Stereochemistry of the Addition of Bromine to trans-Cinnamic Acid
Introduction:
In this experiment, you will add bromine across the double bond of transcinnamic acid to form 2,3-dibromo-3-phenylpropanoic acid. The product of this
reaction contains two asymmetric centers, therefore there are four possible
stereoisomers that could possibly be formed in the reaction. The stereoisomers
that are actually formed in the reaction depend on the mechanism of the reaction.
O
CHBrCHBrCOOH
Br2
OH
trans-cinnamic acid
m.p. 136 oC; m.w. 148.2 g/mol
2,3-dibromo-3-phenylpropanoic acid
m.w. 308.0 g/mol
The stereoisomers that are actually formed in the reaction depend on the
mechanism of the reaction. Through analysis of your product mixture, you will be
able to draw conclusions about the mechanism of bromine addition to a double
bond.
The Fischer projections for the four possible reaction products are shown
below.
COOH
H
Br
H
Br
Ph
COOH
Br
H
Br
H
Ph
COOH
H
Br
Br
H
Ph
COOH
Br
H
H
Br
Ph
The first two structures, where the like groups are on the same side in the
Fischer projections, are known as the erythro isomers. The second two
structures, where like groups are on opposite sides in the Fischer projection, are
known as the threo isomers. The erythro-threo nomenclature is used to describe
configurations of compounds having two stereocenters but no plane of symmetry.
In this reaction, your product will be either a racemic mixture of the two erythro
enantiomers, or a racemic mixture of the two threo enantiomers. The relationship
between the erythro and threo isomers is that they are diastereomers. Since
diastereomers have different physical properties, you will be able to determine
which pair of enantiomers you have formed by taking the melting point of your
product.
The question you are trying answer with this experiment is whether the
addition of bromine across a double bond is a syn addition, or an anti addition.
Syn addition means that the two substituents have added to the same side of
the double bond, anti addition means that the two substituents have added to
opposite sides of the double bond.
Procedure:
To a medium test tube add 150 mg of trans-cinnamic acid and 0.6 mL of
glacial acetic acid. Place the test tube in a 50 oC water bath until all the transcinnamic acid is dissolved. Add 1.0 mL of a 1.0 M bromine in acetic acid solution
to the test tube. Add the solution dropwise over a period of not less than 10 min.
Failure to do so will usually result in a failed reaction. Continue heating at 50 oC
until the red-brown color of the bromine fades to light orange, then continue to
heat for 15 minutes more. If the mixture becomes colorless (or nearly so) during
this period, add more of the bromine solution dropwise until color just persists. If
the mixture has a distinct orange color at the end of the reaction period, add a
drop or two of cyclohexene to turn it light yellow.
Cool the reaction mixture in an ice-water bath for 10 minutes or more,
scratching the sides of the vial to induce crystallization if necessary. Collect the
product by vacuum filtration on a Buchner funnel and wash the solid with several
portions of ice-cold water, until the acetic acid odor is hardly noticeable.
Purify the product by recrystallization from 50% ethanol. Dry the product,
take the weight and measure the melting point to determine whether you have
the erythro (m.p. 204 oC) or threo isomer (m.p. 95 oC).
Safety Considerations:
• Take care when using the concentrated acid or the bromine solution to
not get it on your skin or clothing.
• There will be a waste container for organic liquid waste
• Dispose of used glass pipettes in the broken glass container
Data, Calculations, and Discussion:
• Obtain the mass of the 2,3-dibromo-3-phenylpropanoic acid product
and calculate theoretical and % yields.
• Record the melting point of the product and use it to determine which
pair of enantiomers, the erythro or the threo, you formed.
• From the isomer formed, decide whether the addition was a syn
addition or an anti addition. Be careful in deciding this as it may not
be as obvious as it seems from looking at the Fischer projections.
Remember that carbon-carbon single bonds are freely rotating, and
that Fischer projections show the stereoisomers in their eclipsed
conformations. You should redraw them as skeletal structures or
perspective formulas in order be to able to clearly see whether the two
bromines were added to the same side or opposite sides of the double
bond.
• Draw and name (using R and S designations) the two enantiomers that
you formed in the reaction.