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Prévia do material em texto

184          CHAPTER 7           
 
 
 
It doesn’t work because it is not possible to have a strong acid (H3O+) and a strong base (t-BuOK) present in the same 
reaction flask at the same time (they would simply neutralize each other). In order to achieve the desired 
transformation, the OH group must first be converted to a tosylate (rather than simply being protonated), and then the 
desired reaction can be performed, as shown here: 
 
 
 
Solutions 
7.1. 
(a) The parent is the longest chain, which is five carbon 
atoms in this case (pentane). There are three substituents 
(bromo, bromo, and chloro), and their locants are 
assigned as 4, 4, and 1, respectively. In this case, the 
parent was numbered from right to left, so as to give the 
lowest number to the first substituent (1,4,4 rather than 
2,2,5). Notice that two locants are necessary (rather than 
one) to indicate the locations of the two bromine atoms, 
even though they are connected to the same position 
(4,4-dibromo rather than 4-dibromo). 
 
 
 
(b) The parent is a six-membered ring (cyclohexane). 
There are two substituents (methyl and bromo), both of 
which are located at the C1 position. Substituents are 
alphabetized in the name (bromo precedes methyl). 
 
 
 
(c) The parent is the longest chain, which is eight carbon 
atoms in this case (octane). There are two substituents 
(ethyl and chloro), both of which are located at the C4 
position. These substituents are alphabetized (chloro 
precedes ethyl). In this case, there is also a chiral center, 
so we must assign the configuration (R), which must be 
indicated at the beginning of the name. 
 
 
 
(d) The parent is the longest chain, which is six carbon 
atoms in this case (hexane). There are three substituents 
(fluoro, methyl, and methyl), and their locants are 
assigned as 5, 2, and 2, respectively. In this case, the 
parent was numbered from right to left, so as to give the 
lowest number to the second substituent (2,2,5 rather 
than 2,5,5). The substituents are arranged alphabetically 
in the name, so fluoro precedes dimethyl (the former is 
“f” and the latter is “m”). In this case, there is also a 
chiral center, so we must assign the configuration (R), 
which must be indicated at the beginning of the name. 
 
 
 
(e) The parent is the longest chain, which is nine carbon 
atoms in this case (nonane). There are three substituents 
(methyl, bromo, and isopropyl), and their locants are 
assigned as 2, 3, and 3, respectively. The substituents 
are arranged alphabetically in the name (note that 
isopropyl is alphabetized as “i” rather than as “p”, so it 
comes before methyl). In this case, there is no chiral 
center (C2 is connected to two methyl groups, and C3 is 
connected to two isopropyl groups). 
 
 
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