Reaction Reagents 6

Question 1

Alkene + HX (HBr/HCl)

Answer 1

Pi bond attack H, X- leaves and attacks carbocation

Carbocation rearrangement

Enantiomers (racemic mixture) formed

Question 2

Alkene + H2O/[H2SO4] (hydration reaction)

Answer 2

H2O + dilute H2SO4 (acid catalyst) = H3O+

Pi bond attacks H on H3O+ making carbocation

Carbocation rearrangement

Water attacks carbocation (either side might may enantiomers)

Water acts as base and removes one H on the O from water

Alcohol product

Question 3

Alcohol + conc. H2SO4

Answer 3

Synthesize Alkene (opposite of hydration)

Question 4

Alkene + 1) BH3 • THF, 2) H2O2, NaOH (hydroboration oxidation)

Answer 4

1) Alkene attacks B on BH3

Makes carbocation

H on BH3 attacks carbocation

(B on least subbed C)

2) BH2 subbed for OH

Alcohol formation!

Stereoselective: H and OH are syn (both enantiomers form if one or two chiral centres on Cs that were in C=C)

Question 5

Alkene + h2/M (Pt) - (hydrogenation)

Answer 5

Stereospecific syn addition of H atom on each C part of C=C with metal catalyst

Metal surface binds H2, Alkene attaches to metal, H2 added syn

Question 6

Alkene + Cl2/Br2 (halogenation)

Answer 6

Stereoselective - only anti (both enantiomers form depending on starting Alkene - cis = both enantiomers, trans = meso)

Br2/Cl2 polarized bond

Alkene nucleophilic attack at one Br/Cl, other Br/Cl is LG

Bromonium/chloronium ion (3 membered ring)

LG Br-/Cl- attacks anti at one C in Sn2 rxn

Question 7

Alkene + Br2/Cl2 + H2O (halohydrin formation)

Answer 7

Pi bond attacks Br2/Cl2 in polarized bond

Bromonium/chloronium ion forms (3 membered ring)

Br-/Cl- LG

Water attacks most subbed C in Bromonium/cloronium ion anti to original Br/Cl

More water acts as base to remove one H on O

Making alcohol

Both enantiomers Made depending on starting compound

??OH and Br want to be anti??

Question 8

Alkene + OsO4 (catalytic) + NMO/alkyl peroxide

Answer 8

Syn dihydroxylation

Adds Oh and OH across pi bond in syn, concerted way

Pi bond attacks one O on OsO4, electrons pass to Os, adjacent O attacks other C on pi bond

Ester formation and 5 membered ring

Add naSO3/H2O to make diol

Question 9

Alkene + KMnO4 + NaOH

Answer 9

Alkene attacks one O on MnO4, electrons transfer to Mn, adjacent O attacks other C on C=C bond

Five membered ring formed

NaOH removes MnO2

Syn-dihydroxylation = diol forms

Question 10

Alkene + 1) peroxyacid (RCO3H), 2) H3O+

Answer 10

1. Pi bond attacks O of peroxyacid (RCO3H) to make epoxide
2. O of epoxide attacks H of H3O+, water attacks one C in epoxide, another water deprotonates to make OH

Anti-dihydroxylation (both enantiomers) - trans diol

Question 11

Alkene + 1) O3, 2) DMS

Answer 11

Pi bond turns into 2 double bonded C=O

Question 12

Benzene + Br2

Answer 12

No reaction