- 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?
Answer: Compound A would show an exact mass of 72.0573, while Compound B would show 72.0936.
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.
- 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?
Answer: Loss of Cl from 2-chlorobutane forms a more stable secondary carbocation.
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).
- 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?
Answer: Ethyl acetate
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.
- Which of the following molecules is most likely to have its molecular ion peak as the base peak in its mass spectrum?
Answer: Naphthalene (C₁₀H₈)
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.
- 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?
Answer: An M⁺, M+2, and M+4 peak with a ~3:4:1 intensity ratio.
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.
- 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?
Answer: m/z = 43
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.
- 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?
Answer: Loss of a methyl radical
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.
- A mass spectrum shows a base peak at m/z = 57. Which of the following parent compounds is LEAST likely to produce this fragment?
Answer: Pentanal
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.
- 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?
Answer: Loss of a propyl radical (C₃H₇) from the molecular ion
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.
- 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?
Answer: 2-Pentanone
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.
- The mass spectrum of ethylbenzene (C₆H₅CH₂CH₃) exhibits a base peak at m/z = 91. This fragment is the result of which process?
Answer: Benzylic cleavage involving the loss of a methyl radical
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.
- 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?
Answer: The molecule contains an odd number of nitrogen atoms.
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.
- 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?
Answer: An intense peak at m/z = 45, corresponding to the [CH₂OCH₃]⁺ fragment.
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.
- 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?
Answer: Butanoic acid
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.
