Unformatted text preview: Copyright 2011 James K Whitesell 141 The starting alkene is unsymmetrical and, therefore, two different radicals are formed by abstraction of a hydrogen. Because these radicals are allylic, each can form two different, regio isomeric product bromides. See the answer in the book for the structures. 142 Unlike the alkene in Exercise 14‐1, the alkenes in parts a) and b) are symmetrical and only a single allylic radical results from hydrogen abstraction. Further, the allylic radical formed in a) is also symmetrical so only one allylic bromide results. The radical formed in b) is not symmetrical and therefore, two allylic bromides result. There are three unique allylic radicals that can be formed by hydrogen abstraction from 1‐methylcyclohexene. One of these is symmetrical and results in only one allylic chloride but each of the other two is unsymmetrical and two allylic chlorides result from each. Note that the answer in the book has excluded the last answer, on the right below. Exocyclic double bonds on a six‐membered ring are unusually unstable (steric interaction with the equatorial hydrogen atoms on the ring). Cl
Cl Cl Cl Cl 143 Because an allylic cation can be formed by loss of chloride ion, this substitution reaction proceeds by an SN1 mechanism. Because the original center of chirality at C3 becomes planar in the cation, the chirality present in the starting material is lost. 144 This exercise is worked out in the book as "Working with Concepts".
145 The answer in the back of the book is adequate.
146 The key here is that initially, the reaction is under kinetic control where chloride ion
adds to the initially formed cation. But over time, acetate adds and the greater strength of
the C—O bond compared to the C—Cl bond makes the acetate the more stable product.
146 The answer in the back of the book is adequate.
147 The answer in the back of the book is adequate.
148 The answer in the back of the book is adequate.
149 The answer in the back of the book is adequate.
1410 The answer in the back of the book is adequate.
1411 The answer in the back of the book is adequate. Remember that for E2 reactions,
the best all around base is KOC(CH3)3.
1412 The answer in the back of the book is adequate.
1413 This exercise is worked out in the book as "Working with Concepts".
1414 The answer in the back of the book is adequate.
1415 The answer in the back of the book is adequate.
1416 The answer in the back of the book is adequate.
1417 The answer in the back of the book is adequate.
1418 The answer in the back of the book is adequate.
1419 This exercise is worked out in the book as "Working with Concepts".
1420 The answer in the back of the book is adequate.
1421 The answer in the back of the book is adequate.
1422 This exercise is worked out in the book as "Working with Concepts".
1423 The answer in the back of the book is adequate.
1424 The answer in the back of the book is adequate. 1425 The answer in the back of the book is adequate.
1426 The answer in the back of the book is adequate. 1427 This exercise is worked out in the book as "Working with Concepts". 1428 The answer in the back of the book is adequate.
1429 All other things being equal, the more double bonds in conjugation, the longer the
wavelength of the absorption. ...
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 Spring '06
 Nefzi
 Organic chemistry, Ion, Chloride

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