Organic Chemistry Quiz: Multi Step Synthesis Functional Group Interconversions

What this quiz covers

This quiz focuses on Multi Step Synthesis Functional Group Interconversions, 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. Treatment with POCl₃ in pyridine at 0°C followed by warming to room temperature (correct answer)
2. Conversion to tosylate followed by E2 elimination with potassium tert-butoxide in DMSO
3. Treatment with concentrated H₂SO₄ at 170°C under standard heating conditions
4. Dehydration with Al₂O₃ at 350°C in the gas phase with short contact time

Explanation: When converting alcohols to alkenes, the key challenge is achieving regioselectivity—forming the desired terminal alkene rather than more thermodynamically stable internal alkenes. Different elimination methods favor different products based on their mechanisms and conditions. The POCl₃/pyridine method (option A) is particularly selective for terminal alkenes because it proceeds through a specific mechanism. POCl₃ first converts the alcohol to a chlorophosphate ester, then pyridine acts as both base and nucleophile to eliminate HCl. The low temperature (0°C) and controlled warming minimizes carbocation rearrangements and favors kinetic control, producing primarily 1-hexene. Option B is incorrect because E2 elimination with a bulky base like potassium tert-butoxide typically favors the less substituted alkene (Hofmann product), but the tosylate formation and DMSO conditions can still lead to significant internal alkene formation through competing pathways. Option C fails because concentrated H₂SO₄ at high temperature (170°C) promotes acid-catalyzed dehydration through carbocation intermediates. These conditions strongly favor thermodynamic control, yielding predominantly internal alkenes (2-hexene, 3-hexene) since they're more stable than 1-hexene. Option D is wrong because high-temperature gas-phase elimination over Al₂O₃, despite short contact time, still operates under thermodynamic control due to the extreme temperature (350°C), favoring internal alkenes. Study tip: For maximum selectivity toward terminal alkenes from primary alcohols, remember that mild, non-acidic conditions with controlled temperatures (like POCl₃/pyridine) prevent carbocation formation and thermodynamic equilibration that would favor internal products.

Question 2

1. Acid-catalyzed hydration followed by oxidation with PCC in dichloromethane
2. Hydroboration-oxidation to form primary alcohol, then oxidation with PCC (correct answer)
3. Oxymercuration-demercuration followed by Jones oxidation in acetone
4. Ozonolysis with reductive workup using dimethyl sulfide at low temperature

Explanation: Hydroboration-oxidation gives anti-Markovnikov addition without rearrangement, forming 3-methyl-1-butanol, then PCC oxidizes the primary alcohol to the aldehyde. Choice A (acid-catalyzed hydration) would give Markovnikov addition to form a secondary alcohol, which cannot be oxidized to an aldehyde. Choice C has the same problem with Markovnikov addition. Choice D (ozonolysis) would cleave the C=C bond entirely, destroying the carbon framework needed for 3-methylbutanal.

Question 3

1. The initial bromination reaction does not occur due to the steric hindrance of the tert-butyl group.
2. A carbocation rearrangement occurs during the bromination step, leading to an incorrect dihalide structure.
3. The intermediate vinyl halide formed after the first elimination cannot undergo the second elimination. (correct answer)
4. The base NaNH₂ is not strong enough to perform the second elimination step on the vinyl halide.

Explanation: Let's trace the reaction. Step 1: Bromination of 3,3-dimethyl-1-butene ((CH₃)₃C-CH=CH₂) correctly yields the vicinal dihalide, 1,2-dibromo-3,3-dimethylbutane ((CH₃)₃C-CH(Br)CH₂(Br)). Step 2 is a double dehydrohalogenation using sodium amide. The first elimination removes HBr to form a vinyl halide. The most likely first elimination removes the proton from C-2 to form (CH₃)₃C-C(Br)=CH₂. For the second E2 elimination to occur to form an alkyne, there must be a proton on the carbon adjacent to the one bearing the leaving group (bromine). In the intermediate (CH₃)₃C-C(Br)=CH₂, the adjacent carbon is the quaternary carbon of the tert-butyl group, which has no protons. Therefore, the second elimination is impossible. Distractor A is incorrect; while steric hindrance may slow the reaction, bromination of hindered alkenes still proceeds. Distractor B is incorrect; the addition of Br₂ to an alkene proceeds via a bridged bromonium ion intermediate, not a free carbocation, so rearrangements do not occur. Distractor D is incorrect; NaNH₂ is an extremely strong base, more than sufficient to dehydrohalogenate a vinyl halide if the required proton were present.

Question 4

1. 1. HBr; 2. Mg, ether
   (correct answer)
2. 1. Br₂, CCl₄; 2. Mg, ether
3. 1. HBr, ROOR; 2. Mg, ether
4. 1. Mg, ether; 2. HBr

Explanation: To form isopropylmagnesium bromide, one needs to react isopropyl bromide (2-bromopropane) with magnesium metal in ether. The task is to synthesize 2-bromopropane from propene. Route A is correct. Step 1 is the electrophilic addition of HBr to propene. This reaction follows Markovnikov's rule, where the bromine atom adds to the more substituted carbon (C-2), yielding 2-bromopropane. Step 2 is the standard formation of a Grignard reagent from an alkyl halide. Route B is incorrect. Bromination of propene yields 1,2-dibromopropane, which would form a di-Grignard reagent or undergo other reactions, but would not produce the desired isopropylmagnesium bromide. Route C is incorrect. The addition of HBr in the presence of peroxides (ROOR) follows an anti-Markovnikov pathway, yielding 1-bromopropane. This would lead to the formation of n-propylmagnesium bromide, not isopropylmagnesium bromide. Route D has the reagents in the wrong order. Magnesium does not react with an alkene, and HBr would react with the Grignard reagent if it were somehow formed.

Question 5

1. Step 1 produces 2-bromo-2-methylpropane due to Markovnikov addition, which then undergoes an Sₙ1 reaction.
2. The intermediate alkyl halide is tertiary and sterically hindered, preventing the Sₙ2 reaction in Step 2.
3. The NaOCH₃ in Step 2 is a strong base, causing the intermediate alkyl halide to undergo significant E2 elimination. (correct answer)
4. A carbocation rearrangement occurs in Step 1, leading to the formation of a different constitutional isomer.

Explanation: Let's analyze the proposed synthesis. Step 1: 2-methylpropene reacts with HBr in the presence of peroxides (ROOR). This is an anti-Markovnikov radical addition, which correctly produces the desired intermediate, 1-bromo-2-methylpropane (a primary alkyl halide). Step 2: 1-bromo-2-methylpropane reacts with sodium methoxide (NaOCH₃). NaOCH₃ is both a strong nucleophile and a strong base. The substrate is a primary alkyl halide, but it is sterically hindered at the β-carbon. Therefore, both Sₙ2 (to give the desired 1-methoxy-2-methylpropane) and E2 (to give 2-methylpropene) reactions will occur. Due to the strength of the base and the steric hindrance, the E2 elimination reaction is a significant competing pathway, which reduces the yield of the desired substitution product. Distractor A is incorrect because HBr with ROOR gives anti-Markovnikov addition, not Markovnikov. Distractor B is incorrect because the intermediate, 1-bromo-2-methylpropane, is a primary alkyl halide, not tertiary. Distractor D is incorrect because the radical addition mechanism in Step 1 does not involve carbocations and is not prone to rearrangement.

Question 6

1. Treatment with NaBr and H₂SO₄ at room temperature
2. Treatment with HBr at elevated temperature in the presence of H₂SO₄
3. Treatment with PBr₃ at room temperature under anhydrous conditions (correct answer)
4. Treatment with NaBr alone at elevated temperature under basic conditions

Explanation: PBr₃ converts primary alcohols to alkyl bromides via an SN2 mechanism with inversion, giving clean conversion without rearrangement or elimination. Choice A (NaBr/H₂SO₄) would not work effectively as NaBr is not sufficiently nucleophilic in acidic conditions. Choice B (HBr/H₂SO₄) at high temperature would favor elimination (E2/E1) over substitution, producing butenes as major products. Choice D is impossible as NaBr alone cannot activate the alcohol leaving group.

Question 7

1. A racemic mixture of (2R,3S)- and (2S,3R)-2-bromo-3-methylpentane
2. (S)-1-bromo-3-methylpentane
3. 3-bromo-3-methylpentane (correct answer)
4. A mixture of (E)- and (Z)-3-methyl-2-pentene

Explanation: Step 1 is the acid-catalyzed hydration of an alkene. The proton adds to the less substituted carbon (C-1) of (S)-3-methyl-1-pentene, generating a secondary carbocation at C-2. This secondary carbocation is adjacent to a tertiary carbon (C-3). A 1,2-hydride shift will occur, moving a hydrogen from C-3 to C-2. This forms a more stable tertiary carbocation at C-3. Water then attacks this planar, achiral carbocation to form 3-methyl-3-pentanol, which is an achiral tertiary alcohol. Step 2 involves the reaction of this tertiary alcohol with PBr₃. This reaction proceeds via an Sₙ1-like mechanism, where the hydroxyl group is converted into a good leaving group and departs to reform the stable tertiary carbocation at C-3. The bromide ion then attacks this carbocation to yield the final product, 3-bromo-3-methylpentane, which is also achiral. Distractor A represents the product that would form if no rearrangement occurred and the reaction of the secondary alcohol with PBr₃ proceeded with inversion and retention. Distractor B would result from an anti-Markovnikov addition followed by substitution. Distractor D represents elimination products.

Question 8

1. 1. NaH on isopropanol; 2. CH₃I
   (correct answer)
2. 1. NaH on methanol; 2. 2-iodopropane
3. Methanol and isopropanol with H₂SO₄, heat
4. 1. PBr₃ on isopropanol; 2. Methanol, heat

Explanation: This is a Williamson ether synthesis. The most effective route pairs a sterically unhindered alkyl halide with an alkoxide. The reaction proceeds via an Sₙ2 mechanism, which is sensitive to steric hindrance. Route A is the best choice. Step 1 deprotonates isopropanol to form the isopropoxide nucleophile. Step 2 reacts this nucleophile with methyl iodide, an unhindered methyl halide. This Sₙ2 reaction proceeds in high yield with minimal competing elimination. Route B is a poor choice. It pairs methoxide (a strong base/nucleophile) with 2-iodopropane (a secondary alkyl halide). This combination will result in a significant amount of E2 elimination product (propene) competing with the desired Sₙ2 reaction. Route C, acid-catalyzed condensation, is inefficient. It would lead to a statistical mixture of three ethers (dimethyl ether, diisopropyl ether, and the desired methyl isopropyl ether), as well as elimination byproducts. Route D is incorrect. Step 1 forms 2-bromopropane. Step 2 is an Sₙ1/E1 reaction with a poor nucleophile (methanol) and a secondary halide. This would be very slow and would favor the E1 elimination product (propene).

Question 9

1. 1. HBr; 2. NaC≡CH
2. 1. Br₂/CCl₄; 2. NaNH₂ (excess), heat
   (correct answer)
3. 1. H₂O, H₂SO₄; 2. H₂SO₄, heat; 3. Br₂
4. 1. Br₂/CCl₄; 2. KOC(CH₃)₃ (excess)

Explanation: To convert an alkene to an internal alkyne, a common strategy is addition of a halogen followed by a double elimination. The double elimination may also involve isomerization of the alkyne. Route B is correct. Step 1: Br₂ adds across the double bond of 1-butene to form 1,2-dibromobutane. Step 2: An excess of a very strong base, sodium amide (NaNH₂), is used to perform a double dehydrohalogenation. This initially forms 1-butyne. However, in the presence of hot NaNH₂, the terminal alkyne will isomerize to the more thermodynamically stable internal alkyne, 2-butyne. Route A is incorrect. Step 1 gives 2-bromobutane. The acetylide anion in Step 2 would act as a base, causing E2 elimination to give 2-butene. Route C is futile. It converts 1-butene to 2-butanol, then back to 2-butene (the major product). Route D is incorrect. While KOC(CH₃)₃ is a strong base, it is generally not strong enough to effectively perform the second elimination (from the intermediate vinyl bromide) to form the alkyne in high yield.

Question 10

1. 1. H₂O, H₂SO₄; 2. PCC
2. 1. O₃; 2. Zn, H₂O
3. 1. BH₃·THF, then H₂O₂, NaOH; 2. H₂CrO₄
4. 1. BH₃·THF, then H₂O₂, NaOH; 2. PCC
   (correct answer)

Explanation: The target molecule is propanal, an aldehyde. Aldehydes can be formed by the mild oxidation of primary alcohols. To get a primary alcohol (1-propanol) from propene requires an anti-Markovnikov hydration. Route D is correct. Step 1 (hydroboration-oxidation) is the classic method for anti-Markovnikov hydration of an alkene, yielding 1-propanol. Step 2 uses pyridinium chlorochromate (PCC), a mild oxidizing agent that oxidizes a primary alcohol to an aldehyde and stops, preventing over-oxidation to a carboxylic acid. Route A is incorrect. Step 1 (acid-catalyzed hydration) is a Markovnikov addition, which would yield 2-propanol (a secondary alcohol). Oxidation of 2-propanol with PCC would give acetone, a ketone. Route B is incorrect. Ozonolysis of propene cleaves the double bond, yielding formaldehyde and acetaldehyde, not propanal. Route C is incorrect. Step 1 correctly forms 1-propanol. However, Step 2 uses chromic acid (H₂CrO₄), a strong oxidizing agent, which would oxidize the primary alcohol all the way to a carboxylic acid (propanoic acid).

Question 11

1. 1. NaOH, heat; 2. HBr
2. 1. KOC(CH₃)₃; 2. HBr, ROOR
3. 1. Mg, ether; 2. H₂O; 3. Br₂, hν
4. 1. KOC(CH₃)₃; 2. HBr
   (correct answer)

Explanation: This transformation involves moving a functional group, which requires an elimination followed by an addition. Route D is the most efficient. Step 1: A strong, bulky base like potassium tert-butoxide (KOC(CH₃)₃) is used to perform an E2 elimination on the primary halide to form propene. Step 2: Addition of HBr to propene follows Markovnikov's rule, adding the bromine to the more substituted carbon (C-2) to give the desired 2-bromopropane. Route A is inefficient. Step 1 with NaOH (a strong nucleophile) on a primary halide would favor Sₙ2 substitution to give 1-propanol over elimination. Route B is incorrect. Step 1 correctly forms propene. However, Step 2 uses HBr with peroxides (ROOR), which causes an anti-Markovnikov addition, yielding 1-bromopropane, the original starting material. Route C is a viable but less efficient route. It converts 1-bromopropane to propane, then performs a radical bromination. Radical bromination of propane favors the secondary position, yielding 2-bromopropane as the major product, but it will also produce some 1-bromopropane, making it less clean than the elimination-addition sequence.

Question 12

1. The synthesis is effective; step 1 performs benzylic bromination and step 2 is an E2 elimination. (correct answer)
2. The synthesis fails because step 1 causes bromination of the aromatic ring instead of the ethyl group.
3. The synthesis fails because KOC(CH₃)₃ is a nucleophile and will cause an Sₙ2 reaction in step 2.
4. The synthesis is inefficient because step 1 produces 2-bromoethylbenzene as the major product.

Explanation: This synthesis is a standard method for creating a double bond adjacent to an aromatic ring. Choice A is correct. Step 1 (Br₂ with light, hν) is the condition for free-radical halogenation. On ethylbenzene, the reaction is highly selective for the benzylic position because the benzylic radical is resonance-stabilized. This correctly forms 1-bromo-1-phenylethane. Step 2 uses potassium tert-butoxide, a strong, bulky base, which is ideal for promoting E2 elimination to form the alkene, styrene, in high yield. Choice B is incorrect. Bromination of the aromatic ring requires a Lewis acid catalyst (like FeBr₃), not light. Choice C is incorrect. KOC(CH₃)₃ is a strong, bulky base. While it has some nucleophilicity, its steric hindrance makes it a very poor nucleophile, and it strongly favors elimination over substitution, especially with a secondary halide. Choice D is incorrect. Radical bromination is highly selective for the benzylic position (C-1 of the ethyl group); very little of the primary halide (2-bromo-1-phenylethane) would be formed.

Question 13

1. 1. HBr (1 eq); 2. Mg, ether; 3. CH₃CH₂Br
2. 1. NaNH₂, NH₃(l); 2. CH₃CH₂I
   (correct answer)
3. 1. BH₃·THF, then H₂O₂, NaOH; 2. CH₃CH₂MgBr
4. 1. H₂O, H₂SO₄, HgSO₄; 2. NaH; 3. CH₃CH₂I

Explanation: The starting material is 1-pentyne (a 5-carbon terminal alkyne). The target is 3-heptyne (a 7-carbon internal alkyne). This requires adding a two-carbon (ethyl) group to the alkyne. The most common method for alkylating a terminal alkyne is to deprotonate it to form a nucleophilic acetylide anion, followed by an Sₙ2 reaction with an alkyl halide. Route B is the correct method. Step 1: Sodium amide (NaNH₂) is a very strong base that deprotonates the terminal alkyne to form the sodium pentynide salt. Step 2: The pentynide anion acts as a nucleophile and attacks ethyl iodide in an Sₙ2 reaction, forming the new carbon-carbon bond and yielding 3-heptyne. Route A is incorrect. It attempts to form a vinyl Grignard reagent, which is not a standard or efficient way to achieve this transformation. Route C is incorrect. Step 1 converts the alkyne to pentanal. A Grignard reagent (Step 2) would react with the aldehyde, but this would lead to a secondary alcohol, not an alkyne. Route D is incorrect. Step 1 converts the alkyne to 2-pentanone. The subsequent steps are not chemically sound for reaching the target.

Question 14

1. 1. H₂, Lindlar's catalyst; 2. OsO₄, NMO
   (correct answer)
2. 1. Na, NH₃(l); 2. KMnO₄, NaOH, cold
3. 1. H₂, Lindlar's catalyst; 2. RCO₃H, then H₃O⁺
4. 1. Na, NH₃(l); 2. H₂O, H₂SO₄, HgSO₄

Explanation: The target molecule is meso-2,3-butanediol, which requires a syn-addition of two hydroxyl groups to cis-2-butene, or an anti-addition to trans-2-butene. Route A is correct: Step 1 (H₂, Lindlar's catalyst) reduces 2-butyne to cis-2-butene. Step 2 (OsO₄, NMO) performs a syn-dihydroxylation. The syn-addition to a cis-alkene results in the desired meso compound. Route B is incorrect: Step 1 (Na, NH₃) produces trans-2-butene. Step 2 (KMnO₄) is a syn-addition. Syn-addition to a trans-alkene yields a racemic mixture of enantiomers, not the meso compound. Route C is incorrect: Step 1 produces cis-2-butene. Step 2 (epoxidation followed by acid-catalyzed opening) is an anti-dihydroxylation. Anti-addition to a cis-alkene yields a racemic mixture. Route D is incorrect: Step 1 produces trans-2-butene. Step 2 is the reagent for alkyne hydration, which would produce 2-butanone, not a diol.

Question 15

1. Hydroboration-oxidation using BH₃·THF followed by H₂O₂/NaOH treatment
2. Acid-catalyzed hydration using dilute H₂SO₄ and water at moderate temperature
3. Oxymercuration-demercuration using Hg(OAc)₂/H₂O followed by NaBH₄ reduction (correct answer)
4. Treatment with concentrated HBr followed by substitution with hydroxide ion

Explanation: Oxymercuration-demercuration provides Markovnikov addition of water across the alkene without rearrangement, converting 2-methylpropene to 2-methylpropan-2-ol. Choice A (hydroboration-oxidation) gives anti-Markovnikov addition, producing 2-methylpropan-1-ol instead. Choice B (acid-catalyzed hydration) would work but is much slower for this substituted alkene and may give rearrangement products. Choice D involves two steps with potential elimination in the SN reaction and would not be regioselective.

Question 16

1. PBr₃, then NaNH₂ in liquid NH₃, followed by elimination with strong base
2. H₂SO₄/heat to form alkene, then Br₂ addition, followed by excess NaNH₂ treatment (correct answer)
3. SOCl₂ to form chloride, then NaCN substitution, followed by hydrolysis and decarboxylation
4. TsCl/pyridine to form tosylate, then NaC≡CH substitution in DMSO solvent

Explanation: The sequence alcohol → alkene (E2) → dibromide (addition) → alkyne (double E2 elimination) efficiently converts 1-pentanol to 1-pentyne. Two equivalents of NaNH₂ remove both HBr molecules. Choice A would not work as NaNH₂ cannot directly eliminate from alkyl bromides to form alkynes. Choice C involves nitrile chemistry which is not a direct route to terminal alkynes. Choice D attempts direct SN2 with acetylide, but primary tosylates with strong nucleophiles can give elimination as a competing reaction.

Question 17

1. Direct SN2 substitution with hydroxide ion in aqueous ethanol at elevated temperature
2. SN1 solvolysis in aqueous acetone followed by purification of the major enantiomer
3. Formation of Grignard reagent followed by quenching with water under anhydrous conditions
4. SN2 substitution with acetate ion, followed by basic hydrolysis of the ester product (correct answer)

Explanation: When you need to convert an alkyl halide to an alcohol while maintaining stereochemical purity, you must carefully consider how each mechanism affects the stereochemistry at the chiral center. The correct approach is D: SN2 substitution with acetate ion followed by basic hydrolysis. This two-step process works because SN2 reactions proceed with complete inversion of stereochemistry. Starting with (R)-2-bromobutane, the acetate ion attacks from the backside, producing (S)-2-butyl acetate with clean inversion. The subsequent basic hydrolysis of the ester occurs at the carbonyl carbon (not the chiral center), preserving the (S) configuration to give pure (S)-2-butanol. Option A fails because direct SN2 with hydroxide often leads to competing elimination reactions, especially at elevated temperatures, reducing yield and purity. Option B won't work because SN1 mechanisms proceed through a planar carbocation intermediate that loses stereochemical information, producing a racemic mixture of both (R) and (S) enantiomers. Option C is problematic because Grignard formation involves radical intermediates that scramble stereochemistry, and the subsequent protonation occurs randomly from either face of the carbanion, again leading to racemization. Study tip: Remember that SN2 reactions are your best friend for stereospecific transformations—they give predictable inversion. When direct SN2 with your desired nucleophile isn't practical, use a "good" nucleophile first (like acetate), then convert the product in a separate step that doesn't disturb the chiral center.

Question 18

1. Concentrated H₂SO₄ at 140°C with continuous removal of water vapor (correct answer)
2. Dilute H₂SO₄ at 180°C with excess alcohol present in solution
3. POCl₃ in pyridine at room temperature under anhydrous conditions
4. Concentrated HCl at 160°C with periodic addition of fresh alcohol

Explanation: Concentrated H₂SO₄ at 140°C with water removal favors E2 elimination over SN2 substitution that leads to ether formation. Removing water drives the equilibrium toward elimination. Choice B uses dilute acid and excess alcohol, promoting ether formation via SN2. Choice C (POCl₃/pyridine) works but is unnecessarily expensive and complex for this simple elimination. Choice D with HCl and added alcohol would favor SN2 substitution, forming significant amounts of cyclohexyl ethers.

Question 19

1. Ozonolysis (O₃) followed by oxidative workup with H₂O₂, then Wolff-Kishner reduction
2. Hydroboration-oxidation to form alcohol, then oxidation with Jones reagent (CrO₃/H₂SO₄)
3. Anti-Markovnikov hydration, then PCC oxidation, followed by haloform reaction with base
4. Radical bromination with NBS, then SN2 with cyanide, followed by acid hydrolysis (correct answer)

Explanation: Allylic bromination with NBS introduces Br at the methyl group, SN2 with CN⁻ forms the nitrile, and acid hydrolysis converts the nitrile to carboxylic acid. Choice A (ozonolysis) would cleave the double bond, destroying the carbon skeleton needed for 2-methylbutanoic acid. Choice B (hydroboration-oxidation) gives anti-Markovnikov alcohol addition, but Jones oxidation cannot convert a secondary alcohol to a carboxylic acid. Choice C has the same problem with anti-Markovnikov addition and PCC only oxidizes to ketones.