As shown in Question 3 of the recitation questions, concentrated sulfuric acid c
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As shown in Question 3 of the recitation questions, concentrated sulfuric acid can convert any 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 the E2 mechanism and not the E1 mechanism. Why does elimination not occur via E1? In your experiment the product is characterized using IR spectroscopy, because that is easy for you to do in the lab. However, NMR 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. Do not attempt to distinguish them based on absolute values of chemical shifts (i.e., don't make statements such a "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. 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? In your reaction, one molecule/mole of reactant theoretically gives 1 molecule/mole of product. In the different reaction shown below (this is a retro- reaction). ONE molecule/mole of the reactant gives TWO molecules/moles of the product. In this case, a percent yield calculation needs to take into account the molar ratio (theoretically should make twice as many molecules/moles of product as reactant). If you started with 0.5 moles of the starting and formed 0.9 moles TOTAL acetone product (i.e. 0.9 is the number of moles of all of the acetone combined), calculate the percent yield for the reaction shown, show your work, all equations and include all units.Explanation / Answer
6.
The E1 pathway (formation of a primary carbocation) is not the most likely pathway here.Primary carbocations tend to be extremely unstable, and hence it’s more likely that the reaction passes through an E2 mechanism where the transition state will be lower in energy.
Here in the dehydration reaction, first we protonate the alcohol to give the good leaving group OH2+ , and then a weak base (could be H2O, —OSO3H, or another molecule of the alcohol) could then break C-H, leading to formation of the alkene.
7.
The molecules are Butanol ( CH3CH2CH2OH) and Butylbromide( Ch3CH2CH2Br)
Butanol has an –OH peak and this –OH appears as a singlet in the 1HNMR spectrum of Butanol
This –OH peak is absent in the 1HNMR spectrum of butyl bromide. If –OH peak is not in the 1HNMR spectrum of the product butyl bromide, it means that the starting material butyl alcohol is completely converted to the product.
8.
It depends on the density. The density of concentrated sulphuric acid layer is higher than that of the organic layer. So the concentrated sulphuric acid layer comes at the bottom.
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