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Explain how a mass spectrometer, capable of distinguishing between ions with \(m / e\) values differing by one part in 50,000 , could be used to tell whether an ion of mass 29 is \(\mathrm{C}_{2} \mathrm{H}_{5}^{+}\) or \(\mathrm{CHO}^{+}\).

Short Answer

Expert verified
A mass spectrometer alone cannot differentiate based on m/e values, but fragmentation patterns may help using advanced techniques.

Step by step solution

01

Understand the Problem

The challenge is determining if a mass spectrometer can differentiate between ions with very close mass-to-charge (m/e) ratios. Specifically, differentiating between two ions: \(\mathrm{C}_{2}\mathrm{H}_{5}^{+}\) and \(\mathrm{CHO}^{+}\) when the mass is 29.
02

Calculate the Theoretical Mass for Each Ion

Firstly, calculate the atomic mass of \(\mathrm{C}_{2}\mathrm{H}_{5}^{+}\):- Two carbon atoms: \(2 \times 12 = 24\)- Five hydrogen atoms: \(5 \times 1 = 5\)The total mass is \(24 + 5 = 29\). Thus, \(\mathrm{C}_{2}\mathrm{H}_{5}^{+}\) has a mass of 29.Now, calculate the atomic mass of \(\mathrm{CHO}^{+}\):- One carbon atom: \(12\)- One hydrogen atom: \(1\)- One oxygen atom: \(16\)The total mass is \(12 + 1 + 16 = 29\). Thus, \(\mathrm{CHO}^{+}\) also has a mass of 29.
03

Understand the Instrument's Resolution

The mass spectrometer can distinguish ions with \(m/e\) values differing by one part in 50,000. This means it can resolve mass differences as small as \(29 / 50,000 = 0.00058\). Since the calculated masses of both ions are exactly 29, this level of resolution would not be able to differentiate the ions if their \(m/e\) values are the same.
04

Check for Structural Isomer Distinction

Since the mass is the same but the structure might differ, a mass spectrometer could potentially use a technique such as tandem mass spectrometry (MS/MS) to perform additional fragmentation analyses. This could differentiate between the ions based on their fragmentation patterns, even though their \(m/e\) values are identical.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass-to-Charge Ratio in Mass Spectrometry
The mass-to-charge ratio, denoted as \( m/e \), is a fundamental concept in mass spectrometry. It represents the mass of an ion divided by its charge. This characteristic helps in the separation and identification of ions in a sample. In mass spectrometry, ions are typically charged with a single positive charge, making \( m/e \) essentially equal to the mass of the ion.

Understanding \( m/e \) is essential for differentiating ions, especially when they have similar masses. The instrument can resolve differences in \( m/e \) values, allowing scientists to identify and analyze compounds even if they differ slightly in mass. High-resolution mass spectrometers can detect minuscule differences in \( m/e \), which is crucial for precise analyses in complex mixtures.
Isomer Differentiation with Mass Spectrometry
Sometimes, ions might have the same \( m/e \) but differ structurally, known as isomers. For example, the ions \( \mathrm{C}_{2}\mathrm{H}_5^+ \) and \( \mathrm{CHO}^+ \) both have a mass of 29, but are structurally distinct.

Mass spectrometers can differentiate these isomers, not by their mass, but through their fragmentation patterns.
  • Each isomer breaks down into fragments uniquely after ionization
  • By analyzing these fragmentation patterns, chemists can identify the different isomers present in a sample
This capability is essential in fields such as pharmaceuticals and environmental science, where identifying subtle differences between molecules is vital.
Understanding Tandem Mass Spectrometry (MS/MS)
Tandem mass spectrometry, or MS/MS, is an advanced technique used to analyze ionic species more deeply. This technique involves two stages of mass analysis: First, the ion of interest is isolated, and then it's further fragmented for a second round of analysis.

MS/MS is particularly useful for:
  • Identifying the presence of structural isomers
  • Providing detailed insight into compound structure through fragmentation patterns
Given that isomers like \( \mathrm{C}_{2}\mathrm{H}_5^+ \) and \( \mathrm{CHO}^+ \) cannot be distinguished based on their \( m/e \) alone, MS/MS offers a powerful method for accurate identification. It allows scientists to examine the subtle fragmentation differences to tell these isomers apart.
Ion Fragmentation in Mass Spectrometry
Ion fragmentation is a critical process that occurs when an ionized molecule breaks into smaller pieces inside the mass spectrometer. This process reveals a 'fingerprint' of the molecule, leading to its identification.

Fragmentation can occur naturally through collision with gas atoms in the spectrometer or be induced for more intentional analysis, as in MS/MS.
  • The pattern of fragmentation gives clues about the structure and connectivity of atoms within a molecule
  • Different molecules and isomers produce unique fragmentation patterns
This method is invaluable for structural determination and differentiating molecules with identical masses.
Spectrometric Resolution: Precision in Mass Analysis
Spectrometric resolution refers to a mass spectrometer's ability to differentiate between two ions with very similar \( m/e \) values. The resolution is expressed as \( \frac{m}{\Delta m} \), where \( \Delta m \) is the smallest mass difference that can be distinguished.

A high-resolution mass spectrometer can detect and distinguish minuscule differences in \( m/e \), essential for:
  • Analyzing complex mixtures
  • Identifying substances with closely spaced \( m/e \) values
For example, in our exercise, the spectrometer's resolution was 50,000, which translates to a capability of distinguishing differences as small as 0.00058 mass units. However, when isomers have the exact mass, structural difference techniques like MS/MS become necessary.

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Most popular questions from this chapter

a. Identify the protons with different chemical shifts in each of the structures shown. Use letter subscripts \(\mathrm{H}_{A}, \mathrm{H}_{B}\), and so on, to designate nonequivalent protons. Use models if necessary. (i) cis- and trans-2-butene (ii) 1,3-butadiene (iii) 1 -chloro-2,2-dimethylbutane (iv) 2-butanol (v) trans-1,2-dibromocyclopropane b.* Why does 3-methyl-2-butanol have three methyl resonances with different chemical shifts in its proton \(\mathrm{nmr}\) spectrum? c. \(^{*}\) For the compounds in Part a designated those protons (if any) that are enantiotopic or diastereotopic.

Sketch the proton \(\mathrm{nmr}\) spectrum and integral expected at \(60 \mathrm{MHz}\), with TMS as standard, for the following substances. Show the line positions in \(\mathrm{Hz}\); neglect spin-spin couplings smaller than 1 to \(2 \mathrm{~Hz}\) and all second-order effects. Remember that chlorine, bromine, and iodine (but not fluorine) act as nonmagnetic nuclei. a. \(\mathrm{CH}_{3} \mathrm{Cl}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) c. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) d. \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) f. \(\mathrm{CHCl}_{2} \mathrm{CHBr}_{2}\) g. \(\mathrm{CH}_{3} \mathrm{CHClCOCH}_{3}\) h. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) i. \(\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}\) j. \(\left(\mathrm{ClCH}_{2}\right)_{3} \mathrm{CH}\)

Sketch the proton chemical shifts in ppm and \(\mathrm{Hz}\) as well as the integral you would expect for each of the following substances at \(60 \mathrm{MHz}\). (The spin-spin splitting of the resonance lines evident in Figures 9-23 and 9-27, but not seen in Figure \(9-31\), can be safely neglected with all of the compounds listed.) a. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OCH}_{3}\) b. \(\mathrm{CH}_{2} \mathrm{COC}\left(\mathrm{CH}_{3}\right)_{3}\) c. \(\mathrm{HCOC}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHO}\) d. e. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CCl}_{2}\) f. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COC} \equiv \mathrm{CH}\) g. \(\left(\mathrm{CH}_{2} \mathrm{Cl}\right)_{3} \mathrm{CCO}_{2} \mathrm{H}\) h.* cis-1-methyl-4-tert-butyl-1,2,2,3,3,4,5,5,6,6-decachlorocyclohexane

Which compound in each group would have the most intense infrared absorption band corresponding to stretching vibrations of the bonds indicated? Give your reasoning. a. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}\) (multiple bond) b. \(\mathrm{CH}_{3}-\mathrm{CH}_{3}, \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}(\mathrm{C}-\mathrm{C}\) vs. \(\mathrm{C}-\mathrm{O})\) c. \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}, \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}\) (multiple bond) d. \(\mathrm{H}-\mathrm{Cl}, \mathrm{Cl}-\mathrm{Cl}\)

a. Calculate the relative intensities of the \((\mathrm{M}+1)^{+}\) and \((\mathrm{M}+2)^{+}\) ions for a molecule of elemental composition \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NO}_{2}\) b. The \(\mathrm{M}^{+},(\mathrm{M}+1)^{+}\), and \((\mathrm{M}+2)^{+}\) ion intensities were measured as \(100,8.84\), and \(0.54\) respectively, and the molecular weight as 120 . What is the molecular formula of the compound? c. In our example of how natural \({ }^{13} \mathrm{C}\) can be used to determine the number of carbon atoms in a compound with \(\mathrm{M}^{+}=86\) and a \((\mathrm{M}+1)^{+} / \mathrm{M}^{+}\) ratio of \(6.6 / 100\), we neglected the possible contribution to the \((\mathrm{M}+1)^{+}\) peak of the hydrogen isotope of mass 2 (deuterium). The natural abundance of deuterium is \(0.015 \%\) For a compound of composition \(\mathrm{C}_{6} \mathrm{H}_{14}\), how much do you expect the deuterium to contribute to the intensity of the \((\mathrm{M}+1)^{+}\) peak relative to the \(\mathrm{M}^{+}\) peak?

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