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David Suraj Experiment 11 Mariela Trejo 232 – 5423 GC Analysis of Elimination PLQ 1. A)

For elimination using t-amyl bromide as the alkyl halide, using the ethanol base ,

the secondary elimination product (Zaitsev) was 80 % of the product while the primary elimination product (Hofmann) was 20%, this mechanism is E1. Using the ethoxide base, the Zaitsev product was shown in 69% of the product, while the Hofmann was 31%, this mechanism is E2. Using the t-butoxide base, the Zaitsev product was shown in 27.5% of the product, while the Hofmann was 72.5% of the product, this mechanism is E2. B)

The Ethanol in the study is very similar to the 1 propanol and 1 butanol in our

own experiment, they are very similar because they are both small bases, and they produce the Zaitsev product, they are also both E1. The Ethoxide and the KOH are also very similar because they are bigger bases that produce the Zaitsev product, but they also produce more Hofmann product compared to the smaller bases, they are also E2. The tbutoxide in the study and the KOtBu in our own experiment are similar as well because they yield the most Hofmann product compared to the other bases, while being a big and bulky base, this is E2. C)

Our GC results do not support the findings in the study. In our own GC

experiments, when we were dealing with KOtBu, we had more Zaitsev product compared to Hofmann product. However, in the study, more Hofmann product was found when t-butoxide was used. This was the main difference between the lab and they study. Both of these are big, bulky bases that undergo E2 reactions. The other bases in our lab were similar to the bases found in the lab. The Ethanol in the study is very similar to the 1 propanol and 1 butanol in our own experiment, they are very similar because they are both small bases, and they produce the Zaitsev product, they are also both E1. The Ethoxide and the KOH are also very similar because they are bigger bases that produce the Zaitsev product, but they also produce more Hofmann product compared to the smaller bases, they are also E2

D)

The results are in accord. Essentially the experiment in table four showed that as

the base becomes more sterically inhibited, it is more likely to be a Hofmann product rather than a Zaitsev product in elimination. T-butoxide is a very sterically inhibited base, thus is makes sense that the major product was a Hofmann product. Ethanol is not very sterically inhibited so the major product for this would be Zaitsev. Ethoxide is a base that is also not sterically inhibited, but it is more sterically inhibited than Ethanol is, thus it still favors Zaitsev, but has more Hofmann product than Ethanol has. E) Ethanol was E1, ethoxide E2, t-butoxide is E2

2.

A) For 2-methyl-2-bromopentane the product was 50% Hofmann. For the 2,4-dimethyl-

2-bromopentane, the product was 54% Hofmann. Finally, for the 2,4,4-trimethyl-2bromopentane, the product was 86% Hofmann. The results indicate that as the substrate becomes more substituted and sterically inhibited, the Hofmann product is more likely to occur. B) These findings do not match the lessons taught in class. In the paper, it is shown that as the substrate has more substituents, it is more likely to form the Hoffman product. However, the base itself is not very sterically inhibited. This is a small base, therefore, even as the substrate has more substituents, it should follow the Zaitsev product because of its small size. If the base was big like t-butoxide, then the Hofmann product would be expected. Therefore, the findings do not necessarily match the lessons taught in class. C) The results were not expected and we expected the Zaitsev product. A plausible explanation is that as the substrate gained more substituents, and the groups had less space in between each other, the base was not able to deprotonate the Zaitsev hydrogen. Therefore, it had to take the Hofmann hydrogen....