Organic Chemistry

(Dana P.) #1
Section 3.6 How Alkenes React • Curved Arrows 121

If atoms have the same atomic number,
but different mass numbers, the one
with the greater mass number has the
higher priority.

In the following structures, for example, one of the carbons is bonded to a
deuterium (D) and a hydrogen (H):

D and H have the same atomic number, but D has a greater mass number, so D
has a higher priority than H. The C’s that are bonded to the other carbon are
bothbonded to C,C, and H, so you must go out to the next set of atoms to break
the tie. The second carbon of the group is bonded to H,H, and H,
whereas the second carbon of the group is bonded to H,H, and C.
Therefore, has a higher priority than

Notice in all these examples that you never count the atom attached to the bond
from which you originate. In differentiating between the and
groups in the last example, you do, however, count the atom attached to a bond from
which you originate. In the case of a triple bond, you count the atoms bonded to both
bonds from which you originate.

PROBLEM 9

Draw and label the Eand Zisomers for each of the following compounds:















PROBLEM 10

Draw the structure of -3-isopropyl-2-heptene.

3.6 How Alkenes React • Curved Arrows


There are many millions of organic compounds. If you had to memorize how each of
them reacts, studying organic chemistry would be a horrendous experience. Fortunate-
ly, organic compounds can be divided into families, and all the members of a family
react in similar ways. What determines the family an organic compound belongs to is
its functional group. The functional groupis a structural unit that acts as the center of
reactivity of a molecule. You will find a table of common functional groups inside the
back cover of this book. You are already familiar with the functional group of an
alkene: the carbon–carbon double bond. All compounds with a carbon–carbon double
bond react in similar ways, whether the compound is a small molecule like ethene or a
large molecule like cholesterol.

(Z)

HOCH 2 CH 2 CCCCH

O CHC(CH 3 ) 3

CH 3 CH 2 C CHCH 2 CH 3

Cl

CH 3 CH 2 C

CH 3 CH 2 CH 2 CH 2

CCH 2 Cl

CHCH 3

CH 3

CH 3 CH 2 CH CHCH 3

p

p

CH(CH 3 ) 2 CH“CH 2

s

CH“CH 2 CH(CH 3 ) 2.

CH“CH 2

CH(CH 3 ) 2

sp^2

CHCH 3
C C

H

CH 3

C C

H

CH 3

DCHCH (^2) D CHCH 3
CH CH 2
the Z isomer the E isomer
sp^2
Tutorial:
Eand ZNomenclature
BRUI03-109_140r4 24-03-2003 11:53 AM Page 121

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