you have access to authentic spectra such as those given above, describe how you
ID: 579886 • Letter: Y
Question
you have access to authentic spectra such as those given above, describe how you would use functional group information from infrared spectroscopy to distinguish between the bromide product and the alcohol starting material, AND ALSO information from the fingerprint region (lots of and frequencies are not required here). 14. (2 pts.) In your experiment the product is characterized using IR spectroscopy, because that is easy for lab. However, NM spectroscopy is usually much better for characterizing organic structures. Describe IN WORDS, how you would distinguish the product bromide form the starting material alcohol using proton NMR spectroscopy. You can use the provided structures in your explanation attempt to distinguish them based on absolute values of chemical shifts (ie., don't make statements such as the chemical shift of protons X should be 3.86 in the bromide and 3.65 in the alcohol"), there is another much simpler way of distinguishing the structures. CH2 CH2 CH2 CH2 7. (3 pts.) In your extraction, the organic layer that contained mainly the butyl bromide was on top and the concentrated sulfuric acid layer was the bottom layer. Why is this? 8. (2 pts.) In your reaction, the bromide anion substitutes for H20 to make the alkyl bromide. The mechanism is Sn2. Why is it not Sw1? Give a brief one sentence explanation in terms of the stabilities of organic carbocations. 9.(2 pts.) Concentrated sulfuric acid can convert any unreacted alcohol into alkenes in an unwanted side reaction. This is an elimination reaction, first the acid protonates the oxygen of the alcohol to convert it into a good leaving group, and then elimination occurs via the E2 mechanism and not the E1 mechanism. Why does elimination not occur via E1? 10. (2 pts.) The wet cotton traps any HBr that escapes the reaction mixture. The trapping reaction is simply what happens when HBr dissolves in water, but it is still an acid/base reaction that is in principle reversible. Complete the acid/base equilbrium started for you below, draw the curved arrow pushing in BOTH directions and identify the stronger and weaker Bronsted acid and base on each side of the equilibrium.Explanation / Answer
Answer 13.
In IR spectrum, a broad band is observed at 3300-3500 cm-1 for alcohol starting material. After reaction the starting alcohol is converted into alkyl bromide. Thus, a broad band at 3300-3500 cm-1 is disappeared in alkyl bromide product because of consumption of alcohol group of starting material and a strong peak is observed at 600-700 cm-1 for Bromide group.
Answer 14.
In the NMR spectrum, the broad peak of alcohol group is observed in the alcohol starting material while this broad peak is disappeared in alkyl bromide. Thus, the product alkyl bromide can easily be distinguished from the alcohol starting material.
Answer 7.
In the extraction, the organic layer having butyl bromide was on top and the concentrated sulfuric acid layer was on bottom layer. This is because the density of concentrated layer is higher than that of organic layer having butyl bromide.
Answer 8.
The reaction is SN2. This is because the starting material is primary alcohol and forms primary carbocation which is less stable. Thus, at primary alcohol, SN2 reaction is favored with good nucleophile Br-.
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