Predict Major Product
Antimarkvonikov addition of Br
H-Br ------ ROOR
Replaces OH with Br
PBr3
Syn addition of D (Same thing as hydrogen just need to show it)
D2 (Deutirum) ---- Pd-C
GRIGNARD REAGENT: Only need on carbon group with Grignard reagent because the reaction occurs twice to kick out the OEt group or the Cl group Followed by H3O+
Form a ketone with two carbon chains from acid chloride or ester group
More electronegative atom adds to more substituted carbon...Markovnikov (Br adds to more substituted carbon)
H-Br
Cleave an ether--> R-X and R-X
H-Br or H-I (2 equivalents)
2 halogens are added (markovnikov is followed). Alkane is formed
H-X (2 Equivalents)
Alkene is reduced to alkane
H2 ----- Pd-c
Cis Alkene
H2 H2 ---- or ------- Lindlars Catalyst Pd/BaSO4
Alkene to alkane
H2/Pd/C
Addition of alcohol group-->Tautomerize to ketone
H20 ------ H2SO4
Results in a carboxylic acid
H2CrO4
Alkene
H2SO4 or TsOH
Forms a Water on more substituted carbon
H30+
Replace OH with Br and create a racemic mixture
HBr Secondary Tertiary Alcohol
Add Cl to one end of double bond
HCl
Add O (and carbon chain) to more substituted carbon participating in C-C double bond
Hg(OAc)2 / CH3CH2OH
Adds alcohol markovnikov (more substituted carbon)
Hg(OAc)2/H20 ----------------- NaBH4
Adds OH markovnikov (more substituted carbon)
Hg(OAc)2/H2O ------------------ NaBH4
1.) Alkyne reduces to enol on most substituted carbon 2.) Tautomerization to ketone
HgSO4, H20 --------------- H2SO4
Becomes ketone but creates a racemic mixture
HgSo4, H20 --------------- H2SO4
Divide double bond in half If carbon is secondary--->Ketone If carbon is primary------>OH
KMnO4 --------- H30 +
Two OH groups form cis (Cis diol)
KMnO4 --------- OH
Results in a Ketone
KMnO4 ---------- H3O+
Reduces everything to an OH group (very strong)
LiAlH4
Epoxide formation
MCPBA
Two OH groups form trans --> racemic mixture
MCPBA ---------- H30 +
Keep double bond and add Br allelic (not alpha carbon next one)
NBS
Trans Alkene
Na or Li ----------- NH3
Reduces ketone or aldehyde ONLY to OH
NaBH4
Use acetyl protecting group (HO ---/ OH) looking structure. Then LiAlH4 to reduce acid, then H30+ and heat to remove the protecting group
Need to reduce Carboxylic Acid but not ketone
GRIGNARD REAGENT: Determine the carbon chain and attach MgBr to it. Then add H3O+ to protonate the oxygen anion
Need to reduce ketone and add carbonyl chain
Divide double bond in half If carbon is secondary---> Ketone If carbon is primary-------> Aldehyde
O3
Mixture of Carboxylic acids Terminal Alkyne Carboxylic acid on substituted end Carbon dioxide on terminal carbon
O3 KMnO4 ------ or ------------ OR H30+
Add two OH groups on either side of double bond
OsO4 -------- H2O2
Replace OH with Br and invert sterochem
PBr3
Results in a ketone
PCC
Results in an aldehyde
PCC
No Reaction
PCC or H2CrO4 On a Tertiary Alcohol
1.) Alkyne reduces to enol on less substituted carbon 2.) Tautomerization to aldehyde
R2BH/THF ------------- H2O2, OH
LiAlH4
Reduce alkyl halides or epoxides
Adds Cl group and inverts sterochem
SOCl2
1.) H2/Pdc-->to alkane 2.)H2/Lindlars-->Cis alkene 3.)Na/NH3-->Trans alkene
What are the three reagents to reduce an alkyne
4 Halogens are added. Alkane is result
X-X (2 Equivalents)
Epoxide appearing structure but with C as connecting point
Zn ---- CH2Cl2
Opens epoxide ring
1.) Nu 2.) H20
Adds OH to most substituted carbon
1.) H2O 2.)H2SO4
Add Alkyl group to alkene
1.) NaNH2 (Creates carboanion) 2.)R-Br
Aldehyde
1.) R2BH 2.) H2O2, HO-
Add the OCH3 group in replace of OH
1.) TsCl, Pyr 2.)NaOCH3
One alcohol adds and a hydrogen adds to an end of the double bond
1.)BH3 2.)H2O2
RCO3H
Alkene to epoxide
RCO3H H2O
Alkene to two groups with OH
Adds OH antimarkovnikov
BH3 ------ THF
Adds alcohol antimarkovnikov (less substituted carbon)
BH3/THF ----------- H202/ OH
More electronegative atom (Cl) goes on more substituted carbon
Br-Cl ------- CH2Cl2
Adds 2 Br on side of double bond
Br2
Reduce alkyne to alkene. Add Br on each side of double bond
Br2 ----- CH2Cl2
Add OH to more substituted carbon Add Br to less substituted carbon
Br2 ----- H2O
Forms an epoxide
Br2 + H2O ------------- NaOH
2 Bromines (or Cl) on either side of double bond TRANS
Br2 Cl2 ----- OR ------- CH2Cl2 CH2Cl2
Adds the O (and the chain attached) to the more substituted carbon Can rearrange
CH3OH --------- H
Looks like an epoxide with carbon in middle instead of O. Then two Cl groups coming off of it
CHCl3 --------- KOH
KI (I=N) NaCN (CN=N) NaN3 (N3=N) NaSH (SH=N)
Common nucleophiles undergoing SN2
Alcohol to Carboxylic acid
CrO3