Exam 4 Flashcards
Nomencalture
_ always have the double bond in the parent chain
Zusammen (Z)
- Cis
- Priorities are on the same side of the double bond
Entgegen (E)
- Trans
- Priorities are on opposite sides of the double bond
Synthesis of Alkenes
E1, E2
H2, Pd/c, CH3OH
Alkyne to Alkane
H2, Lindlars catalyst
Alkyne to a cis alkene
Na, NH3
Alkyne to trans alkene
H-X
Mark addition of a hydrogen and a halogen
1. H added to the less substituted carbon
2. Hydrid shift available for Br to attack
H-X, H2O2
anti mark addition of a hydrogen and a Br
X2
trans addition of two halogens
1. one br makes a bromonium ion
2. second br attacks below to release the first br up
X2, ROH
Trans and anti-mark (Br) addition of a halogen and a polar protic molecule
1. one br makes a bromonium ion
2. the polar protic molecule attacts the more substituted carbon releasing the Br
3. the second Br removes the extra hydrogen
H2SO4, H2O
mark addition of a hydrogen and a polar protic molecule (OH)
1. Acid donates a hydrogen
2. hydride or methyl shift
3. the polar protic solvent attacks the more substituted side
4. acid comes back to take off the extra hydrogen
- Hg(OAC)2
- NaBH3
Mark addition of a hydrogen and an OH group
- BH3, THF
- H2O2, NaOH
anti-mark addition of a hydrogen and an OH group
- CH2N2, Light
- CH2I2, Zn, CuCl
makes a Carbene (:CH2)
HCX3, NaOH
makes a carbene with halogens (:CX2)
R-N3, light
makes a nitrene
mCPBA
Makes an epoxide
OsO4, H2O2
syn addition of two OH groups
KMnO4 (Cold), H2O
Syn addition of two OH groups
- mCPBA
- H2SO4, H2O
Trans addition of two Oh groups
1. acid adds a hydrogen to the epoxide
2. H2O attacks the bottom releasing the epoxide upwards
3. acid comes back to take the extra OH
- KMnO4 (Warm/Hot)
- H3O+
Cleaves double and triple bonds, replacing them with a double bonded oxygen adn any lone hydrogens become an OH
- O3
- (CH3)2-S
Cleaves double bonds and replaces it with a double bonded oxygen, Leaves lone Hydrogens alone
- NaNH2
- R-Br
Makes the Alkyne chain longer
1. :NH2 removes the end hydrogen leaving the electrons
2. the electrons attach to the R group and kicks the Br off
KOH (2 equivalents)
Alkane to alkyne by removing hydrogens and halogens
- NaNH2 (xs)
- H3O+
able to move the triple bond from the center of the carbon to the end
KOH (xs)
able to move the triple bond to the center of the molecule
Alkyne: H-X
Mark and syn addition of a hydrogen and a halogen
Alkyne: X2
trans addition of two halogens
Alkyne: HgSO4, H2SO4, H2O
makes a ketone:
1. acid makes the mark product
2. a hydrogen from the acid attacks the remaining double bond
the charge makes a double bond with the OH through resonance
3. the leftover acid takes the extra hydrogen back
- (Sia)2BH
- H2O2, NaOH
Alkyne to antimark Aldehyde
- O3
- H2O
Cleves the triple bond leaving a double bonded oxygen and an OH group
