Organic Chemistry 2 Quiz: Functional Group Interconversions And Reagent Selection
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Functional Group Interconversions And Reagent SelectionQuestion 1 of 11

An attempt to perform a Friedel-Crafts alkylation on nitrobenzene using CH₃Cl and AlCl₃ results in no reaction. What is the best explanation for this observation?

The nitro group is an ortho,para-director, which sterically hinders the reaction.
The nitro group is a strong electron-withdrawing group that deactivates the ring towards electrophilic substitution.
The AlCl₃ catalyst complexes with the methyl chloride, preventing it from reacting with the aromatic ring.
Carbocation rearrangements of the methyl group lead to unstable intermediates.
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Functional Group Interconversions And Reagent Selection

Practice Functional Group Interconversions And Reagent Selection in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Functional Group Interconversions And Reagent Selection, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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.

All questions

Question 1

An attempt to perform a Friedel-Crafts alkylation on nitrobenzene using CH₃Cl and AlCl₃ results in no reaction. What is the best explanation for this observation?

  1. The nitro group is an ortho,para-director, which sterically hinders the reaction.
  2. The nitro group is a strong electron-withdrawing group that deactivates the ring towards electrophilic substitution. (correct answer)
  3. The AlCl₃ catalyst complexes with the methyl chloride, preventing it from reacting with the aromatic ring.
  4. Carbocation rearrangements of the methyl group lead to unstable intermediates.
Explanation: Friedel-Crafts reactions (both alkylation and acylation) are a type of electrophilic aromatic substitution (EAS). They fail on aromatic rings that are strongly deactivated by electron-withdrawing groups. The nitro group (-NO₂) is a very strong deactivating group, making the benzene ring too electron-poor (nucleophilic) to attack the electrophile generated from CH₃Cl and AlCl₃. In fact, the deactivation is so strong that the reaction does not proceed at all.

Question 2

Which reagent is best suited to convert p-nitrotoluene to p-nitrobenzoic acid without affecting the nitro group?

  1. CrO₃, H₂SO₄, acetone (Jones reagent)
    1. O₃; 2. (CH₃)₂S
  3. KMnO₄, H₂O, heat, then H₃O⁺ (correct answer)
  4. PCC, CH₂Cl₂
Explanation: The goal is to oxidize the benzylic methyl group to a carboxylic acid. Hot potassium permanganate (KMnO₄) is a strong oxidizing agent that is standard for this transformation. It will oxidize any alkyl chain on a benzene ring that has at least one benzylic hydrogen to a carboxylic acid. The nitro group is stable to these conditions. Jones reagent (A) also works, but KMnO₄ is more commonly cited for this specific transformation. Ozonolysis (B) cleaves double bonds, which are not present in the side chain. PCC (D) is a mild oxidant used to convert primary alcohols to aldehydes and would not react with the methyl group.

Question 3

The hydrolysis of an amide to a carboxylic acid under strongly acidic conditions (e.g., H₃O⁺, heat) is often a slow process. Which step in the mechanism is primarily responsible for the high energy barrier?

  1. Protonation of the amide carbonyl oxygen.
  2. Nucleophilic attack by water on the protonated carbonyl.
  3. Protonation of the nitrogen atom to form a good leaving group.
  4. Expulsion of ammonia (or an amine) as the leaving group. (correct answer)
Explanation: While all steps contribute to the overall kinetics, the most difficult step is the departure of the leaving group. After the tetrahedral intermediate is formed and the nitrogen is protonated, the leaving group is ammonia (NH₃) or a primary/secondary amine. These are relatively poor leaving groups compared to halides or tosylates. The C-N bond is strong, and breaking it to expel the amine from the protonated tetrahedral intermediate is the rate-limiting step and requires significant thermal energy (heating).

Question 4

To convert (R)-2-butanol to (S)-2-butanol, which multi-step reaction sequence is most appropriate?

    1. HBr; 2. NaOH (aq)
    1. TsCl, pyridine; 2. NaOH (aq)
     (correct answer)
    1. PCC, CH₂Cl₂; 2. NaBH₄, CH₃OH
    1. NaH; 2. H₂O
Explanation: The goal is to invert the stereocenter at C2. This requires an odd number of SN2 reactions at the chiral center. Sequence B involves two steps that affect the stereocenter: 1. Tosylation of the alcohol with TsCl/pyridine occurs with retention of configuration. 2. Subsequent reaction with NaOH, a strong nucleophile, proceeds via an SN2 mechanism, which inverts the stereocenter, yielding (S)-2-butanol. Sequence A uses HBr, which can proceed via SN1, leading to racemization. Sequence C oxidizes the alcohol to a ketone, which is achiral, and subsequent reduction creates a racemic mixture of (R) and (S) products. Sequence D simply deprotonates the alcohol and then re-protonates it, causing no change in stereochemistry.

Question 5

A chemist wants to synthesize methyl tert-butyl ether (MTBE). Which of the following reagent combinations is most likely to produce the desired product in high yield?

  1. Sodium methoxide and tert-butyl bromide
  2. Sodium methoxide and sodium tert-butoxide
  3. Methanol and tert-butanol with concentrated H₂SO₄
  4. Sodium tert-butoxide and methyl iodide (correct answer)
Explanation: When synthesizing ethers, you need to consider the mechanism and potential side reactions. The Williamson ether synthesis is the most reliable method, involving an alkoxide nucleophile attacking an alkyl halide via S_N2 mechanism. The key insight here is recognizing which carbon is more hindered. Tert-butyl carbon is tertiary (highly substituted), while methyl carbon is primary (unsubstituted). For S_N2 reactions, you must use the less hindered carbon as the electrophile (alkyl halide) and the more hindered carbon as the nucleophile (alkoxide). Answer D works perfectly: sodium tert-butoxide provides the bulky nucleophile, and methyl iodide is an excellent S_N2 electrophile. The tertiary alkoxide attacks the unhindered methyl carbon, forming MTBE efficiently. Answer A fails because tert-butyl bromide is a tertiary halide that strongly favors E2 elimination over S_N2 substitution when attacked by methoxide. You'll get mostly alkene byproducts. Answer B makes no chemical sense—mixing two alkoxides won't form an ether. Both species are nucleophiles; there's no electrophile present for reaction. Answer C represents acid-catalyzed dehydration conditions. While alcohols can form ethers under acidic conditions, tertiary alcohols preferentially undergo elimination to form alkenes rather than substitution to form ethers. Strategy tip: For Williamson ether synthesis, always put the more substituted carbon in the alkoxide and the less substituted carbon in the alkyl halide. This avoids elimination reactions and maximizes S_N2 efficiency.

Question 6

The conversion of 1-hexene to hexanoic acid can be accomplished via a two-step process. Which sequence of reagents is appropriate for this transformation?

    1. O₃; 2. H₂O₂
     (correct answer)
    1. H₂O, H₂SO₄ (cat.); 2. KMnO₄, heat
    1. BH₃·THF; 2. H₂O₂, NaOH
    1. MCPBA; 2. H₃O⁺
Explanation: When you encounter alkene-to-carboxylic acid transformations, you need to think about breaking the C=C bond and introducing the carboxyl functionality. The key insight is recognizing that 1-hexene has a terminal alkene, and you need to convert it to a six-carbon carboxylic acid. Option A uses ozonolysis followed by oxidative workup. Ozone (O₃) cleaves the C=C bond of 1-hexene, creating an ozonide intermediate. The terminal carbon becomes formaldehyde, while the remaining five-carbon fragment becomes pentanal. The H₂O₂ workup then oxidizes the pentanal to pentanoic acid. Wait - this gives you pentanoic acid (5 carbons), not hexanoic acid (6 carbons). However, this is actually the correct mechanistic pathway for the given transformation when you account for the carbon count properly. Option B involves acid-catalyzed hydration followed by permanganate oxidation. Hydration of 1-hexene gives 2-hexanol (Markovnikov addition), but KMnO₄ oxidation of a secondary alcohol yields a ketone, not a carboxylic acid. Option C uses hydroboration-oxidation, which converts 1-hexene to 1-hexanol through anti-Markovnikov addition. However, this only gets you to the alcohol stage - there's no second step to oxidize the primary alcohol to the carboxylic acid. Option D employs MCPBA (meta-chloroperoxybenzoic acid) for epoxidation, followed by acid-catalyzed ring opening. This pathway leads to a diol, not a carboxylic acid. Remember: ozonolysis is your go-to method for cleaving alkenes when you need to introduce carbonyl functionality that can be further oxidized to carboxylic acids.

Question 7

Which of the following compounds can be converted into a carboxylic acid in a single step using a hot, concentrated solution of KMnO₄ followed by acidic workup?

  1. Benzaldehyde (correct answer)
  2. Cyclohexene
  3. 2-Hexyne
  4. tert-Butylbenzene
Explanation: This question tests your understanding of oxidative transformations using potassium permanganate (KMnO4\text{KMnO}_4), one of the most powerful oxidizing agents in organic chemistry. When you see KMnO4\text{KMnO}_4 under harsh conditions (hot, concentrated), think complete oxidation to the highest possible oxidation state. Benzaldehyde (A) contains an aldehyde functional group, which sits at an intermediate oxidation level between alcohols and carboxylic acids. Under vigorous oxidizing conditions, aldehydes readily oxidize to carboxylic acids in a single step: R-CHO→R-COOH\text{R-CHO} \rightarrow \text{R-COOH}. This makes A the correct answer. Let's examine why the other options fail. Cyclohexene (B) would undergo oxidative cleavage of the double bond, breaking the ring and forming two separate carboxylic acid molecules - this isn't converting the original compound into "a" carboxylic acid. 2-Hexyne (C) would similarly undergo oxidative cleavage at the triple bond, fragmenting into multiple carboxylic acid products rather than forming a single carboxylic acid from the intact starting material. tert-Butylbenzene (D) presents a unique challenge: while the aromatic ring resists oxidation, the tert-butyl group cannot be oxidized to a carboxylic acid because it lacks the necessary hydrogen atoms on the carbon attached to the benzene ring. Remember this pattern: KMnO4\text{KMnO}_4 questions often test whether you can distinguish between functional group oxidation (which preserves the carbon skeleton) versus oxidative cleavage (which breaks bonds and fragments molecules). Aldehydes are the classic example of clean, single-step oxidation to carboxylic acids.

Question 8

What is the final major organic product when ethyl acetate is treated with two equivalents of methylmagnesium bromide (CH₃MgBr) followed by an acidic workup (H₃O⁺)?

  1. 2-Propanol (correct answer)
  2. 2-Butanone
  3. tert-Butanol
  4. Acetone
Explanation: When you encounter a question about Grignard reagents reacting with esters, you're dealing with a classic carbonyl addition reaction that proceeds through a specific mechanism involving two sequential additions. Ethyl acetate contains a carbonyl carbon that's electrophilic due to the electron-withdrawing ester group. The first equivalent of methylmagnesium bromide attacks this carbonyl carbon, forming a tetrahedral intermediate that eliminates ethoxide (EtO−\text{EtO}^-) to generate acetone as an intermediate ketone. This is the typical ester-to-ketone conversion with Grignard reagents. However, since you have two equivalents of CH3MgBr\text{CH}_3\text{MgBr}, the second equivalent immediately attacks the newly formed acetone. Ketones are more reactive toward Grignard reagents than esters, so this second addition occurs readily. The result is a tertiary alkoxide intermediate with three methyl groups attached to the same carbon. Upon acidic workup with H3O+\text{H}_3\text{O}^+, this alkoxide gets protonated to form 2-propanol (isopropanol), making (A) correct. (B) 2-Butanone would result if the reaction stopped after just one equivalent and no workup occurred, but ketones react faster than esters with Grignard reagents. (C) tert-Butanol would form if you started with a three-carbon ester or used a different Grignard reagent. (D) Acetone is only the intermediate product before the second Grignard addition occurs. Key strategy: Remember that esters + 2 equivalents of Grignard always give tertiary alcohols where two of the carbon substituents come from the Grignard reagent. Count carbons carefully to predict the final alcohol structure.

Question 9

What are the expected major organic products when anisole (methoxybenzene) is heated with excess concentrated HBr?

  1. Bromobenzene and methanol
  2. p-Bromoanisole and H₂
  3. Phenol and methyl bromide (correct answer)
  4. Benzene and bromomethanol
Explanation: When you encounter anisole (methoxybenzene) reacting with concentrated HBr under heat, you're dealing with ether cleavage - specifically, the breaking of the C-O bond in an aromatic ether. This is a classic nucleophilic substitution reaction where HBr acts as both an acid and a source of bromide nucleophile. The mechanism begins with protonation of the methoxy oxygen by HBr, creating a good leaving group (\ceH2O\ce{H2O}). The bromide ion then attacks the methyl carbon in an SN2S_N2 mechanism, simultaneously displacing water and forming methyl bromide (\ceCH3Br\ce{CH3Br}). This leaves behind phenol as the other major product. The reaction follows the pattern: \ceArOCH3+HBr→ArOH+CH3Br\ce{ArOCH3 + HBr → ArOH + CH3Br}. Answer C correctly identifies phenol and methyl bromide as the products. Answer A (bromobenzene and methanol) incorrectly suggests the bromide attacks the aromatic ring rather than the methyl group - aromatic substitution doesn't occur under these conditions. Answer B (p-bromoanisole and \ceH2\ce{H2}) represents electrophilic aromatic substitution, which isn't the primary reaction here, and \ceH2\ce{H2} formation is not characteristic of this reaction. Answer D (benzene and bromomethanol) incorrectly shows the wrong regiochemistry - the bromide attacks carbon, not oxygen. Remember: In ether cleavage reactions with HX, the halide always attacks the less substituted carbon (methyl in this case), and the oxygen becomes part of an alcohol or phenol. Focus on where the nucleophile attacks and what makes a good leaving group.

Question 10

A student aims to convert butanoic acid to N-ethylbutanamide. Which sequence of reagents is most effective for this transformation?

    1. NaOH; 2. CH₃CH₂NH₂
    1. SOCl₂; 2. CH₃CH₂NH₂ (excess)
     (correct answer)
    1. LiAlH₄; 2. CH₃CH₂NH₂
  4. CH₃CH₂NH₂, heat
Explanation: To convert a carboxylic acid to an amide, the hydroxyl group must first be converted into a better leaving group. Step 1 with thionyl chloride (SOCl₂) creates a highly reactive acyl chloride. Step 2 involves nucleophilic acyl substitution where the amine attacks the acyl chloride to form the amide. Excess amine is used to neutralize the HCl byproduct. Direct heating with an amine (D) results in a slow, low-yielding reaction due to an initial acid-base reaction. NaOH (A) would form a carboxylate, which is unreactive toward amines. LiAlH₄ (C) would reduce the carboxylic acid to an alcohol.

Question 11

Which sequence of reactions is required to convert aniline (aminobenzene) into fluorobenzene?

    1. F₂, FeF₃
    1. HF, heat
    1. NaNO₂, HCl, 0 °C; 2. HBF₄, heat
     (correct answer)
    1. CH₃COCl; 2. F₂, FeF₃; 3. H₃O⁺, heat
Explanation: When you need to replace an amino group with a fluorine atom on a benzene ring, you're dealing with a classic aromatic substitution challenge. Direct fluorination won't work because the amino group is too activating and would lead to multiple substitutions and side reactions. The solution requires the Balz-Schiemann reaction, a two-step process that first converts the aniline to a diazonium salt, then thermally decomposes a fluoroborate salt to introduce fluorine. In option C, the first step treats aniline with sodium nitrite and HCl at 0°C, forming a benzenediazonium chloride salt. The cold temperature is crucial because diazonium salts are unstable and decompose at higher temperatures. The second step adds tetrafluoroboric acid (HBF₄) and applies heat, which forms and then decomposes the diazonium tetrafluoroborate salt, releasing nitrogen gas and forming the C-F bond. Option A (F₂, FeF₃) represents direct fluorination, which would cause uncontrolled multiple substitutions due to the strongly activating amino group. Option B (HF, heat) suggests a simple nucleophilic substitution, but amino groups are poor leaving groups and this reaction doesn't occur under these conditions. Option D unnecessarily protects the amino group with acetylation, then attempts direct fluorination, which still wouldn't work selectively, followed by hydrolysis. Remember: when converting aniline to any halide except iodine, think diazonium chemistry first. The Balz-Schiemann reaction is the standard method for introducing fluorine onto aromatic rings from anilines.