E1 mechanism with carbocation rearrangement
2) In questions 2 and 3, only the proton trans to the leaving group can eliminate.
Due to the presence of the bulky t-butyl group, the ring is practically locked up in the most stable conformation with the bulky group being equatorial.
Of the two isomers, the cis is the only one that fulfills the anticoplanar arrangement for E2, where the leaving group and adjacent proton must be anti to each other and in the same plane.
The atoms shown in red fulfill the anticoplanar requirement for E2. Elimination is possible and fast
The atoms shown in red cannot fulfill the anticoplanar requirement. Elimination is slower or not possible
6) The small, unhindered base ethoxide yields the more stable alkene (Saytzeff’s product, i.e. the more highly substituted alkene). When the bulky t-butoxide base is used, the most accessible hydrogen is removed. This results in the least highly substituted alkene (Hoffman’s product).
7) III, IV, and V.
8) In elimination reactions, the most highly substituted alkene predominates.
11) E2. The molecule must rotate around the central cabon-carbon bond to aquire the anticoplanar arrangement required for E2. This is a stereospecific reaction that results in formation of the product where the phenyl groups are cis to each other.
12) Strong base and bulky substrate favor E2. Only carbon 6 has protons trans to the leaving group. The pi bond can only form between carbons 1 and 6. Can you name the product by IUPAC rules?
13) Strong base and bulky substrate favor E2 with preferential formation of Saytzeff’s product.
14) Weak nucleophile (ethanol, the solvent) and bulky substrate favor Sn1 as shown below. Can you tell which is the rate-determining step? Can you tell what type of reaction is involved in the last step?
16) E1 and E2
19) The reacton is E2. See question 12 for a similar case
25) First step is protonation of the alcohol by the strong acid to form a potential water molecule as a leaving group. Next is departure of the leaving group with formation of a carbocation. Next is a carbocation rearrangement from secondary to tertiary. The last step is an elimination step where the water abstracts the acidic proton next to the positive charge to form the alkene.
28) Triethylamine. Amines can be nucleophiles or bases. Increasing their steric bulk near the nitrogen atom diminishes their nucleophilicity while retaining the basicity. Since the substrate is sterically accessible to nucleophilic attack, a bulky base is needed to promote elimination.
30) This is a typical example of a simple, multistep synthesis (in this case only two steps). This tests your ability to use previously learned reactions (e.g. from ch. 4) to design a synthesis towards a particular product, in this case cyclopentene. The last step is an elimination reaction. Any strong base combination will serve the same purpose as NaOH and acetone.
31) Similar to the previous problem, but this time Hoffman’s product is desired. A bulky base must be used in the last step, such as t-butoxide ion.
32) Same as above, but this time Sayteff’s product is desired. A small base must be used in the last step.