Organic Chemistry Quiz: Reaction Conditions And Reagent Recognition

What this quiz covers

This quiz focuses on Reaction Conditions And Reagent Recognition, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. A racemic mixture of (trans)-1,2-dimethylcyclopentane.
2. A single achiral meso compound, (cis)-1,2-dimethylcyclopentane.
3. A mixture of four different stereoisomers.
4. A racemic mixture of (cis)-1,2-dimethylcyclopentane. (correct answer)

Explanation: Catalytic hydrogenation involves the syn-addition of two hydrogen atoms across the double bond from the surface of the catalyst. This means both hydrogens add to the same face of the ring, resulting exclusively in the cis-1,2-dimethylcyclopentane product. Since the starting material is planar and achiral, attack from the top face and bottom face are equally likely, producing the (1R,2S) and (1S,2R) enantiomers in equal amounts. This pair of enantiomers is a racemic mixture. Unlike the six-membered ring analogue, cis-1,2-dimethylcyclopentane is chiral.

Question 2

1. $H_2O$ with catalytic $H_2SO_4$ followed by workup
2. $BH_3 \cdot THF$ followed by $H_2O_2, NaOH$ (correct answer)
3. $Hg(OAc)_2, H_2O$ followed by $NaBH_4$
4. $OsO_4$ followed by $NaHSO_3$ aqueous workup

Explanation: Hydroboration-oxidation ($BH_3 \cdot THF$ followed by $H_2O_2, NaOH$) provides anti-Markovnikov hydration with syn stereochemistry. The boron adds to the less substituted carbon, and both B-H bonds add from the same face. Choice A gives Markovnikov hydration through carbocation intermediates. Choice C (oxymercuration-demercuration) gives Markovnikov hydration despite being syn. Choice D (osmium tetroxide) is used for dihydroxylation to form diols, not simple alcohols.

Question 3

1. 1. LiAlH₄; 2. H₂O
2. Peroxyacetic acid (CH₃CO₃H)
3. Sodium methoxide (NaOCH₃) in methanol (correct answer)
4. 1. CH₃MgBr; 2. H₃O⁺

Explanation: The desired product, 2-methoxy-1-methylcyclohexan-1-ol, results from the nucleophilic attack of methoxide on the less sterically hindered carbon (C2) of the epoxide. This regioselectivity is characteristic of epoxide opening under basic or nucleophilic conditions (SN2-like). Sodium methoxide provides a strong methoxide nucleophile that will attack the less substituted carbon.

Question 4

1. A racemic mixture of (R)- and (S)-2-iodopentane.
2. (R)-2-iodopentane.
3. The meso form of 2-iodopentane.
4. (S)-2-iodopentane. (correct answer)

Explanation: This reaction is a Finkelstein reaction, which proceeds via an SN2 mechanism. A hallmark of the SN2 mechanism is the inversion of configuration at the stereocenter due to backside attack by the nucleophile (iodide). Therefore, the (R) starting material will be converted to the (S) product. Racemization (A) would imply an SN1 mechanism, which is not favored by the conditions (strong nucleophile, polar aprotic solvent).

Question 5

1. 1. PBr₃; 2. Potassium tert-butoxide (KOtBu)
   (correct answer)
2. 1. TsCl, pyridine; 2. Sodium ethoxide (NaOEt)
3. POCl₃, pyridine, at high temperature
4. Concentrated HBr, followed by heating with NaOH

Explanation: To avoid rearrangement and favor the Hofmann product (3-methyl-1-butene), a two-step E2 pathway is required. First, the alcohol is converted to a good leaving group without rearrangement (e.g., using PBr₃ to form 2-bromo-3-methylbutane). Second, a bulky base like potassium tert-butoxide is used to promote E2 elimination by abstracting a proton from the less sterically hindered carbon, yielding the desired Hofmann product.

Question 6

1. A Lewis acid catalyst like FeBr₃
2. A small amount of water as a co-solvent
3. An organic peroxide like dibenzoyl peroxide (ROOR) (correct answer)
4. A phase-transfer catalyst like a quaternary ammonium salt

Explanation: The addition of HBr to an alkene proceeds via a carbocation intermediate (Markovnikov's rule) to give the more substituted halide. However, in the presence of peroxides (ROOR), the mechanism switches to a free-radical chain reaction. The regiochemistry of the radical addition of HBr is anti-Markovnikov, meaning the bromine atom adds to the less substituted carbon of the double bond. This effect is specific to HBr.

Question 7

1. Adding a radical initiator, such as benzoyl peroxide.
2. Using N-bromosuccinimide (NBS) in place of Br₂.
3. Conducting the reaction with water present as the solvent or co-solvent. (correct answer)
4. Using exactly one equivalent of HBr instead of Br₂.

Explanation: In halohydrin formation, after the initial formation of the cyclic bromonium ion intermediate, a nucleophile attacks to open the ring. If water is present in high concentration (as the solvent), it outcompetes the bromide ion as the nucleophile, leading to the addition of -Br and -OH across the double bond.

Question 8

1. $Br_2$ in $CCl_4$ at room temperature in the dark (correct answer)
2. $HBr$ in the presence of peroxides under UV light
3. $NBS$ in $CCl_4$ with catalytic benzoyl peroxide
4. $Br_2$ in aqueous solution with $NaCl$ present

Explanation: $Br_2$ in $CCl_4$ (an inert, nonpolar solvent) proceeds through a bromonium ion intermediate, resulting in anti addition to give 1,2-dibromocyclohexane with anti stereochemistry. The reaction occurs readily at room temperature without light. Choice B would give monobromination through radical addition. Choice C (NBS with peroxides) is used for allylic bromination, not vicinal dibromination. Choice D would lead to bromohydrin formation due to water competition, not dibromination.

Question 9

1. $Pd/C$ with $H_2$ under atmospheric pressure
2. $Na$ in liquid $NH_3$ at $-78°C$
3. Lindlar catalyst ($Pd/CaCO_3/Pb$) with $H_2$ (correct answer)
4. $LiAlH_4$ in anhydrous ether at room temperature

Explanation: Lindlar catalyst (poisoned palladium) selectively reduces alkynes to alkenes with syn addition, giving Z stereochemistry. The lead poisoning prevents over-reduction to alkanes. Choice A ($Pd/C$) would reduce completely to the alkane. Choice B ($Na/NH_3$) gives E alkenes through a trans-reduction mechanism. Choice D ($LiAlH_4$) is not effective for alkyne reduction and is primarily used for carbonyl reductions.

Question 10

1. $Br_2$ with UV light at high temperature ($400°C$)
2. $Br_2$ with UV light at room temperature
3. $NBS$ with benzoyl peroxide in refluxing $CCl_4$ (correct answer)
4. $HBr$ with AIBN initiator at $80°C$

Explanation: NBS (N-bromosuccinimide) with peroxide initiation provides controlled radical bromination with selectivity for the most substituted (tertiary) positions due to radical stability. The low concentration of $Br_2$ generated in situ minimizes polybromination. Choice A uses harsh conditions leading to multiple bromination and decomposition. Choice B with $Br_2$ directly gives less selectivity and multiple products. Choice D uses $HBr$, which doesn't provide the radical source needed for substitution and would require an alkene substrate for addition.

Question 11

1. $CH_3OH$ as solvent with $NaBr$ at room temperature
2. $t-BuOK$ in $t-BuOH$ at elevated temperature (correct answer)
3. $H_2O$ as solvent with $NaCl$ at room temperature
4. $CH_3CH_2OH$ with catalytic $H^+$ at room temperature

Explanation: $t-BuOK$ in $t-BuOH$ at high temperature strongly favors elimination. The bulky, strong base ($t-BuO^-$) preferentially removes protons rather than acting as a nucleophile, the protic solvent stabilizes the leaving group, and elevated temperature kinetically favors elimination. Since 2-methyl-2-butanol is tertiary, it will undergo $E1$ elimination readily. Choices A and C use weak nucleophiles that won't effectively promote reaction. Choice D uses acidic conditions at low temperature, which would favor $S_N1$ over elimination.

Question 12

1. $BH_3 \cdot THF$ followed by $H_2O_2, OH^-$
2. $H_2O$ with catalytic $H_2SO_4$ (correct answer)
3. $Hg(OAc)_2, H_2O$ followed by $NaBH_4$
4. $OsO_4$ followed by $Me_2S$ workup

Explanation: Acid-catalyzed hydration ($H_2O/H_2SO_4$) proceeds through carbocation intermediates, allowing for rearrangement. The primary carbocation initially formed from 3-methyl-1-butene will rearrange to the more stable tertiary carbocation, leading to 3-methyl-2-butanol after water addition. Choice A (hydroboration-oxidation) avoids carbocations, giving anti-Markovnikov addition without rearrangement. Choice C (oxymercuration-demercuration) gives Markovnikov addition but suppresses rearrangements. Choice D forms diols, not alcohols.

Question 13

1. 1. OsO₄ (catalytic), NMO; 2. NaHSO₃/H₂O
2. 1. CH₃CO₃H (mCPBA); 2. H₃O⁺
   (correct answer)
3. Cold, dilute KMnO₄, NaOH
4. 1. O₃; 2. NaBH₄

Explanation: The target molecule is a trans-diol, which results from an anti-dihydroxylation. This is achieved by a two-step process: first, epoxidation of the alkene with a peroxyacid like mCPBA, followed by acid-catalyzed ring-opening of the epoxide. The acid-catalyzed opening involves a backside attack by water, resulting in the trans configuration. Options A and C result in syn-dihydroxylation, yielding the cis-diol.

Question 14

1. H₂ gas with a platinum (Pt) catalyst.
2. Sodium metal (Na) in liquid ammonia (NH₃).
3. H₂ gas with Lindlar's catalyst. (correct answer)
4. H₂ gas with a palladium (Pd/C) catalyst.

Explanation: The conversion of an internal alkyne to a (Z)- or cis-alkene requires partial reduction with syn-stereochemistry. Lindlar's catalyst (palladium poisoned with lead acetate and quinoline) is specifically designed for this transformation, stopping the reduction at the alkene stage and delivering both hydrogen atoms to the same face of the triple bond.

Question 15

1. Dimethyl sulfoxide (DMSO)
2. Methanol (CH₃OH) (correct answer)
3. Acetone (CH₃COCH₃)
4. Acetonitrile (CH₃CN)

Explanation: SN2 reactions with anionic nucleophiles are fastest in polar aprotic solvents (A, C, D) which solvate the cation but leave the anion relatively 'naked' and reactive. Polar protic solvents like methanol (B) form strong hydrogen bonds with the azide anion (N₃⁻), creating a solvent shell that stabilizes the nucleophile, reduces its energy, and sterically hinders it from attacking the electrophile, thus slowing the reaction rate significantly.

Question 16

1. Sodium hydroxide (NaOH) (pKa of H₂O ≈ 15.7)
2. Sodium hydride (NaH) (pKa of H₂ ≈ 35) (correct answer)
3. Sodium bicarbonate (NaHCO₃) (pKa of H₂CO₃ ≈ 6.4)
4. Ammonia (NH₃) (pKa of NH₄⁺ ≈ 9.2)

Explanation: For deprotonation to be essentially complete, the base used must have a conjugate acid with a much higher pKa than the alcohol being deprotonated. The pKa of H₂, the conjugate acid of hydride (H⁻), is about 35, which is much higher than ethanol's pKa of 16. This large difference drives the acid-base reaction to completion. Sodium hydroxide (A) is a poor choice because its conjugate acid (water) has a pKa similar to ethanol, leading to an equilibrium mixture. Bicarbonate and ammonia are far too weak to deprotonate ethanol.

Question 17

1. Thionyl chloride (SOCl₂) in pyridine
2. Sodium chloride (NaCl) in DMSO
3. Phosphorus trichloride (PCl₃)
4. Cold, concentrated hydrochloric acid (HCl) (correct answer)

Explanation: Tertiary alcohols react readily with concentrated hydrogen halides via an SN1 mechanism. The strong acid protonates the hydroxyl group, converting it into an excellent leaving group (H₂O). Departure of water forms a stable tertiary carbocation, which is then rapidly trapped by the chloride ion. Reagents like SOCl₂ (A) and PCl₃ (C) are typically used for primary and secondary alcohols and are less effective for tertiary alcohols, where elimination can be a problem. NaCl (B) is not reactive enough as the hydroxyl group is a poor leaving group.

Question 18

1. $NaF$ in acetone under reflux conditions
2. $AgF$ in methanol with heating
3. $HF$ in aqueous solution at room temperature
4. $KF$ with 18-crown-6 in acetonitrile (correct answer)

Explanation: This question tests your understanding of nucleophilic substitution reactions and the challenges of halogen exchange, particularly the preparation of alkyl fluorides. The key insight is recognizing that fluoride ion ($F^-$) is a poor nucleophile in typical solvents due to its high charge density and strong solvation. Option D is correct because $KF$ with 18-crown-6 in acetonitrile creates optimal conditions for halogen exchange. The crown ether complexes with the $K^+$ ion, effectively separating it from $F^-$ and creating a "naked" fluoride anion with enhanced nucleophilicity. Acetonitrile is an aprotic solvent that doesn't heavily solvate the fluoride ion, further increasing its reactivity. Option A fails because $NaF$ in acetone provides poor solubility for the fluoride salt and insufficient activation of the fluoride nucleophile. Option B using $AgF$ in methanol is problematic because methanol is protic and will extensively hydrogen-bond to fluoride, reducing its nucleophilicity. Additionally, $AgF$ can lead to competing oxidation reactions. Option C with $HF$ in aqueous solution won't work because $HF$ is a weak acid that doesn't dissociate significantly in water, providing very low concentrations of $F^-$ ions. Remember this principle: when you need to use fluoride as a nucleophile, look for conditions that minimize solvation and maximize the "nakedness" of the fluoride ion. Crown ethers with aprotic solvents are classic combinations for activating poorly nucleophilic anions like fluoride.

Question 19

1. Low temperature ($0°C$) with added $CH_3ONa$
2. Low temperature ($0°C$) in DMSO solvent
3. High temperature ($65°C$) with added $CH_3ONa$
4. Room temperature with methanol as the only nucleophile (correct answer)

Explanation: When you encounter substitution reactions, you need to determine whether conditions favor $S_N1$ (unimolecular) or $S_N2$ (bimolecular) mechanisms. The key factors are substrate structure, nucleophile strength, temperature, and solvent properties. 2-bromo-2-methylbutane is a tertiary alkyl halide, which strongly favors $S_N1$ due to the stable tertiary carbocation it forms. However, reaction conditions can still influence which mechanism predominates. Option D creates ideal $S_N1$ conditions because methanol acts as both solvent and a weak nucleophile. Weak nucleophiles favor $S_N1$ since they don't compete effectively in the direct attack required for $S_N2$. Room temperature provides sufficient energy for carbocation formation without being so high that elimination reactions dominate. Option A is wrong because $CH_3ONa$ is a strong nucleophile that would push toward $S_N2$, and low temperature slows carbocation formation needed for $S_N1$. Option B fails because DMSO is an aprotic solvent that enhances nucleophilicity and favors $S_N2$, while low temperature again disfavors carbocation formation. Option C combines strong nucleophile ($CH_3ONa$) with high temperature - while heat might favor $S_N1$, the strong base would likely cause elimination reactions (E2) to compete heavily. Remember this pattern: $S_N1$ is favored by weak nucleophiles, protic solvents, and moderate temperatures with tertiary substrates. Strong nucleophiles and aprotic solvents push toward $S_N2$, even with tertiary substrates.

Question 20

1. $HBr$ in the presence of $H_2SO_4$ at elevated temperature
2. $PBr_3$ in anhydrous conditions at room temperature (correct answer)
3. $NaBr$ in acetone with catalytic $KI$ under reflux
4. $Br_2$ in $CCl_4$ under UV light irradiation

Explanation: $PBr_3$ under anhydrous conditions is the best choice for converting primary alcohols to alkyl bromides. This reagent proceeds via an $S_N2$ mechanism with inversion, giving clean conversion without carbocation rearrangements or elimination side reactions. Choice A ($HBr/H_2SO_4$) would promote elimination and potential rearrangements at high temperature. Choice C represents an $S_N2$ reaction but requires a good leaving group already present (not applicable to alcohols). Choice D is for radical bromination of alkanes, not alcohol conversion.