Organic Chemistry 2 Quiz: Mass Spectrometry Molecular Ion And Fragmentation
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Mass Spectrometry Molecular Ion And FragmentationQuestion 1 of 14

The mass spectrum of ethylbenzene (C₆H₅CH₂CH₃) exhibits a base peak at m/z = 91. This fragment is the result of which process?

Loss of a hydrogen atom from the molecular ion
A McLafferty rearrangement involving the aromatic ring
Benzylic cleavage involving the loss of a methyl radical
Alpha cleavage resulting in the formation of a C₆H₅⁺ ion
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Mass Spectrometry Molecular Ion And Fragmentation

Practice Mass Spectrometry Molecular Ion And Fragmentation 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 Mass Spectrometry Molecular Ion And Fragmentation, 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

The mass spectrum of ethylbenzene (C₆H₅CH₂CH₃) exhibits a base peak at m/z = 91. This fragment is the result of which process?

  1. Loss of a hydrogen atom from the molecular ion
  2. A McLafferty rearrangement involving the aromatic ring
  3. Benzylic cleavage involving the loss of a methyl radical (correct answer)
  4. Alpha cleavage resulting in the formation of a C₆H₅⁺ ion
Explanation: The molecular weight of ethylbenzene is 106. The fragment at m/z = 91 corresponds to a loss of 15 mass units (106 - 91 = 15), which is a methyl group (•CH₃). This fragmentation is a classic example of benzylic cleavage, where the bond between the α and β carbons of the side chain breaks. This process is highly favorable because it forms the very stable benzyl cation (C₆H₅CH₂⁺), which rearranges to the even more stable tropylium ion, both having an m/z of 91.

Question 2

A mass spectrum shows a base peak at m/z = 57. Which of the following parent compounds is LEAST likely to produce this fragment?

  1. 2,2-Dimethylpropane
  2. Butanoyl chloride
  3. 2-Methylbutane
  4. Pentanal (correct answer)
Explanation: The fragment at m/z = 57 typically corresponds to C₄H₉⁺, most commonly the stable tert-butyl cation. 2,2-Dimethylpropane readily loses a methyl radical to form (CH₃)₃C⁺ at m/z = 57. 2-Methylbutane can fragment and rearrange to form the same stable tert-butyl cation. Butanoyl chloride can lose various fragments and rearrange to produce C₄H₉⁺. However, pentanal primarily undergoes McLafferty rearrangement (m/z = 44) and α-cleavage (m/z = 29, CHO⁺). There is no favorable fragmentation pathway for pentanal to produce a significant C₄H₉⁺ fragment at m/z = 57.

Question 3

High-resolution mass spectrometry (HRMS) is used to distinguish between two compounds, A and B, which have the same nominal molecular weight of 72. Compound A is an aldehyde (C₄H₈O) and Compound B is an alkene (C₅H₁₂). Given the exact masses H=1.0078, C=12.0000, O=15.9949, how would HRMS differentiate them?

  1. Compound A would show an exact mass of 72.0573, while Compound B would show 72.0936. (correct answer)
  2. Compound A would show an exact mass of 72.0936, while Compound B would show 72.0573.
  3. Both compounds would show the same exact mass, but have different fragmentation patterns.
  4. HRMS cannot distinguish between these compounds; ¹³C NMR would be required.
Explanation: HRMS measures the exact mass of an ion to several decimal places. For Compound A (C₄H₈O): 4(12.0000) + 8(1.0078) + 1(15.9949) = 48.0000 + 8.0624 + 15.9949 = 72.0573. For Compound B (C₅H₁₂): 5(12.0000) + 12(1.0078) = 60.0000 + 12.0936 = 72.0936. The difference of 0.0363 mass units is easily resolved by HRMS, allowing unambiguous molecular formula determination.

Question 4

The fragmentation of cyclohexanone in a mass spectrometer produces several characteristic peaks. A prominent peak at m/z = 55 is often observed. This fragment is most likely formed by which of the following pathways?

  1. Loss of a propyl radical (C₃H₇) from the molecular ion (correct answer)
  2. Successive loss of two water molecules
  3. Cleavage of the ring followed by loss of a neutral propene molecule
  4. Ring opening followed by alpha-cleavage and loss of a neutral ketene molecule
Explanation: The molecular ion of cyclohexanone is at m/z = 98. A fragment at m/z = 55 represents a loss of 43 mass units, corresponding to C₃H₇. In cyclohexanone, α-cleavage can occur adjacent to the carbonyl group, followed by ring fragmentation and hydrogen rearrangement to eliminate a propyl radical. This produces a stable acylium-type ion at m/z = 55. The other fragmentation pathways either involve incorrect mass losses or are not characteristic of ketone fragmentation patterns.

Question 5

An organic compound with the formula C₅H₁₀O gives a strong absorption in its IR spectrum at 1715 cm⁻¹. Its mass spectrum shows a molecular ion peak at m/z = 86, and other prominent peaks at m/z = 71, 58, and 43. What is the structure of the compound?

  1. Pentanal
  2. 2-Pentanone (correct answer)
  3. 3-Pentanone
  4. 3-Methyl-2-butanone
Explanation: The IR absorption at 1715 cm⁻¹ indicates a ketone or aldehyde. The molecular formula C₅H₁₀O is consistent with this. Let's analyze the mass spectrum fragments for the ketone isomers. The M⁺ peak is at m/z = 86. The peak at m/z = 58 is highly characteristic of a McLafferty rearrangement for a ketone or aldehyde with a γ-hydrogen. 2-Pentanone has γ-hydrogens and would give a McLafferty fragment at m/z = 58. It would also give α-cleavage fragments at m/z = 71 (loss of •CH₃) and m/z = 43 (loss of •C₃H₇). This matches all the data. 3-Pentanone would give an α-cleavage fragment at m/z = 57, not seen. 3-Methyl-2-butanone has no γ-hydrogens and cannot undergo a McLafferty rearrangement, so it would not have a peak at m/z = 58.

Question 6

An unknown compound has the molecular formula C₄H₈O₂. Its mass spectrum shows a base peak at m/z = 43 and another strong peak at m/z = 45. Which of the following is the most likely structure of the compound?

  1. Butanoic acid
  2. Ethyl acetate (correct answer)
  3. Methyl propanoate
  4. Isopropyl formate
Explanation: The molecular weight of C₄H₈O₂ is 88. Ethyl acetate (CH₃COOCH₂CH₃) fragments by α-cleavage to lose an ethoxy radical (OC₂H₅•) forming the acetylium ion (CH₃CO⁺) at m/z = 43, which is typically the base peak for acetate esters. The peak at m/z = 45 corresponds to the ethoxy cation (C₂H₅O⁺) formed by alternative cleavage. This fragmentation pattern is characteristic of ethyl acetate.

Question 7

In the mass spectrum of 2-pentanone (CH₃COCH₂CH₂CH₃), two major fragmentation pathways are α-cleavage and McLafferty rearrangement. Which m/z value corresponds to the fragment that is most likely the base peak?

  1. m/z = 86
  2. m/z = 71
  3. m/z = 58
  4. m/z = 43 (correct answer)
Explanation: The molecular ion peak (M⁺) for 2-pentanone is at m/z = 86. The McLafferty rearrangement yields a fragment at m/z = 58. There are two possible α-cleavages: (1) loss of a propyl radical (•C₃H₇) to form the acetylium ion (CH₃CO⁺) at m/z = 43, and (2) loss of a methyl radical (•CH₃) to form the butyrylium ion (C₃H₇CO⁺) at m/z = 71. The acetylium ion (m/z = 43) is particularly stable and its formation involves the loss of a more stable primary radical compared to a methyl radical. For methyl ketones, the m/z = 43 peak is almost always the base peak. Therefore, m/z = 43 is the most likely base peak.

Question 8

The mass spectrum of a compound containing only carbon, hydrogen, nitrogen, and oxygen exhibits a molecular ion peak at an m/z value of 121. Based on the Nitrogen Rule, which statement can be concluded with the highest certainty?

  1. The molecule contains exactly one nitrogen atom.
  2. The molecule contains an odd number of nitrogen atoms. (correct answer)
  3. The molecule has the formula C₇H₇NO.
  4. The molecule must contain a phenyl group.
Explanation: The Nitrogen Rule states that a molecule with an odd molecular weight must contain an odd number of nitrogen atoms. Since the molecular ion peak is at m/z = 121 (an odd number), the compound must have an odd number of nitrogens (1, 3, 5, etc.). While a molecule with one nitrogen atom (choice A) would fit, we cannot be certain it is exactly one without more information. Similarly, choice C (benzamide, MW=121) is a possible structure, but other formulas like C₆H₁₅N₃ are also possible. Choice D is a structural feature that cannot be determined from the molecular weight alone. The most certain conclusion is the general rule stated in choice B.

Question 9

Which of the following molecules is most likely to have its molecular ion peak as the base peak in its mass spectrum?

  1. 2,2,4-Trimethylpentane
  2. 2-Heptanol
  3. Naphthalene (C₁₀H₈) (correct answer)
  4. Heptanal
Explanation: The molecular ion (M⁺) peak is the base peak when the molecular ion is very stable and resistant to fragmentation. Aromatic compounds like naphthalene have delocalized π-systems that make their radical cations exceptionally stable. In contrast, branched alkanes like 2,2,4-trimethylpentane fragment readily to form stable tertiary carbocations. Alcohols like 2-heptanol readily lose water and undergo α-cleavage, often having a very weak or absent M⁺ peak. Aldehydes like heptanal also fragment easily via α-cleavage and McLafferty rearrangement.

Question 10

Which statement best explains why the mass spectrum of 2-chlorobutane shows a more intense M-35 peak (loss of Cl) than the spectrum of 1-chlorobutane?

  1. The C-Cl bond in 2-chlorobutane is weaker than in 1-chlorobutane.
  2. 1-chlorobutane preferentially loses HCl instead of a chlorine radical.
  3. Loss of Cl from 2-chlorobutane forms a more stable secondary carbocation. (correct answer)
  4. The M+2 isotope peak is more abundant for 2-chlorobutane.
Explanation: The intensity of a fragment peak in a mass spectrum is directly related to the stability of the fragment ion formed. When 2-chlorobutane loses a chlorine radical (Cl•), it forms a secondary butyl carbocation. When 1-chlorobutane loses a chlorine radical, it forms a primary butyl carbocation. Secondary carbocations are significantly more stable than primary carbocations. Because a more stable fragment is formed, this fragmentation pathway is more favorable for 2-chlorobutane, leading to a more intense peak corresponding to the C₄H₉⁺ ion (M-35 or M-37).

Question 11

The mass spectrum of 1-methoxybutane (CH₃OCH₂CH₂CH₂CH₃) is compared to that of its isomer, diethyl ether (CH₃CH₂OCH₂CH₃). Which unique and prominent peak would help to identify 1-methoxybutane?

  1. A base peak at m/z = 31, corresponding to the methoxy cation (CH₃O⁺).
  2. An intense peak at m/z = 45, corresponding to the [CH₂OCH₃]⁺ fragment. (correct answer)
  3. An intense peak at m/z = 59, corresponding to the [CH₂OCH₂CH₃]⁺ fragment.
  4. A molecular ion peak at m/z = 88 that is absent for diethyl ether.
Explanation: Both isomers have a molecular weight of 88. The key is α-cleavage. For 1-methoxybutane, α-cleavage can occur by loss of a propyl radical to form the [CH₂OCH₃]⁺ fragment. This fragment has an m/z of (14+15+16) = 45. This is a very common and intense peak for methyl ethers. Diethyl ether would fragment by loss of a methyl radical to give a [CH₂OCH₂CH₃]⁺ fragment at m/z = 59. Therefore, a prominent peak at m/z = 45 is characteristic of 1-methoxybutane and would distinguish it from diethyl ether.

Question 12

A compound is known to contain both one bromine atom and one chlorine atom. Which pattern of peaks would be expected for its molecular ion cluster in a low-resolution mass spectrometer?

  1. An M⁺ peak and an M+2 peak with a 3:1 intensity ratio.
  2. An M⁺ peak and an M+2 peak with a 1:1 intensity ratio.
  3. An M⁺, M+2, and M+4 peak with a ~9:6:1 intensity ratio.
  4. An M⁺, M+2, and M+4 peak with a ~3:4:1 intensity ratio. (correct answer)
Explanation: This requires considering the isotopic abundances of both halogens. Chlorine has isotopes ³⁵Cl (~75%) and ³⁷Cl (~25%), a 3:1 ratio. Bromine has isotopes ⁷⁹Br (~50%) and ⁸¹Br (~50%), a 1:1 ratio. The M⁺ peak will correspond to the lightest combination: ³⁵Cl and ⁷⁹Br. The M+2 peak will arise from two combinations: ³⁷Cl + ⁷⁹Br and ³⁵Cl + ⁸¹Br. The M+4 peak will correspond to the heaviest combination: ³⁷Cl and ⁸¹Br. Let's approximate the relative probabilities: M⁺ (³⁵Cl, ⁷⁹Br) = 0.75 * 0.50 = 0.375. M+2 = (³⁷Cl, ⁷⁹Br) + (³⁵Cl, ⁸¹Br) = (0.25 * 0.50) + (0.75 * 0.50) = 0.125 + 0.375 = 0.500. M+4 (³⁷Cl, ⁸¹Br) = 0.25 * 0.50 = 0.125. The ratio of intensities is approximately 0.375 : 0.500 : 0.125, which simplifies to 3:4:1.

Question 13

The mass spectrum of triethylamine, N(CH₂CH₃)₃, has a molecular ion peak at m/z = 101. The base peak is found at m/z = 86. This base peak corresponds to a fragment formed by which process?

  1. Loss of a hydrogen atom
  2. Loss of an ethyl radical
  3. Loss of an ethene molecule
  4. Loss of a methyl radical (correct answer)
Explanation: The dominant fragmentation pathway for amines is α-cleavage, which is the breaking of a C-C bond adjacent to the nitrogen atom. In triethylamine, this involves breaking the bond between the α-carbon and β-carbon of one of the ethyl groups. This results in the loss of a methyl radical (•CH₃, mass 15). The molecular ion is at m/z = 101, so 101 - 15 = 86. The resulting fragment, [(CH₃CH₂)₂N=CHCH₃]⁺, is a resonance-stabilized iminium ion, making this a very favorable process and explaining why m/z = 86 is the base peak.

Question 14

A compound with the formula C₄H₈O₂ is analyzed by mass spectrometry. A significant peak is observed at m/z = 60. This observation is most consistent with which of the following structures?

  1. Ethyl acetate
  2. 2-Hydroxybutanal
  3. Butanoic acid (correct answer)
  4. Methyl propanoate
Explanation: A peak at m/z = 60 is characteristic of the McLafferty rearrangement in carboxylic acids. Butanoic acid (CH₃CH₂CH₂COOH, MW = 88) has γ-hydrogens that allow McLafferty rearrangement, losing ethene (C₂H₄, mass 28) to produce a fragment at m/z = 60. This fragment corresponds to the acetic acid radical cation. Ethyl acetate lacks the required γ-hydrogen for McLafferty rearrangement and fragments primarily by α-cleavage to m/z = 43. The other options do not show dominant peaks at m/z = 60.