Question 4 (10 points) 1. How do you explain that the MS spectrum does not show a fragment of m/z 29 even though the los

[answerhappygod]

Question 4 10 Points 1 How Do You Explain That The Ms Spectrum Does Not Show A Fragment Of M Z 29 Even Though The Los 1 (30.95 KiB) Viewed 46 times

Question 4 10 Points 1 How Do You Explain That The Ms Spectrum Does Not Show A Fragment Of M Z 29 Even Though The Los 2 (32.04 KiB) Viewed 46 times

Question 4 10 Points 1 How Do You Explain That The Ms Spectrum Does Not Show A Fragment Of M Z 29 Even Though The Los 3 (56.71 KiB) Viewed 46 times

Question 4 (10 points) 1. How do you explain that the MS spectrum does not show a fragment of m/z 29 even though the loss of a fragment of that mass is suggested elsewhere? The loss of the ethyl group occurs through McLafferty rearrengement, which produces a fragment that carries the charge on the C next to a single-bonded O. The loss of the ethyl group occurs through alpha-cleavage, which produces a fragment that carries the charge on the next to a single-bonded O. There actually is no ethyl group, otherwise the spectrum would show a peak at m'z 29.
Question 5 (10 points) Based on the correct answer to the previous question, what is the most likely feature of the structure of the molecule that explain the observed fragmentation pattern? Justify your choice. O It has a straight chain. The loss of an ethyl group and the formation of the fragment of m/z 43 leaves only one that can be accounted in the fragment of m/z 73 It has a straight chain, because it does not have a fragment of m/z 29. It has a branching chain, because it does not have a fragment of m/z 29.
The formula of the molecule responsible for the attached HNMR spectrum is CSH1002 11 10 9 8 7 2 1 0 6 5 ppm HSP-00-520 What information does the multiplet around 4.1 ppm on the H NMR spectrum provide? Justify your answer. It belongs to H's on a C located next to a CH3 group to account for the 3 components of the multiplet The carrying the resonating H's is bound to an atom with high electronegativity such as Catom to account for the most upfield position of the triplet It belongs to H's on a C located next to a CH2 group to account for the 3 components of the multiplet: The C carrying the resonating H's is boutid to an atom with high electronegativity such as atom to account for the most downfield position of the triplet It belongs a H attached to the O atom to account for the relatively downfield position of the triplet