Organic Chemistry 2 Quiz: Diazonium Chemistry
20 questions · exam conditions
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Diazonium ChemistryQuestion 1 of 20

What is the major product when an aryl diazonium salt reacts with CuBr under Sandmeyer conditions?

Bromobenzene (Ar-Br)
Fluorobenzene (Ar-F)
Phenylamine oxide (Ar-NHO)
Benzaldehyde (Ar-CHO)
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Diazonium Chemistry

Practice Diazonium Chemistry 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 Diazonium Chemistry, 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

What is the major product when an aryl diazonium salt reacts with CuBr under Sandmeyer conditions?

  1. Bromobenzene (Ar-Br) (correct answer)
  2. Fluorobenzene (Ar-F)
  3. Phenylamine oxide (Ar-NHO)
  4. Benzaldehyde (Ar-CHO)
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, the Sandmeyer reaction with CuBr replaces the diazonium group with bromine, yielding bromobenzene. The correct answer highlights bromobenzene, ensuring the student understands the halogen-specific outcome. A common distractor might suggest fluorobenzene, which is a frequent mistake when confusing with other halides. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 2

Which functional group transformation does diazotization of an aniline derivative primarily accomplish?

  1. Ar-NH2_2 to Ar-N2+_2^+ conversion (correct answer)
  2. Ar-NO2_2 to Ar-NH2_2 reduction
  3. Ar-Br to Ar-MgBr formation
  4. Ar-CHO to Ar-CO2_2H oxidation
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, diazotization transforms the amino group of an aniline derivative into a diazonium group, setting up for further substitutions. The correct answer highlights Ar-NH2 to Ar-N2+ conversion, ensuring the student understands the primary transformation. A common distractor might suggest nitro to amino reduction, which is a frequent mistake when reversing the sequence. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 3

Which reagent combination is incorrect for diazonium formation from aniline at 00-5 ∘C5\,^{\circ}\mathrm{C}?

  1. NaNO2_2 + HCl in water
  2. NaNO2_2 + H2_2SO4_4 in water
  3. NaNO2_2 + NaOH in water (correct answer)
  4. NaNO2_2 + HBr in water
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, diazotization requires acidic conditions to generate nitrous acid, so basic combinations like NaNO2 + NaOH are incorrect. The correct answer highlights NaNO2 + NaOH as incorrect, ensuring the student understands the need for acidity. A common distractor might overlook that H2SO4 or HBr can be used, but the question identifies the basic one as wrong. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 4

In a diazotization, why is acidic pH required during formation of the diazonium salt?

  1. Acid generates the nitrosating agent from nitrite (correct answer)
  2. Acid deprotonates aniline to a stronger nucleophile
  3. Acid prevents formation of nitrous acid entirely
  4. Acid forces diazonium to act as nucleophile
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, acidic pH is necessary to protonate nitrite and generate the nitrosating agent for reaction with aniline. The correct answer highlights that acid generates the nitrosating agent, ensuring the student understands the role of acidity. A common distractor might suggest acid deprotonates aniline, which is a frequent mistake in mechanistic confusion. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 5

What is the major product when an aryl diazonium salt is warmed in water without added nucleophiles?

  1. Aryl alcohol (phenol) formation (correct answer)
  2. Aryl chloride formation exclusively
  3. Aryl ketone formation directly
  4. Aryl amide formation directly
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, warming an aryl diazonium salt in water leads to hydrolysis, forming phenol as the major product. The correct answer highlights aryl alcohol (phenol) formation, ensuring the student understands this default reaction pathway. A common distractor might suggest aryl chloride formation, which is a frequent mistake without the Sandmeyer catalyst. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 6

Which statement best describes the role of the diazonium group in substitution reactions on aromatic rings?

  1. It acts as an excellent leaving group (correct answer)
  2. It acts as a strong nucleophile
  3. It permanently deactivates the aromatic ring
  4. It converts benzene into an alkene
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, the diazonium group serves as an excellent leaving group, facilitating nucleophilic substitutions on the aromatic ring. The correct answer highlights this role, ensuring the student understands its mechanistic importance. A common distractor might suggest it acts as a nucleophile, which is a frequent mistake in role reversal. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.

Question 7

The synthesis of an azo dye involves coupling N,N-dimethylaniline with benzenediazonium chloride. What reaction condition is critical for maximizing the yield of this electrophilic aromatic substitution?

  1. Strongly acidic conditions (pH < 2) to fully protonate the diazonium ion, increasing its electrophilicity.
  2. Mildly acidic conditions (pH 4-6) to maintain the diazonium ion concentration while keeping the amine nucleophilic. (correct answer)
  3. Strongly basic conditions (pH > 10) to deprotonate the N,N-dimethylaniline, making it a stronger nucleophile.
  4. Anhydrous conditions with a Lewis acid catalyst to promote the electrophilic attack on the aromatic ring.
Explanation: Azo coupling is an electrophilic aromatic substitution where the diazonium ion is the electrophile and an activated aromatic ring is the nucleophile. For coupling with anilines, the pH must be carefully controlled. Under strongly acidic conditions (pH < 2), the nitrogen of the N,N-dimethylamino group becomes protonated, which converts it into a strongly deactivating group and stops the reaction. Under basic conditions (pH > 7), the diazonium ion is converted to a non-electrophilic diazohydroxide or diazotate anion. Therefore, mildly acidic conditions (pH 4-6) are optimal as they preserve a high concentration of the electrophilic diazonium ion without deactivating the nucleophilic coupling partner.

Question 8

A chemist prepares the diazonium salt from 3,5-dimethylaniline. Which of the following molecules could not be synthesized in a single, high-yield step from this diazonium salt using standard methods?

  1. 3,5-Dimethyliodobenzene
  2. 3,5-Dimethylphenol
  3. 3,5-Dimethylbenzaldehyde (correct answer)
  4. 3,5-Dimethylbenzonitrile
Explanation: Diazonium salts are versatile intermediates. They can be converted to an iodoarene with KI (A), a phenol with H₂O/heat (B), and a benzonitrile with CuCN (a Sandmeyer reaction) (D). However, there is no standard, single-step reaction to convert an aryl diazonium salt directly to a benzaldehyde (C). While reactions like the Gatterman reaction can introduce a formyl group, they do not start from a diazonium salt. Therefore, 3,5-dimethylbenzaldehyde cannot be made in one step from the corresponding diazonium salt.

Question 9

A student attempts to synthesize p-nitrophenol. They start with p-nitroaniline, treat it with NaNO₂ and HCl at 0 °C, and then warm the resulting solution to 50 °C. What is the major organic product of this sequence?

  1. p-Dinitrobenzene
  2. p-Chloronitrobenzene
  3. p-Nitrophenol (correct answer)
  4. 4,4'-Dinitroazobenzene
Explanation: The sequence described is the standard synthesis of a phenol from an aniline via a diazonium salt. First, p-nitroaniline is converted to p-nitrobenzenediazonium chloride at 0 °C. The second step, warming the aqueous solution, causes the hydrolysis of the diazonium salt. Water acts as a nucleophile, attacking the carbon atom of the C-N bond and displacing the excellent leaving group, N₂ gas. The result is the substitution of the diazonium group with a hydroxyl group, yielding p-nitrophenol.

Question 10

What is the major product formed when benzenediazonium chloride is coupled with 2-naphthol (β-naphthol) under mildly basic conditions?

  1. 1-(phenylazo)-2-naphthol (correct answer)
  2. 3-(phenylazo)-2-naphthol
  3. 4-(phenylazo)-2-naphthol
  4. 8-(phenylazo)-2-naphthol
Explanation: This is an electrophilic aromatic substitution (azo coupling) on a naphthalene ring system. The hydroxyl group at the C2 position is a strong activating group and an ortho, para-director. In the naphthalene system, the 'ortho' position is C1 and the 'para' position is C4. Both are activated. However, electrophilic attack at the C1 position is kinetically favored over the C3 position. Attack at C4 is sterically hindered by the hydrogen at C5 (a peri interaction). Therefore, the electrophile (benzenediazonium ion) will preferentially attack the C1 position, yielding 1-(phenylazo)-2-naphthol.

Question 11

If aniline is treated with sodium nitrite containing an isotopic label (Na¹⁵NO₂) and HCl, and the resulting diazonium salt is then allowed to decompose to nitrogen gas and a phenyl cation, where will the ¹⁵N label be located?

  1. The ¹⁵N atom remains bonded to the phenyl cation.
  2. The ¹⁵N atom is exclusively found in the N₂ gas product.
  3. The ¹⁵N atom is equally distributed between the phenyl cation and the N₂ gas.
  4. The ¹⁵N atom is found as the terminal nitrogen in the Ar-N≡¹⁵N⁺ ion, which is then lost as part of the N₂ molecule. (correct answer)
Explanation: The mechanism of diazotization involves the nitrogen atom of the aniline attacking the nitrogen atom of the nitrosonium ion (⁺N=O), which is formed from nitrous acid (HNO₂). The nitrous acid is generated from NaNO₂. Therefore, the nitrogen atom originally from the aniline remains directly attached to the ring (the alpha nitrogen), while the nitrogen atom from the NaNO₂ becomes the terminal nitrogen (the beta nitrogen). The structure of the diazonium ion is Ar-Nₐ≡Nₑ⁺. In this case, it would be Ar-N≡¹⁵N⁺. Upon decomposition, both nitrogen atoms are lost together as N₂ gas. Thus, the label is found on the terminal nitrogen, which departs as part of the N₂ molecule.

Question 12

The use of an amino group as a temporary director, which is later removed via diazotization followed by H₃PO₂, is a powerful synthetic strategy. This approach is most crucial for synthesizing aromatic compounds where:

  1. all substituents are meta-directors, and a 1,2,4-substitution pattern is desired in the final product.
  2. the desired substituents are all ortho,para-directors, but they are arranged in a meta-relationship to each other. (correct answer)
  3. a single, sterically hindered position must be substituted, which is only accessible when the ring is highly activated.
  4. the final product is highly unstable, and the amino group provides a stabilizing effect during intermediate steps.
Explanation: This strategy's main purpose is to overcome the limitations of standard directing effects. For example, to synthesize m-bromotoluene, both the methyl and bromo groups are o,p-directors, making a direct synthesis impossible. The strategy involves using an amino group (a powerful o,p-director) to control the position of a new substituent, creating an intermediate where the groups have the desired relative positions (e.g., placing a bromine ortho to the amino group on p-toluidine makes the bromine and methyl group meta to each other). The amino group is then removed, leaving behind the 'difficult' substitution pattern. Thus, it is ideal for arranging o,p-directing groups in a meta relationship.

Question 13

A student isolates the aryl diazonium tetrafluoroborate salt as a crystalline solid in preparation for a Schiemann reaction. Upon heating the solid, a violent decomposition occurs instead of a smooth conversion to the aryl fluoride. What is the most plausible procedural error leading to this outcome?

  1. The student overheated the solid far above the required decomposition temperature.
  2. The student failed to thoroughly dry the isolated diazonium salt before heating. (correct answer)
  3. The student used HBF₄ with a concentration below 50%, which yields an unstable salt.
  4. The student omitted the copper catalyst that is required for thermal decomposition.
Explanation: Diazonium salts are high-energy compounds that can be explosive, particularly in the solid state. While diazonium tetrafluoroborates are among the most stable, they must be handled with extreme care. A critical safety precaution is to ensure the salt is completely dry before heating. The presence of moisture or other impurities can lead to a rapid, uncontrolled, and sometimes violent decomposition rather than the desired smooth conversion to the aryl fluoride and N₂ gas. The Schiemann reaction does not use a copper catalyst.

Question 14

In some copper-catalyzed Sandmeyer reactions, a symmetric biaryl (e.g., biphenyl from benzenediazonium chloride) is formed as a byproduct. This is thought to occur via a radical pathway. What is the key reactive intermediate that leads to the formation of this biaryl byproduct?

  1. A phenyl cation (C₆H₅⁺)
  2. A copper(III) organometallic complex
  3. A phenyl radical (C₆H₅•) (correct answer)
  4. A dinitrogen radical anion (N₂⁻•)
Explanation: The accepted mechanism for the Sandmeyer reaction involves a single-electron transfer from the copper(I) catalyst to the diazonium ion. This generates an aryl diazonium radical, which rapidly loses a molecule of N₂ to form an aryl radical (in this case, a phenyl radical, C₆H₅•). While this radical typically proceeds to react with the halide to form the desired product, it can also dimerize with another phenyl radical to form the biphenyl byproduct. The phenyl radical is therefore the key intermediate leading to biaryl formation.

Question 15

Diazotization of primary aromatic amines must be carried out at low temperatures (0–5 °C). If the reaction mixture is allowed to warm significantly above this range, what is the major side product and why does it form?

  1. An azo compound, which forms because the higher temperature increases the rate of coupling with unreacted aniline.
  2. A phenol, which forms because water acts as a nucleophile and displaces the N₂ group at an appreciable rate at higher temperatures. (correct answer)
  3. A biphenyl derivative, formed because thermal energy promotes homolytic cleavage of the C–N bond, leading to radical dimerization.
  4. The starting aniline, which is reformed because the diazotization equilibrium is exothermic and thus disfavored by heat.
Explanation: Aryl diazonium salts are thermally unstable. The diazonium group is an excellent leaving group because its departure forms stable N₂ gas. Even a weak nucleophile like water can attack the carbon atom attached to the diazonium group. At 0–5 °C, this reaction is very slow. However, if the solution is allowed to warm up, the rate of this nucleophilic substitution increases significantly, leading to the formation of a phenol as the major byproduct. This decomposition pathway is a primary reason for the strict temperature control required for diazotization reactions.

Question 16

Aniline is successfully converted into benzonitrile via diazotization followed by a Sandmeyer reaction with CuCN. Which of the following spectral changes would be most consistent with this transformation?

  1. In the IR spectrum, a broad peak at ~3400 cm⁻¹ disappears and a strong, sharp peak appears at ~2230 cm⁻¹. (correct answer)
  2. In the ¹H NMR spectrum, a multiplet in the aromatic region is replaced by a single sharp peak at ~7.5 ppm.
  3. In the IR spectrum, a peak for the N-H bend at ~1600 cm⁻¹ is replaced by a stronger peak for a C=N bond at ~1680 cm⁻¹.
  4. In the ¹³C NMR spectrum, the number of aromatic signals changes from four to six due to the removal of symmetry.
Explanation: The transformation converts an amino group (-NH₂) to a nitrile group (-C≡N). In the IR spectrum, the starting material, aniline, shows characteristic N-H stretching vibrations as a broad peak (or pair of peaks) around 3300-3400 cm⁻¹. The product, benzonitrile, lacks these N-H bonds but has a nitrile group, which gives a very characteristic strong, sharp absorption at ~2230 cm⁻¹. Therefore, the disappearance of the N-H stretch and the appearance of the C≡N stretch is the most definitive spectral change.

Question 17

While Sandmeyer reactions introducing -Cl or -Br require a copper(I) catalyst, the analogous reaction to introduce -I proceeds efficiently with just KI. What is the best explanation for this difference?

  1. The C–I bond is significantly weaker than C–Cl or C–Br, making the reaction highly exothermic and kinetically fast without a catalyst.
  2. Iodide (I⁻) is a potent nucleophile and can displace the N₂ group via a direct nucleophilic attack, a pathway not effective for the less nucleophilic Cl⁻ or Br⁻. (correct answer)
  3. Copper(I) iodide is insoluble in the typical aqueous reaction media used for diazotization, preventing its use as a catalyst.
  4. The mechanism for iodination is a high-temperature SₙAr process, whereas the copper-catalyzed reactions proceed through a lower-temperature radical pathway.
Explanation: The key difference lies in the nucleophilicity of the halides. Iodide (I⁻) is a significantly stronger nucleophile than chloride (Cl⁻) or bromide (Br⁻) in aqueous solution. Its high nucleophilicity allows it to directly attack the carbon bearing the diazonium group and displace the N₂ leaving group without the need for a catalyst. Chloride and bromide are weaker nucleophiles and require the copper(I) catalyst to facilitate the reaction, which is believed to proceed through a single-electron transfer and radical mechanism rather than direct nucleophilic attack.

Question 18

The Gattermann reaction uses CuCl/HCl and HCN to formylate an activated aromatic ring, while the Sandmeyer reaction uses CuCN to convert a diazonium salt to a nitrile. A student incorrectly attempts to synthesize p-tolualdehyde from p-toluidine by diazotizing it and then treating it with CuCN. What product will they actually form?

  1. p-Tolualdehyde
  2. p-Methylbenzonitrile (correct answer)
  3. p-Cresol
  4. p-Chlorotoluene
Explanation: This question tests the student's ability to distinguish between different named reactions and predict the outcome of a standard transformation. The diazotization of p-toluidine (p-methylaniline) correctly forms the p-methylbenzenediazonium salt. The subsequent treatment with copper(I) cyanide (CuCN) is a Sandmeyer reaction. This reaction specifically replaces the diazonium group (-N₂⁺) with a cyano group (-CN). The product will therefore be p-methylbenzonitrile, not the aldehyde. The student has confused the Sandmeyer reaction with other reactions that install carbonyl-containing groups.

Question 19

Which sequence of reactions is most appropriate for the synthesis of 1,3,5-tribromobenzene from aniline?

    1. NaNO₂, HCl, 0°C; 2. CuBr; 3. Br₂/FeBr₃ (2 eq.)
    1. Br₂/FeBr₃ (3 eq.); 2. NaNO₂, HCl, 0°C; 3. H₃PO₂
    1. H₂O, heat; 2. Br₂(aq), excess; 3. PBr₃
    1. Br₂(aq), excess; 2. NaNO₂, HCl, 0°C; 3. H₃PO₂
     (correct answer)
Explanation: The synthesis of 1,3,5-tribromobenzene takes advantage of the powerful activating and directing effects of the amino group, which is then removed. The amino group in aniline is so strongly activating that it directs bromination to all three ortho and para positions without a Lewis acid catalyst, using aqueous bromine. This forms 2,4,6-tribromoaniline. The amino group is then converted to a diazonium salt and subsequently removed (deaminated) by reaction with hypophosphorous acid (H₃PO₂), yielding the target 1,3,5-tribromobenzene. Choice D correctly outlines this standard, efficient procedure.

Question 20

In a Sandmeyer chlorination, which copper reagent is used to replace N2+\mathrm{N_2^+} with Cl?

  1. CuCl (copper(I) chloride) (correct answer)
  2. CuCl2_2 (copper(II) chloride)
  3. CuO (copper(II) oxide)
  4. CuSO4_4 (copper(II) sulfate)
Explanation: This question tests knowledge of diazonium chemistry within the context of amines and nitrogen-containing functional groups. Diazonium salts are formed from primary aromatic amines and are highly useful intermediates in organic synthesis due to their ability to undergo substitution reactions. In the specific scenario provided, the Sandmeyer reaction for chlorination uses copper(I) chloride to facilitate the replacement of the diazonium group with Cl. The correct answer highlights CuCl, ensuring the student understands the specific catalyst required. A common distractor might suggest CuCl2, which is a frequent mistake when confusing oxidation states of copper. To help students master diazonium chemistry, emphasize the importance of reaction conditions and the versatility of diazonium salts in synthetic applications. Encourage practice with reaction mechanisms and the prediction of reaction products based on the structure of the starting materials.