Organic Chemistry 2 Quiz: Integrating Ir Nmr Ms To Determine Structures
15 questions · exam conditions
0:00
Integrating Ir Nmr Ms To Determine StructuresQuestion 1 of 15

An organic compound has the formula C₇H₅N. Its IR spectrum shows a prominent, sharp peak at 2230 cm⁻¹. Its ¹H NMR spectrum shows a multiplet at δ 7.6 (2H) and a multiplet at δ 7.4 (3H). If this compound is treated with 1) LiAlH₄, then 2) H₂O, what is the structure of the resulting product?

Benzylamine
Aniline
Benzamide
Benzonitrile
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Integrating Ir Nmr Ms To Determine Structures

Practice Integrating Ir Nmr Ms To Determine Structures 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 Integrating Ir Nmr Ms To Determine Structures, 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 organic compound has the formula C₇H₅N. Its IR spectrum shows a prominent, sharp peak at 2230 cm⁻¹. Its ¹H NMR spectrum shows a multiplet at δ 7.6 (2H) and a multiplet at δ 7.4 (3H). If this compound is treated with 1) LiAlH₄, then 2) H₂O, what is the structure of the resulting product?

  1. Benzylamine (correct answer)
  2. Aniline
  3. Benzamide
  4. Benzonitrile
Explanation: The formula C₇H₅N has 7 - (5/2) + (1/2) + 1 = 6 degrees of unsaturation. The IR peak at 2230 cm⁻¹ is characteristic of a nitrile (C≡N). The ¹H NMR shows a monosubstituted benzene ring. Therefore, the starting material is benzonitrile (C₆H₅CN). The reagent LiAlH₄ is a powerful reducing agent that reduces nitriles to primary amines. The workup protonates the intermediate. Thus, benzonitrile is reduced to benzylamine (C₆H₅CH₂NH₂). Benzamide is an amide, aniline is C₆H₅NH₂, and benzonitrile is the starting material.

Question 2

Treatment of 1-butanol with PCC (pyridinium chlorochromate) in CH₂Cl₂ yields compound Z. The spectra of Z are obtained. Which of the following spectral features would be absent in the spectra of Z but present in the spectra of 1-butanol?

  1. A strong, sharp IR absorption near 1725 cm⁻¹.
  2. A ¹H NMR signal near δ 9.7.
  3. A broad IR absorption in the 3200-3600 cm⁻¹ region. (correct answer)
  4. A molecular ion peak in the mass spectrum.
Explanation: PCC is an oxidizing agent that converts primary alcohols to aldehydes. Therefore, 1-butanol (CH₃CH₂CH₂CH₂OH) is converted to butanal (CH₃CH₂CH₂CHO). The question asks for a feature present in the starting material (1-butanol) but absent in the product (butanal). 1-butanol is an alcohol and has a characteristic broad O-H stretch in its IR spectrum in the 3200-3600 cm⁻¹ region. Butanal is an aldehyde and lacks this O-H bond. Therefore, this feature is absent in Z. Choices A and B describe features of the product, butanal (a C=O stretch and an aldehyde proton signal), which would be absent in the starting material. Choice D is incorrect because both molecules would show a molecular ion peak in their mass spectra.

Question 3

An unknown compound A has the molecular formula C₅H₁₀O. Its IR spectrum shows a strong, sharp absorption at 1715 cm⁻¹. Its ¹H NMR spectrum consists of three signals: a septet at δ 2.5, a doublet at δ 1.1, and a singlet at δ 2.1. Compound A is treated with 1) NaBH₄, followed by 2) H₃O⁺ workup to yield product B. What is the structure of product B?

  1. 3-Methyl-2-butanone
  2. 2-Methoxy-3-methylbutane
  3. 3-Methyl-2-butanol (correct answer)
  4. 2,3-Dimethyl-2-butanol
Explanation: First, the structure of compound A must be determined. The formula C₅H₁₀O has one degree of unsaturation. The IR peak at 1715 cm⁻¹ indicates a ketone. The ¹H NMR shows an isopropyl group (septet/doublet) and a methyl group (singlet). The only structure consistent with this is 3-methyl-2-butanone (isopropyl methyl ketone). The reaction sequence (1. NaBH₄, 2. H₃O⁺) is the reduction of a ketone to a secondary alcohol. Therefore, product B is 3-methyl-2-butanol.

Question 4

An aromatic compound with the formula C₉H₁₀O₂ shows a strong IR absorption at 1720 cm⁻¹. Its ¹H NMR spectrum is: δ 7.9 (d, 2H), δ 7.2 (d, 2H), δ 3.9 (s, 3H), δ 2.4 (s, 3H). Which of the following is the correct structure?

  1. Ethyl benzoate
  2. 4-Methoxyacetophenone
  3. Phenyl propanoate
  4. Methyl 4-methylbenzoate (correct answer)
Explanation: The formula C₉H₁₀O₂ has five degrees of unsaturation, and the IR peak at 1720 cm⁻¹ suggests a carbonyl, likely an ester or ketone, attached to a benzene ring. The two doublets in the aromatic region of the ¹H NMR indicate a 1,4-disubstituted (para) benzene ring. The two singlets at 3.9 and 2.4 ppm correspond to a methoxy group and a methyl group, respectively. This eliminates A and C. Between B and D, 4-methoxyacetophenone would have aromatic protons at ~7.9 ppm (ortho to acetyl) and ~6.9 ppm (ortho to methoxy, shielded). Methyl 4-methylbenzoate has protons ortho to the deactivating ester group at ~7.9 ppm and ortho to the weakly activating methyl group at ~7.2 ppm, which matches the data perfectly.

Question 5

An unknown compound, C₆H₁₂O, shows a strong IR absorption at 1715 cm⁻¹ and a prominent peak in its mass spectrum at m/z = 57. Its ¹H NMR spectrum includes a triplet at δ 2.4 and a singlet at δ 1.0. Which structure is most consistent with all data?

  1. 2-Hexanone
  2. 3-Hexanone (correct answer)
  3. 4-Methyl-2-pentanone
  4. 3,3-Dimethyl-2-butanone (Pinacolone)
Explanation: The IR peak indicates a ketone. The m/z = 57 peak can correspond to a tert-butyl cation or a propanoyl cation. Pinacolone (D) would give a tert-butyl fragment at m/z = 57, but its ¹H NMR consists of only two singlets, which is inconsistent with the observed triplet. 3-Hexanone (B) can undergo alpha-cleavage to produce a propanoyl cation (CH₃CH₂CO⁺) at m/z = 57. Its ¹H NMR would show triplets and quartets for the ethyl and propyl groups, including a triplet around δ 2.4 for the CH₂ group of the propyl chain adjacent to the carbonyl. This matches the NMR data. 2-Hexanone and 4-Methyl-2-pentanone would not produce a major peak at m/z = 57.

Question 6

An unknown, C₅H₁₀O₂, displays a very broad IR absorption from 3300-2500 cm⁻¹ and a strong absorption at 1710 cm⁻¹. Its ¹H NMR spectrum shows a broad singlet at δ 11.5 (1H), a septet at δ 2.6 (1H), and a doublet at δ 1.2 (6H). Which of the following is the unknown compound?

  1. 2-Methylpropanoic acid (correct answer)
  2. Ethyl propanoate
  3. Methyl butanoate
  4. Pentanoic acid
Explanation: The spectral data are characteristic of a carboxylic acid. The IR shows the classic broad O-H stretch and a C=O stretch. The ¹H NMR shows a deshielded, broad proton for the carboxylic acid hydrogen at δ 11.5. The remaining signals, a septet and a 6H doublet, are indicative of an isopropyl group [-CH(CH₃)₂]. Combining these fragments gives 2-methylpropanoic acid. The other options are either esters (B, C) which would lack the broad O-H signals, or a constitutional isomer (D) that would have a different alkyl splitting pattern in the NMR.

Question 7

An unknown, C₉H₈O, exhibits strong, sharp IR absorptions at 3300 cm⁻¹ and 2120 cm⁻¹. Its ¹H NMR spectrum shows a multiplet near δ 7.4 (5H), a doublet at δ 4.8 (1H), a doublet at δ 2.7 (1H), and a broad singlet near δ 2.0 (1H). Which isomer is consistent with this data?

  1. 3-Phenyl-2-propyn-1-ol
  2. Cinnamaldehyde
  3. 1-Phenyl-2-propyn-1-ol (correct answer)
  4. Phenyl vinyl ketone
Explanation: The formula C₉H₈O has six degrees of unsaturation. The IR peaks at 3300 and 2120 cm⁻¹ are characteristic of a terminal alkyne C≡C-H. The broad singlet in the NMR and the C-H stretch near 3300 cm⁻¹ in the IR also suggest an O-H group. The structure must be an alcohol containing a phenyl group and a terminal alkyne. This eliminates B and D. Between A and C, 3-phenyl-2-propyn-1-ol (Ph-C≡C-CH₂OH) would have a 2H singlet for the CH₂ group. 1-Phenyl-2-propyn-1-ol (Ph-CH(OH)-C≡CH) has a methine proton (CH-OH) and a terminal alkyne proton. The methine proton would be a doublet coupled to the OH proton (or a singlet if exchange is fast), and the terminal alkyne proton would be a singlet or a long-range coupled small multiplet. The data given (two doublets for single protons) is most consistent with the benzylic methine proton coupled to the OH proton, and the terminal alkyne proton being present, fitting structure C.

Question 8

An unknown, C₁₀H₁₂O₂, is known to be an ester. Its IR spectrum shows a strong C=O stretch at 1718 cm⁻¹ and C-O stretches around 1280 and 1120 cm⁻¹. Its ¹H NMR spectrum is: δ 7.9 (d, 2H), δ 7.0 (d, 2H), δ 4.3 (q, 2H), δ 3.8 (s, 3H), δ 1.4 (t, 3H). What is the structure of this compound?

  1. Ethyl 4-methoxybenzoate (correct answer)
  2. Methyl 4-ethoxybenzoate
  3. 4-Ethoxyphenyl acetate
  4. Propyl 4-hydroxybenzoate
Explanation: The ¹H NMR spectrum shows a 1,4-disubstituted (para) benzene ring (two doublets). It also shows an ethyl group (quartet at 4.3 ppm, triplet at 1.4 ppm) and a methyl singlet at 3.8 ppm. The formula is C₁₀H₁₂O₂. The structure must contain an ethyl group, a methyl group, and a para-substituted ring. The deshielded quartet at 4.3 ppm suggests an ethyl group attached to an ester oxygen (-COOCH₂CH₃). The singlet at 3.8 ppm is characteristic of a methoxy group (-OCH₃) on the benzene ring. Combining these pieces, the structure is Ethyl 4-methoxybenzoate. This fits all the data, including the formula.

Question 9

An unknown compound has the formula C₇H₅ClO. Its IR spectrum has a strong absorption at 1770 cm⁻¹. Its ¹H NMR spectrum shows a complex multlet between δ 7.2-7.5 corresponding to 5 protons. What is the product when this compound is treated with an excess of ammonia (NH₃)?

  1. Benzoyl chloride
  2. Benzylamine
  3. Phenylacetamide
  4. Benzamide (correct answer)
Explanation: The unknown compound has formula C₇H₅ClO with degree of unsaturation = 4. The IR at 1770 cm⁻¹ is characteristic of an acid chloride C=O stretch. The ¹H NMR shows a monosubstituted benzene ring (5H multiplet). This identifies the compound as benzoyl chloride (C₆H₅COCl). When acid chlorides react with ammonia, they undergo nucleophilic acyl substitution to form primary amides. The chloride is displaced by ammonia as the nucleophile. Therefore, benzoyl chloride + NH₃ → benzamide (C₆H₅CONH₂) + HCl. The other options represent either the starting material (A) or incorrect product types (B, C).

Question 10

An aromatic compound, C₇H₇NO₂, has ¹H NMR signals at δ 8.1 (d, 2H), δ 7.5 (d, 2H), and δ 2.6 (s, 3H). This compound is subjected to bromination (Br₂/FeBr₃). Which structure represents the major organic product?

  1. 4-Bromo-3-nitrotoluene
  2. 2-Bromo-4-nitrotoluene (correct answer)
  3. 1-(Bromomethyl)-4-nitrobenzene
  4. 3-Bromo-4-methylbenzoic acid
Explanation: First, identify the starting material. The ¹H NMR shows a para-disubstituted ring (two doublets) with a methyl group (singlet). The deshielded aromatic protons suggest a strong deactivating group is present. The formula C₇H₇NO₂ indicates a nitro group. The structure is 4-nitrotoluene. In electrophilic aromatic substitution (bromination), we must consider the directing effects of both substituents. The methyl group is an ortho,para-director and an activator. The nitro group is a meta-director and a deactivator. The activating methyl group controls the regiochemistry, directing the incoming electrophile (Br⁺) to the position ortho to itself. This position is also meta to the nitro group. Therefore, the major product is 2-bromo-4-nitrotoluene.

Question 11

An unknown starting material A (C₈H₈O) is treated with excess CH₃MgBr followed by an aqueous workup to produce compound B. The IR spectrum of A shows a strong peak at 1685 cm⁻¹, and its ¹H NMR spectrum shows signals at δ 7.95 (m, 2H), δ 7.50 (m, 3H), and δ 2.60 (s, 3H). What is the structure of the final product B?

  1. 1-Phenylethanol
  2. 2-Phenyl-2-propanol (correct answer)
  3. Acetophenone
  4. Propiophenone
Explanation: First, identify compound A. The formula C₈H₈O, IR peak at 1685 cm⁻¹ (aromatic ketone), and ¹H NMR signals for a monosubstituted phenyl group and a methyl singlet indicate that A is acetophenone (Ph-CO-CH₃). The reaction is the addition of a Grignard reagent to a ketone. The nucleophilic methyl group from CH₃MgBr attacks the carbonyl carbon. Since a ketone is the electrophile, the product is a tertiary alcohol. The addition of one methyl group to acetophenone yields 2-phenyl-2-propanol. Choice A is the product of reduction. Choice C is the starting material. Choice D has the wrong carbon skeleton.

Question 12

A compound with formula C₅H₁₀O gives a positive Tollens' test. Its ¹H NMR spectrum is: δ 9.6 (d, 1H), δ 2.4 (m, 1H), δ 1.1 (d, 6H). The compound is then treated with LiAlH₄ followed by H₃O⁺. What is the major organic product of this reaction?

  1. 3-Methylbutanoic acid
  2. 3-Methyl-1-butanol (correct answer)
  3. 2-Methyl-1-butanol
  4. 3-Methyl-2-butanol
Explanation: First, identify the starting material. A positive Tollens' test and the aldehyde proton signal at δ 9.6 (doublet) in the ¹H NMR indicate an aldehyde. The multiplet and 6H doublet indicate an isopropyl group. The structure is 3-methylbutanal. LiAlH₄ reduces aldehydes to primary alcohols. Therefore, the reduction of 3-methylbutanal yields 3-methyl-1-butanol. The other choices represent an incorrect starting material or an incorrect reaction outcome (oxidation or reduction of a ketone).

Question 13

A compound with the formula C₄H₈O₂ exhibits a strong IR absorption at 1740 cm⁻¹. Its ¹H NMR spectrum is as follows: δ 4.1 (q, 2H), δ 2.0 (s, 3H), δ 1.2 (t, 3H). Which of the following structures is most consistent with this data?

  1. Methyl propanoate
  2. Butanoic acid
  3. Ethyl acetate (correct answer)
  4. Isopropyl formate
Explanation: The IR absorption at 1740 cm⁻¹ suggests an ester. The ¹H NMR data shows an ethyl group (quartet at 4.1 ppm coupled to a triplet at 1.2 ppm) and a methyl singlet at 2.0 ppm. The deshielded quartet (4.1 ppm) is characteristic of a CH₂ group attached to an ester oxygen. The singlet at 2.0 ppm is characteristic of a methyl group attached to a carbonyl. This pattern perfectly matches ethyl acetate (CH₃COOCH₂CH₃). Butanoic acid would show a broad OH peak in the IR and NMR. Methyl propanoate and isopropyl formate would have different NMR splitting patterns and integrations.

Question 14

Compound X (C₄H₁₀O) exhibits a very broad IR absorption from 3200-3600 cm⁻¹. When X is treated with Jones reagent (CrO₃/H₂SO₄), it is converted to compound Y (C₄H₈O₂), which shows a broad IR peak from 2500-3300 cm⁻¹ and a sharp peak at 1715 cm⁻¹. The ¹H NMR of X shows a doublet, a multiplet, and a triplet. Which of the following is compound X?

  1. 1-Butanol (correct answer)
  2. 2-Butanol
  3. tert-Butanol
  4. Diethyl ether
Explanation: Compound X is an alcohol (broad IR OH stretch). It is oxidized by Jones reagent to Y, a carboxylic acid (characteristic IR). This means X must be a primary alcohol. Of the options, only 1-butanol is a primary alcohol. 2-Butanol is secondary and would oxidize to a ketone. tert-Butanol is tertiary and does not oxidize. Diethyl ether is not an alcohol and would not react. The ¹H NMR of 1-butanol is complex but is consistent with the description of a doublet (from OH coupling), multiplet, and triplet, confirming the structure.

Question 15

A hydrocarbon with the formula C₁₀H₁₄ has a ¹H NMR spectrum with a multiplet corresponding to 5H near δ 7.2, a septet (1H) at δ 2.9, and a doublet (6H) at δ 1.2. Its broadband-decoupled ¹³C NMR spectrum shows exactly 4 signals in the aromatic region (120-150 ppm). What is the structure of this hydrocarbon?

  1. n-Butylbenzene
  2. sec-Butylbenzene
  3. tert-Butylbenzene
  4. Isopropylbenzene (Cumene) (correct answer)
Explanation: The formula C₁₀H₁₄ has four degrees of unsaturation, consistent with a benzene ring. The ¹H NMR shows a monosubstituted benzene ring (~5H multiplet) and an isopropyl group (septet and 6H doublet). This strongly suggests isopropylbenzene. The ¹³C NMR data confirms this. Due to symmetry in isopropylbenzene, there are only 4 unique aromatic carbons: the ipso-carbon, the para-carbon, and the equivalent pairs of ortho- and meta-carbons. tert-Butylbenzene would have a 9H singlet in the ¹H NMR. n-Butylbenzene and sec-Butylbenzene would have more complex alkyl signals and 6 aromatic signals in the ¹³C NMR due to lack of symmetry.