Organic Chemistry

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
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