Organic Chemistry 2 Quiz: Multi Step Synthesis Planning
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Multi Step Synthesis PlanningQuestion 1 of 6

Which of the following sequences is the best method for preparing p-bromobenzyl alcohol from toluene?

  1. KMnO₄, H₂O, heat; 2. Br₂, FeBr₃; 3. LiAlH₄, Et₂O; 4. H₃O⁺ workup
  1. NBS, ROOR, heat; 2. Br₂, FeBr₃; 3. NaOH, H₂O
  1. Br₂, FeBr₃; 2. NBS, ROOR, heat; 3. NaOH, H₂O
  1. Br₂, FeBr₃; 2. KMnO₄, H₂O, heat; 3. SOCl₂; 4. NaBH₄, EtOH
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Multi Step Synthesis Planning

Practice Multi Step Synthesis Planning 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 Multi Step Synthesis Planning, 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

Which of the following sequences is the best method for preparing p-bromobenzyl alcohol from toluene?

    1. KMnO₄, H₂O, heat; 2. Br₂, FeBr₃; 3. LiAlH₄, Et₂O; 4. H₃O⁺ workup
    1. NBS, ROOR, heat; 2. Br₂, FeBr₃; 3. NaOH, H₂O
    1. Br₂, FeBr₃; 2. NBS, ROOR, heat; 3. NaOH, H₂O
     (correct answer)
    1. Br₂, FeBr₃; 2. KMnO₄, H₂O, heat; 3. SOCl₂; 4. NaBH₄, EtOH
Explanation: The key to this synthesis is the order of operations. The methyl group of toluene is an ortho,para-director. Therefore, electrophilic aromatic substitution (EAS) with Br₂/FeBr₃ should be done first to install the bromine at the para position, forming p-bromotoluene. Next, the benzylic methyl group is functionalized using N-bromosuccinimide (NBS) via a free-radical mechanism to form p-bromobenzyl bromide. Finally, the benzyl bromide is converted to the target alcohol via an SN2 reaction with aqueous NaOH.

Question 2

A synthesis of p-propylaniline is planned starting from benzene. Which reaction sequence is most likely to succeed?

    1. HNO₃, H₂SO₄; 2. CH₃CH₂COCl, AlCl₃; 3. Zn(Hg), HCl; 4. H₂, Pd/C
    1. CH₃CH₂COCl, AlCl₃; 2. HNO₃, H₂SO₄; 3. H₂, Pd/C
    1. CH₃CH₂CH₂Cl, AlCl₃; 2. HNO₃, H₂SO₄; 3. Fe, HCl
    1. CH₃CH₂COCl, AlCl₃; 2. Zn(Hg), HCl, heat; 3. HNO₃, H₂SO₄; 4. Fe, HCl
     (correct answer)
Explanation: To get a para-substituted pattern, an ortho,para-director must be on the ring before the second group is added. Route D is correct. First, a Friedel-Crafts acylation adds a propanoyl group. This is preferred over alkylation (Choice C) to avoid carbocation rearrangements and polyalkylation. Second, the ketone is reduced to an alkyl group using Clemmensen reduction (Zn(Hg), HCl), forming propylbenzene. The propyl group is an o,p-director. Third, nitration adds a nitro group, with the para isomer being the major product. Finally, the nitro group is reduced to an amine using Fe/HCl.

Question 3

Which set of reagents is most appropriate for the high-yield synthesis of 2-methylcyclohexanone from cyclohexanone?

    1. NaOH, H₂O; 2. CH₃I
    1. LDA, THF, -78 °C; 2. (CH₃)₃CBr
    1. CH₃MgBr, Et₂O; 2. H₂O
    1. LDA, THF, -78 °C; 2. CH₃I
     (correct answer)
Explanation: This transformation is an α-alkylation of a ketone. To achieve efficient mono-alkylation, a strong, non-nucleophilic, sterically hindered base is required to irreversibly and completely convert the ketone to its enolate. Lithium diisopropylamide (LDA) at low temperature is the ideal reagent for this, forming the kinetic enolate. A weak base like NaOH creates an equilibrium that can lead to polyalkylation and aldol side-products. The alkylating agent must be a good electrophile for SN2, like methyl iodide (CH₃I). A bulky halide like t-butyl bromide ((CH₃)₃CBr) would undergo elimination. A Grignard reagent (CH₃MgBr) would act as a nucleophile and add to the carbonyl carbon.

Question 4

To synthesize tert-butyl ethyl ether from ethanol and tert-butanol, which synthetic route is superior and why?

  1. Route 1: Convert ethanol to sodium ethoxide; react with tert-butyl bromide. This route is favored because SN2 reactions are fast on tertiary halides.
  2. Route 2: Convert tert-butanol to potassium tert-butoxide; react with ethyl bromide. This route is favored as it pairs a strong, bulky base with a primary halide, promoting SN2 over E2. (correct answer)
  3. Route 1: Convert ethanol to sodium ethoxide; react with tert-butyl bromide. This route is favored because ethoxide is a small, unhindered nucleophile.
  4. Route 2: Convert tert-butanol to potassium tert-butoxide; react with ethyl bromide. This route is disfavored because the bulky tert-butoxide will primarily cause elimination of ethyl bromide.
Explanation: This is a classic Williamson ether synthesis problem. To form the ether link, an alkoxide must act as a nucleophile on an alkyl halide. The reaction can be SN2 or E2. To favor SN2 and ether formation, the alkyl halide should be unhindered (methyl or primary). Route 2 correctly uses the primary halide (ethyl bromide) and the tertiary alkoxide (potassium tert-butoxide). While tert-butoxide is a strong base, it will act as a nucleophile with an unhindered primary halide. In contrast, Route 1 pairs a strong nucleophile/base (ethoxide) with a tertiary halide (tert-butyl bromide), a combination that will result almost exclusively in E2 elimination to form isobutylene, not the desired ether.

Question 5

What is the most efficient laboratory synthesis of N,N-diethylbenzamide from benzoic acid?

  1. Mix benzoic acid with excess diethylamine and heat at reflux for several hours.
    1. SOCl₂, pyridine; 2. excess (CH₃CH₂)₂NH
     (correct answer)
    1. CH₃OH, H⁺ catalyst; 2. excess (CH₃CH₂)₂NH
    1. LiAlH₄, Et₂O; 2. H₃O⁺; 3. PBr₃; 4. excess (CH₃CH₂)₂NH
Explanation: Directly reacting a carboxylic acid with an amine (Choice A) results in an acid-base reaction to form a stable salt, requiring very high temperatures to form an amide. The most efficient method is to first convert the carboxylic acid into a more reactive acyl derivative, such as an acid chloride. Thionyl chloride (SOCl₂) is a standard reagent for this conversion. The highly electrophilic benzoyl chloride then reacts readily with the nucleophilic diethylamine to form the target amide. Converting to an ester first (Choice C) is a viable but less direct two-step process. Route D is an incorrect sequence that reduces the acid and attempts substitution.

Question 6

Which combination of reactants is most suitable for the synthesis of N-ethylcyclohexylamine via reductive amination?

  1. Cyclohexanone, ethylamine, and NaBH₃CN (correct answer)
  2. Cyclohexanol, ethylamine, and H₂, Pd/C
  3. Cyclohexyl bromide, ethylamine, and a non-nucleophilic base
  4. N-ethylcyclohexanamide and LiAlH₄
Explanation: Reductive amination is a two-step process (often performed in one pot) that converts a ketone or aldehyde into an amine. First, the carbonyl compound (cyclohexanone) reacts with a primary amine (ethylamine) to form an imine intermediate. Then, a mild reducing agent, sodium cyanoborohydride (NaBH₃CN), reduces the imine to the target secondary amine (N-ethylcyclohexylamine). NaBH₃CN is selective for reducing the protonated imine over the ketone, which is crucial for the reaction's success. Choice D describes the reduction of an amide, which is another way to make amines, but is not reductive amination. Choice C describes an SN2 alkylation, which often leads to over-alkylation.