ACS Organic Chemistry Practice Test: 50 Free Questions

Correct answer: D. 2-butene (E2, Zaitsev product)

2-Bromobutane is secondary, and methoxide is a strong, non-bulky base in a protic solvent, so E2 with the Zaitsev alkene (2-butene) predominates. The Hofmann product (1-butene) would require a bulky base.

From practice test 5 of our ACS Organic study guide (question 22). Original question, not an official exam item.

Correct answer: C. 2-chloro-2,3-dimethylhexane

Protonation forms a secondary carbocation at C3, and because C2 has no hydrogen, a methyl shift from C2 gives a tertiary carbocation at C2. Chloride adds at C2, yielding 2-chloro-2,3-dimethylhexane; the direct product 3-chloro-2,2-dimethylhexane would require no rearrangement.

From practice test 11 of our ACS Organic study guide (question 16). Original question, not an official exam item.

Correct answer: D. A methyl radical

A loss of 15 m/z units from the molecular ion is the characteristic loss of a methyl radical, CH3. Ethyl radical loss would be 29, oxygen atom loss would be 16, and hydrogen atom loss would be 1.

From practice test 2 of our ACS Organic study guide (question 5). Original question, not an official exam item.

Correct answer: D. The thermodynamic stability of the conjugate addition product and its enolate

Because the Michael addition is reversible under the reaction conditions, the system relaxes to the thermodynamically most stable adduct. Kinetic factors such as attack rate and reagent concentration do not set the equilibrium product distribution.

From practice test 8 of our ACS Organic study guide (question 58). Original question, not an official exam item.

Correct answer: D. 2-methylpropan-2-ol

Propanone is a ketone; Grignard addition adds a methyl group to the carbonyl carbon, creating a tertiary alcohol, 2-methylpropan-2-ol. Butan-2-ol and pentan-2-ol are secondary alcohols, and 2-methylpropan-1-ol is primary.

From practice test 3 of our ACS Organic study guide (question 15). Original question, not an official exam item.

Correct answer: A. Aqueous acid, such as dilute HCl in THF/water

Acetal deprotection is the reverse of acetal formation and requires aqueous acid to hydrolyze the acetal back to the carbonyl. Fluoride, methoxide, and hydrogenation do not effect this cleavage.

From practice test 6 of our ACS Organic study guide (question 36). Original question, not an official exam item.

Correct answer: A. Chlorine withdraws electron density by induction, deactivating the ring, but its lone pair donates by resonance to stabilize the ortho and para sigma complexes.

Chlorine's electronegativity produces a net inductive withdrawal, so the ring is less reactive than benzene. The meta-directing explanation is incorrect because halogen lone pairs are resonance donors, not resonance acceptors.

From practice test 12 of our ACS Organic study guide (question 23). Original question, not an official exam item.

Correct answer: B. Several adjacent C-H bonds are eclipsed, and the flagpole hydrogens at the two ends are forced close together.

The boat conformation has eclipsed C-H bonds on adjacent carbons and steric crowding between the flagpole hydrogens, both of which raise its energy. Its bond angles remain close to 109.5 degrees, so angle strain is not the main problem.

From practice test 5 of our ACS Organic study guide (question 7). Original question, not an official exam item.

Correct answer: C. R

With group 4 pointing away, the sequence 1 to 2 to 3 is read directly. Left to up to right is clockwise, which corresponds to the R configuration. No reversal is needed because the lowest priority group is already in the back.

From practice test 4 of our ACS Organic study guide (question 54). Original question, not an official exam item.

Correct answer: C. Oxygen reacts with cyclohexyl radicals to form peroxy radicals that are too stable to abstract Br from Br2.

Molecular oxygen is a diradical and reacts rapidly with carbon radicals to form peroxy radicals (ROO·), which are too unreactive to abstract bromine from Br2 and therefore break the propagation cycle. The other options do not describe oxygen's known radical-scavenging behavior.

From practice test 5 of our ACS Organic study guide (question 50). Original question, not an official exam item.

Correct answer: A. Acid X is stronger, and its conjugate base is weaker than the conjugate base of acid Y

Lower pKa means a stronger acid; the stronger acid has the weaker conjugate base. So acid X (pKa 5) is stronger and its conjugate base is weaker than that of acid Y.

From practice test 5 of our ACS Organic study guide (question 40). Original question, not an official exam item.

Correct answer: D. Pyrrole is a very weak base because its nitrogen lone pair is delocalized into the aromatic ring, while pyridine's lone pair is in an sp2 orbital that is available for protonation.

In pyrrole, the nitrogen lone pair is in a p orbital and participates in the six pi electron aromatic sextet, so it is not available for protonation. In pyridine, the lone pair is in an sp2 orbital in the plane of the ring and can bind a proton, making pyridine the stronger base.

From practice test 7 of our ACS Organic study guide (question 16). Original question, not an official exam item.

Correct answer: D. increase by a factor of 6

The E2 rate law is rate = k[alkyl halide][base]. Doubling the halide and tripling the base changes the rate by a factor of 2 x 3 = 6.

From practice test 4 of our ACS Organic study guide (question 29). Original question, not an official exam item.

Correct answer: A. A mixture of regioisomeric alcohols because the two alkene carbons are similarly substituted, so boron has no strong regiochemical preference

Hydroboration-oxidation is concerted and sterically controlled; for internal alkenes with similar substitution the regioselectivity is weak, so a mixture forms. It does not involve a carbocation and is anti-Markovnikov, not Markovnikov.

From practice test 6 of our ACS Organic study guide (question 8). Original question, not an official exam item.

Correct answer: D. It is at a lower wavenumber than the ketone carbonyl

A carboxylic acid carbonyl absorbs near 1710 cm-1, lower than the saturated ketone value near 1715 cm-1. The adjacent O-H group donates electron density to the carbonyl, weakening the C=O bond and lowering the stretching frequency.

From practice test 4 of our ACS Organic study guide (question 36). Original question, not an official exam item.

Correct answer: D. Keto-enol tautomers are constitutional isomers because a hydrogen atom has moved, while resonance forms differ only in electron placement with all atoms fixed.

Tautomers are constitutional isomers because an atom, typically hydrogen, has changed position; resonance forms differ only in electron placement with fixed nuclei. This is why keto-enol tautomerism involves bond breaking and forming, whereas resonance forms do not.

From practice test 2 of our ACS Organic study guide (question 51). Original question, not an official exam item.

Correct answer: D. Excess water with a catalytic amount of strong acid

Acetal hydrolysis is the reverse of acetal formation and requires aqueous acid to protonate the acetal oxygen and promote loss of the alcohol groups. Base cannot catalyze this hydrolysis because the leaving group needs protonation.

From practice test 5 of our ACS Organic study guide (question 25). Original question, not an official exam item.

Correct answer: B. Jones reagent (CrO3, H2SO4, H2O)

Jones reagent is a strong oxidant that oxidizes primary alcohols through the aldehyde to the carboxylic acid under aqueous acidic conditions. PCC and Dess-Martin periodinane stop at the aldehyde, while NaBH4 reduces rather than oxidizes.

From practice test 10 of our ACS Organic study guide (question 52). Original question, not an official exam item.

Correct answer: A. Much slower, because the CF3 group withdraws electron density and destabilizes the sigma complex in the rate-determining step.

The trifluoromethyl group is strongly electron withdrawing by induction, so it destabilizes the arenium ion and raises the activation energy. Substituent effects therefore change the reaction rate as well as the product position.

From practice test 4 of our ACS Organic study guide (question 61). Original question, not an official exam item.

Correct answer: C. Oxygen has two lone pairs, while nitrogen has one, so the two lone pairs on oxygen compress the bonding pairs more.

Both central atoms are sp3 hybridized, but water has two lone pairs and ammonia has one. The greater number of lone pairs on oxygen creates stronger lone-pair-bonding-pair repulsions, which compresses the H-O-H angle to 104.5 degrees, smaller than the H-N-H angle of 107 degrees.

From practice test 9 of our ACS Organic study guide (question 43). Original question, not an official exam item.

Correct answer: C. Horizontal bonds point toward the viewer; vertical bonds point away

The Fischer convention fixes horizontal bonds as coming out of the page and vertical bonds as going into the page. This is the opposite of many other projections, so it is the defining rule for drawing Fischer structures.

From practice test 4 of our ACS Organic study guide (question 67). Original question, not an official exam item.

Correct answer: C. Chlorination is faster but much less selective, so it gives a nearly statistical mixture of primary, secondary, and tertiary chlorides.

Chlorine's high reactivity makes its abstraction transition state early and unselective, giving 9:8:5 for 2-methylbutane rather than a large tertiary preference. Bromine's slower, more selective abstraction (1:80:1600) is required if the tertiary bromide is the target.

From practice test 4 of our ACS Organic study guide (question 38). Original question, not an official exam item.

Correct answer: C. Ethanol is a polar protic solvent that stabilizes the carbocation and the anionic leaving group, favoring ionization to a carbocation intermediate

Ethanol, as a polar protic solvent, stabilizes the developing carbocation and the anionic leaving group through ion-dipole interactions and hydrogen bonding, promoting SN1 and E1 pathways. Ethanol is too weak a base for E2 and too hindered for direct substitution on a tertiary center.

From practice test 2 of our ACS Organic study guide (question 47). Original question, not an official exam item.

Correct answer: D. Heating a cyclohexene derivative to give a diene and a dienophile

A retro-Diels-Alder is the reverse of a [4+2] cycloaddition: heating a cyclohexene derivative cleaves it into a diene and a dienophile. The option describing addition of a diene to a dienophile is the forward reaction.

From practice test 5 of our ACS Organic study guide (question 54). Original question, not an official exam item.

Correct answer: A. A racemic mixture of enantiomers

SN1 proceeds through a planar carbocation intermediate that can be attacked from either face with equal probability. The former stereocenter therefore loses its fixed configuration, giving a racemic mixture. Complete inversion would occur in SN2, not SN1.

From practice test 11 of our ACS Organic study guide (question 69). Original question, not an official exam item.

Correct answer: B. C1, because it gives the more stable secondary carbon radical

In the radical addition, the bromine radical adds to the terminal carbon to form the more stable secondary radical at the internal carbon. Adding to the internal carbon would form a primary radical, which is disfavored. Thus the initial addition is at C1.

From practice test 5 of our ACS Organic study guide (question 12). Original question, not an official exam item.

Correct answer: D. 1800 cm-1

The electron-withdrawing chlorine raises the carbonyl frequency to about 1800 cm-1. Simple ketones absorb near 1715 cm-1 and esters near 1735 cm-1.

From practice test 12 of our ACS Organic study guide (question 51). Original question, not an official exam item.

Correct answer: B. Phenol

Phenol is a fully conjugated enol whose keto tautomer is far less stable, so the enol form dominates. Simple ketones such as cyclohexanone, 2-butanone, and acetone exist overwhelmingly as keto tautomers.

From practice test 2 of our ACS Organic study guide (question 29). Original question, not an official exam item.

Correct answer: D. 4-Methyl-3-penten-2-one

Acetone first forms the beta-hydroxy ketone 4-hydroxy-4-methyl-2-pentanone, which dehydrates on heating to give the alpha,beta-unsaturated ketone 4-methyl-3-penten-2-one. 4-Hydroxy-4-methyl-2-pentanone is the addition intermediate, not the final product.

From practice test 8 of our ACS Organic study guide (question 46). Original question, not an official exam item.

Correct answer: B. A reagent used for one functional group transformation may react with the other functional group if it is already present.

Order matters because most reagents are not perfectly chemoselective, and a reagent that transforms one group can attack another group already present. This is why protecting groups are sometimes needed. The other statements are false.

From practice test 3 of our ACS Organic study guide (question 14). Original question, not an official exam item.

Correct answer: A. -NH2

The amino group has a lone pair on the nitrogen that is attached to the ring, making it a strong resonance donor and an ortho/para director. The nitro, cyano, and trifluoromethyl groups are meta directors.

From practice test 3 of our ACS Organic study guide (question 19). Original question, not an official exam item.

Correct answer: C. -al

Aldehydes are named with the suffix -al when they are the principal functional group. Ketones use -one, alcohols use -ol, and carboxylic acids use -oic acid.

From practice test 6 of our ACS Organic study guide (question 64). Original question, not an official exam item.

Correct answer: D. 16 kJ/mol

The fully eclipsed conformer of butane, in which the methyl groups eclipse each other, is approximately 19 kJ/mol higher in energy than the anti conformer. The gauche conformer is about 3.8 kJ/mol higher, and the 16 kJ/mol value is associated with the methyl-hydrogen eclipsed conformer.

From practice test 11 of our ACS Organic study guide (question 15). Original question, not an official exam item.

Correct answer: D. It reacts with HBr that forms during the reaction, keeping the Br2 concentration very low.

NBS reacts with HBr produced during the radical chain to form a low steady concentration of Br2. This low concentration favors allylic substitution over ionic addition to the double bond.

From practice test 6 of our ACS Organic study guide (question 37). Original question, not an official exam item.

Correct answer: A. ethane less than ethylene less than acetylene

Acidity increases with the s character of the carbon hybrid orbital: sp carbon (acetylene, pKa 25) holds the conjugate base lone pair closer to the nucleus than sp2 (ethylene, pKa 44) or sp3 (ethane, pKa 50). Thus the correct increasing order is ethane, ethylene, acetylene.

From practice test 8 of our ACS Organic study guide (question 48). Original question, not an official exam item.

Correct answer: B. Six molecular orbitals, three bonding and three antibonding

The number of pi molecular orbitals equals the number of p orbitals that combine. A six-carbon conjugated chain has six p orbitals, giving exactly six pi molecular orbitals, three bonding and three antibonding.

From practice test 2 of our ACS Organic study guide (question 1). Original question, not an official exam item.

Correct answer: D. 1-butene

Potassium tert-butoxide is a bulky base that cannot easily approach the carbon for SN2 and is too hindered to access the more substituted beta hydrogen. It removes the more accessible terminal beta hydrogen to give the Hofmann product, 1-butene.

From practice test 2 of our ACS Organic study guide (question 7). Original question, not an official exam item.

Correct answer: C. A racemic mixture of trans-1,2-dibromocyclopentane enantiomers

Bromine adds to cyclopentene through a cyclic bromonium ion, and the second bromide attacks from the opposite face, giving anti addition. The trans-1,2-dibromocyclopentane product is chiral and formed as a racemic mixture; cis products would require syn addition and are not observed.

From practice test 8 of our ACS Organic study guide (question 60). Original question, not an official exam item.

Correct answer: C. The integration ratio is 3:2:1, corresponding to the methyl, methylene, and hydroxyl protons.

Ethanol has three chemically distinct proton environments: the methyl group (3 H), the methylene group (2 H), and the hydroxyl proton (1 H), so the integration ratio is 3:2:1.

From practice test 8 of our ACS Organic study guide (question 18). Original question, not an official exam item.

Correct answer: A. The carbon-fluorine bond is unusually strong, making fluoride a very poor leaving group

Carbon-fluorine bonds are among the strongest in organic chemistry, and fluoride is a very poor leaving group, so SN2 attack on an alkyl fluoride is extremely slow. The other options do not explain the observed lack of reactivity.

From practice test 4 of our ACS Organic study guide (question 68). Original question, not an official exam item.

Correct answer: D. A racemic mixture of enantiomeric tertiary alcohols

The planar carbonyl can be attacked from either face with equal probability, and the product is a tertiary alcohol with four different substituents at the former carbonyl carbon, so it is chiral. Equal attack from both faces gives a 1:1 mixture of enantiomers, not a single stereoisomer.

From practice test 2 of our ACS Organic study guide (question 17). Original question, not an official exam item.

Correct answer: C. To prevent the carbonyl group from reacting with the nucleophile during the SN2 displacement

The cyclic acetal masks the ketone, preventing nucleophilic attack at the carbonyl while the SN2 displacement occurs at the bromide. Without protection, the nucleophile would also react with the ketone.

From practice test 6 of our ACS Organic study guide (question 24). Original question, not an official exam item.

Correct answer: C. nitrobenzene

All three rings are deactivated, but the nitro group withdraws electron density more powerfully than the carbonyl or cyano groups, so nitrobenzene nitrates slowest. It therefore remains in the greatest amount when the nitrating agent is limiting.

From practice test 6 of our ACS Organic study guide (question 22). Original question, not an official exam item.

Correct answer: B. The phenoxide ion is stabilized by resonance, which delocalizes the negative charge over the aromatic ring.

Resonance delocalizes the negative charge of the phenoxide ion onto ortho and para carbons of the aromatic ring, substantially stabilizing the conjugate base. Cyclohexanol forms a simple alkoxide with no resonance stabilization, so phenol is the stronger acid.

From practice test 4 of our ACS Organic study guide (question 51). Original question, not an official exam item.

Correct answer: D. Conformers rapidly interconvert by rotation around single bonds and cannot be separated as distinct compounds.

Conformers interconvert freely by rotation about single bonds, making them different shapes of the same compound rather than separable isomers. Constitutional and configurational isomers are separable because interconversion requires breaking a bond.

From practice test 2 of our ACS Organic study guide (question 37). Original question, not an official exam item.

Correct answer: B. The ethyl radical reacts with a chlorine molecule to form chloroethane and a chlorine radical.

This is the second propagation step: the alkyl radical attacks a halogen molecule, producing the alkyl halide and regenerating the halogen radical. Combining the ethyl radical directly with a chlorine radical is termination, not propagation.

From practice test 2 of our ACS Organic study guide (question 21). Original question, not an official exam item.

Correct answer: A. Because both directions are fast, equilibrium is reached, and the pKa difference between the two acids controls the product ratio

Large rate constants in both directions allow the system to reach equilibrium quickly, so thermodynamic control applies and the pKa difference sets the equilibrium constant. Kinetic control would apply only if one direction were very slow.

From practice test 5 of our ACS Organic study guide (question 55). Original question, not an official exam item.

Correct answer: D. 36

The resonance energy of benzene is about 36 kcal/mol, far larger than the delocalization energy of an acyclic conjugated diene. The other values are too small.

From practice test 6 of our ACS Organic study guide (question 53). Original question, not an official exam item.

Correct answer: D. The secondary carbocation is less stable than the tertiary carbocation because it has fewer alkyl groups donating electron density.

The rate-determining step in SN1 is ionization to a carbocation. A secondary carbocation has only two alkyl groups stabilizing it by hyperconjugation and induction, so it forms more slowly than a tertiary carbocation. Backside attack is an SN2 consideration, and the leaving group is the same bromide in both substrates.

From practice test 12 of our ACS Organic study guide (question 64). Original question, not an official exam item.

Correct answer: C. Disiamylborane (Sia2BH)

Disiamylborane is a bulky dialkylborane, and its steric bulk prevents the second hydroboration, leaving the alkenylborane intermediate for oxidation. BH3-THF would continue to hydroborate the alkenylborane, while sodium borohydride and lithium aluminum hydride do not act as electrophilic borane reagents for alkyne hydroboration.

From practice test 3 of our ACS Organic study guide (question 68). Original question, not an official exam item.