Description
I need a discussion and conclusion based on the data provided on the doc using the tables to talk about the results. Talk about the fragments of each one. Analyze each compound and the mass spectra be sure to use the tables to support the argument talk about the base peaks of each one.
Unformatted Attachment Preview
Daniela Arpi
Dr. Yamaguchi
Instrumental
03/12/2024
Illustrating Gas Chromatography and Mass Spectrometry
Introduction
Gas Chromatography and Mass Spectrometry is used to separate and identify different
components with complex mixtures. The GC technique separates the volatile compounds based
on their distribution between the two stationary phase and mobile phase. Upon injection into
the GC system, the sample undergoes vaporization and is transported by an inert gas along a
lengthy, spiral-shaped column that contains the stationary phase. Once the different
compounds interact with the stationary phase, they travel through the column at different
rates. Mass Spectrometry is used to determine the molecular weight and structural information
of the compound. MS measures the mass to charge ratio of ions formed in the sample. A
detector finds the ions and records their abundance at each m/z value to create a mass
spectrum. When it comes to a compound’s activity within the chromatographic system, the
retention time in GC and the mass spectrum offer useful information for identification. The GCMS offers more specificity and sensitivity than others technique. The objective of this
experiment is to gain experience using the GCMS instrument and to interpret the mass spectra
of the compounds. The isomeric ketones and esters used in this experiment were methyl
hexanoate, methyl n-octanote, 2-heptanone, ethyl butyrate, 5-methyl-2-hexanone, diethyl
succinate, ethyl hexanoate and methyl laurate. The parent ion (M) identifies the mass and
molecular formula of each ketone component. The molecular ion is usually the most common
ion in the mass spectrum of GCMS, having the highest m/z value. The compound’s chemical
formula, C₆H₁₂O, is indicated if the parent ion is at m/z 100. This m/z value represents the
molecular ion’s loss of one carbon atom, which is usually the result of ionization-related
fragmentation processes. The compound’s chemical formula, , C7H14O, is indicated if the
parent ion is at m/z 114. Similarly, this m/z ratio indicates a slightly larger ketone molecule
since it correlates to the molecular ion losing one carbon atom. When isomeric ketones with
identical molecular weights exist, they can be distinguished from one another by the presence
of the parent ion at different m/z values.
Data Analysis
Isomeric Ketones
5-methyl-2-hexanone
Retention Time: 10 minutes
Base Peak: 74 m/z
2-heptanone
Retention Time: ~4.2 minutes
Base Peaks: 43 and 58 m/z
Esters
Methyl hexanoate
Retention Time: Around 7 minutes
Base Peak: 43 m/z
methyl n-octanoate
Retention Time: 10 minutes
Base Peak: 74 m/z
ethyl butyrate
Retention Time: ~ 2.7 minutes
Base Peaks: 43 & 71 m/z
diethyl succinate
Retention Time: 11.5 minutes
Base Peak: 101 m/z
ethyl hexanoate
Retention Time: 7 minutes
Base Peak: 43 m/z
Methyl Laurate
Retention Time: 18 minutes
Base Peaks: 74 & 87 m/z
Discussion
We analyze the samples using the fragmentation table from above where we used the
retention time and mass spectra were used to identify the isomeric ketones and esters. Once
the samples were ran through we used the library tool to verify the compatibible. The first
compound that was analyzed was 5-methyl-2-hexanone, the retention time was of 10 minutes
The most prevalent fragment ion in the 5-methyl-2-hexanone mass spectrum is represented by
the base peak, which has a mass-to-charge ratio (m/z) of 74. When comparing the base peak at
m/z 74 for 5-methyl-2-hexanone with table 11.11 some neutral fragments could be
Acetaldehyde, Butyl radical or Propionaldehyde. In the Methyl hexanoate case the retention
time was about 7 minutes and a base peak of 43m/z. Comparing the base peak to the mass
spectra fragmentation table we are able to see that it might correspond to ketone or propyl
compound. Also other peaks we can notice is 88m/z which corresponds to a carboxyl group
since you add the 43+45=88m/z.
Purchase answer to see full
attachment
