Download Carboxylic Acids and the Acidity of the O—H Bond

Carboxylic Acids and the Acidity of the O—H Bond
Structure and Bonding
• Carboxylic acids are compounds containing a carboxy
group (COOH).
• The structure of carboxylic acids is often abbreviated as
RCOOH or RCO2H, but keep in mind that the central
carbon atom of the functional group is doubly bonded to
one oxygen atom and singly bonded to another.
1
Structure and Bonding
• The C—O single bond of a carboxylic acid is shorter
than the C—O bond of an alcohol.
• This can be explained by looking at the hybridization of
the respective carbon atoms.
• Because oxygen is more electronegative than either
carbon or hydrogen, the C—O and O—H bonds are polar.
2
Nomenclature—The IUPAC System
• In the IUPAC system, carboxylic acids are identified
by a suffix added to the parent name of the longest
chain with different endings being used depending
on whether the carboxy group is bonded to a chain
or a ring.
If the COOH is bonded to a chain, find the longest
chain containing the COOH, and change the “e” ending
of the parent alkane to the suffix “oic acid”.
If the COOH is bonded to a ring, name the ring and add
the words “carboxylic acid”.
Number the carbon chain or ring to put the COOH
group at C1, but omit this number from the name.
Apply all the other usual rules of nomenclature.
3
4
• Greek letters are used to designate the location of substituents
in common names.
• The carbon adjacent to the COOH is called the  carbon,
followed by the  carbon, followed by the  carbon, the  carbon
and so forth down the chain.
• The last carbon in the chain is sometimes called the  carbon.
• The  carbon in the common system is numbered C2 in the
IUPAC system.
5
• Compounds containing two carboxy groups are called
diacids. Diacids are named using the suffix –dioic acid.
• Metal salts of carboxylate anions are formed from
carboxylic acids in many reactions. To name the metal
salt of a carboxylate anion, put three parts together:
6
Figure 19.2
Naming the metal salts of
carboxylate anions
7
Physical Properties
• Carboxylic acids exhibit dipole-dipole interactions
because they have polar C—O and O—H bonds.
• They also exhibit intermolecular hydrogen bonding.
• Carboxylic acids often exist as dimers held together by
two intermolecular hydrogen bonds.
Figure 19.3
Two molecules of acetic acid
(CH3COOH) held together by
two hydrogen bonds
8
Spectroscopic Properties
• Carboxylic acids have very characteristic IR and NMR
absorptions.
• In the IR:
-The C=O group absorbs at ~ 1710 cm-1.
-The O—H absorption occurs from 2500-3500 cm-1.
• In the 1H NMR:
-The O—H proton absorbs between 10-12 ppm.
-The  protons absorb between 2-2.5 ppm.
• In the 13C NMR, the C=O appears at 170-210 ppm.
9
Figure 19.4
The IR spectrum of butanoic
acid, CH3CH2CH2COOH
10
Preparation of Carboxylic Acids
[1] Oxidation of 1° alcohols
[2]
Oxidation of alkyl benzenes
11
[3]
Oxidative cleavage of alkynes
12
Reactions of Carboxylic Acids
The most important reactive feature of a carboxylic acid is
its polar O—H bond, which is readily cleaved with base.
13
• The nonbonded electron pairs on oxygen create electron-rich
sites that can be protonated by strong acids (H—A).
• Protonation occurs at the carbonyl oxygen because the resulting
conjugate acid is resonance stabilized (Possibility [1]).
• The product of protonation at the OH group (Possibility [2])
cannot be resonance stabilized.
14
• The polar C—O bonds make the carboxy carbon
electrophilic. Thus, carboxylic acids react with
nucleophiles.
• Nucleophilic attack occurs at an sp2 hybridized carbon
atom, so it results in the cleavage of the  bond as well.
15
Carboxylic Acids—Strong Organic BrØnsted-Lowry
Acids
• Carboxylic acids are strong organic acids, and as such,
readily react with BrØnsted-Lowry bases to form
carboxylate anions.
16
• An acid can be deprotonated by a base that has a conjugate acid
with a higher pKa.
• Because the pKa values of many carboxylic acids are ~5, bases
that have conjugate acids with pKa values higher than 5 are
strong enough to deprotonate them.
17
18
• Carboxylic acids are relatively strong acids because
deprotonation forms a resonance-stabilized conjugate
base—a carboxylate anion.
• The acetate anion has two C—O bonds of equal length
(1.27 Å) and intermediate between the length of a C—O
single bond (1.36 Å) and C=O (1.21 Å).
19
• Resonance stabilization accounts for why carboxylic acids
are more acidic than other compounds with O—H bonds—
namely alcohols and phenols.
• To understand the relative acidity of ethanol, phenol and
acetic acid, we must compare the stability of their
conjugate bases and use the following rule:
- Anything that stabilizes a conjugate base A:¯ makes the
starting acid H—A more acidic.
20
• Ethoxide, the conjugate base of ethanol, bears a negative
charge on the O atom, but there are no additional factors to
further stabilize the anion. Because ethoxide is less stable than
acetate, ethanol is a weaker acid than acetic acid.
• Phenoxide, the conjugate base of phenol, is more stable than
ethoxide, but less stable than acetate because acetate has two
electronegative O atoms upon which to delocalize the negative
charge, whereas phenoxide has only one.
21
Figure 19.7
Summary: The relationship
between acidity and conjugate
base stability for acetic acid,
phenol, and ethanol
• Note that although resonance stabilization of the conjugate base
is important in determining acidity, the absolute number of
22
resonance structures alone is not what is important!
The Inductive Effect in Aliphatic Carboxylic Acids
23
24
Substituted Benzoic Acids
Recall that substituents on a benzene ring either donate or
withdraw electron density, depending on the balance of their
inductive and resonance effects. These same effects also
determine the acidity of substituted benzoic acids.
[1] Electron-donor groups destabilize a conjugate base, making an
acid less acidic—The conjugate base is destabilized because
electron density is being donated to a negatively charged
carboxylate anion.
25
[2] Electron-withdrawing groups stabilize a conjugate base,
making an acid more acidic. The conjugate base is stabilized
because electron density is removed from the negatively
charged carboxylate anion.
26
Figure 19.8
How common substituents
affect the reactivity of a
benzene ring towards
electrophiles and the acidity of
substituted benzoic acids
27
Sulfonic Acids
• Sulfonic acids have the general structure RSO3H.
• The most widely used sulfonic acid is p-toluenesulfonic acid.
• Sulfonic acids are very strong acids because their conjugate
bases (sulfonate anions) are resonance stabilized, and all the
resonance structures delocalize negative charge on oxygen.
28
Introduction to Carbonyl Chemistry; Organometallic
Reagents; Oxidation and Reduction
Introduction
Two broad classes of compounds contain the carbonyl
group:
[1]
Compounds that have only carbon and hydrogen atoms
bonded to the carbonyl
[2]
Compounds that contain an electronegative atom bonded to
the carbonyl
• The presence or absence of a leaving group on the
carbonyl determines the type of reactions the carbonyl
compound will undergo.
• Carbonyl carbons are sp2 hybridized, trigonal planar, and
have bond angles that are ~1200. In these ways, the
carbonyl group resembles the trigonal planar sp2
hybridized carbons of a C=C.
• In one important way, the C=O and C=C are very
different.
• The electronegative oxygen atom in the carbonyl group
means that the bond is polarized, making the carbonyl
carbon electron deficient.
• Using a resonance description, the carbonyl group is
represented by two resonance structures.
General Reactions of Carbonyl Compounds
Carbonyls react with nucleophiles.
Aldehydes and ketones react with nucleophiles to form
addition products by a two-step process: nucleophilic
attack followed by protonation.
• The net result is that the  bond is broken, two new 
bonds are formed, and the elements of H and Nu are
added across the  bond.
• Aldehydes are more reactive than ketones towards
nucleophilic attack for both steric and electronic
reasons.
Carbonyl compounds with leaving groups react with nucleophiles
to form substitution products by a two-step process: nucleophilic
attack, followed by loss of the leaving group.
The net result is that Nu replaces Z, a nucleophilic substitution
reaction. This reaction is often called nucleophilic acyl
substitution.
• Nucleophilic addition and nucleophilic acyl substitution involve
the same first step—nucleophilic attack on the electrophilic
carbonyl carbon to form a tetrahedral intermediate.
• The difference between the two reactions is what then happens
to the intermediate.
• Aldehydes and ketones cannot undergo substitution because
they do not have a good leaving group bonded to the newly
formed sp3 hybridized carbon.
Preview of Oxidation and Reduction
• Carbonyl compounds are either reactants or products in
oxidation-reduction reactions.
The three most useful oxidation and reduction reactions of
carbonyl starting materials can be summarized as follows:
Reduction of Aldehydes and Ketones
• The most useful reagents for reducing aldehydes and ketones
are the metal hydride reagents.
• Treating an aldehyde or ketone with NaBH4 or LiAlH4, followed by
H2O or some other proton source affords an alcohol.
• The net result of adding H:¯ (from NaBH4 or LiAlH4) and H+ (from
H2O) is the addition of the elements of H2 to the carbonyl  bond.
• Catalytic hydrogenation also reduces aldehydes and ketones
to 1° and 2° alcohols respectively, using H2 and a catalyst.
• When a compound contains both a carbonyl group and a
carbon—carbon double bond, selective reduction of one
functional group can be achieved by proper choice of the
reagent.
A C=C is reduced faster than a C=O with H2 (Pd-C).
A C=O is readily reduced with NaBH4 and LiAlH4, but a
C=C is inert.
• Thus, 2-cyclohexenone, which contains both a C=C and a
C=O, can be reduced to three different compounds depending
upon the reagent used.
The Stereochemistry of Carbonyl Reduction
• Hydride converts a planar sp2 hybridized carbonyl carbon to a
tetrahedral sp3 hybridized carbon.
Enantioselective Carbonyl Reductions
• Selective formation of one enantiomer over another can occur if
a chiral reducing agent is used.
• A reduction that forms one enantiomer predominantly or
exclusively is an enantioselective or asymmetric reduction.
• An example of chiral reducing agents are the enantiomeric CBS
reagents.
• CBS refers to Corey, Bakshi and Shibata, the chemists who
developed these versatile reagents.
• One B—H bond serves as the source of hydride in this reduction.
• The (S)-CBS reagent delivers H:- from the front side of the C=O. This
generally affords the R alcohol as the major product.
• The (R)-CBS reagent delivers H:- from the back side of the C=O. This
generally affords the S alcohol as the major product.
• These reagents are highly enantioselective. For example,
treatment of propiophenone with the (S)-CBS reagent
forms the R alcohol in 97% ee.
Reduction of Carboxylic Acids and Their Derivatives
• LiAlH4 is a strong reducing agent that reacts with all
carboxylic acid derivatives.
• Diisobutylaluminum
hydride
([(CH3)2CHCH2]2AlH,
abbreviated DIBAL-H, has two bulky isobutyl groups
which makes this reagent less reactive than LiAlH4.
• Lithium tri-tert-butoxyaluminum hydride,
LiAlH[OC(CH3)3]3, has three electronegative O atoms
bonded to aluminum, which makes this reagent less
nucleophilic than LiAlH4.
• Acid chlorides and esters can be reduced to either aldehydes or 1°
alcohols depending on the reagent.
• In the reduction of an acid chloride, Cl¯ comes off as the
leaving group.
• In the reduction of the ester, CH3O¯ comes off as the
leaving group, which is then protonated by H2O to form
CH3OH.
• The mechanism illustrates why two different products are
possible.
• Carboxylic acids are reduced to 1° alcohols with LiAlH4.
• LiAlH4 is too strong a reducing agent to stop the reaction
at the aldehyde stage, but milder reagents are not strong
enough to initiate the reaction in the first place.
• Unlike the LiAlH4 reduction of all other carboxylic acid
derivatives, which affords 1° alcohols, the LiAlH4
reduction of amides forms amines.
• Since ¯NH2 is a very poor leaving group, it is never lost
during the reduction, and therefore an amine is formed.
Oxidation of Aldehydes
• A variety of oxidizing agents can be used, including
CrO3, Na2Cr2O7, K2Cr2O7, and KMnO4.
• Aldehydes can also be oxidized selectively in the
presence of other functional groups using silver(I) oxide
in aqueous ammonium hydroxide (Tollen’s reagent).
Since ketones have no H on the carbonyl carbon, they
do not undergo this oxidation reaction.
Organometallic Reagents
• Other metals in organometallic reagents are Sn, Si, Tl, Al, Ti, and
Hg. General structures of the three common organometallic
reagents are shown:
• Since both Li and Mg are very electropositive metals,
organolithium (RLi) and organomagnesium (RMgX) reagents
contain very polar carbon—metal bonds and are therefore very
reactive reagents.
• Organomagnesium reagents are called Grignard reagents.
• Organocopper reagents (R2CuLi), also called organocuprates,
have a less polar carbon—metal bond and are therefore less
reactive. Although they contain two R groups bonded to Cu, only
one R group is utilized in the reaction.
• In organometallic reagents, carbon bears a - charge.
• Organolithium and Grignard reagents are typically
prepared by reaction of an alkyl halide with the
corresponding metal.
• With lithium, the halogen and metal exchange to form
the organolithium reagent. With Mg, the metal inserts in
the carbon—halogen bond, forming the Grignard
reagent.
• Grignard reagents are usually prepared in diethyl ether
(CH3CH2OCH2CH3) as solvent.
• It is thought that two ether O atoms complex with the Mg
atom, stabilizing the reagent.
• Organocuprates are prepared from organolithium
reagents by reaction with a Cu+ salt, often CuI.
• Acetylide ions are another example of organometallic
reagents.
• Acetylide ions can be thought of as “organosodium
reagents”.
• Since sodium is even more electropositive than lithium,
the C—Na bond of these organosodium compounds is
best described as ionic, rather than polar covalent.
• An acid-base reaction can also be used to prepare sp
hybridized organolithium compounds.
• Treatment of a terminal alkyne with CH3Li affords a
lithium acetylide.
• The equilibrium favors the products because the sp
hybridized C—H bond of the terminal alkyne is more
acidic than the sp3 hybridized conjugate acid, CH4, that
is formed.
• Organometallic reagents are strong bases that readily abstract a
proton from water to form hydrocarbons.
• Similar reactions occur with the O—H proton of alcohols and
carboxylic acids, and the N—H protons of amines.
• Since organolithium and Grignard reagents are
themselves prepared from alkyl halides, a two-step
method converts an alkyl halide into an alkane (or other
hydrocarbon).
• Organometallic reagents are also strong nucleophiles
that react with electrophilic carbon atoms to form new
carbon—carbon bonds.
• These reactions are very valuable in forming the carbon
skeletons of complex organic molecules.
Examples of functional
organometallic reagents:
group
transformations
involving
[1] Reaction of R—M with aldehydes and ketones to afford
alcohols
[2] Reaction of R—M with carboxylic acid derivatives
[3] Reaction of R—M with other electrophilic functional groups
Reaction of Organometallic Reagents with Aldehydes
and Ketones.
• Treatment of an aldehyde or ketone with either an
organolithium or Grignard reagent followed by water
forms an alcohol with a new carbon—carbon bond.
• This reaction is an addition because the elements of R’’
and H are added across the  bond.
• This reaction follows the general mechanism for
nucleophilic addition—that is, nucleophilic attack by a
carbanion followed by protonation.
• Mechanism 20.6 is shown using R’’MgX, but the same
steps occur with RLi reagents and acetylide anions.
Note that these reactions must be carried out under anhydrous
conditions to prevent traces of water from reacting with the
organometallic reagent.
• This reaction is used to prepare 1°, 2°, and 3°
alcohols.
Protecting Groups
• Addition of organometallic reagents cannot be used with molecules
that contain both a carbonyl group and N—H or O—H bonds.
• Carbonyl compounds that also contain N—H or O—H bonds
undergo an acid-base reaction with organometallic reagents, not
nucleophilic addition.
Solving this problem requires a three-step strategy:
[1] Convert the OH group into another functional group that does
not interfere with the desired reaction. This new blocking
group is called a protecting group, and the reaction that
creates it is called “protection.”
[2] Carry out the desired reaction.
[3] Remove the protecting group. This reaction is called
“deprotection.”
A common OH protecting group is a silyl ether.
tert-Butyldimethylsilyl ethers are prepared from alcohols by
reaction with tert-butyldimethylsilyl chloride and an amine
base, usually imidazole.
The silyl ether is typically removed with a fluoride salt such as
tetrabutylammonium fluoride (CH3CH2CH2CH2)4N+F¯.
The use of tert-butyldimethylsilyl ether as a protecting
group makes possible the synthesis of 4-methyl-1,4pentanediol by a three-step sequence.
Figure 20.7
General strategy for using a
protecting group
Reaction of Organometallic Reagents with Carboxylic
Acid Derivatives.
• Both esters and acid chlorides form 3° alcohols when
treated with two equivalents of either Grignard or
organolithium reagents.
• To form a ketone from a carboxylic acid derivative, a less
reactive organometallic reagent—namely an organocuprate—is
needed.
• Acid chlorides, which have the best leaving group (Cl¯) of the
carboxylic acid derivatives, react with R’2CuLi to give a ketone
as the product.
• Esters, which contain a poorer leaving group (¯OR), do not react
with R’2CuLi.
Reaction of
Compounds
Organometallic
Reagents
with
Other
• Grignards react with CO2 to give carboxylic acids after
protonation with aqueous acid.
• This reaction is called carboxylation.
• The carboxylic acid formed has one more carbon atom than the
Grignard reagent from which it was prepared.
• The mechanism resembles earlier reactions of
nucleophilic Grignard reagents with carbonyl groups.
• Like other strong nucleophiles, organometallic
reagents—RLi, RMgX, and R2CuLi—open epoxide rings
to form alcohols.
• The reaction follows the same two-step process as opening
of epoxide rings with other negatively charged
nucleophiles—that is, nucleophilic attack from the back side
of the epoxide, followed by protonation of the resulting
alkoxide.
• In unsymmetrical epoxides, nucleophilic attack occurs at the
less substituted carbon atom.
,-Unsaturated Carbonyl Compounds
• ,-Unsaturated carbonyl compounds are conjugated molecules
containing a carbonyl group and a C=C separated by a single 
bond.
• Resonance shows that the carbonyl carbon and the  carbon
bear a partial positive charge.
• This means that ,-unsaturated carbonyl compounds can react
with nucleophiles at two different sites.
• The steps for the mechanism of 1,2-addition are exactly
the same as those for the nucleophilic addition of an
aldehyde or a ketone—that is, nucleophilic attack,
followed by protonation.
• Consider the conversion of a general enol A to the carbonyl
compound B. A and B are tautomers: A is the enol form and B is the
keto form of the tautomer.
• Equilibrium favors the keto form largely because the C=O is much
stronger than a C=C. Tautomerization, the process of converting
one tautomer into another, is catalyzed by both acid and base.
Summary
Reagents
of
the
Reactions
of
Organometallic
[1] Organometallic reagents (R—M) attack electrophilic
atoms, especially the carbonyl carbon.
[2]
After an organometallic reagent adds to the carbonyl group,
the fate of the intermediate depends on the presence or
absence of a leaving group.
[3]
The polarity of the R—M bond determines the reactivity of the
reagents:
—RLi and RMgX are very reactive reagents.
—R2CuLi is much less reactive.
Synthesis
Figure 20.8
Conversion of 2–hexanol into
other compounds
1) Give the IUPAC name for each compound.
a)
O
3,3’-dimethylhexanoic acid
OH
b)
4-chloropetanoic acid
Cl
O
OH
c)
O
OH
2,4-diethylhexanoic acid
d)
O
OH
4-isopropyl-6,8-dimethylnonanic
acid
2) Draw the structure corresponding to the IUPAC name.
O
a) 2-bromobutanoic acid
OH
Br
b) 2,3-dimethylpentanoic acid
O
OH
c) 3,3’,4-trimethylheptanoic acid
OH
O
d) 2-secbutyl-4,4’-diethylnonanoic acid
O
OH
e) 3,4-diethylcyclohexanecarboxylic acid
O
OH
f) 1-isopropylcyclobutanecarboxylic acid
O
OH
5) Give the IUPAC name for each metal salt.
a) C6H5CO2-+Li
Lithium benzoate
b) HCO2-+Na
Sodium methanoate
c) (CH3)2CHCO2-+K
Potassium 2-methylpropanoate
d) (CH3CH2)2CHCH2CHBrCH2CH2CO2-+Na
Sodium 4-bromo-6-ethyloctanoate
7) Explain how you would distinguish between these
compounds using IR spectroscopy.
O
O
H3CH2CH2CH2C
OH
H3CH2CH2CH2C
OH
OCH3
O
2 peaks
1 peak
1 peak
C=O 1710
C=O 1700
-OH 32003600
-OH 2500-3500
8) Propose a compound with the formula, C4H8O2
and the following date:
0.95 (triplet, 3H)
1.65 (multiplet, 2H)
2.30 (triplet, 2H)
11.8 (singlet, 1H)
O
OH
11)Identify the starting material in each reaction.
O
a)
Na2Cr2O7
OH
b)
OH
H2SO4, H2O
O
H3C
CH3
O3
2
H2O
OH
c)
O2N
O
KMnO4
OH
O2N
d)
O
OH
OH
CrO3
O
H2SO4, H2O
OH
19.13) Which of the following bases are strong enough to
deprotonate CH3COOH?
a) F-
pka of CH3COOH is 4.8.
b) (CH3)3CO-
b) has a pka of 18
c) has a pka of 50
c) CH3-
d) has a pka of 38
d) NH2e) Cl-
pka of conjugate acid must be greater
than that of the carboxylic acid being
deprotonated.
19.14) Rank the labeled protons in order of increasing
acidity.
Ha
H
OHb
OHc
O
Ha<Hb<Hc
The more stable the conjugate base the more acidic the
proton.
19.15) Match each pka value with each carboxylic
acid.(3.2, 4.9 and 0.2)
a) CH3CH2COOH
4.9
b) CF3COOH
0.2
c) ICH2COOH
3.2
Electron withdrawing groups make acids more
acidic.
19.16) Why is formic acid more acidic than
acetic acid?
H
OH
O
OH
O
The methyl group is electron donating and stabilizes
the acid while destabilizing the conjugate base thus
making it less acidic.
19.17) Rank the compounds within each group in order
of decreasing acidity.
a) CH3COOH, HSCH2COOH, HOCH2COOH
3
2
1
b) ICH2COOH, I2CHCOOH, ICH2CH2COOh
2
1
3
19.18) Rank each group of compounds in order of
decreasing acidity.
a)
CO2H
CO2H
CO2H
Cl
2
1
3
CO2H
b)
CO2H
CO2H
O
H3CO
2
1
3
19.19) Is the following compound more or less
acidic than phenol?
OH
HO
R
The more electron donating groups present, the less
acidic a compound is. This compound has an
additional hydroxy and alkyl group, both electron
donating. So it is less acidic.
19.22) Comparing CF3SO3H and CH3SO3H, which
has the weaker conjugate base? Which conjugate
base is the better leaving group? Which of these
acids has the higher pka?
CF3SO3H is the weaker conjugate base.
CF3SO3H is the better leaving group because it
is the weaker conjugate base.
CH3SO3H, with the electron donating methyl
group, has the higher pka and is thus a weaker
acid.
20.1) What type of orbitals make up the indicated
bonds? And in what orbitals do the lone pairs on the
oxygen lie?
O
a
b
c
a. sp3-sp2
b. sp2-sp2, p-p
c. sp3-sp2
The lone pairs lie in sp2 hybridized
orbitals.
20.2) Which compounds undergo nucleophilic
addition and which substitution?
O
a)
O
b)
H3CH2CH2C
addition
Cl
substitution
O
c)
O
d)
H
H3C
OCH3
substitution
addition
20.3) Which compound in each pair is more reactive
toward nucleuphilic attack?
a)
H3CH2CH2C
H
H3C(H3C)HC
CH2CH3
H3CH2C
O
O
O
c)
H3CH2C
H
O
O
H3CH2C
d)
H3CH2CH2C
H3CH2CH2C
O
b)
O
O
O
Cl
O
OCH3
Cl
O
O
OCH3
H3CH2C
NHCH3
OCH3
20.4) What alcohol is formed when each
compound is treated with NaBH4 in MeOH?
a)
O
OH
NaBH4
H3CH2CH2C
MeOH
H
H3CH2CH2C
H
H
b)
OH
O
NaBH4
MeOH
c)
NaBH4
MeOH
O
OH
20.5) What aldehyde or ketone is needed to synthesize
each alcohol by metal hydride reduction?
a)
b)
OH
OH
O
O
c)
OH
O
20.6) Why can’t 1-methylcyclohexanol be prepared from
a carbonyl by reduction?
OH
Tertiary alcohols can not be made by reduction of a
carbonyl because there are no hydrogens on the
carbon with the -OH.
20.7) Draw the products of the following reactions?
a)
O
OH
LiAl4
H2O
O
b)
OH
NaBH4
MeOH
c)
O
O
H2 (1 equiv.)
Pd-C
d)
O
OH
H2 (excess)
Pd-C
e)
O
OH
NaBH4 (excess)
MeOH
f)
O
NaBD4
MeOH
D
OH
20.8) Draw the products when the following compounds
are treated with NaBH4 in MeOH.
a)
O
HO
NaBH4
H
H
MeOH
b)
+
NaBH4
O
OH
MeOH
c)
(H3C)3C
O
NaBH4
MeOH
(H3C)3C
OH
(H3C)3C
+
OH
OH
20.9) What reagent is needed to carry out the reaction
below?
O
HO
Cl
H
Cl
Two reagents are needed to carry out this
reaction. First, the (S)-CBS reagent to
produce the R-enantiomer. Followed by H2O
to protonate the alcohol.
20.10) Draw a stepwise mechanism for the following
reaction.
O
LiAlH4
OH
H2O
Cl
O
O
O
Cl
Cl
H
H
H3Al
H
+ Cl
+ AlH3
O
O
H
OH
OH
H
H
H
H
H3Al
H
+ AlH3
H
+ OH
20.11) Draw an acid chloride and an ester that can be
used to produce each product.
a)
O
O
Cl
OCH3
CH2OH
Cl
OCH3
b)
O
OH
O
O
c)
OH
H3CO
H3CO
Cl
O
H3CO
OCH3
20.12) Draw the products of LiAlH4 reduction of each
compound.
O
a)
OH
OH
O
b)
NH2
NH2
c)
O
N(CH3)2
N(CH3)2
d)
O
NH
NH
20.13) What amide will form each of the following
amines when treated with LiAlH4?
O
a)
NH2
NH2
O
b)
N
c)
N
O
N
H
N
H
20.14) Predict the products of these compounds
when treated with the following reagents.
O
O
OH
a)
LiAlH4
OCH3
H2O
OH
OH
O
NaBH4
MeOH
b)
O
OCH3
O
LiAlH4
H3CO
OH
H2O
NaBH4
MeOH
HO
OH
No reaction
c)
H3CO
H3CO
O
OH
LiAlH4
H2O
NaBH4
MeOH
H3CO
OH
20.15) Predict the products in the following
reactions.
a)
OH
Ag2O
No Reaction
NH4OH
O
Na2Cr2O7
H2SO4, H2O
b)
OH
O
OH
OH
O
Ag2O
NH4OH
OH
Na2Cr2O7
O
O
H2SO4, H2O
OH
20.16) Predict the products of the compound below when
OH
reacted with each reagent.
O
HO
a)
OH
NaBH4
OH
MeOH
HO
b)
LiAlH4
OH
H2O
OH
HO
OH
O
HO
c)
O
PCC
O
O
OH
d)
O
Ag2O
OH
NH4OH
HO
O
e)
O
CrO3
OH
H2SO4, H2O
HO
O
20.17) Write out the rea tions needed to
convert CH3CH2Br to each of the following
reagents.
a) H3CH2C
H3CH2C
b)
Li
H3CH2C
H3CH2C
c)
+ 2 Li
Br
Li
H3CH2C
MgBr
+ Li
Br
MgBr
+ Mg
Br
H3CH2C
CuLi
H3CH2C
2 H3CH2C
H3CH2C
Br
Li
+ 2 Li
H3CH2C
Li
Li
Cu
+ CuI
H3CH2C
+ Li
+ Li
CH2CH3
Br
I
20.18) 1-octyne reacts readily with NaH, forming a gas
that bubbles out of the reaction mixture. 1-octyne also
reacts with CH3MgBr and a different gas is produced.
Write out balanced equations for each reaction.
HC
CCH2CH2CH2CH2CH2CH3
+ NaH
NaC
HC
CCH2CH2CH2CH2CH2CH3
CCH2CH2CH2CH2CH2CH3
+ H2
+ CH3MgBr
BrMgC
CCH2CH2CH2CH2CH2CH3
+ CH4
20.19) Draw the product of the following reactions.
a)
Li
+ H2O
+ LiOH
b)
MgBr
c)
+ H2O
MgBr
+ H2O
+ HOMgBr
+ HOMgBr
d)
LiC
CCH2CH3
+ H2O
HC
CCH2CH3
+ LiOH
20.20) Draw the product formed when each compound
is treated with C6H5MgBr followed by H2O.
a)
H
O
OH
H
b)
H
H
O
CH2CH3
OH
H3CH2C
CH2CH3
CH2CH3
O
c)
CH2CH3
H3CH2C
OH
H
H
d)
OH
O
20.21)Draw the products of each reaction.
a)
H3CH2CH2C
Li
+ LiOH
H2O
O
HO
b)
CH2CH2CH3
Li
H
O
+ LiOH
HO
H
H2O
H
H
OH
c)
C6H5Li
O
H2 O
+ LiOH
d)
CNa
H2C
H2
C
O
C
H2O
OH
+ NaOH
20.22) Draw the products (including stereochemistry) of
the following reactions.
a)
O
H
H3CH2C
H3C
b)
H
OH
H
MgBr
OH
+
H2O
CH2CH3
H3CH2C
O
Li
OH
H2O
+
CH2CH3
OH
20.23) What Grignard and carbonyl are needed to
prepare each alcohol?
a)
O
OH
+
H3C
MgBr
H
MgBr
O
OH
b)
+
H
H
O
OH
c)
+
H3CH2C
MgBr
or
O
MgBr
|+
d)
OH
O
H3C
+
Br
or
O
MgBr
+
20.24) Tertiary alcohols with three different R groups on
the carbon attached to the OH can be prepared in three
different ways using the Grignard reagent. Show them.
a)
OH
O
H3C
CH2CH3
CH2CH3
CH2CH2CH3
+ H3C
MgBr
+ H CH C
3
2
MgBr
H3CH2CH2C
O
H3C
CH2CH2CH3
O
+ H CH CH C
3
2
2
H3C
CH2CH3
MgBr
b)
OH
O
+
H3C
MgBr
MgBr
O
+
MgBr
O
+
c)
OH
O
MgBr
+
MgBr
O
+
+ H3CH2C
O
MgBr
20.25) Show the steps for the following reaction.
CH2CH2CH2CH3
HO
O
HO
OH
TBDMS-Cl
HO
O
TBDMSO
H
N
O
N
BrMg
CH2CH2CH2CH3
H2O
CH2CH2CH2CH3
FN(CH2CH2CH2CH3)4
HO
OH
CH2CH2CH2CH3
TBDMSO
H2O
OH