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| Getting CN as substituent? |
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WAY 1: NH2 + HONO and H2SO4 --> makes nitrile then CuCN
WAY 2: Add NaCN to a carbonyl, makes OH and CN
WAY 3: 1,4 addition |
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| Getting OMe as a substituent? |
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Get Cl on there, then do NaOMe and HOMe and heat
(this applies to getting OH on as well just use NaOH and H2O and heat) |
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| Removing NO2 from a ring? |
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1) H2/Ni
2)HONO/H2SO4
3)H3PO2 |
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| Getting OR as a substituent from OH? |
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| Carboxylic acid pops off as CO2 |
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Imidazole is more basic than pyridine
One of its N's is has a parallel lone pair and an H, like pyridine, but another has a perpendicular lone pair that is not involved in aromaticity |
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| add carboxylic acid and then H2O2, makes separation in charge, get NO2 on at C2 or C4, base restores aromaticity, and again separation in charge, then PCl3 gets rid of O on N and you have POCl3 and NO2 on pyridine |
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| NaNH2/ NaNH2 NH3 to the C2, then the H will drop down and restore aromaticity, it will pull off an H from NH3 and you will end with pyridine with NH2, H2, and NH2- |
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| Pyrole is more reactive to EAS as it is electron donating |
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Oxalic- 0 c's
Malonic- 1c
Succinic- 2c's
Gluatnic- 3c's
Adipic- 4c's
Pinelic- 5c's
Suberic- 6c's
Azelic- 7c's
Sebacic- 8c's
Oh! MSG! Ady Produced Super Amazing Seasalt! |
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R- right, S-left (do reverse of what it is in fischer b/c H is up)
D- OH on right, L- OH on left
alpha- OH is axial, beta- OH is equitorial |
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| Making hydroxy carboxylic acid |
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FROM CA:
Way 1:
PBr3/Br2 and 2 H2O
Way 2:
1) ROH, H+
2) OR- and aldol
3) H3O+
4) 1) NaBH4, 2) H20
FROM BETA KETO ALDEHYDE
Way 3:
1) CN-
2)H2SO4/H2O
3) One OH drops down to kick off NH3+
FROM ALDEHYDE
Way 4:
Aldol then Ag+ for selective oxidation of carbonyl |
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| Breaking a sugar with HIO4? |
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There is a cut in between every 2 oxidized carbons. Everything gets oxidized.
If one alcohol on the chain is cut on both sides, it will be oxidized twice (to a carboxylic acid). |
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| Oxidize the anomeric center only |
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| Oxidize both ends of the sugar chain |
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| Reduce the anomeric center |
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Breaks the ring
NaBH4 MeOH or H2O
ONLY if you have an alcohol at the anomeric position |
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| Make all of the alcohols on the sugar change |
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excess acetic anhydride
*these change them to CH3-(O)*
excess of CH3-(O)-Cl
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excess of CH3-(O)-O-(O)-CH3 |
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| Only change anomeric on cyclic and go back |
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ROH, H+
then Acid H2O
NOT BASE! NR WITH BASE! |
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| Change all but anomeric on cyclic |
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Do acid and EtOH
then add NaOH, MEI
OR
Do NaOH and MEI on all, anomeric is not stable in acid like this, so add acid and water, and anomeric will go back to OH |
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CN-, makes 2 enantiomers, then H2SO4/H2O
you are left with a carboxylic acid on top instead of aldehyde |
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H2N-OH
Ac2O
Base
leaves you with CN- (driving force), OAc- |
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