march 3oth, 2011 goals for the day: 1)naming amines 2)physical properties of amines 3)amine...
TRANSCRIPT
March 3oth , 2011Goals for the day:
1) Naming Amines
2) Physical Properties of Amines
3) Amine reactions
* Conversion of an alkyl halide
* Gabriel Synthesis
* Hoffman Elimination
4) IR/NMR of Amines
5) Multiple Choice Questions…
Chapter 24 AminesName the groups (up to 3) that are attached.Di & tri prefixes are used if groups are the same
NH3 ammonia
methyl amine
isopropyl methyl amine
NH2
NH
Degree of substitution…Substitution on the nitrogen is important …
The number of carbons directly attached to the nitrogen is whatYou want to look at
NH2NH N
Primary secondary tertiary
Bonding to N is similar to that in ammoniaN is sp3-hybridizedC–N–C bond angles are close to 109°
tetrahedral value
24.2 Properties of Amines
Amines with fewer than five carbons are water-soluble
Primary and secondary amines form hydrogen bonds, increasing their boiling points
Amines Form H-Bonds
The lone pair of electrons on nitrogen makes amines basic and nucleophilic
They react with acids to form acid–base salts and they react with electrophiles
24.3 Basicity of Amines
Amides (RCONH2) in general are not proton acceptors except in very strong acid
The C=O group is strongly electron-withdrawing, making the N a very weak base
Addition of a proton occurs on O but this destroys the double bond character of C=O as a requirement of stabilization by N=
Amides
Ammonia and other amines are good nucleophiles
SN2 Reactions of Alkyl Halides
Primary, secondary, and tertiary amines all have similar reactivity, the initially formed monoalkylated substance undergoes further reaction to yield a mixture of products
Problem: Uncontrolled Multiple Alkylation (i.e. can’t stop the reaction! ).
A phthalimide alkylation for preparing a primary amine from an alkyl halide
The N-H in imides (–CONHCO–) can be removed by KOH followed by alkylation and hydrolysis
Gabriel Synthesis of Primary Amines
Converts amines into alkenesNH2
is very a poor leaving group so it is converted to an alkylammonium ion, which is a good leaving group
Hofmann Elimination
Exchanges hydroxide ion for iodide ion in the quaternary ammonium salt, thus providing the base necessary to cause elimination
The Elimination Step
We would expect that the more highly substituted alkene product predominates in the E2 reaction of an alkyl halide (Zaitsev's rule)
However, the less highly substituted alkene predominates in the Hofmann elimination due to the large size of the trialkylamine leaving group
The base must abstract a hydrogen from the most sterically accessible, least hindered position
Orientation in Hofmann Elimination
Steric Effects Control the Orientation
Polycyclic Heterocycles
Infrared Spectroscopy:Characteristic N–H stretching absorptions
3300 to 3500 cm1 Amine absorption bands are sharper and less
intense than hydroxyl bandsProtonated amines show an ammonium band in
the range 2200 to 3000 cm1
24.10 Spectroscopy of Amines -Infrared
Examples of Infrared Spectra
NMRHydrogens on a carbon attached to an oxygen (or
nitrogen) are around 3-4 ppm.Hydrogens on a carbon next to a carbonyl are around
2 ppm.
Putting it all together…Reactions for exam three are on the following
pages.
Note that reactions with a *** you need to know the mechanism…
Oxidation of primary alcohols
Reagents can be H2CrO4, CrO3, [O], PCCPrimary alcohols can be oxidized to carboxylic acids
Reaction of CO2 with a Grignard reagent ***
OH
OCrO3
OH
H3C OH
O1) CO2
2) H3O+CH3 : MgBr
Reactions that Carboxylic Acids do…
React with bases… ***
Makes acid halides(do the same reactions asacids, but are faster)…
Reduces to a primary alcohol(goes through an aldehyde)…
OH
O
O
ONaOH
Na
OH
O SOCl2
Cl
O
OHOH
O1) LAH
2) H3O+
H H
Two main Carboxylic Acid Reactions(need to be acid or base catalyzed)…
Esterification…***
Important in fats
Amide formation…***
Important in proteins (amide bonds)
OH
O ROH
OR
O
H + or OH -
OH
O RNH2
NHR
O
H + or OH -
Note: You will need to know all of the mechanisms, but the good newsis that they are very similar! (We will review them with the acid derivatives…)
Reactions of Acid Derivatives…
Cl
O ROH
OR
O
Cl
O RNH2
NHR
O
Ester and amide formation using an acid halide
More reactions of Acid Derivatives…
X = halogen, OR, or NHR
X
O H2O
OH
O
may needH+ or OH-
OHX
O1) LAH
2) H3O+
H H
2) H3O+
CH3 : MgBr
X
O
OH
H3C CH3
Hydrolysis ***(very important for bothesters and amides)
Reduction using LAH(gives an amine if starting
with an amide)
Goes through a ketone ***
Summary of the reactions thus far…
O OKOH Br2 O
Br
Br2
OH
O
xs
Haloform test
O
Br
H2O
alpha bromination
alkylation
***
***
Next three (Aldol, acetoacetate synthesis & claisen)…
O
OO NaOEt H3O +
O
Br
O
O
1) NOEt, ethanol
2) H3O+
2)
ClaisenCondensation
Acetoacetate Synthesis
O 1) KOH
OH-
O O
aldol ***
Chain lengthening of amines (from an alkyl halide)…
Gabriel Synthesis…
H3O+
RNH2
Hoffman Elimination…
***
Which of the following is not an acyl derivative?
O
O1.
Cl
O2.
HN
O
3.
O
O
O
4.
OO
5.
Answer
5. OO
What is the IUPAC name for the following carboxylic acid?
1. 3-bromo-4-methylbenzoic acid2. 4-methyl-3-bromobenzoic acid3. phthalic acid4. 4-carboxy-2-bromotoluene5. 5-carboxy-2-methyl-1-bromobenzene
OH
O
Br
Answer
1. 3-bromo-4-methylbenzoic acid
What is the IUPAC name for the following carboxylic acid?
1. 2-propylhexanoic acid2. 4-propylhexanoic acid3. 2-butylpentanoic acid4. 4-carboxyoctane5. 5-carboxyoctane
OH
O
Answer
1. 2-propylhexanoic acid
The structure for oleic acid is shown below. What is the IUPAC name of this compound?
a. (Z)-octadec-9-enoic acidb. (E)- octadec-9-enoic acid
OH
O
Answer
a. (Z)-octadec-9-enoic acid
Rank the following molecules in acidity from least acidic to most acidic.
OH
O
A
OH
O
B
OH
OCl
Cl
ClC
1. A, C, B2. B, A, C3. C, B, A4. B, C, A5. C, A, B
Answer
2. B, A, C
Chloroacetic acid is a stronger acid than acetic acid. Which is the best explanation?
1. More resonance structures can be drawn for chloroacetic acid than for acetic acid.
2. More resonance structures can be drawn for chloroacetate ion than for acetate ion.
3. Because of its high electronegativity, chlorine is able to donate electrons to the chloroacetate ion by the inductive effect, thereby stabilizing this ion.
4. Because of its high electronegativity, chlorine is able to withdraw electrons from the chloroacetate ion by the inductive effect, thereby stabilizing this ion.
5. Chlorine is larger than hydrogen and can better hold a negative charge.
H3C COOH H3C COOH2C COOH
Cl
H2C COO
Cl
acetic acid acetate chloroacetic acid chloroacetate
Answer
4. Because of its high electronegativity, chlorine is able to withdraw electrons from the chloroacetate ion by the inductive effect, thereby stabilizing this ion.
p-Chlorobenzoic acid is more acidic than p-methylbenzoic acid.
1. True2. False
OH
O
Cl
OH
O
Answer
1. True
Which of the following is not an ester?
O
O1. O
O2.
OO3.
O
O4.
Answer
2. O
O
Determine the IUPAC name for the following molecule.
1. 3,4-dimethylphenyl pentanoate2. 3,4-dimethylpentyl benzenoate3. 3,4-dimethylpentyl benzoate4. benzyl 3,4-dimethylpentanoate5. phenyl 3,4-dimethylpentanoate
O
O
Answer5. phenyl 3,4-dimethylpentanoate
Which of the statements is true concerning the following two carboxylic acid derivatives?
1. Only molecule A can be hydrolyzed.2. Only molecule B can be hydrolyzed.3. Both molecules can be hydrolyzed, but A
will react faster than B.4. Both molecules can be hydrolyzed, but B
will react faster than A.5. A and B can be hydrolyzed at roughly the
same rate.
NH
O
O
O
A B
Answer4. Both molecules can be
hydrolyzed, but B will react faster than A.
Predict the outcome of the following reaction.
OH1.
H
O2.
Cl
O3.
SOH
O4.
S
O
O
OH
5.
OH
O
SOCl2
Answer
3. Cl
O
Which of the following steps occurs first in the mechanism of Fischer esterification?
1. attack of the nucleophile on the carbonyl carbon
2. protonation of the carbonyl oxygen3. loss of water from the tetrahedral carbonyl
addition intermediate4. protonation of the alcohol oxygen5. formation of the tetrahedral carbonyl
addition intermediate
Answer2. protonation of the carbonyl oxygen
Which of the following is the tetrahedral intermediate that appears in the Fischer esterification of ethanol and benzoic acid?
O
HO OH1.
O
O OH2.
HO OH3.
O
O O4.
HO O
O OH5.
Answer1.
O
HO OH
1.
2.
3.
4.
5. None of these
Predict the product of the following reaction.
NH2
OH
NH2
HO OH
NH2 NH2+
OH OH
NH2
O 1) LiAlH4
2 ) H2O
Answer1. NH2
What is the predominant enol form of the following molecule?
OH OH1. OH OH2.
OH O3.
OH O4. None of these is favored over the others.
5.
O O
Answer
4. OH O
Which of the following will occur when the following optically active compound is placed in dilute acid?
1. It will form an acetal.2. It will form a diol.3. It will become an alcohol.4. It will lose its optical activity.5. none of these
O
Answer
4. It will lose its optical activity.
Identify the expected major product of the following reaction.
O
Br
1. Br
O2.
O
Br
3.
OH
O
Br
4.Br
O
Br Br
5.
O Br2
acetic acid
Answer
3. O
Br
How many protons with a pKa < 22 exist in the following molecule?
1. 02. 13. 24. 35. 4
Answer
5. 4
Which of the following reagents will not form an enolate upon reaction with a ketone?
1. 2.
3.
4.5.
LiAlH4
Answer
4. LiAlH4
Which of the above carbonyl compounds is most acidic?
1. A2. B3. C4. D
O
O
O
O
O
O
O
A B C D
Answer
3. C
Which of the following statements explains why the following aldehyde will not undergo an aldol reaction with itself?
1. The benzene ring makes the carbonyl group unreactive towards aldol reactions.
2. A carbonyl group must be connected to two alkyl groups in order to undergo an aldol reaction.
3. The molecule does not possess any hydrogens α to the carbonyl group.
4. Electrophilic aromatic substitution competes favorably with the aldol reaction.
5. Nucleophilic acyl substitution competes favorably with the aldol reaction.
Answer
3. The molecule does not possess any hydrogens α to the carbonyl group.
Predict the aldol reaction product of the following ketone.
O OH
1.O OH
2.
O
HO
3.
O OH
4. OH
O
5.
O NaOH
Ethanol
Answer
2. O OH
Which starting material(s) will produce the following aldol reaction product?
1. 2.
3.
4.5.
Answer
2.
Select the correct aldol reaction product for the following reaction.
1. 2.
3.
4. 5.
Answer
4.
Which pair of compounds would be required to prepare the following aldol product?
H
O
+O
1.
H
O
+O
O2.
H
O
+
O3.
H
O
+O
4.H
O
+O
5.
O
Answer
3. H
O
+
O
Select the correct Claisen condensation product for the following reaction.
1. 2.
3.
4. 5.
Answer
4.
What type of reaction has occurred in the above biological process?
1. Claisen condensation2. aldol reaction3. nucleophilic acyl substitution4. β-elimination5. both Claisen condensation and
nucleophilic acyl substitution
Answer
2. aldol reaction
Select the best classification for the following molecule:
1. 1˚ aliphatic amine2. 2˚ aliphatic amine3. 3˚ aliphatic amine4. aromatic amine5. heterocyclic aromatic amine
Answer
2. 2˚ aliphatic amine
Determine the IUPAC name for the following molecule:
1. ethylpropylphenylamine2. N-phenyl-N-ethylpropanamine3. N-ethyl-N-phenylpropanamine4. N-ethyl-N-propylaniline5. N-phenyl-N-ethylaniline
Answer
4. N-ethyl-N-propylaniline
The major alkaloid present in tobacco leaves is nicotine, whose structure is shown below. Which will be the major ammonium ion formed when nicotine is treated with one equivalent of a strong acid?
1. 2.
N
N
CH3
nicotine
Answer
2.
Arrange the above from strongest to weakest base:
1. A, B, C2. B, C, A3. C, A, B4. A, C, B5. B, A, C
Answer
3. C, A, B
Amine A is more basic than amine B.
1. True2. False
Answer
1. True
Which of the following statements is true regarding the following two molecules?
1. Both A and B are aromatic.2. Both A and B are aliphatic amines.3. A is more basic than B.4. B is more basic than A.5. Both A and B are planar molecules.
Answer
4. B is more basic than A.
Select the most acidic compound from the choices provided.
1. 2.
3.
4. 5.
Answer
4.
If a protonated amine with a pKa of 10 is placed in a solution of pH 12, the predominant form of the amine in solution will be the protonated form.
1. True2. False
Answer
2. False