Organic Chemistry 2 Quiz: Benzylic Reactions Oxidation Radical Bromination
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Benzylic Reactions Oxidation Radical BrominationQuestion 1 of 13

Which sequence of reagents would be most effective for converting ethylbenzene into styrene (phenylethene)?

  1. Br₂, FeBr₃; 2. NaOH, heat
  1. KMnO₄, H₃O⁺, heat; 2. LiAlH₄
  1. NBS, hν; 2. KOtBu, heat
  1. H₂SO₄, SO₃; 2. NaOH, fusion
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Benzylic Reactions Oxidation Radical Bromination

Practice Benzylic Reactions Oxidation Radical Bromination 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 Benzylic Reactions Oxidation Radical Bromination, 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 sequence of reagents would be most effective for converting ethylbenzene into styrene (phenylethene)?

    1. Br₂, FeBr₃; 2. NaOH, heat
    1. KMnO₄, H₃O⁺, heat; 2. LiAlH₄
    1. NBS, hν; 2. KOtBu, heat
     (correct answer)
    1. H₂SO₄, SO₃; 2. NaOH, fusion
Explanation: This transformation is an elimination reaction. To achieve this, a leaving group must first be installed at the benzylic position. Step 1: Benzylic bromination using NBS and light (hν) converts ethylbenzene to 1-bromo-1-phenylethane. Step 2: A strong, bulky base like potassium tert-butoxide (KOtBu) is used to promote E2 elimination to form the alkene (styrene) as the major product, minimizing the competing SN2 reaction.

Question 2

A pure sample of (S)-1-phenylethanol is first converted to (S)-1-chloro-1-phenylethane using SOCl₂ with pyridine. The resulting chloride is then treated with NBS and UV light. What is the stereochemical outcome for the major organic product containing bromine?

  1. A racemic mixture of (R)- and (S)-1-bromo-1-chloro-1-phenylethane (correct answer)
  2. Only (S)-1-bromo-1-chloro-1-phenylethane
  3. Only (R)-1-bromo-1-chloro-1-phenylethane
  4. A mixture of diastereomeric products resulting from bromination of the methyl group
Explanation: The first step produces (S)-1-chloro-1-phenylethane with retention of configuration. The second step is a benzylic radical bromination. The benzylic carbon is also the stereocenter. The mechanism involves abstraction of the benzylic hydrogen to form a planar, achiral radical intermediate. The bromine radical can then attack this planar intermediate from either face with equal probability, leading to the formation of a racemic mixture of the (R) and (S) enantiomers.

Question 3

The radical bromination of toluene with NBS is highly selective for the benzylic position over any C-H bond on the aromatic ring. What is the primary reason for this selectivity?

  1. The benzylic radical intermediate is stabilized by resonance, lowering the activation energy for its formation. (correct answer)
  2. The bromine radical (Br•) is too large to approach the aromatic ring due to steric hindrance.
  3. The aromatic ring is electron-rich and repels the incoming electrophilic bromine radical.
  4. Breaking an sp² C-H bond on the ring is energetically much more difficult than breaking an sp³ C-H bond.
Explanation: While breaking an sp² C-H bond is indeed difficult (making D a true statement, but not the primary reason for selectivity), the key factor driving the high selectivity is the stability of the intermediate. The rate-determining step is hydrogen abstraction. Abstraction of a benzylic hydrogen leads to a benzylic radical, which is significantly stabilized by delocalizing the unpaired electron over the adjacent aromatic ring via resonance. This stabilization lowers the transition state energy for this pathway, making it much faster than any other possible hydrogen abstraction.

Question 4

A compound with formula C₉H₁₂ reacts with NBS/light to give a single achiral monobrominated product. The ¹³C NMR spectrum of the starting material shows only 6 distinct signals. What is the identity of the starting material?

  1. n-Propylbenzene
  2. Isopropylbenzene (correct answer)
  3. 1,3,5-Trimethylbenzene
  4. p-Ethyltoluene
Explanation: The formation of a single monobrominated product strongly suggests a high degree of symmetry or a single reactive site. Isopropylbenzene has a single unique benzylic C-H, and its bromination product is achiral. Its ¹³C NMR spectrum has 6 signals (ipso, ortho, meta, para aromatic carbons, plus the benzylic CH and the two equivalent methyl carbons). n-Propylbenzene would give a chiral product and has 7 ¹³C signals. 1,3,5-Trimethylbenzene would give an achiral product but only has 3 ¹³C signals due to its high symmetry. p-Ethyltoluene would yield a mixture of two bromination products.

Question 5

Toluene undergoes electrophilic chlorination with Cl₂/AlCl₃, and the major para isomer is isolated. This product is then subjected to vigorous oxidation with hot, acidic Na₂Cr₂O₇. What is the final major product?

  1. Benzoic acid
  2. 4-chlorobenzoic acid (correct answer)
  3. 4-chlorobenzyl alcohol
  4. No reaction occurs in the second step.
Explanation: This is a two-step synthesis problem. In the first step, electrophilic aromatic substitution occurs. Toluene's methyl group is an ortho,para-director, so chlorination yields a mixture of ortho- and para-chlorotoluene. The para isomer is isolated. In the second step, the isolated p-chlorotoluene is oxidized. Strong oxidizing agents like Na₂Cr₂O₇ convert any alkyl group with a benzylic hydrogen to a carboxylic acid. The methyl group is oxidized to -COOH, and the chloro group on the ring is unaffected by the oxidant. The final product is 4-chlorobenzoic acid.

Question 6

An equimolar mixture of isopropylbenzene and tert-butylbenzene is subjected to vigorous oxidation with excess hot, acidic KMnO₄. After the reaction is complete and neutralized, what is the major organic compound isolated from the mixture?

  1. Benzoic acid (correct answer)
  2. A mixture of benzoic acid and p-tert-butylbenzoic acid
  3. Only unreacted starting materials
  4. A mixture of acetophenone and 2,2-dimethyl-1-phenylpropan-1-one
Explanation: Strong oxidation of an alkylbenzene with reagents like KMnO₄ requires the presence of at least one benzylic hydrogen. Isopropylbenzene has a benzylic hydrogen and will be oxidized completely to benzoic acid. Tert-butylbenzene has no benzylic hydrogens (the benzylic carbon is quaternary), so it will not react under these conditions. Therefore, the only major organic product formed is benzoic acid, which can be isolated from the unreacted tert-butylbenzene.

Question 7

A sample of toluene in which the methyl group is fully deuterated (C₆H₅-CD₃) is reacted with NBS and light alongside an identical concentration of normal toluene (C₆H₅-CH₃). Which statement accurately describes the relative rates of reaction?

  1. Both samples will react at the same rate because C-H bond cleavage is not rate-determining.
  2. Both samples will react at the same rate because radicals are too reactive to distinguish between isotopes.
  3. The deuterated sample will react significantly faster due to a more stable C-D bond.
  4. The deuterated sample will react significantly slower due to a primary kinetic isotope effect. (correct answer)
Explanation: When you encounter questions about deuterated compounds in radical reactions, you're dealing with kinetic isotope effects—a key concept in understanding how isotopic substitution affects reaction rates. In this NBS (N-bromosuccinimide) bromination reaction, the rate-determining step involves breaking a C-H or C-D bond during hydrogen atom abstraction by a bromine radical. This is where the isotope effect becomes crucial. Deuterium forms stronger bonds than hydrogen due to its greater mass, which affects the zero-point vibrational energy. The C-D bond requires more energy to break than the corresponding C-H bond, making the deuterated sample react significantly slower. This primary kinetic isotope effect typically shows rate ratios (kH/kDk_H/k_D) between 2-8 for reactions involving C-H bond cleavage. Answer A incorrectly assumes C-H bond cleavage isn't rate-determining, but in radical halogenation with NBS, the hydrogen abstraction step is indeed rate-limiting. Answer B misunderstands radical selectivity—while radicals are highly reactive, they still exhibit measurable selectivity, especially regarding bond strengths. The kinetic isotope effect demonstrates this selectivity clearly. Answer C gets the direction completely wrong, suggesting deuterated samples react faster due to C-D bond stability, when that very stability actually slows the reaction. Remember this pattern: whenever you see deuterium substitution at a position where bond-breaking occurs in the rate-determining step, expect the deuterated compound to react slower due to the primary kinetic isotope effect. This concept frequently appears in mechanistic organic chemistry problems.

Question 8

A chemist wants to synthesize benzyl bromide from toluene. Which of the following sets of reagents and conditions would be the most successful?

  1. Br₂ with FeBr₃ in the dark
  2. N-bromosuccinimide with UV light (correct answer)
  3. HBr with a peroxide initiator
  4. Aqueous Br₂ at room temperature
Explanation: The target product, benzyl bromide, results from substitution at the benzylic position of the methyl group. This requires a radical chain mechanism. N-bromosuccinimide (NBS) in the presence of a radical initiator like UV light or AIBN is the standard, high-yield method for selective benzylic bromination. Choice A would result in electrophilic aromatic substitution on the ring. Choice C (HBr/peroxide) is for anti-Markovnikov addition to alkenes. Choice D would be very slow and non-selective.

Question 9

An optically pure sample of (R)-2-phenylbutane is treated with NBS and benzoyl peroxide. Assuming monobromination is the major pathway, what is the best description of the product mixture?

  1. An equal mixture of (R)- and (S)-2-bromo-2-phenylbutane (correct answer)
  2. A single enantiomer, (S)-2-bromo-2-phenylbutane
  3. A single enantiomer, (R)-2-bromo-2-phenylbutane
  4. A mixture of four stereoisomers due to bromination at C2 and C3
Explanation: The starting material has a stereocenter at the benzylic carbon (C2). Radical bromination proceeds via abstraction of the benzylic hydrogen, which is the most favorable position. This creates a trigonal planar, sp²-hybridized radical at C2. This intermediate is achiral. The subsequent attack by a bromine atom can occur from either face of the planar radical with equal probability. This results in the formation of both (R) and (S) enantiomers in equal amounts, producing a racemic mixture.

Question 10

What is the expected outcome when 2,4,6-trinitrotoluene (TNT) is treated with hot, acidic potassium dichromate?

  1. No reaction occurs due to the strong deactivating effect of the nitro groups.
  2. The aromatic ring is cleaved by the strong oxidizing conditions.
  3. The nitro groups are reduced to amino groups, yielding 2,4,6-triaminotoluene.
  4. The methyl group is oxidized to a carboxylic acid, yielding 2,4,6-trinitrobenzoic acid. (correct answer)
Explanation: When you encounter oxidation reactions in organic chemistry, focus on identifying which functional groups are susceptible to the oxidizing agent. Potassium dichromate (K2Cr2O7K_2Cr_2O_7) in hot, acidic conditions is a powerful oxidizing agent that selectively targets certain organic functional groups. In TNT, you have a benzene ring with three nitro groups and one methyl group. The key insight is understanding what each functional group does under these conditions. The methyl group attached to the aromatic ring (a benzylic position) is highly susceptible to oxidation by dichromate. This oxidation proceeds through several steps: methyl → primary alcohol → aldehyde → carboxylic acid. The strong oxidizing conditions drive this reaction to completion, converting the −CH3-CH_3 group to −COOH-COOH. Answer D correctly identifies this transformation, yielding 2,4,6-trinitrobenzoic acid. The nitro groups remain unchanged because they're already in a highly oxidized state. Answer A is incorrect because while nitro groups are strongly deactivating for electrophilic aromatic substitution, they don't prevent oxidation of the methyl group. Answer B misunderstands the selectivity—dichromate oxidizes specific functional groups but doesn't cleave stable aromatic rings under these conditions. Answer C confuses oxidation with reduction; dichromate is an oxidizing agent, not a reducing agent, so it cannot convert nitro groups (−NO2-NO_2) to amino groups (−NH2-NH_2). Remember: Strong oxidizing agents like dichromate target electron-rich sites like alkyl groups attached to aromatics, while leaving electron-poor groups like nitro substituents untouched.

Question 11

Consider the compound 4-(chloromethyl)toluene. It is treated with excess hot KMnO₄ followed by an acidic workup. What is the major product?

  1. p-Toluic acid chloride
  2. 4-chlorobenzoic acid (with retention of chlorine)
  3. 4-methylbenzoic acid
  4. Terephthalic acid (correct answer)
Explanation: When you see a question about treating an aromatic compound with hot KMnO₄, you're dealing with oxidative cleavage of alkyl side chains. Permanganate is a powerful oxidizing agent that converts alkyl groups attached to benzene rings into carboxylic acids, regardless of the alkyl chain length. Let's trace what happens to 4-(chloromethyl)toluene. This compound has two substituents on the benzene ring: a methyl group (-CH₃) and a chloromethyl group (-CH₂Cl), positioned para to each other. Under hot KMnO₄ conditions, both alkyl side chains undergo complete oxidation. The methyl group becomes -COOH, and the chloromethyl group also becomes -COOH after the chlorine is displaced during oxidation. This produces terephthalic acid (1,4-benzenedicarboxylic acid), making D correct. Looking at the wrong answers: A (p-toluic acid chloride) incorrectly assumes only one group oxidizes and forms an acid chloride rather than a carboxylic acid. B (4-chlorobenzoic acid) makes the common mistake of thinking chlorine survives the harsh oxidative conditions—it doesn't. The chloromethyl group still gets fully oxidized to carboxylic acid. C (4-methylbenzoic acid) wrongly assumes only the chloromethyl group reacts while the methyl group remains unchanged. Study tip: Remember that hot KMnO₄ is ruthless—it oxidizes ALL alkyl groups on aromatic rings to carboxylic acids, regardless of what other functional groups are present. Count the alkyl substituents to predict how many -COOH groups you'll end up with.

Question 12

Which of the following transformations is best accomplished using pyridinium chlorochromate (PCC) in CH₂Cl₂?

  1. Toluene to benzaldehyde
  2. Toluene to benzoic acid
  3. Benzyl alcohol to benzaldehyde (correct answer)
  4. Benzyl alcohol to benzoic acid
Explanation: PCC is a mild oxidizing agent used to oxidize primary alcohols to aldehydes and secondary alcohols to ketones. It is not strong enough to oxidize an alkyl group like in toluene. To convert benzyl alcohol (a primary benzylic alcohol) to benzaldehyde, a mild oxidant like PCC is ideal because it stops the oxidation at the aldehyde stage. Stronger oxidants like KMnO₄ or Na₂Cr₂O₇ would oxidize benzyl alcohol all the way to benzoic acid (Choice D).

Question 13

The compound 1-tert-butyl-4-methylbenzene is heated with excess potassium permanganate in a basic solution, followed by acidification. What is the major organic product?

  1. Terephthalic acid (benzene-1,4-dicarboxylic acid)
  2. 4-tert-butylbenzoic acid (correct answer)
  3. Benzoic acid
  4. No reaction occurs.
Explanation: Benzylic oxidation with KMnO₄ requires that the alkyl group has at least one benzylic hydrogen. In 1-tert-butyl-4-methylbenzene, the methyl group has three benzylic hydrogens and will be oxidized to a carboxylic acid. The tert-butyl group, however, has a quaternary benzylic carbon with no benzylic hydrogens. Therefore, the tert-butyl group remains intact during the reaction. The final product is 4-tert-butylbenzoic acid.