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21 Cards in this Set
- Front
- Back
- 3rd side (hint)
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R-Br + ammonia (excess)
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R-NH2
(ammonia is the nucleophile in an SN2 mechanism) |
An alkylation of an amine by an alkyl halide
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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.
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Carboxylic acid to an acid chloride?
R-CO-OH -(?)-> R-CO-Cl |
SOCl2 (Thionyl chloride)
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Thionyl....
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amine + sulfonyl chloride?
R'-NH2 + R-SOO-Cl --> ? |
sulfonamide
R-SOO-NHR' |
SO works like CO
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amide -(?)-> amine + c. acid
R2N-CO-R' -> R2N-H + R'COOH |
H3O+ and heat (hydrolysis)
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Activate the carbonyl, attack with water, lose the protonated amine.
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R-NH2 + NaNO2 + 2HCl-->
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diazonium salt
R-N2+Cl- |
Makes an excellent l.g.
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Ar-N2+ to Ar-OH
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H3O+
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Ar-N2+ to Ar-Br
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CuBr
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Ar-N2+ to Ar-Cl
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CuCl
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Ar-N2+ to Ar-CN
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CuCN
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Ar-N2+ to Ar-F
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HBF4
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Ar-N2+ to Ar-I
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KI
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Ar-N2+ to Ar-H
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H3PO2
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Ar-N2+ to Ar-N=N-Ar'
(diazo coupling) |
H-Ar' (Ar' is activated)
ex. phenol |
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Which is more basic, pyridine or piperidine?
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Piperidine.
The sp3 electrons of piperidine are held more losely than the sp2 e- of pyridine. (1/4th vs. 1/3rd s-character) |
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Which is more basic, pyrrole or pyridine?
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Pyridine. The electrons in pyrrole are involved in the aromatic ring and are unavailable to be donated.
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E2 elimination of a quarternary ammmonium hydroxide will give the ____ ___________ alkene.
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least substituted / terminal
(the Hofmann product) |
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The reagents to convert a tertiary amine into an amine oxide for a Cope elimination.
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1) Peroxide
(H2O2 or MCPBA or R-COOOH) 2) heat |
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Steriochemistry for a Cope elimination.
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Syn. The cyclic transition state puts the N+ and H in the same plane on the same side.
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Steriochemistry for a Hofman elimination.
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Anti. The anti-coplanar arrangement of the N+ and H is required for E2.
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Amine + HCl -->
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ammonium salt (now soluble in the aqueous layer)
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