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9. Addition of X2 (X = Cl, Br) – ANTI addition

X X2

X So: Let’s look at bond connections first: Step 1: the e- rich approaches the X-X bond, causing the X-X to polarize, creating an for the nucleophilic alkene to react with:

+ δ δ- X-X X

In the next step, the halide attacks:

X X X X

Now – go back and add in stereochemistry… No, seriously – go back and add in the wedges/dashes…

The Side View: top face blocked!

Br Br Br Br Br H H H H H H Br

10. Formation – Addn of OH, X to alkene Markovnikov, ANTI Ex. X X2 H O 2 OH Alternatively, if you substitute an alcohol for the and make an ether instead: Ex. X X2

CH3OH OCH3

1 Draw the Products: Ex.

Br2 H O 2

Ex.

Cl2 H O 2

Ex.

Br2

CH3OH

Mechanism:

OH Br2 H O 2 Br

Top View: H H O O Br Br H H Br Br

Br

H O

Br

Side View: Look at that ANTI addition occurring…

2 Br Br Br Br H-O-H H CH H 3 CH3 H CH3 O H H

Br

Br

H CH3 O H So – Why “ANTI”?

One face is blocked by halonium , just like in the addition of just X2:

Br H CH 3 .

And Now - Why “Markovnikov”?

First, recall than any positive charge, full or partial, is more stable on a more substituted carbon. Markovnikov means the reaction proceeds through the most stable intermediate, full or partial charge. Which end is more substituted?

Br

H CH3

The more substituted end of the halonium ion breaks the C-X bond more easily because that carbon atom forms a partial positive charge that would be more stable (lower in energy, lower activation barrier), and thus the addition of water occurs there, faster, just like in oxymercuration, with a mercurinium ion. Br

H CH3

Nuc The more substituted side would form a more stable partial charge. More stable intermediate = Markovnikov

And now the last reaction type: 11. Oxidative Cleavage of – requires the breaking of both the pi AND sigma bonds of alkenes, thus fracturing the carbon skeleton into pieces.

3

R1 R3 R1 R3 O + O R R R2 R4 2 4

There are two sets of reagents you must know how to work with: a. Ozonolysis – (with reductive work-up) -Conversion of alkenes to aldehydes and/or ketones

-Uses (O3) – formed from in an ozonator e- O2 O3

O O O O O O

Reagent: 1. O3 + 2. Zn, H3O

Aldehydes and Ketones form with ozonolysis:

R1 H R1 H O + O R R R2 R4 2 4

R1 H R1 H O + O R H R2 H 2

Draw the products (BOTH): Ex.

1. O3 2. Zn, H O+ 3

Ex.

1. O3 2. Zn, H O+ 3

4 Ex.

1. O3 + 2. Zn, H3O

+ b. KMnO4, H2O or H3O

• Conversion of alkenes to ketones/carboxylic acids/CO2 (H2CO3) • pH dependent reaction – if the reaction pH becomes too HIGH, dihydroxylation occurs!

General Equations:

R1 R3 R1 R3 O + O R R R2 R4 2 4

R1 R3 R1 R3 O + O R2 H R2 H

R3 O OH

R1 H R1 H O + O R2 H R2 H

OH O OH

Draw the products: Ex.

KMnO4, H2O

5 Ex.

+ KMnO4, H3O

Ex.

+ KMnO4, H3O

Now: Both of them…

Ex.

1. O3 2. Zn, H O+ 3

Ex.

+ KMnO4, H3O

Ex.

KMnO4,NaOH, H2O

Finally…What about:

1. O3 + 2. Zn, H3O

KMnO4, H2O

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