Organic Chemistry 2 Quiz: Spectroscopic Verification And Purity Checks
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Spectroscopic Verification And Purity ChecksQuestion 1 of 16

A Friedel-Crafts acylation of anisole (methoxybenzene) is expected to yield 4-methoxyacetophenone as the major product. A student suspects their purified product is contaminated with unreacted anisole. How many signals would be expected in the aromatic region (110-165 ppm) of the broadband-decoupled ¹³C NMR spectrum of this mixture?

4 signals
6 signals
8 signals
10 signals
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Spectroscopic Verification And Purity Checks

Practice Spectroscopic Verification And Purity Checks 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 Spectroscopic Verification And Purity Checks, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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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.

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Question 1

A Friedel-Crafts acylation of anisole (methoxybenzene) is expected to yield 4-methoxyacetophenone as the major product. A student suspects their purified product is contaminated with unreacted anisole. How many signals would be expected in the aromatic region (110-165 ppm) of the broadband-decoupled ¹³C NMR spectrum of this mixture?

  1. 4 signals
  2. 6 signals (correct answer)
  3. 8 signals
  4. 10 signals
Explanation: The product 4-methoxyacetophenone has 4 unique aromatic carbon environments due to symmetry. Anisole also has 4 unique aromatic carbon environments. However, some signals overlap between the two compounds. Specifically, the carbons with similar electronic environments (some C-H carbons ortho/meta to substituents) will have nearly identical chemical shifts. Analysis shows that 2 anisole signals are sufficiently distinct from the 4 product signals, giving 6 total signals in the aromatic region.

Question 2

To monitor a Fischer esterification of benzoic acid with methanol, a student takes a ¹H NMR spectrum of the crude reaction mixture. Key integrations are recorded: a singlet for the product's OCH₃ group at 3.9 ppm integrates to 2.4H, and a broad singlet for the starting material's COOH proton at 12.5 ppm integrates to 0.4H.

Based on this ¹H NMR integration data, what is the approximate molar ratio of product (methyl benzoate) to starting material (benzoic acid) in the mixture?

  1. 1:1
  2. 2:1 (correct answer)
  3. 6:1
  4. 8:1
Explanation: To find the molar ratio, the integration value for each signal must be normalized by the number of protons it represents. The product's OCH₃ signal represents 3 protons, so its relative molar amount is 2.4H / 3H = 0.8. The starting material's COOH signal represents 1 proton, so its relative molar amount is 0.4H / 1H = 0.4. The ratio of product to starting material is therefore 0.8 : 0.4, which simplifies to 2:1.

Question 3

A chemist deprotects an alcohol that was protected as a TBDMS (tert-butyldimethylsilyl) ether using tetra-n-butylammonium fluoride (TBAF). The progress of the reaction is monitored using IR spectroscopy. Which pair of spectral changes provides the most conclusive evidence that the desired deprotection is occurring?

  1. Appearance of a broad O-H stretch around 3400 cm⁻¹ and disappearance of C-H stretches around 2900 cm⁻¹.
  2. A significant shift of the C-O stretch from ~1100 cm⁻¹ to ~1250 cm⁻¹.
  3. Appearance of a strong C=O stretch around 1720 cm⁻¹ indicating cleavage of the ether.
  4. Disappearance of strong Si-O and Si-C stretches (1100-1250 cm⁻¹) and appearance of a broad O-H stretch around 3400 cm⁻¹. (correct answer)
Explanation: When monitoring deprotection reactions by IR spectroscopy, you need to identify which bonds are breaking and forming. TBDMS deprotection involves cleaving silicon-oxygen and silicon-carbon bonds while regenerating the free alcohol's O-H bond. The correct answer is D because it identifies the two most diagnostic changes: disappearance of strong Si-O and Si-C stretches (1100-1250 cm⁻¹) from the TBDMS group, and appearance of the characteristic broad O-H stretch around 3400 cm⁻¹ from the newly formed free alcohol. These changes directly correspond to the bond-breaking and bond-forming events in the deprotection mechanism. Option A is partially correct about the O-H stretch appearance, but wrong about C-H stretches disappearing around 2900 cm⁻¹. The TBDMS group does have C-H bonds, but so does the rest of the molecule—these stretches don't provide conclusive evidence of deprotection. Option B describes an unrealistic shift in C-O stretching frequency. The C-O bond character doesn't change dramatically enough during deprotection to cause such a significant shift from 1100 to 1250 cm⁻¹. Option C suggests formation of a C=O stretch around 1720 cm⁻¹, which would indicate oxidation of the alcohol to a ketone or aldehyde—not deprotection. This represents a completely different reaction outcome. Study tip: For protecting group removal reactions, focus on the bonds that are actually breaking (protecting group bonds) and forming (functional group bonds). The most reliable IR evidence combines loss of protecting group signals with appearance of the target functional group signal.

Question 4

A student plans to convert 1-bromobutane into butylamine via Sₙ2 reaction with sodium amide (NaNH₂). To confirm the identity of the purified product and rule out unreacted starting material, which spectroscopic technique would provide the most unambiguous data?

  1. Mass spectrometry, by confirming the molecular ion peak shifts from m/z=136/138 to m/z=73. (correct answer)
  2. IR spectroscopy, by observing the appearance of two medium N-H stretches around 3300-3400 cm⁻¹.
  3. ¹H NMR spectroscopy, by observing the downfield shift of the protons on the carbon adjacent to the functional group.
  4. ¹³C NMR spectroscopy, by observing the change in chemical shift for the C1 carbon from ~33 ppm to ~42 ppm.
Explanation: When analyzing reaction products spectroscopically, you need to choose the technique that provides the most definitive, unambiguous evidence of structural change. Each method offers different types of information about molecular identity. Mass spectrometry provides the most conclusive evidence here because it directly measures molecular weight, giving you an unequivocal molecular fingerprint. The conversion of 1-bromobutane (C₄H₉Br) to butylamine (C₄H₉NH₂) involves replacing bromine (atomic mass ~80) with an amino group (mass 16), creating a dramatic mass difference. The molecular ion peak shifts from m/z 136/138 (showing the characteristic bromine isotope pattern) to a single peak at m/z 73, providing unmistakable confirmation of the product identity and absence of starting material. Option B is problematic because while IR would show N-H stretches for the amine product, it cannot definitively rule out trace amounts of unreacted starting material since C-Br stretches are weaker and might be missed. Option C fails because ¹H NMR chemical shifts for -CH₂Br versus -CH₂NH₂ are relatively similar (both around 3-4 ppm), making differentiation challenging, especially with overlapping signals. Option D is unreliable because the predicted ¹³C chemical shift change is modest and could be obscured by other factors or impurities. Study tip: When comparing spectroscopic techniques for product confirmation, mass spectrometry typically provides the most unambiguous structural evidence because molecular weight is a fundamental molecular property. Always consider which technique gives you the clearest "before and after" comparison.

Question 5

A student attempts to hydrolyze benzamide to benzoic acid using aqueous NaOH, followed by acidification. The IR spectrum of the isolated crude solid shows a very broad peak from 2500-3300 cm⁻¹, a strong C=O peak at 1700 cm⁻¹, and another strong C=O peak at 1660 cm⁻¹. What is the most likely interpretation of this spectrum?

  1. The reaction went to completion, and the peak at 1660 cm⁻¹ is due to aromatic C=C stretching.
  2. The reaction failed, and the spectrum shows only unreacted benzamide.
  3. The product is a mixture of benzoic acid and unreacted benzamide. (correct answer)
  4. The product is contaminated with benzoic anhydride, which has two carbonyl peaks.
Explanation: The very broad peak from 2500-3300 cm⁻¹ and the C=O stretch at 1700 cm⁻¹ are characteristic of a carboxylic acid (benzoic acid). The strong C=O stretch at 1660 cm⁻¹ is characteristic of a primary amide (benzamide). The presence of both sets of peaks indicates that the hydrolysis did not go to completion, and the isolated product is a mixture of the starting material and the desired product.

Question 6

A crude product containing a desired compound (P) and a more polar impurity (I) is purified by column chromatography. Fractions are analyzed by TLC. The crude mixture shows spots at Rf=0.5 (P) and Rf=0.2 (I). Which set of collected fractions should be combined to obtain the purest sample of P?

  1. The first fractions collected, which show only a spot at Rf=0.5. (correct answer)
  2. The middle fractions, which show spots at both Rf=0.5 and Rf=0.2.
  3. The last fractions collected, which show only a spot at Rf=0.2.
  4. The first fractions collected, which show only a spot at Rf=0.2.
Explanation: In normal-phase chromatography (e.g., silica gel), less polar compounds elute faster and have a higher Rf value. The desired product (P) has a higher Rf (0.5) than the impurity (I, Rf=0.2), meaning P is less polar. Therefore, P will be collected in the first fractions from the column. The purest sample of P would be obtained by combining the initial fractions that show only the spot for P.

Question 7

A student synthesizes a ketone via the hydration of an alkyne and purifies it. The mass spectrum shows the expected molecular ion at m/z = 86. The student needs to confirm that the product is 2-pentanone and not the isomeric 3-pentanone. The presence of a prominent peak at which m/z value would most strongly support the 2-pentanone structure?

  1. m/z = 86
  2. m/z = 71
  3. m/z = 58 (correct answer)
  4. m/z = 57
Explanation: 2-Pentanone possesses γ-hydrogens (the C4 protons), making it capable of undergoing the McLafferty rearrangement. This specific fragmentation pathway involves the transfer of a γ-hydrogen to the carbonyl oxygen, followed by cleavage of the α-β bond, resulting in the loss of a neutral alkene (ethene) and formation of a radical cation with m/z = 58. 3-Pentanone does not have γ-hydrogens and cannot undergo this rearrangement. Therefore, a peak at m/z=58 is a definitive marker for 2-pentanone.

Question 8

During the nitration of benzaldehyde with a mixture of HNO₃ and H₂SO₄, a student isolates a product. The ¹H NMR spectrum shows the expected signals for m-nitrobenzaldehyde, but also includes a very broad singlet far downfield at 12.1 ppm. What is the most likely identity of the impurity that gives rise to this signal?

  1. m-nitrobenzoic acid. (correct answer)
  2. Unreacted benzaldehyde.
  3. Residual nitric acid.
  4. Phenol, from decomposition.
Explanation: When analyzing ¹H NMR spectra of organic reactions, extremely downfield signals (around 10-13 ppm) are characteristic of highly deshielded protons, particularly those involved in hydrogen bonding or on electron-withdrawing groups like carboxylic acids. The broad singlet at 12.1 ppm is a telltale sign of a carboxylic acid proton (COOH). During nitration reactions, aldehydes can undergo oxidation as a side reaction, converting the aldehyde group (-CHO) to a carboxylic acid group (-COOH). This occurs because the nitrating mixture contains strong oxidizing agents. The broadness of the signal results from rapid proton exchange and hydrogen bonding typical of carboxylic acids. Looking at the wrong answers: (B) Unreacted benzaldehyde would show signals around 7-10 ppm for the aldehyde proton, not 12.1 ppm. (C) Residual nitric acid wouldn't produce a discrete signal at 12.1 ppm in organic NMR solvents and would likely exchange rapidly. (D) Phenol protons appear around 4-7 ppm, much more upfield than 12.1 ppm, and phenol formation would require breaking the aromatic C-C bond, which is unlikely under these conditions. The correct answer is (A) m-nitrobenzoic acid, formed by oxidation of the starting benzaldehyde during the harsh nitrating conditions. Study tip: Remember that signals above 10 ppm in ¹H NMR almost always indicate carboxylic acid protons. When you see such extreme downfield shifts in reaction mixtures, consider whether oxidation of aldehydes or alcohols might have occurred as a side reaction.

Question 9

A student performs an oxidation of cyclohexanol to cyclohexanone using pyridinium chlorochromate (PCC). The crude product is isolated and an IR spectrum is obtained. The spectrum shows a strong, sharp absorption at 1715 cm⁻¹ and a noticeable, broad absorption centered around 3300 cm⁻¹. Which of the following is the most accurate conclusion about the reaction outcome?

  1. The reaction was unsuccessful, and the spectrum shows only unreacted cyclohexanol.
  2. The reaction proceeded to form the desired ketone, but the crude product is contaminated with unreacted cyclohexanol. (correct answer)
  3. The reaction is complete and the product is pure cyclohexanone, with the broad peak resulting from absorbed atmospheric moisture.
  4. The reaction produced an acidic byproduct, likely from over-oxidation, which is responsible for the broad peak.
Explanation: The strong, sharp absorption at 1715 cm⁻¹ is characteristic of the C=O stretch of a ketone, confirming that the product, cyclohexanone, was formed. The broad absorption around 3300 cm⁻¹ is characteristic of the O-H stretch of an alcohol. Its presence indicates that some of the starting material, cyclohexanol, remains in the crude product, meaning the reaction is incomplete or the product requires purification.

Question 10

To probe a reaction mechanism, benzaldehyde is reduced with sodium borodeuteride (NaBD₄) followed by an H₂O workup, yielding Ph-CHD-OH. Which NMR spectroscopic feature provides the most unambiguous evidence that a single deuterium atom has been incorporated at the benzylic position?

  1. In the ¹H NMR, the benzylic proton signal integrates to 1H.
  2. In the ¹H NMR, the hydroxyl proton signal is split into a doublet.
  3. In the proton-decoupled ¹³C NMR, the signal for the benzylic carbon appears as a 1:1:1 triplet. (correct answer)
  4. In the proton-decoupled ¹³C NMR, the signal for the ipso-aromatic carbon is significantly shifted.
Explanation: In a proton-decoupled ¹³C NMR spectrum, a carbon bonded to deuterium (spin I=1) is split into a 2nI+1 = 2(1)(1)+1 = 3 line pattern, a 1:1:1 triplet. This C-D coupling is not removed by proton decoupling. This splitting pattern is a highly specific and unambiguous confirmation of the C-D bond, directly indicating successful incorporation of deuterium at that carbon. While changes in ¹H NMR integration are also evidence, they can be subject to error, and splitting patterns can be complex or unresolved.

Question 11

A student isolates a product after column chromatography that used a 9:1 mixture of hexanes and ethyl acetate as the eluent. The ¹H NMR spectrum in CDCl₃ shows all the expected product signals, but also a quartet at 4.12 ppm, a singlet at 2.05 ppm, and a triplet at 1.25 ppm. The singlet for residual CHCl₃ at 7.26 ppm is also visible. What is the most likely source of these three additional signals?

  1. Residual hexanes from the chromatography eluent.
  2. Decomposition of the product on the silica gel column.
  3. An unreacted starting material that co-eluted with the product.
  4. Residual ethyl acetate from the chromatography eluent. (correct answer)
Explanation: The pattern of a quartet (~4.1 ppm), a singlet (~2.0 ppm), and a triplet (~1.2 ppm) is the classic ¹H NMR signature for ethyl acetate (CH₃COOCH₂CH₃). It is a very common impurity when used as a chromatography solvent because it has a higher boiling point than hexanes and can be difficult to remove completely under reduced pressure.

Question 12

Electrophilic nitration of bromobenzene produces a mixture of o-bromonitrobenzene and p-bromonitrobenzene. To determine the isomeric ratio, a ¹H NMR of the crude product is taken. Which feature of the spectrum would be the most straightforward to use for this quantification?

  1. The chemical shift difference between the most downfield signals of the two isomers.
  2. The splitting pattern of the proton ortho to the bromine in the ortho isomer compared to the para isomer.
  3. The total integration of the entire aromatic region, which corresponds to 8 protons total for a 1:1 mixture.
  4. The ratio of the integrations of the two well-separated doublets from the para isomer versus the complex multiplets from the ortho isomer. (correct answer)
Explanation: When analyzing isomeric ratios using ¹H NMR, you need signals that are well-separated and easily integrated. The key is finding protons that appear in distinctly different chemical environments between the two isomers. In p-bromonitrobenzene, the molecule has perfect symmetry. The two protons ortho to bromine are equivalent to each other, and the two protons ortho to the nitro group are also equivalent. This creates two clean doublets due to ortho coupling between adjacent aromatic protons. These doublets will be well-separated because protons near electron-withdrawing nitro groups are significantly more downfield than those near bromine. In contrast, o-bromonitrobenzene has no symmetry. Each aromatic proton sits in a unique chemical environment, creating a complex multiplet pattern where individual signals overlap and are difficult to integrate accurately. Choice D correctly identifies that you can easily measure the ratio by comparing the integration of the para isomer's clean doublets against the ortho isomer's overlapping multiplets. Choice A focuses on chemical shifts rather than integration, which doesn't help determine ratios. Choice B mentions splitting patterns, but these help identify the isomers, not quantify them. Choice C contains a fundamental error—regardless of the ratio, you always have 4 aromatic protons per molecule, so the total integration doesn't reveal the isomeric ratio. Strategy tip: For NMR quantification problems, always look for well-separated, easily integrated signals. Symmetrical molecules typically give cleaner spectra that are easier to integrate than asymmetrical ones.

Question 13

To determine the yield of an aldol condensation without isolating the product, a student dissolves the crude reaction mixture in CDCl₃. They add 16.8 mg of 1,4-dinitrobenzene (MW = 168 g/mol) as an internal standard. In the ¹H NMR, the singlet for the 4 protons of the standard at 8.4 ppm integrates to 1.00. A doublet for one of the vinyl protons of the product (MW = 208 g/mol) at 7.8 ppm integrates to 1.25. What is the approximate mass of aldol product in the sample?

  1. 26 mg
  2. 52 mg
  3. 104 mg (correct answer)
  4. 208 mg
Explanation: First, calculate the moles of the internal standard: 16.8 mg / 168 g/mol = 0.100 mmol. The signal for the standard (4 protons) has an integral of 1.00, so the response factor is 0.100 mmol / 1.00 integral = 0.100 mmol per unit of integration. The product signal (1 proton) has an integral of 1.25. Moles of product protons = 1.25 * 0.100 mmol = 0.125 mmol. Since the signal represents 1 proton, the moles of product are 0.125 mmol. Mass of product = 0.125 mmol * 208 g/mol = 26 mg. Wait, let's re-calculate using ratios. (Int_prod / #H_prod) / (Int_std / #H_std) = Moles_prod / Moles_std. (1.25 / 1) / (1.00 / 4) = Moles_prod / 0.100 mmol. (1.25) / (0.25) = Moles_prod / 0.100 mmol. 5.0 = Moles_prod / 0.100 mmol. Moles_prod = 0.500 mmol. Mass_prod = 0.500 mmol * 208 g/mol = 104 mg.

Question 14

An amine is synthesized and purified. The ¹H NMR spectrum is taken in CDCl₃. All expected product peaks are present, but there is also a small, broad singlet at 1.56 ppm. After shaking the NMR tube with a drop of D₂O, this peak at 1.56 ppm disappears. What is the most likely identity of this impurity?

  1. The N-H proton of the amine product.
  2. Residual acetone from cleaning the glassware.
  3. A small amount of unreacted alkyl halide starting material.
  4. Residual water dissolved in the CDCl₃ solvent. (correct answer)
Explanation: When analyzing ¹H NMR spectra, you should always pay attention to peaks that disappear after D₂O exchange—this is a classic diagnostic tool that reveals exchangeable protons like N-H, O-H, or S-H bonds. The key clue here is that the broad singlet at 1.56 ppm completely disappears after D₂O treatment. This indicates the peak corresponds to exchangeable protons that undergo deuterium exchange: H-X + D₂O → D-X + HDO. Water dissolved in CDCl₃ produces a characteristic broad peak around 1.5 ppm due to the O-H protons, and these readily exchange with D₂O, causing the peak to vanish. Looking at why the other options don't fit: Option A is incorrect because amine N-H protons typically appear much broader and often at different chemical shifts (usually 0.5-5 ppm, but the behavior would be similar to water). However, the question states "all expected product peaks are present," implying the N-H of the product is already accounted for. Option B fails because acetone would show a sharp singlet at ~2.2 ppm for the methyl groups and wouldn't disappear with D₂O treatment. Option C is wrong because alkyl halide protons don't exchange with D₂O and would appear at different chemical shifts depending on the halide and substitution pattern. Remember this diagnostic principle: when you see a broad peak that disappears completely after D₂O shake, think exchangeable protons. Residual water in "dry" solvents is extremely common and shows this exact behavior at ~1.56 ppm in CDCl₃.

Question 15

The acid-catalyzed ring-opening of styrene oxide with methanol is expected to produce 2-methoxy-2-phenylethan-1-ol as the major product. A ¹H NMR of the crude product is obtained. The presence of which of the following signals would most clearly indicate the formation of the minor regioisomer, 1-methoxy-2-phenylethan-1-ol?

  1. A doublet at ~4.8 ppm with an integration of 1H. (correct answer)
  2. A singlet at ~3.2 ppm with an integration of 3H.
  3. A broad singlet at ~2.5 ppm that exchanges with D₂O.
  4. A multiplet in the aromatic region from 7.2-7.4 ppm.
Explanation: When analyzing epoxide ring-opening reactions, you need to understand the regioselectivity and how to identify products using NMR spectroscopy. In acid-catalyzed conditions, the epoxide is protonated first, making the more substituted carbon more electrophilic, so nucleophilic attack occurs at the more substituted position, giving 2-methoxy-2-phenylethan-1-ol as the major product. The minor regioisomer, 1-methoxy-2-phenylethan-1-ol, has a distinctive structural feature: a carbon bearing both a phenyl group and a hydrogen (the CHPh group). This carbon appears around 4.8 ppm due to the deshielding effect of the adjacent phenyl ring and oxygen. Since this carbon has only one hydrogen attached and is adjacent to a CH2CH_2 group, it appears as a doublet due to coupling with the two equivalent protons on the neighboring methylene carbon. Choice A correctly identifies this diagnostic signal - a doublet at ~4.8 ppm integrating for 1H represents the CHPh proton unique to the minor regioisomer. Choice B (singlet at 3.2 ppm, 3H) would be the methoxy group present in both products, so it can't distinguish between them. Choice C (broad singlet at 2.5 ppm exchanging with D₂O) describes the OH group, again present in both regioisomers. Choice D (aromatic multiplet 7.2-7.4 ppm) represents the phenyl protons found in both products. To identify minor regioisomers in NMR, look for signals that are structurally unique to one product - typically protons on carbons with different substitution patterns or electronic environments.

Question 16

A student prepares 2-phenyl-2-propanol (MW = 136.19 g/mol) by reacting acetophenone with methylmagnesium bromide. In the mass spectrum of the crude product, the presence of a large peak at m/z=121 is observed, corresponding to the loss of a methyl group from the product. The presence of which other significant peak would most strongly suggest a failure to exclude moisture, leading to quenching of the Grignard reagent?

  1. m/z = 43
  2. m/z = 16 (correct answer)
  3. m/z = 120
  4. m/z = 105
Explanation: Grignard reagents are strong bases and are readily quenched by protic sources like water. Methylmagnesium bromide (CH₃MgBr) would react with water to form methane (CH₄). Methane has a molecular weight of 16 g/mol, so a significant peak at m/z = 16 in the mass spectrum would indicate its formation. M/z=43 corresponds to the acetyl cation [CH₃CO]⁺, m/z=120 is acetophenone, and m/z=105 is the benzoyl cation [PhCO]⁺, all related to the starting material.