Reactions Flashcards
(27 cards)
Basicity of Amines
Pka ~ 10
W/ Aromatic ring ~ 6 or less so not very basic
2nd Activating group on ring makes it more basic up to 6 pka
2nd deactivating aka withdrawing groups make it weaker Base 1-4 pka
Amide
Carboxylic Acid with a NH2 instead of OH
Basic? Why or why not?
Not the lone pair resonates with double bond O and lone pairs there
Reduction of Nitriles
Reagents?
R- C triple bond N
- LiAlH4, ether
- H2O
->
R-CH2-NH2
Reduction of
Amide
Carboxylic Acid with a NH2 instead of OH
- LiAlH4, ether
- H2O
R-CH2- NH2
NaBH4 won’t reduce
Benzene- NO2 to NH2
- H2 Pd
Benz- NH2
Or
- SnCl2, H3O+
- NaOH
Benz- NH2
NH3 + R-X
- SN2
—> RN+H3 X- - NaOH
—-> RNH2 primary amine
Will keep reacting with more R-X - SN2
—> RN+H3 X- - NaOH
—-> R2NH 2ndary amine
Will react and make R3N which will react with R-X again and make a quaternary salt
ammonium
R4N+ X-
Always keeps going
In reality make trace of salt and and R-N
Most makes RNH2 and RNH
Synthesis of RNH2 using Azide method
R-X
- NaN3, ethanol
—-> R-NNN
- LiAlH4, ether
- H2O or H3O
—-> RNH2
Reductive Amination
Aldehyde or Ketone to Amine
Reagents?
- NH3 or R-NH2 or R2NH and acid
- H2/Ni or NaBH4 or LiAlH4
—> NH2 replace carbonyl double bond O
- Protonate Carbonyl O with acid H3O
- NH3 nuc attack C attached to O
- Proton shift from N to O
- Lone pair from N moves in and forms double bond and H2O leaves
- H2O take H from NH2 leaving lone pair behind
- H- from reduction agent attacks carbon and double bond jumps up to make 2 lone pairs on N
- One lone pair steals H from H3O forming a C-NH2 bond in place of Carbonyl O
Amino acid coupling
DCC structure
R-N:=C=:N-R
1. N: deprotonates the -OH on C term Carbox acid
2. O- that was deprotonated nuc attacks DCC carbon forming bond to DCC
3. N: term of amino acid attacks C term carbonyl carbon making carbonyl double bond electrons jump up to O making it O-
4. O electrons jump down and reform double bond kicking out O- DCC
5 Randon B: deprotonates the N leaving lone pair behind
Amino acids protonation state
Protonated if pH is below pKa
Deprotonated if pH is above pKa
At 7.3 pH which amino acid are protonated?
Lysine and arginine are protonated and positively charge
His is usually unprotonated since pKa is 6 also can coordinate metals
Tyrosine OH pKa is 10 so protonated and Ser OH pKa is super high so always protonated yet neutral charge
Glu and Asp are 4.2 and 3.9 so unprotonated and negatively charged
Edman Degradation
Removes peptide from chain from the N term end
Uses Acid and PITC
C6H5-N=C=S
1. N: from N term attacks C =S electrons from double bond takes H from acid
2. S lone pair attacks carbonyl C term C that is still attached to peptide chain forming ring while double bond carbonyl electrons take another H from acid forming OH
3 Lone pair from that O moves back down in and reforms double bond breaking the bond from that C to the N term of peptide chain causing the ring to leave
4. ATZ rearranges ring to make PTH
FMoc protecting group
Protects N terminus
Added with Fluoromethyloxycarbonyl-chloride aka FMoc and (CH3CH2)3N
Taken off by base
Boc protecting group
Amino acid plus reagents
Di-tert-butyl dicarbonate
(CH3CH2)3N
Add to N term and is removed by strong acid
such as CF3CO2H
- Peptide N term lone pair nuc attacks a carbonyl carbon on Boc, double bond breaks and electrons jump up to O making O-
- Electrons come back down and ester bond to other half of Boc molecule breaks leaving
- (CH3CH2)3N: deprotonates the amino acid N terminus
Making Boc-NH-amino acid
Removing a Boc protecting group?
Acid
- Lone pair on ether bond on Boc takes H from acid leaving positive charge O
- Bond from O to the tert-butyl group on Boc breaks leaving lone pair with O making a carbon cation which forms an alkene biproduct
- Cl- from earlier takes H off the ether O making a double bond to carbonyl carbon and the electron from carbonyl C to N from amino acid break bond and grab and H from acid to make NH2
R-NH2 amino acid unprotected
CO2
And (CH3)2CCH2 byproduct
Protecting Carbox Acid end of amino acid?
Method 1
Alcohol/CH3OH and HCl to add CH3 to O from OH on Carbox acid making an ester
- Protonate carbonyl C
2 O on alcohol nuc attacks carbonyl C breaking double bond kicking electrons up to O - Proton transfer from alcohol to Existing OH
4 the top carbonyl carbon is deprotonated causing double bond to reform kicking out water as leaving group
Benz alcohol adds similar
To remove
Method 1 use saponification
- NaOH
- H3O+
Remove OCH3 group and replace with OH from base
- Base OH- nuc attacks carbonyl carbon, double bond breaks and electrons jump up to O making it -
- Electrons come back down and kick out OCH3 group but OCH3 group often deprotonates the OH so need a acid step to protonate both
Or use acid 1 protonate carbonyl O 2 water nuc attacks breaks double bond 3 Water deprotonates carbonyl O reforming double bond and and kicking out RO ester group Leaving Carbox acid
Or remove Benz alcohol w/ H2/Pd hydrogenation
Merrifield peptide synthesis
Nuc addition C end to polymer fix benzene-CH2-Cl
Remove protection from N term
Add DCC and protected N term of next peptide
Wash to remove protection and excess
Repeat
Citrate Synthase
Acetyl CoA
“A reaction waterfall”
Base deprotonates the C next to carbonyl C causing electrons to form double bond between the 2 carbons
Which make double bond to O to break and those electrons to grab H from acid side chain which will take a proton from another side chain leaving an exposed base B:
Which B: takes his proton back causing acid to take their proton back reforming carbonyl O=C double bond breaking the C=C double bond whose electrons attack the carbonyl C on Oxaloacetate breaking the C=O carbonyl double bond which grabs a proton from a his side chain which grabs a proton from an acid
H2O nuc attacks thioester bond on Acetyl CoA breaking it leaving citrate and HSCoA
Sugar ring plus Acid Anhydride and Pyridine
(CH3CO)2O
What does it do?
Adds to CH3CO to all alcohols on sugar making ethers
OH is deprotonated then that O- nuc attacks one of the carbonyl carbons on anhydride
Breaks double bond which jump up to O making it O-, when the electrons come back down reform the double bond half of the anhydride with ether O leaves O- COCH3
Sugar ring plus alkyl halide R-X and AgO
Adds R to all alcohols making ethers
Base deprotonates OH’s on sugar then O- attacks C attached to halide causing halide to leave
The hemi acetal aka R-O-C-OH when treat with alcohol R-O-H and acid makes?
An acetal
The reaction involves the R then corner O- anomeric carbon - OH which is a hemi acetal.
The RO from the alcohol adds to anomeric carbon and OH that is there is protonated by acid and leaves as water.
Mixed alpha beta product
Reducing Sugars?
Aldoses only
Ketoses don’t reduce things
Reduction of monosaccharides
What reagent reduces CHO end to CH2OH
NaBH4 and H2O
Had to be in open form
Reduces aldehyde carbonyl end only
What oxidizes aldehyde carbonyl end to carboxylic acid?
Br2, H2O
pH of 6
Other reagents will oxidize every OH on sugar ring
