INTOCHEMISTRY · ORIGINAL FREE MCAT-STYLE CHEMISTRY PRACTICE · SET 2 50 brand-new questions. Six lab passages. Step-by-step explanations. Continue your General and Organic Chemistry practice with 50 new, original questions—not repeats of IntoChemistry’s first free set. The questions emphasize experimental interpretation, chemical reasoning, units, mechanisms, and spectroscopy.
25 General Chemistry 25 Organic Chemistry 30 passage-linked 20 independent Free self-scoring
How to use this set: Aim for about 75–80 minutes with paper and no calculator, then grade your work. Each question has four choices and a hidden explanation. The final Grade my work button reports your General and Organic Chemistry raw scores and reveals explanations for missed or unanswered questions. No login, purchase, or submission is required. If JavaScript is unavailable, answers remain reviewable using the disclosure sections.
General Chemistry Organic Chemistry Grade my work Original 50-question set Study guide Part I: General Chemistry (G26–G50) 10 new independent questions G26 · Mass percent and solution preparation A technician dissolves 0.250 g of a nonvolatile solute and adds water until the total solution mass is 50.0 g. What is its mass percent?
A. 0.500% B. 20.0% C. 0.0500% D. 5.00% Reveal answer and detailed explanation Correct answer: A. Mass percent = (0.250 g / 50.0 g) × 100% = 0.500%. The denominator is the final solution mass, not just the solvent mass.
G27 · Formal charge in nitrite In a valid Lewis resonance contributor of NO2 − , the central nitrogen has one lone pair, one N–O single bond, and one N=O double bond. What is nitrogen’s formal charge?
A. 0 B. −1 C. +1 D. +2 Reveal answer and detailed explanation Correct answer: A. Nitrogen has 5 valence electrons, 2 nonbonding electrons, and 6 bonding electrons: FC = 5 − 2 − 6/2 = 0. The singly bonded oxygen has the −1 charge.
G28 · Transition-metal electron configurations Which ground-state electron configuration describes Fe3+ after removing electrons from neutral iron, Fe = [Ar]4s2 3d6 ?
A. [Ar]4s1 3d4 B. [Ar]3d3 C. [Ar]3d5 D. [Ar]4s2 3d3 Reveal answer and detailed explanation Correct answer: C. Forming transition-metal cations removes 4s electrons before 3d electrons. Remove two 4s electrons and one 3d electron to get [Ar]3d⁵.
G29 · Sigma and pi bonds How many sigma (σ) and pi (π) bonds are present in one molecule of ethyne, H–C≡C–H?
A. 3 σ and 2 π B. 5 σ and 0 π C. 2 σ and 3 π D. 4 σ and 1 π Reveal answer and detailed explanation Correct answer: A. Each C–H bond is one σ bond; the C≡C bond has one σ plus two π bonds. Total: 3 σ and 2 π.
G30 · Changing an equilibrium expression For A(aq) ⇌ B(aq), the equilibrium constant is K = 4.0 at a given temperature. What is K for 2A(aq) ⇌ 2B(aq) at the same temperature?
A. 2.0 B. 4.0 C. 8.0 D. 16 Reveal answer and detailed explanation Correct answer: D. Multiplying all reaction coefficients by 2 raises the equilibrium constant to the second power: Knew = (4.0)² = 16.
G31 · Gas collected over water A gas sample is collected over water. The total pressure is 760 torr, and water vapor has a partial pressure of 24 torr. What is the partial pressure of the dry gas?
A. 736 torr B. 784 torr C. 24 torr D. 760 torr Reveal answer and detailed explanation Correct answer: A. Dalton’s law gives Ptotal = Pdry gas + PH₂O; therefore Pdry gas = 760 − 24 = 736 torr.
G32 · Lewis acids and bases Which species acts as a Lewis acid when it forms an adduct with ammonia by accepting a lone pair?
A. Cl− B. BF3 C. NH3 D. OH− Reveal answer and detailed explanation Correct answer: B. BF₃ is electron deficient at boron and accepts the lone pair from NH₃. NH₃, OH⁻, and Cl⁻ are electron-pair donors in the usual contexts.
G33 · Henry’s law At constant temperature, a dilute solution obeys Henry’s law, with dissolved gas concentration directly proportional to the gas’s partial pressure. If pressure doubles, what happens to the equilibrium dissolved concentration?
A. It doubles. B. It is halved. C. It increases fourfold. D. It remains unchanged. Reveal answer and detailed explanation Correct answer: A. Under the stated Henry’s-law regime, c = kH × P. Doubling P doubles dissolved-gas concentration.
G34 · Molality A sample contains 0.300 mol of solute dissolved in 0.750 kg of solvent. What is the solution’s molality?
A. 0.300 m B. 2.50 m C. 0.225 m D. 0.400 m Reveal answer and detailed explanation Correct answer: D. Molality = mol solute / kg solvent = 0.300 / 0.750 = 0.400 mol/kg.
G35 · Photon energy and wavelength For two photons in vacuum, photon B has twice the wavelength of photon A. What is the ratio EB /EA ?
A. 4 B. 1/2 C. 1/4 D. 2 Reveal answer and detailed explanation Correct answer: B. Photon energy E = hc/λ is inversely proportional to wavelength, so doubling λ halves E.
Three new experimental passages · 15 questions Passage G4 — UV–visible calibration of a metal complex Spectrophotometry · Dilution · Beer–Lambert law
A chemist measures the absorbance of a colored coordination complex at its absorption maximum using a 1.00 cm cell. The calibration is linear and passes through the origin after proper blank correction. The solution contains one major absorbing species. An unknown sample is diluted by a factor of 5.00 (one part original sample to five parts final volume) and yields a corrected absorbance of 0.150 .
Assume A = εbc, where ε is molar absorptivity, b is optical path length, and c is molar concentration. A separate raw reading is 0.170 before subtraction of a blank absorbance of 0.020.
G36 · Determine diluted concentration What is the concentration of the diluted unknown based on its corrected absorbance?
A. 1.50 × 10−5 M B. 3.00 × 10−5 M C. 6.00 × 10−5 M D. 1.50 × 10−4 M Reveal answer and detailed explanation Correct answer: B. The calibration gives A/c = 0.100/(2.00×10⁻⁵) = 5.00×10³ M⁻¹. Thus c = 0.150/(5.00×10³) = 3.00×10⁻⁵ M.
G37 · Reconstruct original concentration What was the concentration in the original unknown before the fivefold dilution?
A. 3.00 × 10−5 M B. 3.00 × 10−4 M C. 1.50 × 10−4 M D. 6.00 × 10−6 M Reveal answer and detailed explanation Correct answer: C. The original was five times as concentrated: (3.00×10⁻⁵ M) × 5.00 = 1.50×10⁻⁴ M.
G38 · Molar absorptivity What is ε for the complex at the measured wavelength?
A. 2.00 × 10−4 M−1 cm−1 B. 5.00 × 102 M−1 cm−1 C. 5.00 × 103 M−1 cm−1 D. 5.00 × 104 M−1 cm−1 Reveal answer and detailed explanation Correct answer: C. ε = A/(bc) = 0.100/[1.00 cm × (2.00×10⁻⁵ M)] = 5.00×10³ M⁻¹ cm⁻¹.
G39 · Effect of path length If the same diluted unknown is measured in a 0.500 cm cell under otherwise identical conditions, what absorbance is predicted?
A. 0.0750 B. 0.0250 C. 0.150 D. 0.300 Reveal answer and detailed explanation Correct answer: A. At constant ε and c, A is proportional to b. Halving 1.00 cm to 0.500 cm halves 0.150 to 0.0750.
G40 · Blank correction A second reading has raw absorbance 0.170 and blank absorbance 0.020. What corrected absorbance should be used for calibration?
A. 0.170 B. 0.190 C. 0.150 D. 0.020 Reveal answer and detailed explanation Correct answer: C. Subtract the blank from the raw measurement: 0.170 − 0.020 = 0.150. Adding the blank or leaving it uncorrected biases concentration upward.
Passage G5 — Preparing and challenging an acetate-like buffer Conjugate pairs · Stoichiometry · Henderson–Hasselbalch
A simplified monoprotic buffer contains an acid HA and its conjugate base A− . At 25 °C, Ka = 1.00 × 10−5 , so pKa = 5.00. A 100. mL buffer portion initially contains 0.0100 mol HA and 0.0200 mol A− . Separate, initially identical portions receive either 0.00500 mol strong acid, 0.00500 mol strong base, or tenfold dilution with water.
Assume each strong-acid or strong-base addition reacts essentially completely with the appropriate buffer component; neglect volume changes from reagent addition, activity corrections, and small shifts in weak-acid dissociation. Use pH ≈ pKa + log10 (nA− /nHA ) after neutralization.
G41 · Initial buffer pH What is the approximate initial buffer pH?
A. 5.70 B. 4.70 C. 5.00 D. 5.30 Reveal answer and detailed explanation Correct answer: D. pH ≈ 5.00 + log₁₀(0.0200/0.0100) = 5.00 + log₁₀(2) ≈ 5.30.
G42 · After adding strong acid What is the approximate pH after adding 0.00500 mol HCl to one initial portion?
A. 5.00 B. 4.70 C. 5.30 D. 5.70 Reveal answer and detailed explanation Correct answer: A. H⁺ converts A⁻ into HA. A⁻ becomes 0.0200−0.00500 = 0.0150 mol and HA becomes 0.0100+0.00500 = 0.0150 mol, so log(1) = 0 and pH ≈ 5.00.
G43 · After adding strong base What is the approximate pH after adding 0.00500 mol NaOH to a fresh initial portion?
A. 5.00 B. 5.30 C. 4.30 D. 5.70 Reveal answer and detailed explanation Correct answer: D. OH⁻ converts HA into A⁻: HA = 0.00500 mol and A⁻ = 0.0250 mol. The ratio is 5, giving pH ≈ 5.00 + log₁₀(5) = 5.70.
G44 · Tenfold dilution What happens to this buffer’s approximate pH upon a tenfold dilution if its composition ratio is preserved and the stated approximations hold?
A. It becomes exactly 7.00. B. It stays approximately the same. C. It decreases by 1.00 pH unit. D. It increases by 1.00 pH unit. Reveal answer and detailed explanation Correct answer: B. Dilution changes both concentrations by the same factor, leaving [A⁻]/[HA] (and hence the logarithm in the buffer equation) unchanged under the stated approximation. The buffer capacity does decrease.
G45 · Designing buffer capacity At a fixed total analytical concentration of HA plus A− , which composition generally gives the strongest resistance to both small acid and base additions near the target pH?
A. Almost entirely HA B. Almost entirely A− C. Pure water without buffer species D. Equal amounts of HA and A− Reveal answer and detailed explanation Correct answer: D. A conjugate acid/base mixture resists both added acid and base. Buffer capacity in the target region is usually highest near [HA] ≈ [A⁻], where pH ≈ pKa.
Passage G6 — Determining a reaction rate law Experimental controls · Initial rates · Reaction orders
Researchers study the reaction A(aq) + B(aq) → products at a fixed temperature. They vary initial concentrations while measuring the initial product formation rate. All experiments use the same reaction conditions except for the concentrations listed. Assume a rate law of the form rate = k[A]m [B]n .
G46 · Order in A What is the experimentally determined reaction order with respect to A?
A. First B. Second C. Third D. Zero Reveal answer and detailed explanation Correct answer: B. Compare Trials 1 and 2: [B] remains fixed while [A] doubles and the rate increases fourfold. Since 2^m = 4, m = 2.
G47 · Order in B What is the experimentally determined order with respect to B?
A. Third B. Zero C. First D. Second Reveal answer and detailed explanation Correct answer: C. Compare Trials 1 and 3: [A] remains fixed while [B] doubles and rate doubles. Thus 2^n = 2 and n = 1.
G48 · Overall reaction order What is the overall order of the measured rate law?
A. First B. Second C. Third D. Fourth Reveal answer and detailed explanation Correct answer: C. The experimentally determined exponents add: overall order = m+n = 2+1 = 3.
G49 · Rate constant and units What is k for these data, including the appropriate units?
A. 2.00 M−1 s−1 B. 0.200 M−2 s−1 C. 20.0 s−1 D. 0.0200 M−2 s−1 Reveal answer and detailed explanation Correct answer: B. Using Trial 1, k = (2.00×10⁻⁴ M/s)/[(0.100 M)²(0.100 M)] = 0.200 M⁻²s⁻¹. A third-order rate law requires concentration⁻² time⁻¹ units.
G50 · Predicting another initial rate At [A] = 0.300 M and [B] = 0.100 M, what initial rate is predicted?
A. 1.20 × 10−3 M/s B. 6.00 × 10−4 M/s C. 9.00 × 10−4 M/s D. 1.80 × 10−3 M/s Reveal answer and detailed explanation Correct answer: D. rate = 0.200(0.300)²(0.100) = 0.00180 M/s = 1.80×10⁻³ M/s.
Part II: Organic Chemistry (O26–O50) 10 new independent questions O26 · Basic epoxide ring opening When 1,2-epoxypropane reacts with methoxide (CH3 O− ) under basic conditions and is then protonated, which major product is expected?
A. CH3 –CO–CH3 B. CH3 –CH(OCH3 )–CH2 OH C. CH3 –CH2 –CH2 OH D. CH3 –CH(OH)–CH2 –OCH3 Reveal answer and detailed explanation Correct answer: D. Under basic conditions methoxide attacks the less substituted terminal epoxide carbon by an SN2-like pathway, breaking its C–O bond. Protonation places OH on the more substituted carbon.
O27 · IR nitrile identification Which sharp absorption is most diagnostic of a nitrile (C≡N) in an organic IR spectrum?
A. Around 2250 cm−1 B. Around 1700 cm−1 C. Around 3000 cm−1 D. Broad band from 3200–3600 cm−1 Reveal answer and detailed explanation Correct answer: A. The C≡N stretching band typically appears around 2210–2260 cm⁻¹. Bands around 1700 cm⁻¹ often indicate carbonyls; broad 3200–3600 cm⁻¹ often indicates O–H.
O28 · E/Z stereoisomerism A drawing of 2-butene has the two CH3 groups on the same side of the C=C bond. What is its correct E/Z designation?
A. Z B. Neither, because there are no substituents C. Both E and Z simultaneously D. E Reveal answer and detailed explanation Correct answer: A. For each alkene carbon, CH₃ has higher CIP priority than H. The two higher-priority CH₃ groups on the same side define Z (zusammen).
O29 · Sodium borohydride selectivity What is the principal product when 2-butanone is reduced with NaBH4 and followed by aqueous workup?
A. 2-Butanol B. Butanal C. 1-Butanol D. Butanoic acid Reveal answer and detailed explanation Correct answer: A. NaBH₄ reduces a ketone C=O to a secondary alcohol. The carbon skeleton and carbonyl carbon position remain unchanged, giving 2-butanol.
O30 · Aromatic substitution directing effects Under ordinary electrophilic nitration conditions, what positions are generally favored on methoxybenzene (anisole) relative to the methoxy substituent?
A. Meta because methoxy withdraws by resonance B. Ortho and para because methoxy donates by resonance C. Only the ipso carbon D. All positions equally because the ring is symmetric Reveal answer and detailed explanation Correct answer: B. The oxygen lone pair donates electron density by resonance into the ring, stabilizing ortho/para sigma-complex contributors. Para can be favored by sterics, but both ortho and para are directing positions.
O31 · Relative acidity Which compound is the strongest Brønsted acid in water among the following?
A. Phenol B. Acetic acid C. Ethyne D. Ethanol Reveal answer and detailed explanation Correct answer: B. Acetic acid (pKa about 4.8) is much more acidic than phenol (~10), ethanol (~16), or a terminal alkyne (~25), because its conjugate carboxylate charge is delocalized over two oxygens.
O32 · Base-promoted ester hydrolysis Methyl acetate (CH3 COOCH3 ) reacts with excess aqueous NaOH. Before any acid workup, which principal organic products are present?
A. Acetic acid and dimethyl ether B. Acetic acid and methane C. Acetaldehyde and methanol D. Acetate ion and methanol Reveal answer and detailed explanation Correct answer: D. Base hydrolysis (saponification) converts an ester into a carboxylate and an alcohol. The carboxylic acid remains deprotonated before acid workup.
O33 · Proton NMR equivalence Ignoring the exchangeable O–H proton, how many distinct 1 H NMR chemical environments are present in pure 2-propanol (CH3 –CHOH–CH3 ) in an achiral medium?
A. One B. Three C. Two D. Four Reveal answer and detailed explanation Correct answer: C. The two methyl groups are equivalent and give one environment; the methine C–H gives the other. The OH proton is excluded by the stem.
O34 · E2 elimination conformation Which geometric arrangement is generally required for an optimal concerted E2 elimination of an acyclic alkyl halide?
A. The substrate must form a planar carbocation first. B. The leaving group must be on a benzene ring. C. The β-C–H bond must be approximately anti-periplanar to C–X. D. The β-C–H bond must be perpendicular to C–X. Reveal answer and detailed explanation Correct answer: C. The anti-periplanar β-C–H and C–X arrangement provides favorable stereoelectronic overlap for concerted E2 elimination, without a carbocation intermediate.
O35 · Mass spectral chlorine isotope pattern For a molecular ion containing one chlorine atom , which approximate ratio of M to M+2 peaks is expected from naturally occurring chlorine isotopes?
A. 9:1 B. 1:3 C. 1:1 D. 3:1 Reveal answer and detailed explanation Correct answer: D. Natural ³⁵Cl:³⁷Cl abundance is approximately 3:1. A molecular ion containing one chlorine gives a characteristic M:M+2 pair near that ratio.
Three new experimental passages · 15 questions Passage O4 — Synthesizing and analyzing acetylsalicylic acid Acylation · Yield · IR · TLC
Students react 1.38 g of salicylic acid (molar mass 138 g/mol) with an excess of acetic anhydride under acidic catalysis. The target is acetylsalicylic acid (aspirin; molar mass 180 g/mol ). The balanced process consumes one mole of salicylic acid per mole of aspirin formed. After isolation and drying, the students obtain 1.44 g of crude product.
On silica gel TLC with the specified eluent, authentic salicylic acid has Rf = 0.25 , and authentic aspirin has Rf = 0.60 . An IR spectrum is collected to help distinguish the starting phenolic group from the new ester functionality. Ignore possible side reactions for theoretical yield.
O36 · Acetic anhydride function What is the role of acetic anhydride in the synthesis?
A. An acetylating reagent that transfers an acyl group B. An oxidant that increases the aromatic ring oxidation state C. A reducing agent that saturates the aromatic ring D. A base that converts phenol into an ether Reveal answer and detailed explanation Correct answer: A. Acetic anhydride provides the acetyl group that esterifies the phenolic OH of salicylic acid. Aspirin also retains its carboxylic acid group.
O37 · Calculate theoretical yield What is the approximate maximum theoretical mass of aspirin from the given salicylic acid?
A. 1.38 g B. 1.80 g C. 2.76 g D. 1.00 g Reveal answer and detailed explanation Correct answer: B. 1.38 g / 138 g/mol = 0.0100 mol salicylic acid. A 1:1 stoichiometry gives 0.0100 mol aspirin × 180 g/mol = 1.80 g.
O38 · Calculate percent yield What is the crude product’s percent yield relative to that theoretical maximum?
A. 96.0% B. 80.0% C. 60.0% D. 72.0% Reveal answer and detailed explanation Correct answer: B. Percent yield = (1.44 g / 1.80 g) × 100% = 80.0%. The crude mass can overstate true chemical yield if impurities remain.
O39 · IR functional-group evidence Which IR observation most directly supports formation of an ester group in the product?
A. A characteristic ester C=O absorption near 1750 cm−1 B. A strong C≡N stretch near 2250 cm−1 C. A broad free-amine N–H envelope near 3300 cm−1 D. Disappearance of every carbonyl absorption Reveal answer and detailed explanation Correct answer: A. The newly formed acetate ester introduces an additional strong ester carbonyl absorption around 1750 cm⁻¹. Aspirin still contains the carboxylic acid carbonyl, so carbonyl signals do not all disappear.
O40 · Interpret TLC behavior Under the stated silica-gel TLC conditions, which interpretation best explains why salicylic acid has a smaller Rf than aspirin?
A. A smaller Rf always indicates a larger molecular radius. B. Aspirin must decompose before leaving the origin. C. Salicylic acid interacts more strongly with the polar stationary phase. D. Salicylic acid must have a higher molar mass. Reveal answer and detailed explanation Correct answer: C. The phenolic OH makes salicylic acid relatively strongly retained by the polar silica stationary phase in the stated solvent, resulting in lower Rf. TLC separation depends on intermolecular interactions, not molar mass alone.
Passage O5 — Kinetics and stereochemistry of azide substitution SN2 mechanisms · Solvent effects · Steric accessibility
A researcher studies the substitution of 1-bromobutane by azide ion (N3 − ) in dry acetone. The isolated main product is 1-azidobutane , and the measured initial rate is proportional to both substrate and azide concentrations. Follow-up experiments compare other alkyl bromides, and a separate enantiopure secondary bromide is reacted under conditions in which substitution occurs almost exclusively through an SN2 pathway.
Assume comparable temperature, reagent identities, and solvent unless a question specifies otherwise. Consider substitution rather than any side reactions for the chiral-substrate experiment.
O41 · SN2 rate law Which rate expression matches the observed substitution of 1-bromobutane?
A. rate = k[1-bromobutane]² B. rate = k[1-bromobutane] C. rate = k[1-bromobutane][N3 − ] D. rate = k[N3 − ] Reveal answer and detailed explanation Correct answer: C. The observed rate depends first-order on each reaction partner, consistent with bimolecular substitution: rate = k[substrate][nucleophile].
O42 · Combining concentration changes If substrate concentration is doubled and azide concentration is tripled, what initial rate change is expected?
A. Twofold B. Threefold C. Sixfold D. Fivefold Reveal answer and detailed explanation Correct answer: C. With rate ∝ [substrate][azide], the factors multiply: 2 × 3 = 6. The temperature and rate constant remain fixed.
O43 · Polar aprotic solvent role Why is acetone a suitable solvent for promoting this substitution with an anionic nucleophile?
A. It converts the primary substrate into a stable tertiary carbocation. B. It removes the leaving group by aromatic substitution. C. It strongly protonates azide before it attacks. D. It generally solvates anions less strongly than protic solvents, favoring nucleophilicity. Reveal answer and detailed explanation Correct answer: D. Polar aprotic acetone solvates cations but does not donate hydrogen bonds to solvate anions as strongly as alcohols or water, helping anionic nucleophiles react via SN2.
O44 · Steric control If 1-bromobutane is replaced by tert-butyl bromide, what happens to the tendency for direct SN2 substitution at the brominated carbon?
A. It increases due to additional methyl inductive stabilization. B. It increases sharply because the carbon is tertiary. C. It remains identical if azide concentration is unchanged. D. It becomes essentially negligible because backside attack is strongly hindered. Reveal answer and detailed explanation Correct answer: D. Backside approach to the tertiary carbon is strongly blocked by three alkyl groups. Other mechanisms can occur under suitable conditions, but direct SN2 is not favorable.
O45 · SN2 stereochemical result If an enantiopure secondary alkyl bromide undergoes an uncomplicated SN2 displacement at its stereogenic carbon, what happens to the local tetrahedral geometry?
A. Retention of spatial arrangement B. Complete loss of stereochemical information via a planar carbocation C. Inversion of the spatial arrangement D. Automatic conversion of the R descriptor to S in all cases Reveal answer and detailed explanation Correct answer: C. Backside attack causes Walden inversion at the reacting center. R/S labels can change differently if CIP priorities change, so the general defensible conclusion is inversion of geometry, not always R→S.
Passage O6 — Identifying and transforming an unknown carbonyl IR + ¹H NMR · Hydride reduction · Cyanohydrin formation
An unknown achiral sample X has the molecular formula C4 H8 O . Its IR spectrum shows a strong absorption near 1715 cm−1 but no broad O–H band. Its 1 H NMR spectrum has three major signals: a 3H singlet near δ 2.1 , a 2H quartet near δ 2.4 , and a 3H triplet near δ 1.0 . Signal areas are proportional to hydrogen count.
Reaction 1: X is treated with NaBH4 followed by aqueous workup. Reaction 2: X is separately treated with HCN under suitable catalytic conditions to form a cyanohydrin. All reaction media are achiral, and the carbonyl group of X is planar before nucleophilic attack.
O46 · Identify the carbonyl compound Which structure is most consistent with all of the spectral information?
A. Tetrahydrofuran (C4 H8 O) B. 2-Butanone (CH3 COCH2 CH3 ) C. 2-Buten-1-ol (C4 H8 O) D. Butanal (CH3 CH2 CH2 CHO) Reveal answer and detailed explanation Correct answer: B. 2-Butanone has a ketone C=O near 1715 cm⁻¹, a methyl directly on the carbonyl giving a 3H singlet, and a neighboring ethyl group giving a 2H quartet plus 3H triplet. Butanal should show an aldehydic proton.
O47 · IR after reduction After Reaction 1 is complete, which IR change most directly supports conversion of the carbonyl into an alcohol?
A. A broad O–H stretching band and loss of the ketone C=O band B. Disappearance of all C–H stretching absorptions C. New sharp nitrile absorption near 2250 cm−1 D. Growth of a carbonyl absorption near 1715 cm−1 Reveal answer and detailed explanation Correct answer: A. Reducing the ketone converts C=O into a secondary alcohol: the ~1715 cm⁻¹ ketone band disappears and a broad O–H band develops, usually around 3200–3600 cm⁻¹.
O48 · Reduction product What is the constitutional identity of the major product from Reaction 1?
A. 2-Butanol B. 1-Butanol C. 2-Butene D. Butanoic acid Reveal answer and detailed explanation Correct answer: A. Hydride addition and protonation reduce the ketone at C2 without changing the C4 skeleton, yielding 2-butanol. In an achiral setting the newly stereogenic product is racemic, but the requested identity is constitutional.
O49 · Cyanohydrin mechanism In Reaction 2, which event forms the new carbon–carbon bond?
A. An H atom migrates from the ethyl group to the carbonyl carbon. B. The carbonyl carbon expels the methyl group as a leaving group. C. The carbonyl oxygen attacks a cyanide nitrogen. D. CN− attacks the electrophilic carbonyl carbon. Reveal answer and detailed explanation Correct answer: D. Cyanide attacks the electrophilic carbonyl carbon and the C=O π electrons move onto oxygen. Protonation of the alkoxide furnishes the cyanohydrin with a new C–CN bond.
O50 · Stereochemical outcome With an achiral medium and no stereoselective catalyst, what stereochemical mixture is expected for the new cyanohydrin from X?
A. An E/Z alkene mixture B. A racemic pair of enantiomers under ideal achiral conditions C. Only one enantiomer because X contains no stereocenter D. Only a meso compound Reveal answer and detailed explanation Correct answer: B. The planar ketone can be attacked on either enantiotopic face with equal probability under achiral conditions. The new OH-bearing center has different substituents and gives enantiomers in a racemic mixture.
Scope, sources, and editorial notes This is an independently developed, unofficial chemistry-only practice set . The full MCAT includes physics, biology, biochemistry, psychology, sociology, and CARS, none of which this chemistry-only set attempts to cover comprehensively. The real Chemical and Physical Foundations section has 59 questions in 95 minutes and uses a different discipline distribution; the equal General–Organic split here is chosen for balanced chemistry review, not to imitate official weighting.
Read the official AAMC section overview and official exam content outline for authoritative exam specifications. MCAT® is a trademark of the Association of American Medical Colleges. IntoChemistry is not affiliated with, endorsed by, or reviewed by the AAMC.
All questions and laboratory scenarios are original educational examples. Numerical data are illustrative and may be idealized as stated in their passages. Questions have four answer choices, verified keys and solution explanations. Last reviewed November 2026.
