Answer Essay

1372 Words Nov 7th, 2015 6 Pages
CHAPTER 15 ORGANIC COMPOUNDS AND
THE ATOMIC PROPERTIES OF CARBON
15.15

a) Octane denotes an eight carbon alkane chain. A methyl group (–CH3) is located at the second and third carbon position from the left.
CH3
CH3

CH

CH

CH2

CH2

CH2

CH2

CH3

CH3

b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically, numbering starts at the top carbon atom of the ring for convenience. The ethyl group (–CH2CH3) is located at position 1 and the methyl group is located at position 3.
CH2

CH3

1
6

2

3
5

CH3

4

c) The longest continuous chain contains seven carbon atoms, so the root name is “hept-.” The molecule
…show more content…
The third carbon has three distinct groups: 1) –Cl, 2) –CH3, 3) two –CH2CH3 groups.

15-1

CH3
CH3

CH2

C

CH2

CH3

Cl

c) This compound is a 4 carbon chain with Br atoms on the first and second carbon atoms and a methyl group on the second carbon. 1,2–dibromo–2–methylbutane is optically active because the 2nd carbon is chiral, bonded to the four groups: 1) –CH2Br, 2) –CH3, 3) –Br, 4) –CH2CH3.
CH3
CH2

15.25

C

Br

chiral carbon
CH2

CH3

Br

a) The structure of propene is CH2=CH–CH3. The first carbon that is involved in the double bond is bonded to two of the same type of group, hydrogen. Geometric isomers will not occur in this case.
b) The structure of 3–hexene is CH3CH2CH=CHCH2CH3. Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur.
CH3

CH2

CH2
C

CH3

CH3

CH2
C

C

H

H

H

C

H

cis-3-hexene

CH2

CH3

trans-3-hexene

c) The structure of 1,1–dichloroethene is CCl2=CH2. Both carbons in the double bond are bonded to two identical groups, so no geometric isomers occur.
d) The structure of 1,2–dichloroethene is CHCl=CHCl. Each carbon in the double bond is bonded to two distinct groups, so geometric isomers do exist.
Cl

Cl
C

H

Cl

C

C
H

H

cis-1,2-dichloroethene

15.35

H
C
Cl

trans-1,2-dichlorethene

a) CH3CHBrCH3 + KI

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