Download - KCN K+ R KOH + H2O

Cyanohydrin formation (nucleophilic addition of CN–)
H(R')
R
KCN, HCl
C
O- K+
NC
O
(R')H
H(R')
-KCN
NC
CN
R
OH
R
H
- Cyanohydrins are more easily formed with aldehydes than ketones; yet, another
way to make C–C bond!
- HCN is formed from acid-base reaction of –CN with HCl.
- the reaction does not occur with HCN alone; cyanide is required. Thus, the
reaction can be made to occur using HCN/catalytic base. Generally, use of HCN
is avoided since it’s poisonous!
Cyanohydrins can be converted back to the carbonyl compound by treatment
with base:
H(R')
NC
H(R')
KOH
OH
-
NC
O
K
+
R
- KCN
+ H2O
C
O
(R')H
R
R
- the cyano group of a cyanohydrin is readily hydrolyzed to a COOH group by
heating with aqueous base or acid:
H(R')
NC
H(R')
H2 O
OH
(H+ or –OH)
R
HOOC
R
!
H3O+, !
OH
OH
OH
COOH
CN
Imines and Enamines
Just as with alcohols, amines can add to ketones and aldehydes. Primary amines
add to give imines while secondary amines give enamines (pronounced eneamines). The reactions are generally catalyzed by a small amount of acid and
need to be buffered to a pH of ~4.5.
a) imines
R
C
O
H(R')
R"-NH2
"RHN
(R')H
C
-H2O
R
C
OH
(R')H
R
NR"
imine
carbinolamine
H(R')
"RH2N
C
R
- H3O+
H3O+
H(R')
O
"RHN
C
R
OH2
-H2O
R
C
(R')H
NR"
H
OH2
1
i) The reactions involve nucleophilic addition to yield a carbinolamine
(hemiaminal) followed by E1 elimination of H2O. Note that this reaction requires
the presence of at least two hydrogens in the starting amine (H+ comes from N;
O is protonated before it leaves)
ii) Equilibrium favors imine formation when R” = hydroxy, alkoxy, or amino
groups; the products called oximes, oxime ethers, or hydrazones, respectively.
When R”= alkyl, equilibrium usually favors the carbonyl compound, but can be
pushed toward imine by removal of H2O.
iii) Fastest at near-neutral pH: acidic pH protonates starting material and slows
first reaction, basic pH slows protonation of O in hemiaminal and prevents
second reaction.
b) Enamines
In the case of secondary amines, we have a lack of protons that can easily be
removed from the amine – the mechanism thus requires that the offending
positive charge to be neutralized by removing a proton from the alkyl group.
Thus, there must be at least one H atom bonded to the carbon alpha to the
carbonyl
R
C
O
R2"NH
R'H2C
CH2R'
R"2N
C
-H2O
R
C
OH
R"2N
enamine
R
carbinolamine
R"2HN
C
R
-H3O+
R
H3O+
CH2R'
CH2R'
O
R"2N
C
R
CHR'
C
-H2O
R"2N
CHR'
H
OH2
OH2
R
C
CH2R'
R"2N
Equilibrium usually favors carbonyl, but can be pushed toward enamine by
removal of H2O.
Clearly, hydrolysis of enamines and imines are simply the reverse of these
reactions (reverse mechanism) and hence form aldehydes and ketones.
De- Ox yge natio n re action s.
In Chapter 14, we saw two general reactions for the complete de-oxygenation of
ketones and aldehydes. The general scheme for de-oxygenation is:
The two methods are the Wolff-Kishner (runs under basic conditions) and the
Clemmensen (under acidic conditions). For examples,
2
The Witt ig Re action is another deoxygenation reaction that’s useful for
making mono-, di-, and trisubstituted alkenes, but not for tetrasubstituted
alkenes. The reaction involves nucleophilic addition to the carbonyl compound
followed by elimination.
R"
R
C
O
+ Ph3P
C
C
R'
R*
R"
Ph3P
R"
R
+ Ph3P=O
R*
R'
Either R' or R" Must be hydrogen
C
R*
i.e.
CH3
H3C
C
O
+ Ph3P
C
H
CH3
H3C
C
CH3
H
+ Ph3P=O
CH3
CH3
CH3
C
O
+ Ph3P
+ Ph3P=O
C
C
H
H
Note that the alkene product must contain at least one hydrogen; another way
to make C–C bond!
In the most general scheme of this reaction, an alkyl halide (usually the bromide)
and an aldehyde or ketone are taken to an alkene.
In a more detailed picture of this reaction, the alkyl bromide is allowed to react
with triphenyl phosphine to form an alkylphosphonium salt:
Ph3P + BrCHR”R* → [Ph3P+–CHR”R*]Br–
- this SN2 works best for CH3X and 1˚ alkyl halides; yields are lower for 2˚ alkyl
hakides
- alkylphosphonium salts are generally quite stable, and can be stored over long
periods of time (i.e. many of these salts are even sold commercially). The
phosphonium salt is then deprotonated (usually with sodium hydride, or
3
butyllithium) to form the ylide. An ylide is simply a charge-separated species, as
shown below:
[Ph3P+–CHR”R*]Br– + BuLi → Ph3P+–C–R”R* <-> Ph3P=CR”R*
This ylide is then allowed to react with a ketone or aldehyde, to form a betaine
intermediate. This intermediate cyclizes to form another intermediate, an
oxaphosphetane:
O
R
O
R
R'
Ph3P
R"
R'
R"
Ph3P
O
Ph3P
R*
R*
R
R'
R"
R*
betaine
oxaphosphetane
The oxaphosphetane decomposes rapidly to form the alkene and
triphenylphosphine oxide:
R
O
R'
Ph3P
R"
O
Ph
P
Ph
+
Ph
R*
R'
R
*R
R"
Thus, half of an alkene can come from an alkyl bromide, and the other half from
a ketone or aldehyde. The main driving force for this reaction is the formation of
the phosphorous-oxygen double bond; this is one of the strongest bonds known,
and its formation drives the reaction to completion.
A limitation of the Wittig reaction is that a mixture of stereoisomers sometimes
forms (below):
H
H3CH2C
C
O
+ Ph3P
H
H
H3CH2C
C
C
(CH2)4CH3
(CH2)4CH3
H
H
H
+
C
(CH2)4CH3
H3CH2C
41%
59%
This approach to alkene formation is better than addition of a Grignard to
aldehyde or ketone followed by dehydration, because dehydration follows
Saytzeff’s rule (thermodynamic ratio of products, more substituted favored)
while Wittig reaction gives only one product:
CH3
C O
1. CH3Li
2. H2O
C
CH3
H3O+
OH
H2O
major
+
H
C
H
minor
versus
H
H
C O + Ph3P C
C
H
+ Ph3P=O
H
4
Retrosynthetic analysis:
RCH=CHR' ⇒ RCHO + BrCH2R' or RCH2Br + R'CHO.
Remember, RBr ⇒ ROH; and we have seen that RCHO or R2CO ⇒ R”CH2OH or
R”2CHOH (oxidation of aldehydes and ketones)
Which starting materials would you use to prepare PhCH=C(CH3)2? PhCHO and
(CH3)2CHBr versus PhCH2Br and (CH3)2CO?
How would you prepare PhCH2Br from PhCOOMe?
How would you prepare PhCHO from PhCOOMe?
Below are a few examples of the Wittig reaction at work:
A simple modification of the Wittig reaction leads to a dibromo-olefin, which is
an excellent precursor to alkynes:
5