18 Chapter 2

18 CHAPTER 2

IDENTIFY WHICH RESONANCE STRUCTURES BELOW ARE SIGNIFICANT AND WHICH ARE INSIGNIFICANT

OOH OOH OOH OOH

SkillBuilder 2.9 Identifying Localized and Delocalized Lone Pairs

IDENTIFY WHETHER THE LONE PAIR ON THE IDENTIFY THE HYBRIDIZATION STATE NITROGEN ATOM BELOW IS DELOCALIZED OF THE NITROGEN ATOM

O O

NH2 NH2

Solutions

2.1. H H HH H C H HCH H H H H C HH H H C H H C HCCCOH H C C C O H HCCO H H C H H H H C H CC H C H C H HCH H H H H a) H b) H H c ) H H

H H H H H H C H H H H H C H C H C H H H H C C O CCH CC H H CC H C H C H C H H H H H C H C C H H C H d) H H e) H H f) H H

H H H H C H H H C C C C OH H H H H H H H H H H H C H H H C C C C CCH H C C C OCH H H g) H h ) H H H H H H i) H H H H

H H H C H H C H H H H C H C H HCOH H C H H H H H C HCCOC H H C H H C H H C H H HCCOH H C H H C H H C H j) H H k) H H l ) H H CHAPTER 2 19

(CH ) COCH (CH ) CHOCH CH 2.2 3 3 3 and 3 2 2 3

2.3 Six

2.4 H2C=CHCH3

2.5. H H H H H H C H CC H H C H C H H H H C H HH HH H C H H C C H H C N C H C C C C C H H CCH H H C H HH O HH a) HH b ) c) H H

CH3 O HH O H H H3C C CH3 C C C H C C H C C H C C OH C C H H H C C CH H C C H 3 3 HHCC C H H d) CH3 e) H H f) HH

OH H O H H C C H H C H H H C C H C H C C C C C C C C H C H H H C H C C C H H H H H g) HH h) H i ) H H

H H HH HH O H H C CC O 2 HH C H C C C C H C C C C H2C C CH2 H C HH C C H H H H C C j) HH k) H H l) H2 H2

2.6 a) decrease (7!6) b) no change (8!8) c) no change (8!8) d) increase (5!7)

2.7 a) increase (12!14) b) decrease (8!6)

2.8. OH

O OH

a) b) c) OH d) 20 CHAPTER 2

OH OH

e) f) g) h) O

NH2 O i) j) O k) l)

Cl m) n) o ) p) O

OH O O q) r)

2.9.

2.10.

O O Br δδδ+ Cl a) b) δδδ+ c) δδδ+ δδδ+ O O δδδ+ H δδδ+

2.11. ester amide OO NH2 ether N O N N O amide aromatic HO O H amine H OH O amine aromatic alcohol carboxylic acid

2.12.

N N a) b) No charge c) d) No charge

CHAPTER 2 21

2.13.

O O H a) b) c) No charge d) O H H

2.14. O O O O O OH O O a) b) c) d) e)

H O H O O H H O O H H O O O R f) g) h) i) j)

2.15. There are no hydrogen atoms attached to the central carbon atom. The carbon atom has four valence electron. Two valence electrons are being used to form bonds, and the remaining two electrons are a lone pair. This carbon atom is using the appropriate number of valence electrons.

2.16.

N N N a) b) H c) d) no lone pairs

H NH2 N N N e) f) g) no lone pairs h)

2.17. H O

N O CN OON a) b) c)

O O O N O CN d) e) NH2 f)

2.18. a) one b) zero c) one d) five 22 CHAPTER 2

2.19 Five lone pairs: O R O NH3

2.20 a) HO

O O N O O N N H N H N H S N S S O O O O O O Troglitazone Rosiglitazone Pioglitazone b) Yes, it contains the likely pharmacophore highlighted above.

2.21 a) Violates second rule by giving a fifth bond to a nitrogen atom. b) Does not violate either rule. c) Violates second rule by giving five bonds to a carbon atom. d) Violates second rule by giving three bonds and two lone pairs to an oxygen atom. e) Violates second rule by giving five bonds to a carbon atom. f) Violates second rule by giving five bonds to a carbon atom. g) Violates second rule by giving five bonds to a carbon atom, and violates second rule by breaking a single bond. h) Violates second rule by giving five bonds to a carbon atom, and violates second rule by breaking a single bond. i) Does not violate either rule. j) Does not violate either rule. k) Violates second rule by giving five bonds to a carbon atom. l) Violates second rule by giving five bonds to a carbon atom.

2.22.

O

2.23.

a) b) CHAPTER 2 23

O O N N O O c) d)

OO OO OH OH N N

e) f)

O O O O N N OO

g) h)

2.24.

OO OO

a) b)

O O

N N c) b)

2.25.

O O

O O

a) b)

O O NH2 NH2

c) d)

24 CHAPTER 2

O O O O O O N N

e) f)

O O N N O O g)

O O

H2N OH H2N OH O O h)

2.26.

a) b)

c)

d)

CHAPTER 2 25

2.27.

NN NN a) b)

OO

c)

2.28.

N N O O

a) b)

O O

c)

2.29.

OH OH

O O

2.30.

O O

26 CHAPTER 2

2.31.

HO OH HO OH

H2N H2N

2.32.

O OO

a)

b)

c)

O O O

O O O

d)

OO

e)

H H H H H O O O O O

f)

CHAPTER 2 27

NN

g)

N NN

h)

NN

i)

O O O

Cl Cl Cl

j)

2.33.

H H H N N N

a)

N NN

b)

NN

c)

OCN OCN OCN d) 28 CHAPTER 2

O O S S e)

f)

OO O

g)

h)

OO

i)

N N C C

j)

N N N C C C

k)

OH OH

l) CHAPTER 2 29

2.34. O δδδ+ δδδ+ δδδ+ δδδ+ δδδ+ δδδ+

2.35. δδδ- OH

δδδ- δδδ- δδδ- δδδ-

2.36. localized O sp2 hybridized geometry not relevant O N (connected to only one atom) H N H H N delocalized H One of these lone pairs is sp2 hybridized delocalized. The oxygen trigonal planar delocalized localized atom is therefore sp2 sp2 hybridized sp3 hybridized hybridized and has bent a) trigonal planar trigonal pyramidal b) geometry.

localized One of these lone pairs is sp2 hybridized delocalized. The oxygen OOH 2 geometry not relevant atom is therefore sp hybridized and has bent (connected to only one atom) H2N NH2 geometry.

localized delocalized sp3 hybridized 2 sp hybridized trigonal pyramidal c) trigonal planar

localized sp2 hybridized N d) bent

delocalized N sp2 hybridized H e) trigonal planar

30 CHAPTER 2

f) localized 2 sp hybridized O geometry not relevant O (connected to only one atom) O

localized One of these lone pairs is sp3 hybridized delocalized. The oxygen bent atom is sp2 hybridized and has bent geometry.

2.37. Both lone pairs are localized and, therefore, both are expected to be reactive.

2.38. localized O (not participating in resonance) localized NH (not participating in resonance) N 2 N H

delocalized localized (participating in resonance) (not participating in resonance)

2.39.

H H H OOH HH H C HH H C N C HH O C H C C C H H C O C C N C C C H H N C C C O C H HO C C C C C O H N H C H H O N H C H H H 2.40.

2.41.

2.42. HO O O HO OH HO OH Vitamin A Vitamin C CHAPTER 2 31

2.43. Twelve (each oxygen atom has two lone pairs)

2.44.

O NO N O N

2.45.

O OO

O O

2.46. a)

C4H10 C6H14 C8H18 C12H26

In each of the compounds above, the number of hydrogen atoms is equal to two times the number of carbon atoms, plus two. b)

C4H8 C7H14 C7H14 C12H24

In each of the compounds above, the number of hydrogen atoms is two times the number of carbon atoms. 32 CHAPTER 2

c)

C6H10 C9H16 C9H16 C7H12

In each of the compounds above, the number of hydrogen atoms is two times the number carbon atoms, minus two.

d) A compound with molecular formula C24H48 must have either one double bond or one ring. It cannot have a triple bond, but it may have a double bond.

e)

2.47. a) an sp2 hybridized atomic orbital b) a p orbital c) a p orbital

2.48.

OH OH OH H H H N N N

a)

O OO b)

c)

2.49. a) (CH3)3CCH2CH2CH(CH3)2 b) (CH3)2CHCH2CH2CH2OH c) CH3CH2CH=C(CH2CH3)2

CHAPTER 2 33

2.50. a) C9H20 b) C6H14O c) C8H16

2.51. (d) is not a valid resonance structure, because it violates the octet rule. The nitrogen atom has five bonds in this drawing, which is not possible, because the nitrogen atom only has four orbitals with which it can form bonds.

2.52. 15 carbon atoms and 18 hydrogen atoms:

HHHH HH C C HH C C H C C C C HH H C C C H C C H H C C H HH

2.53.

O O a) b)NN c) d) N

2.54. Cl Cl Cl Cl

2.55.

OO O O

a) b)

NN c) d)

34 CHAPTER 2

e)

H H H H HHH H H H N N N N N

f)

g)

OOOO

O O O

O O O

O O O h)

CHAPTER 2 35

i)

OH OH OH

OH

j)

O O O

O O

2.56. These structures do not differ in their connectivity of atoms. They differ only in the placement of electrons, and are therefore resonance structures.

2.57. a) constitutional isomers b) same compound c) different compounds that are not isomeric d) constitutional isomers

2.58. OH O a) b) c)

Br O d) O e) f)

36 CHAPTER 2

2.59. The nitronium ion does not have any significant resonance structures because any attempts to draw a resonance structure will either 1) exceed an octet for the nitrogen atom or 2) generate a nitrogen atom with less than an octet of electrons, or 3) generate a structure with three charges. The first of these would not be a valid resonance structure, and the latter two would not give significant resonance structures.

2.60.

O O O OOO

2.61. Both nitrogen atoms are sp2 hybridized and trigonal planar, because in each case, the lone pair participates in resonance.

2.62.

OH OH OH

H H H

H H H H H H HO HO HO

OH OH

H H

H H H H HO HO

OH OH OH

H H H

H H H H H H O O O

CHAPTER 2 37

2.63. a) The molecular formula is C3H6N2O2 b) There are two sp3 hybridized carbon atoms c) There is one sp2 hybridized carbon atom d) There are no sp hybridized carbon atoms e) There are six lone pairs (each nitrogen atom has one lone pair and each oxygen atom has two lone pairs) f) O localized delocalized HNH N 2 localized O localized

g) not relevant (only connected to one other atom) trigonal planar O trigonal planar

HNH N 2 bent O trigonal pyramidal

tetrahedral tetrahedral

h)

O O O

HNH HNH HNH N 2 N 2 N 2 O O O

2.64. a) The molecular formula is C16H21NO2 b) There are nine sp3 hybridized carbon atoms c) There is seven sp2 hybridized carbon atoms d) There are no sp hybridized carbon atoms e) There are five lone pairs (the nitrogen atom has one lone pair and each oxygen atom has two lone pairs) f) The lone pairs on the oxygen of the C=O bond are localized. One of the lone pairs on the other oxygen atom is delocalized. The lone pair on the nitrogen atom is delocalized. g) All sp2 hybridized carbon atoms are trigonal planar. All sp3 hybridized carbon atoms are tetrahedral. The nitrogen atom is trigonal planar. The oxygen atom of the C=O bond does not have a geometry because it is connected to only one other atom, and the other oxygen atom has bent geometry.

38 CHAPTER 2

2.65.

O OH δδδ+ δδδ+ δδδ+ H δδδ- δδδ- H

O OH O OH O OH O OH

H H H H H H H H

O OH O OH O OH

H H H H H H

2.66. a) Compound B has one additional resonance structure that Compound A lacks, because of the relative positions of the two groups on the aromatic ring. Specifically, Compound B has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge:

O O

O O

Compound B

Compound A does not have a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge. That is, Compound A has fewer resonance structures than Compound B. Accordingly, Compound B has greater resonance stabilization. b) Compound C is expected to have resonance stabilization similar to that of Compound B, because Compound C also has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge:

O O OO

Compound C

CHAPTER 2 39

2.67. The single bond mentioned in this problem has some double bond character, as a result of resonance:

Each of the carbon atoms of this single bond uses an atomic p orbital to form a conduit (as described in Section 2.7):

H C H CH3 C C H H

Rotation about this single bond will destroy the overlap of the p orbitals, thereby destroying the resonance stabilization. This single bond therefore exhibits a large barrier to rotation.