1 (222)

Prévia do material em texto

214          CHAPTER 7           
 
(c) Given the location of the  bond, we consider the 
following two possible alkyl halides as potential starting 
materials. 
 
 
Compound A has three  positions, but only two of them 
bear protons, and those two positions are identical. 
Deprotonation at either location will result in the desired 
alkene. In contrast, compound B has two different  
positions that bear protons. Therefore, if compound B 
undergoes an E2 elimination, there will be two possible 
regiochemical outcomes, so more than one alkene will be 
formed. 
 
(d) Given the location of the  bond, we consider the 
following two possible alkyl halides as potential starting 
materials. 
 
 
 
Compound A has only one  position, giving rise to only 
one alkene. In contrast, compound B has more than one 
 position, giving rise to more than one alkene. 
 
 
7.66. Hydroxide is both a strong nucleophile and a 
strong base. The substrate is tertiary, so we expect an E2 
process only. In the transition state, the hydroxide ion is 
in the process of removing the proton, the double bond is 
in the process of forming, and the leaving group is in the 
process of leaving. We use dotted lines to indicate the 
bonds that are in the process of being formed or broken, 
and we use  symbols to indicate the distribution of 
charge. Note that the negative charge is in the process of 
being transferred from the oxygen atom to the chlorine 
atom, and the  symbols indicate that each location 
bears partial negative character in the transition state. 
Finally, brackets are drawn, together with the symbol 
that indicates that this is a transition state: 
 
 
 
 
7.67. 
(a) The Zaitsev product is desired, so we must a use a 
base that is not sterically hindered. Sodium ethoxide is 
the correct choice, because potassium tert-butoxide is a 
sterically hindered base. 
(b) The Hofmann product is desired, so a sterically 
hindered base should be used. The correct choice is 
potassium tert-butoxide. 
 
 
7.68.  bonds cannot be formed at the bridgehead of a 
bicyclic compound, unless one of the rings is large (at 
least eight carbon atoms). This rule is known as Bredt’s 
rule. 
 
 
 
7.69. For substituted cyclohexanes, an E2 reaction will 
occur if the leaving group and the  proton can achieve 
antiperiplanarity. In order to achieve this, one must be 
on a wedge and the other must be on a dash. The leaving 
group (Brˉ) is on a wedge. Therefore, we are looking for 
a  proton that is on a dash. In this case, there is only 
one such proton, highlighted below, so there is only one 
possible regiochemical outcome. In this case, the 
Hofmann product is formed regardless of the choice of 
base. 
 
 
 
7.70. 
(a) In acidic conditions, the OH group is protonated, which converts it from a bad leaving group to a good leaving 
group. In aqueous sulfuric acid, the acid that is present in solution is H3O+ (because of the leveling effect, as explained 
in Section 3.6), so we use H3O+ as the proton source in the first step of the mechanism. The next two steps of the 
mechanism constitute the core steps of an E1 process: (i) loss of a leaving group (H2O), which requires one curved 
arrow, and (ii) proton transfer, which requires two curved arrows, as shown: 
 
www.MyEbookNiche.eCrater.com