Organic Chemistry

Section 1.4 Representation of Structure 17

Kekulé structure Condensed structures

Two or more identical groups considered bonded to the “first”atom on the left can be shown (in parentheses) to the left of that

atom, or hanging from the atom.

An oxygen doubly bonded to a carbon can be shown hanging off the carbon or to the right of the carbon.

O

CH 3 CH 2 CCH 3 or CH 3 CH 2 COCH 3 or CH 3 CH 2 C( O)CH 3

O

CH 3 CH 2 CH 2 CH or CH 3 CH 2 CH 2 CHO or CH 3 CH 2 CH 2 CH O

O

CH 3 CH 2 COH or CH 3 CH 2 CO 2 H or CH 3 CH 2 COOH

O

CH 3 CH 2 COCH 3 or CH 3 CH 2 CO 2 CH 3 or CH 3 CH 2 COOCH 3

HHC

H

(CH 3 ) 2 NCH 2 CH 2 CH 3 or CH 3 NCH 2 CH 2 CH 3

CH 3

C

H H

H

C

H

HH HH

H

C

C

H

N

HHC (CH 3 ) 2 CHCH 2 CH 2 CH 3 or CH 3 CHCH 2 CH 2 CH 3

CH 3

H

HH HH

H

C

C

H H

H

C

H

H

C

H

C

PROBLEM 10 SOLVED

Draw the Lewis structure for each of the following:

a. d. g. j. NaOH

b. e. h. k.

c. f. i. l.

SOLUTION TO 10a The only way we can arrange one N and three O’s and avoid

single bonds is to place the three O’s around the N. The total number of valence

electrons is 23 (5 for N, and 6 for each of the three O’s). Because the species has one neg-

ative charge, we must add 1 to the number of valence electrons, for a total of 24. We then

use the 24 electrons to form bonds and fill octets with lone-pair electrons.

When all 24 electrons have been assigned, we see that N does not have a complete octet. We

complete N’s octet by using one of oxygen’s lone pairs to from a double bond. (It doesn’t

make any difference which oxygen atom we choose.) When we check each atom to see

whether it has a formal charge, we find that two of the O’s are negatively charged and the N

is positively charged, for an overall charge of

O

N+

O

−

O

−

• 
  
  
  
  1. 
     
  
  
  
  

O

N

O O

O¬O

NO 2 - N 2 - CH 3 Na 2 CO 3

NO 2 + HCO 3 - +C 2 H 5 NH 4 Cl

NO 3 - CO 2 CH 3 NH 3 +

Table 1.5 (continued)