Organic Chemistry Ch. 19 of Wade (sec. 1-18, amines)
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- R-Br + ammonia (excess)
-
R-NH2
(ammonia is the nucleophile in an SN2 mechanism) -
acid chloride + amine?
R-CO-Cl + R'-NH2 --> ? -
amide
R-CO-NHR'
Don't forget to deprotonate the N+ with another amine. -
Carboxylic acid to an acid chloride?
R-CO-OH -(?)-> R-CO-Cl - SOCl2 (Thionyl chloride)
-
amine + sulfonyl chloride?
R'-NH2 + R-SOO-Cl --> ? -
sulfonamide
R-SOO-NHR' -
amide -(?)-> amine + c. acid
R2N-CO-R' -> R2N-H + R'COOH - H3O+ and heat (hydrolysis)
- R-NH2 + NaNO2 + 2HCl-->
-
diazonium salt
R-N2+Cl- - Ar-N2+ to Ar-OH
- H3O+
- Ar-N2+ to Ar-Br
- CuBr
- Ar-N2+ to Ar-Cl
- CuCl
- Ar-N2+ to Ar-CN
- CuCN
- Ar-N2+ to Ar-F
- HBF4
- Ar-N2+ to Ar-I
- KI
- Ar-N2+ to Ar-H
- H3PO2
-
Ar-N2+ to Ar-N=N-Ar'
(diazo coupling) -
H-Ar' (Ar' is activated)
ex. phenol - Which is more basic, pyridine or piperidine?
-
Piperidine.
The sp3 electrons of piperidine are held more losely than the sp2 e- of pyridine. (1/4th vs. 1/3rd s-character) - Which is more basic, pyrrole or pyridine?
- Pyridine. The electrons in pyrrole are involved in the aromatic ring and are unavailable to be donated.
- E2 elimination of a quarternary ammmonium hydroxide will give the ____ ___________ alkene.
-
least substituted / terminal
(the Hofmann product) - The reagents to convert a tertiary amine into an amine oxide for a Cope elimination.
-
1) Peroxide
(H2O2 or MCPBA or R-COOOH)
2) heat - Steriochemistry for a Cope elimination.
- Syn. The cyclic transition state puts the N+ and H in the same plane on the same side.
- Steriochemistry for a Hofman elimination.
- Anti. The anti-coplanar arrangement of the N+ and H is required for E2.
- Amine + HCl -->
- ammonium salt (now soluble in the aqueous layer)