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The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.
Answer:
See figures 1 and 2
Explanation:
For this question, we can analyze each compound:
<u>a. (4Z,3S)-2,4,7-trimethyl-3-amino-4-octene</u>
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In this case, we have to start with a <u>carbon chain of 8 carbons</u>. Then in carbons 2, 4 and, 7, we have to put three methyl groups (). In carbon 4 we have to put an amino group (
). When we put the amino group in carbon 4 we will have a <u>chiral carbon</u>. So, we have to indicate the orientation in the bond of the amino group. If we want an "S" configuration the atoms must have a <u>counterclockwise orientation</u>. So, the amino group must have a dashed bond. Finally, in carbon 4 we have to put a double bond, the configuration of this double bond is <u>"Z"</u>. So, the most important groups must be on the <u>same side</u>, in this case.
See figure 1 for further explanations
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<u>b. (3S,6R)-6-(2,4-dinitrophenyl)-3-amino-1-heptyne</u>
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In this case, we have to start with a <u>chain of 7 carbons</u>. In carbon 3 we have to put an amino group (). When we put this group in carbons we will have a chiral carbon is the name we have an <u>"S" configuration</u> for this carbon. So, we have to use the dashed bond in order to have a <u>counterclockwise orientation</u> for all the atoms bonded on carbon 3. In carbon 6 we have to put a big group "2,4-dinitrophenyl". In this group, we have a bond with the main carbon chain in carbon 6 (of benzene), in this same benzene ring we have to put nitro groups (
) in carbons 2 and 4. Additionally, carbon 6 is a chiral carbon, and the configuration of this carbon is "R". Therefore we have to use a wedge bond for all the "2,4-dinitrophenyl" in order to have a clockwise orientation for all the atoms bonded on carbon 6. Finally, we have to put a triple bond between carbons 2 and 3.
See figure 2 to further explanations
