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

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CHEMICAL HIGHLIGHT 8-3

Transition Metal-Catalyzed Cross-Coupling Reactions
The general coupling reaction of a haloalkane, containing a
positively polarized carbon, with an alkylmetal, containing a
negatively polarized carbon, is quite exothermic.

� �R
��

M
��

R�
��

X
��

O O R MXR�O

Yet, in the case of Li and Mg, such couplings are either
too slow at room temperature or lead to product mixtures

on heating. As this process constitutes one of the most
fundamental C – C bond-forming reactions, it is not surpris-
ing that synthetic chemists have devoted considerable
effort to the solution of the problem, an effort that is
ongoing. An early solution was provided by copper salts
as catalysts. Catalysts enable reactions to proceed faster,
through lower-energy transition states and mechanisms
(Section 3-3).

The method has been applied on an industrial scale in
the manufacture of muscalure, the sex attractant of the
housefl y. In conjunction with a toxic ingredient, it is

� CH3(CH2)5CH2MgCl CH3(CH2)8OCH2CH3O
Br 5% CuI

� MgBrCl
82%

�

H H

CuI
� MgBr2

CH3(CH2)7 CH3(CH2)3CH2MgBr(CH2)7CH2Br

H H

CH3(CH2)7 (CH2)12CH3

Muscalure

80%

used for pest control, particularly in poultry, swine,
beef and dairy cattle facilities, and stables (see also
 Section 12-17).

The mechanism of these reactions proceeds through organo-
copper species, also called cuprates (see Section 18-10),
which can be generated and used stoichiometrically, for
example, from alkyllithium reagents.

Fatal attraction to
muscalure spells doom to
the common housefl y.

Let us begin with a few examples in which we predict reactivity on mechanistic grounds.
Then we shall turn to synthesis — the making of molecules. How do chemists develop new
synthetic methods, and how can we make a “target” molecule as effi ciently as possible?
The two topics are closely related. The second, known as total synthesis, usually requires
a series of reactions. In studying these tasks, therefore, we will also be reviewing much of
the reaction chemistry that we have considered so far.

Mechanisms help in predicting the outcome of a reaction
First, recall how we predict the outcome of a reaction. What are the factors that let a par-
ticular mechanism go forward? Here are three examples.

How to Predict the Outcome of a Reaction on Mechanistic Grounds

Example 1. What happens when you add I2 to FCH2CH2CH2Br?

ICH2CH2CH2Br
Not formed

FCH2CH2CH2Br FCH2CH2CH2I
I� I�

Explanation. Bromide is a better leaving group than fl uoride.

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Example 2. How does a Grignard reagent add to a carbonyl group?

A

A
CH3CMgBr CH3MgBr�C

OOCH3

H
A

A
CH3CCH3

OMgBr
��

H
Not formed

(CH3CH2)2O

H3C H

(CH3CH2)2OB

HE

��

����
��

Explanation. The positively polarized carbonyl carbon forms a bond to the negatively
polarized alkyl group of the organometallic reagent.

Example 3. What is the product of the radical bromination of methylcyclohexane?

�

Not formed

Br2, hv

CH3 H3CCH3 CH2Br Br
Br2, hv

Br

�Other bromides

More recently, a number of variations on this theme
have been explored, employing M 5 Zn, Sn, Al, and
others, in the presence of catalysts based on Ni, Pd, Fe,
and Rh, just to name a few. The aim is to improve not

only effi ciency, but also functional group tolerance. For
example, unlike alkyllithium and Grignard reagents, the
corresponding Zn compounds do not attack carbonyl
 functions.

�

O
Ni catalyst

52%

O

I ZnI

�
Ni catalyst

62%

O

O

O

OI BrZn

2 CH3CH2CH2CH2Li 1 CuI ¡ (CH3CH2CH2CH2)2CuLi 1 LiI
 Lithium dibutylcuprate

� 3 (CH3CH2CH2CH2)2CuLiCH3S
O

O

CH3OCH2CH2O CH3O(CH2)5CH3
90%

In these cases, the mechanism is not direct nucleophilic
substitution, but rather assembly of the two fragments of

R and R9 around the catalyst, as schematized in a simplifi ed
manner below.

R
� Ni R�ZnX

� ZnX2
XO R Ni NiXO O R

� Ni RO O R�R� O

The development of transition metal – catalyzed C – C bond-
forming reactions has seen an explosive growth during the
past decade, and widely used methods employing metals in

the coupling to alkenes and alkynes will be discussed in
Chemical Highlights 12-4 and 13-1, and in Sections 13-9,
18-10, and 20-2.

8 - 9 S t r a t e g i e s t o C o m p l e x A l c o h o l s

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312 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

Explanation. The tertiary C – H bond is weaker than a primary or secondary C – H bond,
and Br2 is quite selective in radical halogenations.

New reactions lead to new synthetic methods
New reactions are found by design or by accident. For example, consider how two different
students might discover the reactivity of a Grignard reagent with a ketone to give an alco-
hol. The fi rst student, knowing about electronegativity and the electronic makeup of ketones,
would predict that the nucleophilic alkyl group of the Grignard species should attach itself
to the electrophilic carbonyl carbon. This student would be pleased by the successful out-
come of the experiment, verifying chemical principles in practice. The second student, with
less knowledge, might attempt to dilute a particularly concentrated solution of a Grignard
reagent with what one might conceive to be a perfectly good polar solvent: acetone. A
violent reaction would immediately reveal that this notion is incorrect, and further investi-
gation would uncover the powerful potential of the reagent in alcohol synthesis.

Exercise 8-17

Working with the Concepts: Using Mechanistic Knowledge to Predict the
Outcome of a Reaction

Predict and explain the outcome of the following reaction on mechanistic grounds.

ClCH2CH2CH2C(CH3)2
CH2Cl

NaOH
H2O

�
A

Strategy
The fi rst step is to identify the functional sites in the two starting materials. Then you can list the
possible modes of reactivity for these functional groups and sort out which ones best apply.
Solution
• The organic component is a dihaloalkane. Hence, it contains two reaction sites that might be
subject to the chemistry described in Chapters 6 and 7: SN2, SN1, E2, and E1.
• The inorganic NaOH is a strong unhindered base and nucleophile. Inspection of Table 7-4 reveals
that hydroxide attacks haloalkanes at primary centers to make alcohols by SN2, but forms alkenes
at more hindered (to nucleophilic attack) positions by E2.
• Turning to the haloalkane, one Cl resides at an unhindered primary center; it should be replaced
by OH through SN2. The second Cl is also bound to a primary carbon; however, it is sterically
hindered by b branching. Such steric hindrance retards nucleophilic attack, resulting in favorable
E2, but only in cases in which a b hydrogen is available for deprotonation. In the present case,
the carbon is neopentyl-like and E2 is not possible. Therefore, no reaction occurs at this center.
Consequently, the product is

HO Cl

Exercise 8-18

Try It Yourself

Predict and explain the outcome of the following reactions on mechanistic grounds.

(a) ClCH2CH2CH2C(CH3)2 CH3CH2OH
Br

�
A

(b) HOCH2CH2CH2C(CH3)2
OH

PCC, CH2Cl2A

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When a reaction has been discovered, it is important to show its scope and its limita-
tions. For this purpose, many different substrates are tested, side products (if any) noted,
new functional groups subjected to the reaction conditions, and mechanistic studies carried
out. Should these investigations prove the new reaction to be generally applicable, it is
added as a new synthetic method to the organic chemist’s arsenal.

Because a reaction leads to a very specifi c change in a molecule, it is frequently useful
to emphasize the general nature of this “molecular alteration.” A simple example is the addi-
tion of a Grignard or alkyllithium reagent to formaldehyde. What is the structural change in
this transformation? A one-carbon unit is added to an alkyl group. The method is valuable
because it allows a straightforward one-carbon extension, also called a homologation.