C2 of one propanal molecule adds to the carbonyl carbon of a second propanal; under mild conditions without heat the beta-hydroxy aldehyde (3-hydroxy-2-methylpentanal) is isolated before dehydration.
Q2Acetophenone is treated with benzaldehyde and NaOH with heating. What is the major isolated product?
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Correct: (D)
This Claisen-Schmidt condensation first forms the aldol adduct (C), but heating drives dehydration to the conjugated enone, 1,3-diphenylprop-2-en-1-one.
Q3Ethyl propanoate (CH3CH2CO2Et) is treated with sodium ethoxide followed by aqueous acid workup. What is the major product?
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Correct: (A)
Claisen condensation between two molecules of ethyl propanoate gives the beta-keto ester CH3CH2COCH(CH3)CO2Et, ethyl 2-methyl-3-oxopentanoate.
Q4Diethyl pimelate (EtO2C-CH2-CH2-CH2-CH2-CH2-CO2Et) undergoes Dieckmann cyclization with NaOEt. What ring size is formed in the major product?
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Correct: (C)
The seven-atom diester chain closes between C2 and C7, giving a six-membered ring (five- and six-membered rings are strongly favored in the Dieckmann reaction).
Q5Ethyl benzoate is treated with 2 equivalents of methylmagnesium bromide followed by aqueous acid workup. What is the major organic product?
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Correct: (C)
The first equivalent of Grignard reagent gives acetophenone after ethoxide expulsion, and the second equivalent adds to that ketone, giving the tertiary alcohol PhC(OH)(CH3)2 after workup.
Q6Which reagent converts ethyl benzoate (PhCO2Et) to benzaldehyde (PhCHO) without reducing it further?
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Correct: (C)
DIBAL-H at low temperature reduces esters only as far as the aldehyde; NaBH4 does not reduce esters at all, and LiAlH4 reduces esters all the way to primary alcohols.
Q7Benzaldehyde reacts with the stabilized ylide Ph3P=CHCO2Et. What is the major alkene product?
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Correct: (A)
Stabilized ylides (with electron-withdrawing groups on the ylide carbon) give predominantly the E-alkene, whereas non-stabilized alkyl ylides favor the Z-alkene.
Q8Cyclohexanone is converted to its pyrrolidine enamine, treated with methyl iodide, then hydrolyzed with aqueous acid. What is the major product?
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Correct: (D)
In the Stork enamine alkylation, the nucleophilic alpha-carbon of the enamine attacks methyl iodide; acidic hydrolysis regenerates the ketone, now methylated at the alpha position.
Q9Methyl 4-oxopentanoate (CH3COCH2CH2CO2CH3) is first protected as the ethylene ketal, then treated with LiAlH4 followed by aqueous acid workup. What is the final major product?
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Correct: (A)
The ketal protects the ketone from LiAlH4, so only the ester is reduced to a primary alcohol; aqueous acid then removes the ketal to reveal CH3COCH2CH2CH2OH.
Q10Which carbonyl compound forms the most stable hydrate in water?
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Correct: (C)
Alkyl groups donate electron density that stabilizes the carbonyl and disfavors hydrate formation; formaldehyde, with no alkyl substituents, has the largest hydration equilibrium constant.
Q11Benzaldehyde is treated with concentrated NaOH. What are the major organic products?
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Correct: (B)
Benzaldehyde has no alpha-hydrogens, so it undergoes the Cannizzaro reaction: one molecule is reduced to benzyl alcohol while another is oxidized to benzoate.
Q123-methylcyclohex-2-en-1-one is treated with NaBH4 and CeCl3 in methanol. What is the major product?
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Correct: (D)
Luche reduction (NaBH4/CeCl3) delivers selective 1,2-reduction of the enone, giving the allylic alcohol while leaving the C=C bond intact.
Q132-methylcyclohexanone is treated with LDA at -78 C, then benzyl bromide. What is the major product?
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Correct: (A)
LDA at -78 C removes the less hindered alpha-proton irreversibly, giving the kinetic (less substituted) enolate at C6, which is then benzylated.
Q14Ethyl acetoacetate is treated with NaOEt, then ethyl iodide, then aqueous NaOH with heating. What is the major final product?
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Correct: (B)
Alpha-alkylation gives CH3COCH(Et)CO2Et; hydrolysis and heating cause decarboxylation of the beta-keto acid, losing CO2 to give CH3COCH2CH2CH3.
⚙ Substitution & Elimination
Q15For the solvolysis of tert-butyl bromide in aqueous acetone, which rate law is observed?
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Correct: (C)
This is an SN1 reaction whose rate-determining step is ionization of the alkyl halide alone, so the rate depends only on the substrate concentration (first order).
Q16Which salt gives the fastest SN2 reaction with methyl iodide in methanol?
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Correct: (A)
In polar protic solvents, small anions are heavily solvated; the largest, most polarizable anion (I-) is least solvated and therefore the most nucleophilic.
Q17Which is the best leaving group?
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Correct: (D)
Leaving-group ability parallels the strength of the conjugate acid: HI is the strongest acid listed, so I- is the weakest base and the best leaving group.
Q182-bromo-2-methylbutane is treated with potassium tert-butoxide (t-BuOK). What is the major alkene product?
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Correct: (B)
The bulky base abstracts the least hindered beta-proton (from a methyl group), giving the less substituted Hofmann alkene rather than the more substituted Zaitsev product.
Q193,3-dimethyl-2-butanol is heated with concentrated H2SO4. What is the major alkene product?
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Correct: (C)
Loss of water gives a secondary carbocation that undergoes a methyl shift to the more stable tertiary carbocation; deprotonation then gives the tetrasubstituted alkene 2,3-dimethyl-2-butene.
Q20(R)-2-bromobutane reacts with NaCN in DMSO. What is the stereochemical outcome at C2?
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Correct: (D)
Cyanide attacks from the back side in an SN2 reaction, so the configuration at C2 inverts from R to S.
Q21(R)-2-bromobutane undergoes solvolysis in aqueous acetone. What is the stereochemical outcome of the substitution product?
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Correct: (A)
Ionization gives a planar, achiral carbocation at C2 that water attacks from either face with equal probability, producing a racemic alcohol.
Q22In the E2 dehydrohalogenation of 2-bromobutane, which arrangement of the C2-Br bond and a C3-H bond reacts fastest?
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Correct: (B)
The anti-periplanar geometry allows optimal overlap between the breaking C-H sigma bond and the C-Br sigma* orbital as the pi bond forms.
Q23Which solvent change most accelerates the SN1 solvolysis of tert-butyl chloride?
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Correct: (C)
Increasing the water content raises the ionizing power of the polar protic medium, which stabilizes the charge-separated SN1 transition state and the carbocation intermediate.
🔬 Stereochemistry
Q24A stereocenter in butan-2-ol is drawn with OH on a solid wedge (up), H on a hashed wedge, CH2CH3 to the left, and CH3 to the right. What is the configuration?
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Correct: (D)
Priorities are OH (1) > CH2CH3 (2) > CH3 (3) > H (4); with H on the hashed wedge, the 1->2->3 sequence (up -> left -> right) is counterclockwise, giving S.
Q25In a Fischer projection of lactic acid (CH3CH(OH)COOH), COOH is on top, CH3 on the bottom, OH on the right, and H on the left. What is the configuration at the stereocenter?
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Correct: (A)
With priorities OH (1) > COOH (2) > CH3 (3), the 1->2->3 order appears counterclockwise (S), but H is on a horizontal bond (forward), so the true assignment is inverted to R.
Q26Which of the following compounds is meso?
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Correct: (C)
The (2R,3S) isomer has an internal plane of symmetry, making it achiral despite having two stereocenters.
Q27Which compound is the enantiomer of (2R,3R)-3-bromo-2-butanol?
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Correct: (B)
The enantiomer has every stereocenter inverted, so (2R,3R) becomes (2S,3S); the other stereoisomers are diastereomers.
Q28A sample of butan-2-ol has an observed specific rotation of +15 degrees; pure (R)-butan-2-ol has a specific rotation of +60 degrees. What is the enantiomeric excess and the percentage of the major enantiomer?
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Correct: (D)
Enantiomeric excess = (15/60) x 100% = 25%; the major enantiomer makes up (100 + 25)/2 = 62.5% of the mixture.
Q29The anti staggered conformation and a gauche staggered conformation of butane (viewed down the C2-C3 bond) are best described as what?
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Correct: (A)
They differ only by rotation about a single bond, interconvert rapidly at room temperature, and have identical connectivity and configuration.
Q30Which of the following is optically active?
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Correct: (B)
The (2R,3R) enantiomer is chiral with no plane of symmetry, so a pure sample rotates plane-polarized light; the meso compound, the racemic mixture, and the achiral cyclopropane do not.
🧬 Alkenes & Alkynes
Q31Ozonolysis of 3-methyl-1-butene followed by reductive workup (Zn, H2O) gives which pair of products?
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Correct: (C)
Reductive ozonolysis cleaves each doubly bonded carbon to a carbonyl: the terminal CH2 becomes formaldehyde and the (CH3)2CH- fragment becomes 2-methylpropanal.
Q32Ozonolysis of 1-methylcyclopentene followed by oxidative workup (H2O2) gives which product?
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Correct: (A)
Oxidative cleavage opens the ring; the substituted ring carbon (no H) becomes a ketone and the CH carbon becomes a carboxylic acid, giving CH3COCH2CH2CH2COOH.
Q33Treatment of 1-methylcyclohexene with (1) BH3-THF, (2) H2O2, NaOH gives which major product?
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Correct: (D)
Hydroboration-oxidation adds H and OH syn to the same face with anti-Markovnikov regiochemistry (OH on the less substituted carbon), placing the methyl and OH groups trans; attack from either face gives the racemate.
Q34Treatment of 3,3-dimethyl-1-butene with (1) Hg(OAc)2, H2O, (2) NaBH4 gives which major product?
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Correct: (B)
Oxymercuration-demercuration gives Markovnikov addition of water through a mercurinium ion with no free carbocation, so no methyl shift occurs.
Q35The major product of the Diels-Alder reaction between 1-methoxy-1,3-butadiene and methyl acrylate is best described as:
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Correct: (A)
The ortho rule places the diene donor (OMe) adjacent to the dienophile acceptor (ester), and the endo rule places the ester cis to the new double bond.
Q36Which reagent converts 3-hexyne to (E)-3-hexene?
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Correct: (C)
Dissolving-metal reduction gives the trans (E) alkene; Lindlar gives the cis (Z) alkene, excess Pt gives hexane, and Hg2+/H2O gives a ketone.
Q372-Methyl-2-butene is treated separately with (i) HBr in the presence of peroxides and (ii) Br2 in water. The major products are:
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Correct: (D)
The peroxide effect reverses HBr regiochemistry (Br adds to the less substituted carbon), while halohydrin formation goes through a bromonium ion opened by water at the more substituted carbon.
💧 Aromatic Chemistry
Q38Which substituent is an ortho/para director but deactivates the ring toward electrophilic aromatic substitution?
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Correct: (B)
Halogens donate electron density by resonance (ortho/para directing) but withdraw inductively, with the inductive effect dominating so the ring is deactivated.
Q39Rank the following from fastest to slowest rate of nitration with HNO3/H2SO4: toluene, benzene, chlorobenzene, nitrobenzene.
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Correct: (D)
The activating methyl group speeds nitration, benzene is the reference, deactivating chlorine slows it, and the strongly deactivating nitro group slows it most.
Q40Which compound reacts fastest with NaOH via nucleophilic aromatic substitution?
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Correct: (A)
SNAr requires nitro groups ortho or para to the leaving group to stabilize the Meisenheimer complex by resonance; two such nitro groups give the fastest reaction, while meta nitro groups provide no resonance stabilization.
Q41Treatment of benzene with 1-chlorobutane and AlCl3 gives as the major product:
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Correct: (C)
The primary butyl cation rearranges by hydride shift to the more stable secondary cation before electrophilic attack on the ring.
Q42Aniline is treated with NaNO2/HCl at 0 C, followed by CuBr. The major product is:
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Correct: (B)
Diazotization forms the benzenediazonium ion, and the Sandmeyer reaction with CuBr replaces the diazonium group with bromine.
Q43Oxidation of isobutylbenzene with hot KMnO4 gives:
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Correct: (B)
Any alkylbenzene with at least one benzylic hydrogen is oxidized completely to benzoic acid, and isobutylbenzene has a benzylic CH2 group.
🔍 Spectroscopy
Q44A compound with formula C4H8O2 shows three 1H NMR signals: a singlet (3H) at 3.7 ppm, a quartet (2H) at 2.3 ppm, and a triplet (3H) at 1.1 ppm. The compound is:
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Correct: (A)
The 3.7 ppm singlet (3H) is characteristic of a methyl ester OCH3 group; ethyl acetate would show its singlet near 2.0 ppm and the acids would show a broad OH signal.
Q45How many signals appear in the proton-decoupled 13C NMR spectrum of 1,3,5-trimethylbenzene?
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Correct: (C)
Symmetry makes the three methyl carbons equivalent, the three substituted ring carbons equivalent, and the three unsubstituted ring carbons equivalent.
Q46Which IR absorption is characteristic of the C-H bond of a terminal alkyne?
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Correct: (B)
The terminal alkyne C-H stretch is a sharp band near 3300 cm-1; 2250 cm-1 is the C triple bond stretch, 1715 cm-1 is C=O, and the broad 2500-3300 cm-1 band is a carboxylic acid O-H.
Q47A compound shows molecular ion peaks at m/z 156 and 158 of nearly equal intensity, and a prominent fragment at m/z 77. The compound is most likely:
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Correct: (A)
Equal-intensity M and M+2 peaks indicate one bromine atom (79Br/81Br), and m/z 77 is the phenyl cation C6H5+.
Q48A compound C5H10O shows a strong IR band at 1725 cm-1 with weak bands at 2720 and 2820 cm-1. Its 1H NMR shows only two singlets: 9H at 1.1 ppm and 1H at 9.6 ppm. The compound is:
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Correct: (D)
The 2720/2820 cm-1 bands and the 9.6 ppm singlet identify an aldehyde, and the 9H singlet identifies a tert-butyl group with no neighboring protons.
📐 Conformations & Bonding
Q49What is the most stable chair conformation of cis-1-tert-butyl-3-methylcyclohexane?
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Correct: (B)
Cis-1,3 substituents cannot both be equatorial; the very bulky tert-butyl group occupies the equatorial position, forcing the methyl group axial.
Q50What is the hybridization of the central carbon atom in allene (CH2=C=CH2)?
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Correct: (A)
The central carbon uses two p orbitals to form two pi bonds, leaving two sp hybrid orbitals for its two sigma bonds.
Q51Which of the following species is aromatic?
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Correct: (D)
The cyclopentadienyl anion has 6 pi electrons in a planar, cyclic, conjugated system (Huckel 4n+2); cyclobutadiene and the cyclopentadienyl cation are antiaromatic (4 pi electrons) and cyclooctatetraene is nonplanar and nonaromatic.
Q52In which compound is the nitrogen atom sp2 hybridized?
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Correct: (C)
The amide nitrogen lone pair is delocalized into the carbonyl pi system, making nitrogen planar (sp2); the amines are pyramidal (sp3) and the nitrile nitrogen is sp.
🧫 Acids & Bases
Q53Rank the following acids from strongest to weakest: acetic acid, p-nitrophenol, phenol, ethanol.
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Correct: (B)
The carboxylic acid (pKa ~4.8) is strongest, the nitro group stabilizes the phenoxide by resonance making p-nitrophenol (~7.2) stronger than phenol (~10), and the alcohol (~16) is weakest.
Q54Rank the following bases from strongest to weakest: the amide ion, the acetylide ion, the hydroxide ion, the acetate ion.
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Correct: (D)
Base strength increases as the conjugate acid gets weaker: ammonia (pKa 35) > acetylene (25) > water (15.7) > acetic acid (4.8).
Q55Based on the pKa values of their conjugate acids, which is the strongest base? (HF: 3.2; methanol: 15.5; acetylene: 25; HCl: -7)
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Correct: (A)
The strongest base has the weakest conjugate acid, i.e., the highest pKa, which is acetylene at 25.
⚡ Radicals
Q56Rank the following carbon radicals from most stable to least stable: tert-butyl radical, isopropyl radical, n-propyl radical, methyl radical.
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Correct: (C)
Radical stability increases with alkyl substitution (tertiary > secondary > primary > methyl) due to hyperconjugation.
Q57Bromination of 2-methylbutane gives almost exclusively the tertiary bromide, while chlorination gives a mixture of all possible chlorides. The best explanation is:
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Correct: (D)
By the Hammond postulate, the endothermic bromine abstraction has a late transition state resembling the carbon radical, so radical stability differences matter greatly; the exothermic chlorine abstraction has an early transition state and is unselective.
Q58In the free-radical chlorination of methane, which of the following is a propagation step?
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Correct: (D)
A propagation step consumes one radical and generates a new one; (A) is initiation, (B) is termination, and (C) does not occur under these conditions.
🔗 Polymers
Q59The polymer with repeat unit -[CH2-CHCl]-n is made from which monomer?
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Correct: (A)
Addition polymerization of vinyl chloride (chloroethene) gives poly(vinyl chloride) with the -[CH2-CHCl]- repeat unit.
Q60Which polymer is formed by condensation polymerization?
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Correct: (C)
Nylon-6,6 forms from a diamine and a diacid with loss of water; the others form by addition across carbon-carbon double bonds.
ANSWER KEY TEST 18: Q1 B, Q2 D, Q3 A, Q4 C, Q5 C, Q6 C, Q7 A, Q8 D, Q9 A, Q10 C, Q11 B, Q12 D, Q13 A, Q14 B, Q15 C, Q16 A, Q17 D, Q18 B, Q19 C, Q20 D, Q21 A, Q22 B, Q23 C, Q24 D, Q25 A, Q26 C, Q27 B, Q28 D, Q29 A, Q30 B, Q31 C, Q32 A, Q33 D, Q34 B, Q35 A, Q36 C, Q37 D, Q38 B, Q39 D, Q40 A, Q41 C, Q42 B, Q43 B, Q44 A, Q45 C, Q46 B, Q47 A, Q48 D, Q49 B, Q50 A, Q51 D, Q52 C, Q53 B, Q54 D, Q55 A, Q56 C, Q57 D, Q58 D, Q59 A, Q60 C
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