Vollhardt  Capítulo 8 (Álcoois)
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Vollhardt Capítulo 8 (Álcoois)


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in the presence of bases.
Br
NH3
\ufffd
Expertise in synthesis, as in many other aspects of organic chemistry, develops largely 
from practice. Planning the synthesis of complex molecules requires a review of the reac-
tions and mechanisms covered in earlier sections. The knowledge thus acquired can then 
be applied to the solution of synthetic problems.
THE BIG PICTURE
Where do we stand now, and where do we go from here? With Chapter 8, we have begun 
to discuss a second important functional class of compounds after the haloalkanes, namely, 
the alcohols. We used the haloalkanes to introduce two major mechanistic pathways: radi-
cal reactions (in order to make haloalkanes; Chapter 3) and ionic pathways (to demonstrate 
their reactivity in substitution and elimination, Chapters 6 and 7). In contrast, we have used 
the alcohols to introduce new types of reactions: oxidations, reductions, and organometallic 
additions to aldehydes and ketones. This discussion has allowed us to examine the idea of 
how to plan a synthesis. It requires us to start with a given product and work backward 
(retrosynthetic analysis) to determine what reactants and what kind of reaction conditions 
we need to obtain this product. As we proceed through the course and build up our knowl-
edge of families of compounds and their chemistry, we shall keep returning to synthetic 
strategies as a way of classifying and applying this information.
This chapter and the next follow the format we shall use to present all future functional 
groups: (a) how to name them; (b) a description of their structures and general properties; 
(c) how to synthesize them; and (d) what kinds of reactions they undergo and how we can 
apply these reactions in various ways.
CHAPTER INTEGRATION PROBLEMS
8-22. Tertiary alcohols are important additives in some industrial processes utilizing Lewis acidic 
metal compounds (Section 2-2) as catalysts. The alcohol provides the metal with a sterically protecting 
and hydrophobic environment (see Figure 8-3; see also Chemical Highlight 8-1), which ensures 
solubility in organic solvents, longer lifetimes, and selectivity in substrate activation. Preparation of 
these tertiary alcohols typically follows the synthetic principles outlined in Section 8-9.
Starting from cyclohexane and using any other building blocks containing four or fewer carbons, 
in addition to any necessary reagents, formulate a synthesis of tertiary alcohol A.
OH
A
SOLUTION
Before we start a random trial-and-error approach to solving this problem, it is better to take an 
inventory of what is given. First, we are given cyclohexane, and we note that this unit shows up as 
a substituent in tertiary alcohol A. Second, a total of seven additional carbons appears in the product, 
so our synthesis will require some additional stitching together of smaller fragments because we cannot 
C h a p t e r I n t e g r a t i o n P r o b l e m s
320 C h a p t e r 8 H y d r o x y F u n c t i o n a l G r o u p : A l c o h o l s
use compounds containing more than four carbons. Third, target A is a tertiary alcohol, which should 
be amenable to the retrosynthetic analysis introduced in Section 8-9 (M 5 metal):
M
A
a
a
b
b
c
c
\ufffd
O
BB
\ufffd
O
BB
M
B C
M
\ufffd
O
BB
OH
Approach a is clearly the one of choice because it breaks up tertiary alcohol A into evenly sized 
fragments B and C.
Having chosen route a as the most suitable way to fi nd direct precursors to A, we proceed to 
work backward further: What are the appropriate precursors to B and C, respectively? Compound B 
is readily traced by retrosynthesis to our starting hydrocarbon, cyclohexane: The precursor to the 
organometallic compound B must be a halocyclohexane, which in turn can be made from cyclohexane 
by free-radical halogenation:
B
M X
M\ufffd X2\ufffd
Ketone C must be broken into two smaller components. The best breakdown would be a \u201cfour 1 
three\u201d carbon combination: It is the most evenly sized solution, and it suggests the use of cyclobutyl 
intermediates. Because the only C \u2013 C disconnection that we know at this stage is that of alcohols, the 
fi rst retrosynthetic step from C is its precursor alcohol (plus a chromium oxidant). Further retrosyn-
thesis then provides the required starting pieces D and E.
M
\ufffd
C D E
H
OHO
BB
O
BB
Now we can write the detailed synthetic scheme in a forward mode, with cyclohexane and pieces 
D and E as our starting materials:
\ufffd
Compound A
OH
Na2Cr2O7
Li
H
Br2, Mg
Br MgBr
h\ufffd
O
BB
O
BB
A fi nal note: In this and subsequent synthetic exercises in this book, retrosynthetic analysis 
requires that you have command of all reactions not only in a forward fashion (i.e., starting material 1 
reagent y product) but also in reverse (i.e., product z starting material 1 reagent). The two sets 
address two different questions. The fi rst asks: What are all the possible products that I can make 
from my starting material in the presence of all the reagents that I know? The second asks: What are all 
 C h a p t e r 8 321
the conceivable starting materials that, with the appropriate reagents, will lead to my product? The 
two types of schematic summaries of reactions that you will see at the end of this and subsequent 
chapters emphasize this point.
8-23. In this chapter we introduced redox reactions interconverting alcohols with aldehydes and 
ketones. The reagents employed were Cr(VI) (in the form of chromates, for example Na2Cr2O7) and H
2 
(in the form of NaBH4 and LiAlH4). Organic chemists usually don\u2019t worry about the inorganic 
products of such processes, because they are routinely discarded. However, for the purposes of 
electron bookkeeping (and in experimental recipes), it is useful (essential) to write balanced equations, 
showing how much of any given starting material is \u201cgoing into\u201d the reaction and how much of any 
possible product is \u201ccoming out.\u201d Most of you have dealt with the problem of balancing equations 
in introductory chemistry, but usually only for redox exchanges between metals. Can you balance the 
following general oxidation of a primary alcohol to an aldehyde?
Na2Cr2O7 Cr2(SO4)3 Na2SO4 H2OH2SO4RCH2OH \ufffd \ufffd \ufffd\ufffd\ufffd RCH
O
B
SOLUTION
It is best to think of this transformation as two separate reactions taking place simultaneously: (1) the 
oxidation of the alcohol, (2) the reduction of the Cr(VI) species to Cr(III). These two parts are called 
half-reactions. In water, which is the common solvent for this and similar redox processes, we balance 
the half-reactions by
a. treating any hydrogen atoms consumed or produced as H3O
1 (or, for simplicity, H1),
b. treating any oxygen atoms consumed or produced as H2O (in acidic solutions) or 
2OH (in basic 
solution),
c. adding electrons explicitly to the side defi cient in negative charge.
Let us apply these guidelines to equation (1), which can be thought of as the removal of two 
hydrogens from the alcohol. These hydrogens are written as protons on the product side (rule a), 
charge-balanced by two added electrons (rule c):
 2 H
\ufffd 2 eRCH2OH \ufffd\ufffdRCH
O
B
 (1)
Turning to the half-reaction of the chromium species, we know that Cr2O7
22 is changed to (two) 
Cr31 ions:
Cr2O7
22 uy 2 Cr31
We note that seven oxygen atoms need to appear on the right side; rule (b) stipulates seven molecules 
of H2O:
Cr2O7
22 uy 2 Cr31 1 7 H2O
This change requires that 14 hydrogens be added to the left side; rule (a) says 14 H1 atoms:
14 H1 1 Cr2O7
22 uy 2 Cr31 1 7 H2O
Is the charge balanced? No, rule (c) tells us to add six electrons to the left side, resulting in 
balanced equation (2).
 14 H1 1 Cr2O7
22 1 6 e uy 2 Cr31 1 7 H2O (2)
Inspection of the two half-reactions shows that, as written, (1) generates 2e (an oxidation, Section 8-6) 
and (2) consumes 6e (a reduction, Section 8-6). Since a chemical equation does not show electrons, 
we need to balance the production and consumption of electrons. We do this by simply multiplying 
(1) by 3:
 6 H
\ufffd 6 e3 RCH2OH \ufffd\ufffd3 RCH
O
B
 (3)
C h