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36 Cards in this Set

  • Front
  • Back

Alkane to alkyne


2NaNH2


Heat

Dissolving metal of an alkyne yields

trans alkene (E)

Alkyne to cis (Z) alkene

H2


(Lindlers catalyst) Pd, CaCO3, Quinoline



Or



Ni2B(P-2)

Pg 13 test



C2H5OH delta

Alkyne to trans (E) alkene


Anti addition

(Li) + Base


(Na)


Disolving metal reduction

Alkene forms cyclopropane

CH2I2 , Zn(Cu)



Retains stereo chem

Alkene Br added to less substituted carbon

HBr , ROOR


Antimarkovnikov

Alkane with X (Br) elimination at less substituted C

t-BuOK , t-BuOH


Bulky base and Neutriphile


Hoffman product

Alkene I added to less sub and Cl added most subbed Carbon

ICl


Alkane Br sub for H (tert)

Br2 hv



Radical rxn

(CH3)2C=CH2 into cyclopropane (dichloro)

HCCl3 + (CH3)3COK

Radical

Intermediate with an unpaired electron

Syn addition

Pg9

Dehydration leaving group

H2O

Geometry of methyl radical

Trigonal planar

Stability of radiacal

3°>2°>1°>ch3

Electrophilic addition of hydrogen halides yields in RDS

Carbocation

Decomposition of diazomethane in heat or light yields

Methelene :CH2


simplest carbene

Rxn only bonds breaking



Only forming

Endothermic



Exothermic

Haloalcohol

Halohydride


Ffhjne

Show synthesis of product given reactant

Start with final product go backwards using to determine reagents used

Chain terminating step

1)No free radical formed


2)process is exothermic


3) energy of activation is 0


4) delta H is negative

Addition of hydrogen chloride to alkene stereo chem


Fill either P orbital


Both enantiomers

ICl type of addition

Markovnikov product


Regioselective

Br2 to alkene selectivity

Stereo slective

Alkene markovnikov addition of h2o without rearrangement

1)Hg(OAc)2 , THF-H2O


2) NaBH4, HO-

Q10 pg5

A

Alcohol that most rapidly dehydrates with H2SO4

Tert, alcohol

Carbocations

1) rearrange to be more stable


2) lose a proton to form an alkene


3) combine with a nucleophile


4) react with an alkene to form larger...

Heat (reflux) favors

E2


Zaytsev product

Elimination favors

1) high temp


2) tert-butoxide ion used


3) 3°(2°) alkyl halides with strong base


Sn1 rxn will be fallowed by

E1 Rxn



pg2

Ir bands 3300cm

Pg14

Elimination of R-X (Br)


Less substituted

NaOEt heat(^)

Cyclohexane to cyclohexene

Br2 hv (adds br)


RO- Heat(^) removes Br forms double bond

Add OH to terminal double bond


Antimarkovnikov

1) BH3/THF


2) H2O2 , -OH