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21 Cards in this Set

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R-Br + ammonia (excess)
R-NH2

(ammonia is the nucleophile in an SN2 mechanism)
An alkylation of an amine by an alkyl halide
acid chloride + amine?
R-CO-Cl + R'-NH2 --> ?
amide
R-CO-NHR'

Don't forget to deprotonate the N+ with another amine.
The amine is a nucleophile attacking the reactive (partial positive) carbonyl.
Carboxylic acid to an acid chloride?

R-CO-OH -(?)-> R-CO-Cl
SOCl2 (Thionyl chloride)
Thionyl....
amine + sulfonyl chloride?
R'-NH2 + R-SOO-Cl --> ?
sulfonamide
R-SOO-NHR'
SO works like CO
amide -(?)-> amine + c. acid
R2N-CO-R' -> R2N-H + R'COOH
H3O+ and heat (hydrolysis)
Activate the carbonyl, attack with water, lose the protonated amine.
R-NH2 + NaNO2 + 2HCl-->
diazonium salt
R-N2+Cl-
Makes an excellent l.g.
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)