Expierement #9 5a7c3f5a5de34531825e9ce379aaaabe
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Expierement #9
1
Expierement #9
1. Using the instruction in the Rubric, calculate the percentage of both 1-
methylcyclohexene and 3-methylcyclohexene in both fractions. How much are they different? Does “Evelyn effect” operate as it is described in the manual?
The 1-Methylcyclohexane is the major product since the peak is larger and it is the most product formed and 3-Methylcyclohexane is a minor product because it always produced in minority compared to the major product.
8 mL Fraction:
3-Methylcyclohexane Area = 10698257
1-Methylcyclohexane Area = 146659909
Total Area = 157705849
3-Methylcyclohexane Percentage = 10698257/157705849 = 7%
1-Methylcyclohexane Perecentage = 146659909/157705849 = 93%
6 mL Fraction:
3-Methylcyclohexane Area = 30706428
1-Methylcyclohexane Area = 124400365
Total Area = 160981817
3-Methylcyclohexane Percentage = 30706428/160981817 = 19%
1-Methylcyclohexane Perecentage = 124400365/160981817 = 77%
Among the two fractions the alkenes varied in ratio/percentage. For instance in the first fraction (8mL) the 3-Methylcyclohexane was 7% and in the second fraction (6mL) it more than doubled to 19%. Oppositley, 1-Methylcyclohexane was originally 93% in the 8mL and then was 77% in the 6mL. The “Eveyln Effect” says that there will be a change in ratio of percent compositions between the alkenes through time because of the
Expierement #9
2
reaction mechanisms that are competing. The Evelyn Effect also states that there will be a lower percentage of the expected (zaitzev product). There is compeition between E1 and E2 reaction which causes the competition between mechanisms and shows how the Evelyn Effect is taking place and causing a change in percent compositions. Furthermore, there were both cis and trans isomers in the reaction, the cis reacts faster and when there is more availability of trans after the cis is reacted, the trans will start reacting more. That creates the change in percetages between the two times.
2. Compare the retention times for each peak (i.e. time corresponding to the maximum of each peak), with the boiling point of each alkene. Does the alkene with lower boiling point really appear first (has lower retention time)?
The lower boiling point does have a lower retention time as it does appear first. For instance, 3-Methylcyclohexane has a boiling point of 104 degrees celcius and it has a shorter peak that happens at about 1.45 minutes. This peak happens first and shows that the lower boiling point come first. Then the major product (1-methylcyclohexane) that has a larger boiling point of 110 degrees celcius comes with the second larger peak. This peak has a retention time around 1.7 minutes. During both of the fractions, 3-
methylcyclohexane occurs first and it has a lower boiling point and the larger boiling point peak occurs after.
3. Consider the dehydration reaction of 4-methylcyclohexanol. By considering the stability of the relevant conformations required for E2 elimination, predict which stereoisomer (
cis
or trans
4-methylcyclohexanol) should undergo faster dehydration if we assume that both E1 and E2 occur concurrently?
Expierement #9
3
Cis-4-methylcyclohexanol will react faster than the trans-4-methylcyclohexanol. This is because the hydrogen is in the anit-perplanar position, meaning it is on the opposite side of the leabing group. This means E2 will react faster. For the trans, the hydrogen is not anti-perplanar meaning it will be much harder for the E2 reaction to occur because the hydrogens on the beta hydrogens are not anti-
perplanar. Furthermore, when place in a chair diagram, the trans is less stable and it does not favor a chair flip, however cis does favor a chair flip and is more stable. This means that cis will occur faster. There is more strain on the trans chair diagram because of steric hinderance that allows it to not be stable and occur slower. Furthermore, in a E1 reaction, the carbocation intermediate is more stable for the cis which means it will also occur faster.
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