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 contains only single bonds, so the suffix is “-ane.” Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other will result in side groups at positions 4 and 5. Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3,4– dimethylheptane. Note that the prefix “di-” is used to denote that two methyl side groups are present in this molecule. d) At first glance, this molecule looks like a 4 carbon ring, but the two –CH3 groups mean that they cannot be bonded to each other. Instead, this molecule is a 4–carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is 2,2–dimethylbutane.

15.21

An optically active compound will contain at least one chiral center, a carbon with four distinct groups bonded to it. a) This compound is a 6 carbon chain with a Br on the third carbon. 3–bromohexane is optically active because carbon #3 has four distinct groups bonded to it: 1) –Br, 2) –H, 3) –CH2CH3, 4) –CH2CH2CH3. chiral carbon
CH3

CH2

CH

CH2

CH2

CH3

Br
b) This compound is a 5 carbon chain with a Cl and a methyl (CH3) group on the third carbon.
3–chloro–3–methylpentane is not optically active because no carbon has four distinct groups. 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  CH3CHICH3 + KBr
b)
Cl
+

Cl2
Cl

15-2

c)
O
CH3

15.37

OH

C

+

CH3

Ni

H2

CH 3

CH

CH 3

a) The structures for chloroethane and methylethylamine are given below. The compound methylethylamine is more soluble due to its ability to form H-bonds with water. Recall that N–H, O–H, or F–H bonds are required for
H-bonding.
H
H
H
H
H
H

C

C

H

Cl

H

H

C

C

N

C

H

H

H

H

H

b) The compound 1–butanol is able to H-bond with itself (shown below) because it contains covalent oxygenhydrogen bonds in the molecule. Diethylether molecules contain no O–H covalent bonds and experience dipoledipole interactions instead of H-bonding as intermolecular forces. Therefore, 1–butanol has a higher melting point because H-bonds are stronger intermolecular forces than dipole-dipole attractions.
CH3

CH2

CH2

CH2

H

O

O

CH2

CH2

CH2

CH3

H hydrogen bond

H

H
H3C

C

O

H

C

CH3

H

diethyl ether
c) Propylamine has a higher boiling point because it contains N–H bonds necessary for hydrogen bonding.
Trimethylamine is a tertiary amine with no N–H bonds, and so its intermolecular forces are weaker.

CH3

CH2

CH2

N

H

CH3

H
Propylamine

N

CH3

CH3
Trimethylamine

15-3

15.45
(a)

O

(b)
CH3

CH

CH

CH2

Cl

OH

Alkene

CH2

C

OH

Alcohol
Carboxylic Acid
Aromatic Ring

Haloalkane
O

(c)

(d)

C

N

CH3

C

CH2

C

CH3

N
Ketone

O
Nitrile

H
Amide
Alkene

(e)

O
C

O

CH2

CH3

Ester

15.48

Aldehydes:
O

O
CH3

CH2

CH2

CH2

C

CH3

H

CH2

CH

C

H

CH3
O
CH3

CH

CH2

C

CH3
H

C

CH3

CH3

O
C

H

CH2

CH3

CH3

Ketones:
CH3

C

CH2

CH2

CH3

CH3

O

CH2

C
O

15-4

CH3

CH

O

15.49

C

CH3

CH3

First, draw all primary amines (formula R–NH2). Next, draw all secondary amines (formula R–NH–R'). There is only one possible tertiary amine structure (formula R3–N). Eight amines with the formula C4H11N exist.
CH3

CH2

CH2

CH2

CH3

CH

CH2

NH2

NH

NH2

CH3

CH3

CH2

CH

NH2

CH3
CH3

CH2

CH2

NH

CH3

CH3

CH3

CH

CH3

CH2

CH3

CH3

CH3
CH3

CH3

N

C

CH3

NH2

CH2

NH

CH2

CH3

CH3

15.51

a) This reaction is a dehydration-condensation reaction.
O
CH3

O

C

O

H

H

N

CH3

CH3

C

H

N

CH3

+

H 2O

H

H2O eliminated

b) An alcohol and a carboxylic acid undergo dehydration-condensation to form an ester.
O
CH3

CH2

CH2

C

CH3
O

H

H

O

CH
CH3

H2O Eliminated

O
CH 3

CH 2

CH 2

C

CH 3
O

CH
CH 3

15-5

+ H2O

c) This reaction is also an ester formation through dehydration-condensation.
CH3

O
H

C

O

H

H

O

CH2

CH

CH3
O

CH3

H 2O Eliminated
H

15.53

C

O

CH

CH3

To break an ester apart, break the –C–O– single bond and add water (–H and –OH) as shown.

O
R

C

O

Add -H
O

R

R'

C

OH

+

Bond Broken
Add -OH
(a)
O
CH3

(CH2)4

OH

+

HO

CH2

CH3

OH

+

HO

CH2

CH2

CH3

H

NH2

C

(b)

O
C

(c)

O
CH3

15.54

CH2

CH2

OH

+ HO

C

CH2

CH2

a)

O
H3C

CH2

C

OH

+

15-6

H

O

R'

+ H2O

b)
CH3
CH3

O

CH

C

+

OH

H

N

CH2

CH3

CH3

c)
O
H

OH +

C

H

N
H

15.56

a) Markovnikov’s rule ;치환기 수가 많은 탄소에 치환기가 붙는다. b)번 제외

H+, H2O
CH3

CH2

CH

CH2

CH3

CH2

CH

CH3

OH
Cr2O72-, H+
CH3

CH2

C

CH3

O
15.81

a) Functional groups in jasmolin II alkene CH2

CH
1
CH3

2
CH

CH3
O

3
CH

4
C

O

5
CH

C
6
C

C

O

CH3

C 11
CH2

O
CH3

C
7
O

ester

alkene ester ketone

b) Carbon atoms surrounded by four electron regions (4 single bonds) are sp3 hybridized. Carbon atoms surrounded by three electron regions (2 single bonds and 1 double bond) are sp2 hybridized.
Carbon 2 is sp3 hybridized.
Carbon 1 is sp2 hybridized.
3
Carbon 3 is sp hybridized.
Carbon 4 is sp2 hybridized.
3
Carbon 6 and 7 are sp2 hybridized.
Carbon 5 is sp hybridized.
c) Carbons 2, 3, and 5 are chiral centers as they are each bonded to four different groups.

15-7