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