Transcript Document
20.8) Draw the products when the following compounds are treated with NaBH 4 in MeOH.
a)
O HO H H OH NaBH 4 MeOH +
b)
O NaBH 4 MeOH OH
c)
(H 3 C) 3 C O NaBH 4 MeOH (H 3 C) 3 C OH + (H 3 C) 3 C OH 1
20.9) What reagent is needed to carry out the reaction below?
O HO H Cl Cl
Two reagents are needed to carry out this reaction. First, the (S)-CBS reagent to produce the R-enantiomer. Followed by H 2 O to protonate the alcohol.
2
20.12) Draw the products of LiAlH 4 compound.
reduction of each a)
O OH OH
b)
O NH 2 NH 2 O
c)
N(CH 3 ) 2 N(CH 3 ) 2 3
d)
O NH NH 4
20.14) Predict the products of these compounds when treated with the following reagents.
O O OH
a)
LiAlH 4 OCH 3 H 2 O OH OH O NaBH 4 MeOH OCH 3
b)
O O H 3 CO OH LiAlH 4 H 2 O HO OH NaBH 4 MeOH
No reaction
5
c)
H 3 CO O LiAlH 4 H 2 O H 3 CO NaBH 4 MeOH H 3 CO OH OH 6
20.15) Predict the products in the following reactions.
a)
OH Ag 2 O NH 4 OH
No Reaction
O Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O OH
b)
OH O Ag 2 O NH 4 OH Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O OH O O O OH OH 7
20.16) Predict the products of the compound below when
OH
reacted with each reagent.
O HO
a)
OH NaBH 4 MeOH OH HO
b)
LiAlH 4 H 2 O OH OH HO 8
c)
PCC O
d)
Ag 2 O NH 4 OH HO
e)
CrO 3 H 2 SO 4 , H 2 O HO O O HO OH O O O OH O OH OH 9 O
20.22) Draw the products (including stereochemistry) of the following reactions.
a)
O H 3 C H H 3 CH 2 C H 2 O MgBr H OH + H OH
b)
H 3 CH 2 C O H 2 O Li CH 2 CH 3 OH + CH 2 CH 3 OH 10
20.26) Draw out the products when each compound is treated with 2 equivalents of CH 3 CH 2 CH 2 CH 2 MgBr followed by H 2 O?
a)
O OH H 3 CH 2 C CH 2 CH 2 CH 2 CH 3 H 3 CH 2 C Cl CH 2 CH 2 CH 2 CH 3 O
b)
HO CH 2 CH 2 CH 2 CH 3 OCH 3 CH 12 CH 2 CH 2 CH 3
c)
H 3 CH 2 CH 2 CH 2 CH 2 C O OCH 2 CH 3 H 3 CH 2 CH 2 CH 2 CH 2 C OH CH 2 CH 2 CH 2 CH 3 CH 2 CH 2 CH 2 CH 3 11
20.28) Tertiary alcohols can also be formed by reacting dimethyl carbonate with excess Grignard reagent. Draw out a mechanism for the following reaction.
OH O C 6 H 5 MgBr + H 2 O C 6 H 5 C 6 H 5 + MeOH + HOMgBr H 3 CO OCH 3 C 6 H 5 O O O + OCH 3 H 3 CO OCH 3 MgBr H 3 CO OCH 3 + MgBr H 3 CO MgBr 12
O + OCH 3 MgBr HO H O + MgBr OH O OCH 3 + MgBr + HOMgBr + HOCH 3 13
20.32) What carboxylic acid is formed from each aqlkyl halide when treated with Mg, CO 2 and H 2 O?
O
a)
Br Mg MgBr CO 2 O H 2 O O OH 14
b)
H 3 CH 2 CH 2 CH 2 C Cl Mg O H 3 CH 2 CH 2 CH 2 C MgCl CO 2 H 3 CH 2 CH 2 CH 2 C H 2 O O O OH
c)
CH 2 Br Mg CH 2 MgBr CO 2 O C O H 2 O O H 3 CH 2 CH 2 CH 2 C C OH OCH 3 OCH 3 OCH 3 OCH 3 15
20.34) Draw the products of each compound when treated with either dimethylcuprate or lithium acetylide followed by water.
O
a)
O CuLi H 2 O CLi HO H 2 O 16
b)
H 3 CH 2 C O CH 2 CH 3 H 2 O CuLi H 3 CH 2 C O CLi H 2 O H 3 CH 2 C OH CH 2 CH 3 CH 2 CH 3 17
c)
O CLi H 2 O CuLi H 2 O HO O 18
20.35) What reagents are needed in each reaction?
a)
O CuLi H 2 O O
b)
O Li H 2 O OH
c)
OH O CuLi H 2 O Li H 2 O 19
20.36) Synthesize each of the following using cyclohexanol, ethanol and any other inorganic reagent.
OH
a)
OH OH HBr or PBr 3 PCC O Br Mg H 2 O MgBr HO MgBr 20
b)
Br HO HBr Br
c)
HO + H 2 SO 4 H 2 Pd-C 21
d)
O OH HBr or PBr 3 OH PCC Br Mg O MgBr O PCC HO H 2 O 22
e)
OH OH H 2 SO 4 MgBr m-ClPBA O + H 2 O OH 23
Aldehydes and Ketones —Nucleophilic Addition
Introduction
Aldehydes and ketones contain a carbonyl group. An aldehyde contains at least one H atom bonded to the carbonyl carbon, whereas the ketone has two alkyl or aryl groups bonded to it.
Two structural features determine the chemistry and properties of aldehydes and ketones.
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Aldehydes and Ketones —Nucleophilic Addition
Introduction
• •
Aldehydes and ketones react with nucleophiles.
As the number of R groups around the carbonyl carbon increases, the reactivity of the carbonyl compound decreases, resulting in the following order of reactivity:
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Nomenclature of Aldehydes
• •
If the CHO is bonded to a chain of carbons, find the longest chain containing the CHO group, and change the –e ending of the parent alkane to the suffix –al. If the CHO group is bonded to a ring, name the ring and add the suffix
–carbaldehyde
.
Number the chain or ring to put the CHO group at C1, but omit this number from the name. Apply all the other usual rules of nomenclature.
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• •
Like carboxylic acids, many simple aldehydes have common names that are widely used.
A common name for an aldehyde is formed by taking the common parent name and adding the suffix -aldehyde.
•
Greek letters are used to designate the location of substituents in common names.
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Nomenclature of Ketones
• • • •
In the IUPAC system, all ketones are identified by the suffix “one”.
Find the longest continuous chain containing the carbonyl group, and change the –e ending of the parent alkane to the suffix -one.
Number the carbon chain to give the carbonyl carbon the lowest number. Apply all of the usual rules of nomenclature.
With cyclic ketones, numbering always begins at the carbonyl carbon, but the “1” is usually omitted from the name.
The ring is then numbered clockwise or counterclockwise to give the first substituent the lower number.
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•
Most common names for ketones are formed by naming both alkyl groups on the carbonyl carbon, arranging them alphabetically, and adding the word “ketone”.
•
Three widely used common names for some simple ketones do not follow this convention:
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•
Sometimes, acyl groups must be named as substituents. The three most common acyl groups are shown below:
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Spectroscopic Properties —IR Spectra
• •
Aldehydes and ketones exhibit a strong peak at ~1700 cm -1 due to the C=O.
The sp 2 hybridized C —H bond of an aldehyde shows one or two peaks at ~2700 –2830 cm -1 .
Figure 21.3
The IR spectrum of propanal, CH 3 CH 2 CHO 31
• •
Most aldehydes have a carbonyl peak around 1730 cm -1 , whereas for ketones, it is typically around 1715 cm -1 .
Ring size affects the carbonyl absorption in a predictable manner.
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•
Conjugation also affects the carbonyl absorption in a predictable manner.
Figure 21.4
The effect of conjugation on the carbonyl absorption in an IR spectrum 33
Spectroscopic Properties —NMR Spectra Aldehydes and ketones exhibit the following NMR absorptions.
1
H and
13
C
•
The sp 2 hybridized C —H proton of an aldehyde is highly deshielded and absorbs far downfield at 9-10 ppm.
Splitting occurs with protons on the
carbon, but the
•
coupling constant is often very small (J = 1-3 Hz).
Protons on the
carbon to the carbonyl group absorb at 2-2.5
ppm.
Methyl ketones, for example, give a characteristic singlet at ~2.1 ppm.
•
In a 13 C NMR spectrum, the carbonyl carbon is highly deshielded, appearing in the 190-215 ppm region.
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Figure 21.5
The 1 H and 13 C NMR spectra of propanal, CH 3 CH 2 CHO 35
36
Preparation of Aldehydes and Ketones
Common methods to synthesize aldehydes:
37
Common methods to synthesize ketones:
38
Aldehydes and ketones are also both obtained as products of the oxidative cleavage of alkenes.
39
Which compound is most reactive toward nucleophilic addition?
O A O H B H O C H
B is most reactive. The less R groups attached to the carbonyl the less sterically hindered it is and thus more reactive.
40
Which is more reactive and why?
O A H O B H
B is more reactive because it isn’t resonance stabilized like A.
41
Which of these compounds is 2-isobutyl-3 isopropylhexanal?
O O A B H O H C
C is the correct answer.
42
Which compound is 5-ethyl-4-methyl-3-octanone
O A C O
B is the correct answer.
O B 43
Which of the following is the name of the compound below?
O
a) 4-methyl-3-hexanone b) Isopropyl ethyl ketone c) 3-methyl-4-hexanone] d) sec-butyl ethyl ketone A and D are the correct answers
44
Which of the following is the name of the compound below?
O
a) 2-methyl-5-ethylhexanal b) 3-ethyl-6-methylcyclohexanecarbaldehyde c) 2-methyl-5-ethylcyclohexanone d) 3-ethyl-5-methylcyclohexanone C is the correct answer.
45
Which compound absorbs at the higher IR frequency?
A O O B
B due to it being unconjugated.
B A O O
A due to ring strain increasing as rings decrease in size.
46
Three compounds with the formula C 4 H 8 O are below. Compound A has 3 peaks in the C-13 NMR. Compound B has a signal at 9.8 ppm in the proton NMR. Compound C has a singlet at 2.1 ppm. Label the compounds below with the correct letter.
O O O H H
B A c
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O OCH 3 DIBAL-H H 2 O PCC OH BH 3 H 2 O / -OH O 3 Zn, H 2 O O H O O O 48
ClCOCH 3 AlCl 3 O O Cl (CH 3 ) 2 CuLi H 2 O H 2 O H 2 SO 4 HgSO 4 O O 49
Reactions of Aldehydes and Ketones —General
[1] Reaction at the carbonyl carbon —the elements of H and Nu are added to the carbonyl group.
[2] Reaction at the
carbon.
50
• •
With negatively charged nucleophiles, nucleophilic addition follows a two-step process —nucleophilic attack followed by protonation, as shown below.
This process occurs with strong, neutral or negatively charged nucleophiles.
51
•
With some neutral nucleophiles, nucleophilic addition only occurs if an acid is present —In this mechanism, protonation precedes nucleophilic attack as shown below.
52
It is important to know what nucleophiles will add to carbonyl groups.
• •
Cl ¯ , Br ¯ and I ¯ reactions at sp 3 are good nucleophiles in substitution hybridized carbons, but they are ineffective nucleophiles in addition.
When these nucleophiles add to a carbonyl, they cleave the C —O
bond, forming an alkoxide. Since X ¯ is a much weaker base than the alkoxide formed, equilibrium favors the starting materials, not the addition product.
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Nucleophilic Addition of H
¯
and R
¯
—A Review
Treatment of an aldehyde or ketone with either NaBH 4 LiAlH 4 or followed by protonation forms a 1 ° or 2° alcohol.
The nucleophile in these reactions is H: ¯.
Hydride reduction occurs via a two-step mechanism.
54
Treatment of an aldehyde or ketone with either an organolithium (R”Li) or Grignard reagent (R”MgX) followed by water forms a 1 ° , 2 ° , or 3 ° alcohol containing a new C —C bond. In these reactions, R”¯ is the nucleophile.
Nucleophilic addition occurs via a two-step mechanism.
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Nucleophilic Addition of
¯
CN
•
Treatment of an aldehyde or ketone with NaCN and a strong acid such as HCl adds the elements of HCN across the C —O
bond, forming a cyanohydrin.
•
The mechanism involves the usual two steps of nucleophilic addition —nucleophilic attack followed by protonation.
56
•
Cyanohydrins can be reconverted to carbonyl compounds by treatment with base. This process is just the reverse of the addition of HCN: deprotonation followed by elimination of ¯ CN.
•
The cyano group of a cyanohydrin is readily hydrolyzed to a carboxy group by heating with aqueous acid or base.
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The Wittig Reaction
•
The Wittig reaction reagent ) to form uses a carbon nucleophile (the Wittig alkenes —the carbonyl group is converted to a C=C.
58
• •
The Wittig reagent is an organophosphorus reagent .
A typical Wittig reagent has a phosphorus atom bonded to three phenyl groups, plus another alkyl group that bears a negative charge.
• •
A Wittig reagent is an ylide , a species that contains two oppositely charged atoms bonded to each other, with both atoms having octets.
Phosphorus ylides are also called phosphoranes .
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• • • •
Since phosphorus is a second-row element, it can be surrounded by more than eight electrons.
Thus, a second resonance structure can be drawn that places a double bond between carbon and phosphorus.
Regardless of which resonance structure is drawn, a Wittig reagent has no net charge.
However, note that in one resonance structure, the carbon bears a net negative charge, making it nucleophilic.
60
Wittig reagents are synthesized by a two-step procedure.
61
To synthesize the Wittig reagent Ph 3 P=CH 2 , use the following two steps: Step [1] Form the phosphonium salt Ph 3 P: and CH 3 Br.
by S N 2 reaction of Step [2] Form the ylide by removal of a proton using as a strong base.
BuLi
62
The currently accepted mechanism of the Wittig reaction involves two steps.
63
One limitation of the Wittig reaction is that a mixture of stereoisomers sometimes forms.
64
• •
An advantage of the Wittig reaction over elimination methods used to synthesize alkenes is that you always know the location of the double bond —the Wittig reaction always gives a single constitutional isomer.
Consider the two methods that can be used to convert cyclohexanone into cycloalkene B.
65
Addition of 1
0
Amines —Formation of Imines
•
Amines are classified as 1 °, 2°, or 3° by the number of alkyl groups bonded to the nitrogen atom.
•
Treatment of an aldehyde or a ketone with a 1 ° amine affords an imine (also called a Schiff base ).
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Addition of 1
°
Amines —Formation of Imines
•
Because the N atom of an imine is surrounded by three groups (two atoms and a lone pair), it is sp 2 hybridized, making the C —N—R bond angle 120°, (not 180°). Imine formation is fastest when the reaction medium is weakly acidic (pH 4-5).
67
68
• •
In imine formation, mild acid is needed for protonation of the hydroxy group in step 3 to form a good leaving group.
Under strongly acidic conditions, the reaction rate decreases because the amine nucleophile is protonated —with no free electron pair, it is no longer a nucleophile, and so nucleophilic addition cannot occur.
69
Addition of 2
°
Amines —Formation of Enamines
•
A 2 ° amine reacts with an aldehyde or ketone to give an enamine . Enamines have a nitrogen atom bonded to a C —C double bond.
70
71
Figure 21.10 The formation of imines and enamines compared 72
Imine and Enamine Hydrolysis
• •
Because imines and enamines are formed by a reversible set of reactions, both can be converted back to carbonyl compounds by hydrolysis with mild acid.
The mechanism of hydrolysis is the exact reverse of the mechanism written for formation of imines and enamines.
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21.14) Draw the products. (include stereoisomers) a)
O H OH NaBH 4 MeOH + H OH
b)
O CHMgBr H 2 O OH + OH
c)
O LiAlH 4 H 2 O H OH + H OH 74
21.15) Draw the products.
a)
O NaCN HCl OH CN
b)
OH CN H 3 O + heat OH COOH 75
21.17) Draw the products.
a)
O + Ph 3 P CH 2
b)
O + Ph 3 P CHCH 2 CH 2 CH 2 CH 3 CHCH 2 CH 2 CH 2 CH 3 76
21.18) Synthesize each reagent.
a)
Ph 3 P CHCH 3 PPh 3 + H 3 CH 2 C Br Ph 3 P + Br CH 2 CH 3 Bu-Li Ph 3 P CHCH 3
b and c follow the same process. The alkyl group doesn’t effect the reaction.
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21.20) Draw the products.(include stereoisomers) a)
O Ph 3 P CHCH 2 CH 3 + CH 2 CH 3 CH 2 CH 3
b and c work the same way. You get an E and Z isomer for each reaction unless your Wittig reagent is Ph 3 P=CH 2 .
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21.21) Draw the starting materials. If there are two routes indicate the preferred one.
a)
CH 2 CH 3 CH 2 CH 3 PPh 3 + O H H or H 3 CH 2 C PPh 3 + O H
The second route is preferred due to the Wittig reagent coming from a less hindered alkyl halide. Both b and c only have one route because both sides of the alkene product are the same.
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21.22) Show two different methods for the following reactions.
O
a)
OH O H 3 C H 2 O MgBr + H 2 SO 4 + minor major E and Z 80
O H 2 C PPh 3
With the two step process you get multiple products with the major one being the most substituted alkene. However with the Wittig you only get one product.
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21.23) Draw the products.
a)
O H 3 CH 2 CH 2 CH 2 C NH 2 H 3 O + NCH 2 CH 2 CH 2 CH 3
Primary amines react with aldehydes or ketones in the presence of a mild acid to form an imine
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21.24) What primary amine and carbonyl are needed to prepare the product.
a)
NCH 2 CH 2 CH 3 O + H 2 N CH 2 CH 2 CH 3 H H
When you have an imine as a product, remove the nitrogen containing end of the double bond and replace it with an oxygen. And add 2 H’s to the nitrogen end to make your primary amine.
83
21.26) what two enamines are formed?
O N + NH + N
The number of enamines formed is determine by the number of different beta carbons with available protons.
84
21.27)Explain why benzaldehyde cannot form an enamine with a secondary amine.
O
This carbon has 4 bonds to carbon and thus no proton which is needed in the formation of the enamine.
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21.28) What carbonyl and amine are produced under hydrolysis conditions?
a)
C H N H 2 O H + CH O + H 2 N
b)
CH 2 N H 2 O H + NH CH 2 + O 86
Addition of H
2
O — Hydration
•
Treatment of a carbonyl compound with H 2 O in the presence of an acid or base catalyst adds the elements of H and OH across the C —O
bond, forming a gem-diol or hydrate .
•
Gem-diol product yields are good only when unhindered aldehydes or aldehydes with nearby electron withdrawing groups are used.
87
•
Increasing the number of alkyl groups on the carbonyl carbon decreases the amount of hydrate at equilibrium. This can be illustrated by comparing the amount of hydrate formed from formaldehyde, acetaldehyde and acetone.
88
Other electronic factors come into play as well.
This explains why chloral forms a large amount of hydrate at equilibrium. Three electron-withdrawing Cl atoms result in a partial positive charge on the
carbon of the carbonyl, destabilizing the carbonyl group, and therefore increasing the amount of hydrate at equilibrium.
89
Both acid and base catalyze the addition of H 2 O to the carbonyl group.
•
With base, the nucleophile is ¯ OH, and the mechanism follows the usual two steps: nucleophilic attack followed by protonation.
•
The reaction rate increases in the presence of base because the base converts H nucleophile.
2 O into ¯ OH, a stronger
90
•
The reaction rate increases in the presence of acid because the acid protonates the carbonyl group, making it more electrophilic and thus more susceptible to nucleophilic attack.
91
Addition of Alcohols —Acetal Formation
• • •
Aldehydes and ketones react with two equivalents of alcohol to form acetals .
Acetal formation is catalyzed by acids, such as Note that acetals are not ethers .
TsOH .
92
•
When a diol such as ethylene glycol is used in place of two equivalents of ROH, a cyclic acetal is formed.
• •
Like gem-diol formation, the synthesis of acetals is reversible, and often, the equilibrium favors the reactants.
In acetal synthesis, since water is formed as a by-product, the equilibrium can be driven to the right by removing H 2 O as it is formed using distillation or other techniques.
Driving an equilibrium to the right by removing one of the products is an application of Le Ch âtelier’s principle.
93
Figure 21.11
A Dean –Stark trap for removing water 94
•
The mechanism for acetal formation can be divided into two parts, the first of which is addition of one equivalent of alcohol to form the hemiacetal .
95
•
The second part of the mechanism involves conversion of the hemiacetal into the acetal.
96
• •
Because conversion of an aldehyde or ketone to an acetal is a reversible reaction, an acetal can be hydrolyzed to an aldehyde or ketone by treatment with aqueous acid.
Since the reaction is also an equilibrium process, it is driven to the right by using a large excess of water for hydrolysis.
97
Acetals as Protecting Groups
• • •
Acetals are valuable protecting groups for aldehydes and ketones.
Suppose we wish to selectively reduce the ester to an alcohol in compound A, leaving the ketone untouched.
Because ketones are more readily reduced, methyl-5 hydroxyhexanoate is formed instead.
98
To solve this problem, we can use a protecting group to block the more reactive ketone carbonyl. The overall process requires three steps.
[1] Protect the interfering functional group —the ketone carbonyl.
[2] Carry out the desired reaction.
[3] Remove the protecting group.
99
Cyclic Hemiacetals
Cyclic hemiacetals containing five- and six-membered rings are stable compounds that are readily isolated.
100
Cyclic hemiacetals are formed cyclization of hydroxy aldehydes.
by intramolecular Such intramolecular reactions to form five- and six membered rings are faster than the corresponding intermolecular reactions. The two reacting functional groups (OH and C=O), are held in close proximity, increasing the probability of reaction.
101
Hemiacetal formation is catalyzed by both acid and base.
102
Intramolecular cyclization of a hydroxy aldehyde forms a hemiacetal with a new stereogenic center, so that an equal amount of two enantiomers results.
Cyclic hemiacetals can be converted to acetals by treatment with an alcohol and acid.
103
104
• • •
In the conversion of hemiacetals to acetals, the overall result is the replacement of the hemiacetal OH group by an OCH 3 group.
This reaction occurs readily because the carbocation formed in step 2 is stabilized by resonance. This fact makes the hemiacetal OH group different from the hydroxy group in other alcohols.
Thus, when a compound with both an alcohol OH and a hemiacetal OH is treated with an alcohol and acid, only the hemiacetal OH reacts to form the acetal.
105
For Monday, 21.31-39.
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