March 3oth , 2011 Goals for the day: Naming Amines Physical Properties of Amines Amine reactions * Conversion of an alkyl halide * Gabriel Synthesis * Hoffman Elimination 4) IR/NMR of Amines 5) Multiple Choice Questions…
Chapter 24 Amines NH3 ammonia methyl amine isopropyl methyl amine Name 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
Degree of substitution… Substitution on the nitrogen is important … The number of carbons directly attached to the nitrogen is what You want to look at Primary secondary tertiary
24.2 Properties of Amines Bonding to N is similar to that in ammonia N is sp3-hybridized C–N–C bond angles are close to 109° tetrahedral value
Amines Form H-Bonds Amines with fewer than five carbons are water-soluble Primary and secondary amines form hydrogen bonds, increasing their boiling points
24.3 Basicity of Amines 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
Amides 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=
SN2 Reactions of Alkyl Halides Ammonia and other amines are good nucleophiles
Problem: Uncontrolled Multiple Alkylation (i. e Problem: Uncontrolled Multiple Alkylation (i.e. can’t stop the reaction! ). Primary, secondary, and tertiary amines all have similar reactivity, the initially formed monoalkylated substance undergoes further reaction to yield a mixture of products
Gabriel Synthesis of Primary Amines 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
Hofmann Elimination Converts amines into alkenes NH2 is very a poor leaving group so it is converted to an alkylammonium ion, which is a good leaving group
The Elimination Step Exchanges hydroxide ion for iodide ion in the quaternary ammonium salt, thus providing the base necessary to cause elimination
Orientation in Hofmann Elimination 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
Steric Effects Control the Orientation
Polycyclic Heterocycles
24.10 Spectroscopy of Amines -Infrared Infrared Spectroscopy: Characteristic N–H stretching absorptions 3300 to 3500 cm1 Amine absorption bands are sharper and less intense than hydroxyl bands Protonated amines show an ammonium band in the range 2200 to 3000 cm1
Examples of Infrared Spectra
NMR Hydrogens 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], PCC Primary alcohols can be oxidized to carboxylic acids Reaction of CO2 with a Grignard reagent ***
Reactions that Carboxylic Acids do… React with bases… *** Makes acid halides (do the same reactions as acids, but are faster)… Reduces to a primary alcohol (goes through an aldehyde)…
Two main Carboxylic Acid Reactions (need to be acid or base catalyzed)… Esterification…*** Important in fats Amide formation…*** Important in proteins (amide bonds) Note: You will need to know all of the mechanisms, but the good news is that they are very similar! (We will review them with the acid derivatives…)
Reactions of Acid Derivatives… Ester and amide formation using an acid halide
More reactions of Acid Derivatives… X = halogen, OR, or NHR Hydrolysis *** (very important for both esters and amides) Reduction using LAH (gives an amine if starting with an amide) Goes through a ketone ***
Summary of the reactions thus far… *** ***
Next three (Aldol, acetoacetate synthesis & claisen)… ***
Chain lengthening of amines (from an alkyl halide)… *** Gabriel Synthesis… Hoffman Elimination…
Which of the following is not an acyl derivative? 1. 2. 3. 4. 5. 1 2 3 4 5
Answer 5.
What is the IUPAC name for the following carboxylic acid? 3-bromo-4-methylbenzoic acid 4-methyl-3-bromobenzoic acid phthalic acid 4-carboxy-2-bromotoluene 5-carboxy-2-methyl-1-bromobenzene
Answer 1. 3-bromo-4-methylbenzoic acid
What is the IUPAC name for the following carboxylic acid? 2-propylhexanoic acid 4-propylhexanoic acid 2-butylpentanoic acid 4-carboxyoctane 5-carboxyoctane
Answer 1. 2-propylhexanoic acid
The structure for oleic acid is shown below The structure for oleic acid is shown below. What is the IUPAC name of this compound? a. (Z)-octadec-9-enoic acid b. (E)- octadec-9-enoic acid
Answer a. (Z)-octadec-9-enoic acid
Rank the following molecules in acidity from least acidic to most acidic. A, C, B B, A, C C, B, A B, C, A C, A, B
Answer 2. B, A, C
Chloroacetic acid is a stronger acid than acetic acid Chloroacetic acid is a stronger acid than acetic acid. Which is the best explanation? More resonance structures can be drawn for chloroacetic acid than for acetic acid. More resonance structures can be drawn for chloroacetate ion than for acetate ion. Because of its high electronegativity, chlorine is able to donate electrons to the chloroacetate ion by the inductive effect, thereby stabilizing this ion. Because of its high electronegativity, chlorine is able to withdraw electrons from the chloroacetate ion by the inductive effect, thereby stabilizing this ion. Chlorine is larger than hydrogen and can better hold a negative charge.
Answer 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. True False
Answer 1. True
Which of the following is not an ester? 1. 2. 4. 3. 1 2 3 4
Answer 2.
Determine the IUPAC name for the following molecule. 3,4-dimethylphenyl pentanoate 3,4-dimethylpentyl benzenoate 3,4-dimethylpentyl benzoate benzyl 3,4-dimethylpentanoate phenyl 3,4-dimethylpentanoate
Answer 5. phenyl 3,4-dimethylpentanoate
Which of the statements is true concerning the following two carboxylic acid derivatives? Only molecule A can be hydrolyzed. Only molecule B can be hydrolyzed. Both molecules can be hydrolyzed, but A will react faster than B. Both molecules can be hydrolyzed, but B will react faster than A. A and B can be hydrolyzed at roughly the same rate.
Answer Both molecules can be hydrolyzed, but B will react faster than A.
Predict the outcome of the following reaction. 1. 2. 3. 4. 5. 1 2 3 4 5
Answer 3.
Which of the following steps occurs first in the mechanism of Fischer esterification? attack of the nucleophile on the carbonyl carbon protonation of the carbonyl oxygen loss of water from the tetrahedral carbonyl addition intermediate protonation of the alcohol oxygen formation of the tetrahedral carbonyl addition intermediate
Answer 2. protonation of the carbonyl oxygen
Which of the following is the tetrahedral intermediate that appears in the Fischer esterification of ethanol and benzoic acid? 1. 2. 3. 4. 5. 1 2 3 4 5
Answer 1.
Predict the product of the following reaction. 1. 2. 3. 4. 5. None of these 1 2 3 4 5
Answer 1.
What is the predominant enol form of the following molecule? 1. 2. 3. 5. None of these is favored over the others. 4. 1 2 3 4 5
Answer 4.
Which of the following will occur when the following optically active compound is placed in dilute acid? It will form an acetal. It will form a diol. It will become an alcohol. It will lose its optical activity. none of these
Answer 4. It will lose its optical activity.
Identify the expected major product of the following reaction. 2. 1. 3. 4. 5. 1 2 3 4 5
Answer 3.
How many protons with a pKa < 22 exist in the following molecule? 1 2 3 4
Answer 5. 4
Which of the following reagents will not form an enolate upon reaction with a ketone? 2. 1. 3. 5. 4. LiAlH4 Choice One Choice Two Choice Three Choice Four Choice Five
Answer 4. LiAlH4
Which of the above carbonyl compounds is most acidic?
Answer 3. C
Which of the following statements explains why the following aldehyde will not undergo an aldol reaction with itself? The benzene ring makes the carbonyl group unreactive towards aldol reactions. A carbonyl group must be connected to two alkyl groups in order to undergo an aldol reaction. The molecule does not possess any hydrogens α to the carbonyl group. Electrophilic aromatic substitution competes favorably with the aldol reaction. Nucleophilic acyl substitution competes favorably with the aldol reaction.
Answer The molecule does not possess any hydrogens α to the carbonyl group.
Predict the aldol reaction product of the following ketone. 2. 1. 3. 4. 5. 1 2 3 4 5
Answer 2.
Which starting material(s) will produce the following aldol reaction product? 1. 2. 3. 5. 4. 1 2 3 4 5
Answer 2.
Select the correct aldol reaction product for the following reaction. 1. 2. 3. 5. 4. 1 2 3 4 5
Answer 4.
Which pair of compounds would be required to prepare the following aldol product? 1. 2. 3. 4. 5. 1 2 3 4 5
Answer 3.
Select the correct Claisen condensation product for the following reaction. 1. 2. 3. 4. 5. 1 2 3 4 5
Answer 4.
What type of reaction has occurred in the above biological process? Claisen condensation aldol reaction nucleophilic acyl substitution β-elimination both Claisen condensation and nucleophilic acyl substitution
Answer 2. aldol reaction
Select the best classification for the following molecule: 1˚ aliphatic amine 2˚ aliphatic amine 3˚ aliphatic amine aromatic amine heterocyclic aromatic amine
Answer 2. 2˚ aliphatic amine
Determine the IUPAC name for the following molecule: ethylpropylphenylamine N-phenyl-N-ethylpropanamine N-ethyl-N-phenylpropanamine N-ethyl-N-propylaniline 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. 1 2
Answer 2.
Arrange the above from strongest to weakest base: A, B, C B, C, A C, A, B A, C, B B, A, C
Answer 3. C, A, B
Amine A is more basic than amine B. True False
Answer 1. True
Which of the following statements is true regarding the following two molecules? Both A and B are aromatic. Both A and B are aliphatic amines. A is more basic than B. B is more basic than A. 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. 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. True False
Answer 2. False