What IR absorption does a:

1. Primary Amine have?

2.  Secondary Amine have?

3.  Alcohol have?

4.  Tertiary Amine have?

1.  2 spikes at around 3200-3500

2.  1 small spike at arond 3200-3500

3.  1 broad peak in the same area

4.  No absorption

N-H shifts appear in the range of 1-4

Absorption decreases/disappears after shaking with D2O.  Alpha C-H are around 2-3 (increasing with more substitution).  Betas are smaller, 1.1-1.8.  Gammas are even smaller.

Alpha carbons show about 30-50.  Betas are about 27.  Gammas are about 11. 

(P 871)

Amine groups are strong activating groups.  Direct Ortho/Para.  Acidic reagents will protonate the NH2, leading it to be a deactivating group.

Draw the mechanism for Electrophilic Aromatic Substitution of Pyridine

(with sigma complexes)

(P892)

Pyridine is basically a strong deactivating group.  Directs meta, requires strong conditions to even work.

Draw the mechanism for Nucleophilic Aromatic Substitution of Pyridine.

(sigma complexes, too)

(P893)

Pyridine is activating towards stuff with a negative charge (in contrast to electrophilic sub., which is deactivating).  Negative charge is delocalized onto the Nitrogen.

Describe the differences between Exhaustive alkylation and Reacting with a large excess of starting material (ammonia).

(include a reaction for both)

Exhaustive - Excess reagent is used, and a tetraalkylammonium salt is formed.  Mild bases are added to deprotonate the intermediate amines and to neutralize the HX formed.

CH₃CH₂CH₂-NH₂ + 3CH₃-I =>(with NaHCO₃)

CH₃CH₂CH₂-N⁺(CH₃)₃   I^-

XS Ammonia - Yields primary amine

10NH₃ + R-CH₂-X =>

R-CH₂-NH3⁺   X^-

(P895)

Don't forget tetrahedral transition state, as well as deprotonation step at end.

Adding H₃O⁺ will get rid of the extra stuff on the NH by protonating it to NH₂.  Dilute HCl also does.

(P897)

No tetrahedral transition state like with Acid Chlorides.  R-NH₂ directly attacks the S, immediately having the X leave.  Gets deprotonated by NaOH.

Note:  Amine group on a benzene must be protected by an amide group (acyl chloride derivative).  If not, the NH₂ will attack the sulfur directly. 

Draw the mechanism for Hofmann Elimation of Amines

(hint: forms alkene)

(P899)

A quarternary ammonium iodide (product of exhaustive alkylation of ammonia) needs to be converted to hydroxide salt first (via Ag₂O and H₂O).  OH^- attacks a close carbon that will form the LEAST substituted, deprotonates it and forms an alkene, expelling the amine as a leaving group.  Heat is needed. 

Oxidation of Amines: The Cope Elimation

Draw the mechanism

(P903)

Amine is first oxidized by H₂O₂.  Cyclic transition state.  Occurs with syn stereochemistry.  LESS hindered Carbon is attacked.

Reactions of Amines with Nitrous Acid

(mechanism of primary, secondary amines)

(P904)

NaNO₂ is added to HCl to form HON=O (nitrous acid). 

Reaction with Primary amine yields R-N=N + Cl^- + 2H₂O + NaCl

Reaction with Secondary amine yields RR-N-N=O

Aliphatic (i.e. cyclohexanimine) diazonium ions rapidly decompose to carbocations

How do Arenediazonium Salts (benzene with N=N group) react with the following:

1.H₃O⁺, warm (generic acid)

2.CuCl(Br)

3.CuCN

4.HBF₄(KI)

5.H₃PO₂

6.H-Ar'

1.Ar-OH

2.Ar-Cl(Br)

3.Ar-CN

4.Ar-F(I)

5.Ar-H

6.Ar-N=N-Ar'

Reductive animation

(synthesis of amines)

(P913)

R-NH₂ + O=C-R'R' => R-N=CR'R' +LiAlH₄ =>

R-NH-CHR'R'

Primary: Ketone + H₂N-OH (with H+) => oxime + reduction => Primary amine (reduc done by H₂/Ni or LiAlH₄ [with water] or Zn/HCl)

Secondary: Same, but with H₂N-R

Tertiary: Same, but with R-NH-R' and the reducing agent used is Na(CH₃COO)₃BH in the presence of CH₃COOH

Acylation-Reduction

(synthesis of Amines)

(P915)

Amine + Acid Cl = Amide + LiAlH4 = alkylated amine

Direct Alkylation

(synthesis)

(P916)

Remember when you added excess ammonia?  yeah another one of the products that wasn't mentioned was ⁺NH₄    X^-

Kaboom, primary amine

(P917)

Phthalimide + KOH (with H₂O) = Phthalimide ion

Phthalimide ion + R-X = N-alkyl phthalimide

N.A.P. + H2N-N2H = Primary amine with R group

(P918)

R-X + NaNNN = R-N-N=N + H₂/Pd = R-NH₂

R-X + CN = R-C=N + H₂/Pd = R-CH₂-NH₂

(P920)

NO2 is reduced by:

H2/Catalyst

Sn,H2SO4

Zn,HCl

LiAlH4

Hofmann rearrangement of amides

Draw mech

(P921)

Amide in presence of strong base and Cl/Br will basically just cleave out the C=O

Important because the alkyl group that gets attached can be 1* or 2* or 3*.