Brominationas a tool for structure

determinationExperimental:

• Dissolve the sample in DCM (1 mL)• Add dropwise a 1% bromine solution of DCM• Positive test, if red color disappears

Conclusion:

The sample contains a multiple bond, which undergoes bromination

Hypothesis:

The sample is an alkeneif the red color

disappears

Br2Br

Br

Br2 consumedred color disappears

Sample 2 also an

alkene

H3CO

OCH3

O

O

O O

CH3H3C OO

dimethyl fumarate dimethyl maleate

O O

CH3H3C OO Br2

red color persists

O O

CH3H3C OO Br2

red color persists

No Reaction?

Why?

Let’s take a look at the mechanism of alkene bromination

Electrophilic Additon

Br BrBr

Electrophilic Additon

Alkene is the Nucleophile

The stronger the nucleophile, the faster is

the reactionAn electron-rich alkene

makes a stronger nucleophile,

What makes an electron rich alkene?

O O

CH3H3C OO

Is a maleate an electron rich alkene?

O O

CH3H3C OO

Let’s take a look at the substituents

O

O

Electron-Withdrawing• Reduces the electron

density of the double bond

• Makes the double bond less nucleophilic

Br Br

O

does not happen

Maleate is not electron-rich enough

(even though it is an alkene)

Conclusion:“False” Negative

What if we “helped” maleate to go over the activation energy

barrier?by heating it

Drawbacks:We might evaporate the whole

solventO O

CH3H3C OO

Br Br

expected product

a White Solid

Sample 3 not an alkene

OCH3

anisole

an aromatic compound

aromatic compounds do not undergo electrophilic addition reactions

BUT OCH3

Br2

red color disappears

OCH3

Br2

Is anisole undergoing an electrophilic addition?

Or could it be an aromatic electrophilic substitution (EAS)?(the first step of which is an electrophilic addition anyway)

Can this happen in the absence of a Lewis Acid?

Check the Substituent

OCH3 Methoxy group,• a strong

activatorwill enable anisole undergo bromination (EAS)in the absence of a Lewis Acid

OCH3

Br Br

OCH3H

Br

BrOCH3

Br

+ HBr

can HBr be detected?

Conclusion:“False” Negative

OCH3

Br2

OCH3

Br

+ HBr

OCH3

Br

+

red color disappears

Further Conclusions (if HBr is detected):Sample is an aromatic compoundwith a strongly activating

substituent