Question

Characterize the following alkene as having the E or Z configuration. Draw the product(s) of bromination...

Characterize the following alkene as having the E or Z configuration. Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any). Use wedge-and-dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one.

Characterize the following alkene as having the E

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Answer #1
Concepts and reason

Alkenes involve in electrophilic addition relations. Where, alkene acts as nucleophile and electrophile attacks on, one of the carbon of a double bond. Draw the major organic product for the reaction between an alkene and bromine.

Fundamentals

If two high priority groups are on the same side of the double bond, then the compound is called as Z-alkene. On the other hand, if the two highest priority groups on the opposite side, then the compound is called as E-alkene.

Alkenes react with dibromine (Br2){\rm{(B}}{{\rm{r}}_{\rm{2}}}) to form 1,2-dibromo substituted compound through formation of a bromonium intermediate. Z-alkenes react with dibromine (Br2){\rm{(B}}{{\rm{r}}_{\rm{2}}}) gives enantiomers as products. E-alkene reacts with dibromine (Br2){\rm{(B}}{{\rm{r}}_{\rm{2}}}) gives meso compound.

(a)

The configuration at double bond is Z.

(b)

The formation of bromonium intermediate is,

Br:
CH3
*
HIM...
....1111CH3
=-*
+
Br— Br
dibromine
Br
H3C
H3C
(Z)-3-methylpent-2-ene
Bromonium intermediate

Stereochemistry to bromonium intermediate can be applied as follows:

In (Z)-3-methylpent-2-ene, the groups away from the observer are shown with dashed bond, and groups towards the observer are shown with wedge bond.

The formation of products from bromonium intermediate is as follows:

Br
:Br:
Hlm.
ICH3
Hmm...R
(R)
H,C
/
Н;С
(2R 3R)-2,3-dibromo-3-methylpentane
:Br:
Br?
н,C,
B-
H
111CH3
...
(B)
(S)
H,C
:Br
CH3

There is an equal probability of attacking the second bromide ion on the two carbon atoms in bromonium intermediate. Hence, the possibility of formation of the two products is equal. Hence, a racemic mixture will form.

Ans: Part a

The configuration of the alkene is Z.

Part b

The products bromination of given alkene are,

:Br:
H3C
S
)
(8)
H
: Br
HEC
(2R,3R)-2,3-dibromo-3-methylpentane
(2S,3S)-2,3-dibromo-3-methylpentane

The products of the reaction are characterized as, (R, R), (S, S) and racemic.

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