CHAPTER 3

An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Representation of Structure

1. a. n-propyl alcohol or propyl alcohol b. dimethyl ether c. z-propylamine or propylamine

Notice that each carbon forms four bonds and each hydrogen and bromine forms one bond.CH3I

CH3CCH3IBr

2.

CH3CH2CH2CH2Br

n-butyl bromideor

butyl bromide

3. a. CH3CHOHICH3

b. CH3CHCH2CH2FICH3

CH3CHCH2CH'II .Br

CH:,!CHCH2Br. ICH3

see-butyl bromide isobutyl bromide tert-butyl bromide

CH3I .c. CH3CH20CH2CH2CH3 e. CH3CNH2

ICH3

d. CH3CH2CHIICH3

f. CH3CH2CH2CH2CH2CH2CH2CH2Br

23

24 Chapter 3

CH'j CH3I - I7. a. #1 CH3CH2CH2CH2CH2CH2CH2CH3 #9 CH3CHCH2CH2CHCH3

octane 2, 5-dimethylhexane

CH'j CH3 CH3I - I - I -#2 CH3CHCH2CH2CH2CH2CH3 #10 CH3CH2CH - CHCH2CH3

2-methylheptane 3,4-dimethylhexane

CH3 CH'j CH:~I - I - I -#3 CH3CH2CHCH2CH2CH2CH3 #11 CH3C-CHCH2CH3

ICH3

3-methylheptane 2,2,3-trimethylpentane

CH'j CH3 CH1I - I 1-#4 CH3CH2CH2CHCH2CH2CH3 #12 CH'jCCH2CHCH3- I

CH3

4-methylheptane 2,2,4-trimethylpentane

CH3 CH'j CH3I - I - I -#5 CH3CCH2CH2CH2CH'j #13 CH3CH-CCH2CH3- I - I

CH3 CH3

2,2-dimethylhexane 2,3,3-trimethylpentane

CH'j CH3 CH3 CH3I - I I - I -#6 CH3CH2CCH2CH2CH'j #14 CH3CH-CH-CHCH3I -

CH3

3,3-dimethy Ihexane 2,3,4-trimethylpentane

CH3 CH3 CH3 CH1I - I - I - I -#7 CH3CH-CHCH2CH2CH3 #15 CH3C-CCH'jI I -

CH3 CH3

2,3-dirnethy lhexane 2,2,3,3-tetrarnethylbutane

CH1 CH3I - I .#8 CH3CHCH2CHCH2CH3 #16 CH3CH2CHCH2CH2CH3I -

CH2CH3

2,4-dimethylhexane 3-ethy lhexane

9.

10.

11.

12.

CH3I

#17 CH3CH2CHCHCH3I -CH2CH3

3-ethy 1-2-methy lpentane

Chapter 3 25

CH"I -#18 CH1CH2CCH2CH3- I -

CH2CH3·3-ethyl- 3-methy lpentane

b. The systematic names are under the compound.

c. #2,#7,#8,#9,#12,#13,#14,#17

d. #3,#8,#10,#11

~ #5,#11,#12,#15

8. a. 2,2,4-trimethylhexane

b. 2,2-dimethylbutane

c. 3,3-diethylhexane

d. 2,5-dimethylheptane

e. 4-isopropyloctane

f. 4-ethyl-2,2,3-trimethylhexane

{3IN ~2

OH

a.~OH

b.~

menthol = CIOH200

a. 1-ethyl-2-methylcyclopentane

b. ethylcyclobutane

c.·~

Br

d··~O/

terpin hydrate = CIOH2002

c. 3,6-dimethyldecane

d. 5-isopropylnonane

b. Both compounds have the same name (1-iodo-2-methylcyclohexane), so they are the samecompound.

26 Chapter 3

13. a. Both compounds have the same name (l-bromo-3-methylhexane), so they are the samecompound.

14. a. see-butyl chloride c. cyclohexyl bromide2-chlorobutane bromocyclohexane

b. isohexyl chloride d. isopropyl fluoridel-chloro-4-methylpentane 2-fluoropropane

15. a tertiary alkyl bromide b. a tertiary alcohol . .a. c. a pnmary amme

16. a. methy lpropy lamine c. diethylaminesecondary secondary

h. trimethylamine d. buty ldimethy laminetertiary tertiary

17. a. eSel c. &CI &CICH3

Cl

18. a. The bond angle is predicted to be similar to the bond angle in water (~104.5°).

b. The bond angle is predicted to be similar to the bond angle in ammonia (~107.3°).

c. The bond angle is predicted to be similar to the bond angle in water (~l 04.5°).

19. Room temperature is about 25°C. Therefore, pentane, with a boiling point of 36 °C is thesmallest alkane that is a liquid at room temperature.

20. a. An 0- H hydrogen bond is longer.

b. Because it is shorter, an 0 - H covalent bond is stronger.

Chapter 3 27

21. a. 1,4, and 5

b. 1,2,4,5, and 6

OH OH OH NH2

22. HO~OH > ~OH > ~ > ~ >

~> ~

23. a. cH3cH2cH2cH2cH2CH2Br > CH3CH2CH2CH2CH2Br > CH3CH2CH2CH2Br

b. CH3CH2CH2CH2CH2CH2CH2CH2CH3 > CH3CH2CH2CH2CH2CH2CH2CH3 >

CH3CHCH2CH2CH2CH2CH". 1 .

CH3

H3C CH3·1 I·> CH3C-CCH3. I I .

H3C CH3

c. CH3CH2CH2CH2CH20H > CH3CH2CH2CH20H > CH3CH2CH2CH2CI > CH3CH2CH2CH2CH3

24. Figure 3.2 in the text shows that even number alkanes have higher melting points than oddnumber alkanes. Therefore, since more energy is required to disrupt the crystal lattice of evennumbered alkanes, we can conclude that even number alkanes pack more tightly.

25. a. HOCH2CH2CH20H > CH3CH2CH20H >CH3CH2CH2CH20H > CH3CH2CH2CH2CI

b.0 OH 0 The amine is more soluble

6 than the alcohol in water,because the amine has two

> > hydrogens that can formhydrogen bonds with waterand the alcohol has only one.

26. Because cyclohexane is a nonpolar compound, it will have the lowest solubility in the most polarsolvent, which, of the solvents given, is ethanol.

CH3CH2CH2CH2CH20Hl-pentanol

CH3CH20CH2CH3diethyl ether .

CH3CH20Hethanol

CH3CH2CH2CH2CH2CH3hexane

28 Chapter 3

27. Hexethal would be expected to be the more effective sedative because it is more nonpolar thanbarbital since hexethal has a hexyl group in place of the ethyl group of barbital. Being morenonpolar, hexethal will be better able to penetrate the nonpolar membrane of a cell.

28.

29.

30.

31.

32.

33.

a.

b. yes c. yes

a. H~H

HyHCH2CH:~

15 4 CH

CH~C 23

3

2 HH HH

b.

Cl axialc0-CI equatorial

At anyone time, there will be more molecules of isopropylcyc1ohexane with the substituent inthe equatorial position because the isopropyl substituent is larger than the ethyl substituent.Because the isopropyl substituent is larger, the axial confomer of isopropylcyclohexane has moresteric strain than the axial conformer of ethylcyclohexane, so the isopropyl group will have agreater preference for the equatorial position.

a. CIS b. C1S c. trans d. trans

a. ~CH2CH3

CH3

c. trans-l-Ethyl-2-methylcyclohexane is more stable because both substituents are in equatorialpositions, whereas cis-l-ethyl-2-methylcyclohexane has one substituent in the equatorialposition and the other in the axial position.

Chapter 3 29

34. Both Kekule and skeletal structures are shown.

~OX

CH31 -

CHCH31

f. CH3CH2CH2CHCHCH2CH2CH2CH3- 1

CHCH31

CH3

CH31

a. CH,CH?CHOCCH3- -I 1CH3 CH3

.Cf CH3

e.CH3

Bf• 1

J. CH3CHCH2CH2CCH2CH2CH3- 1 1 -

CH3 Bf

Cf Br Br

30 Chapter 3

35. a. 5-bromo-2-methyloctane d. 3,3-diethylpentane

b. 2,2,6-trimethylheptane e. isopropylcyclohexane

c. 2,3,5-trimethylhexane f. N,N-dimethy J eye lohexanamine

36. a. 3 b. 6 c. 3

CH1 CH3 CHlI . I . I .37. a. CH3CHCH2CH2CH2CH3 b. CH3CH-CCH:~I .

CH32-methy Ihexane 2,2,3-trimethylbutane

38. The first conformation is the most stable because the three substituents are more spread out, soits gauche interactions will not be as large - the Cl in the first conformation is between a CH3

and an H, whereas the Cl in the other two conformers is between two CH3 groups.

39. a. diethylpropylamine c. isopenty lpropy lamineTERTIARY SECONDARY

b. see-butylisobutylamine d. cyclohexylamineSECONDARY PRIMARY

CH3CH3

0 I I40. a. b. CH3C-CCH1 c. CH3CHCH2CHCH3I I . I I

CH3CH3 CH~ CH3.'

41. a. l-ethoxypropane f. 2-bromo-2-methylbutaneethyl propyl ether tert-pentyl bromide

b. 4-methyl-l-pentanol g. cyclohexanolisohexyl alcohol cyclohexyl alcohol

c. 2-butanamine h. bromocyclopentanesee-butylamine cyclopentyl bromide

d. 2-chlorobutane i. 3-propanaminesee-butyl chloride isopropylamine

e. 2-methylpentane j. 3-methyl-N-propyl-l-butanamineisohexane see-butylethylamine

42. a. I-bromohexane c. butyl alcohol e. hexaneb. pentyl chloride d. hexyl alcohol f. pentyl alcohol

Chapter 3 31

43.a. ~f~-rru

CH3 CH3

b. CH3H3C~CH3

H3C~CH3H3C

44. Ansaid is more soluble in wat,er. It has a fluoro substituent that can hydrogen bond to water.Motrin has a nonpolar isobutyl substituent in place of Ansaid's fluoro substituent.

45. Only one is named correctly.

a. 3-ethyl-2-methyloctaneb. 4-ethyl-2,2-dimethylheptanec. l-bromo-3-methylbutane

d. correcte. 2,5-dimethylheptanef. 2,3,3-trimethyloctane·

46. B has the highest energy (is the least stable). They are all diaxial-substituted cyclohexanes, soeach one has four 1,3-diaxial interactions. Only B has a 1,3-diaxial interaction between CH3 andCI, which will be greater than a 1,3-diaxial interaction between CH3 and H or a 1,3-diaxialinteraction between CI and H.

47. The only one is 2,2,3-trimethylbutane.

CH3CH3I 'ICH3C-CHCH'l, I '

CH3>y

48. a.

boQ49. a. l-bromopentane

b. butyl alcohol

c. octane

c. .r:d.

d. isopentyl alcohol (it has stronger hydrogen bonds)

e. hexylamine (it has more hydrogen bonds)

50. Alcohols with low molecular weights are more water soluble than alcohols with high molecularweights because, as a result of having fewer carbons, they have a smaller nonpolar componentthat has to be dragged into water.

32 Chapter 3

51.

CH,I -

a. H~H~

HyHCH2CH3

most stable

b.

c. Rotation can occur about all the C-C bonds. There are six carbon-carbon bonds in thecompound, so there are five other carbon-carbon bonds, in addition to the C3-C4 bond, aboutwhich rotation can occur.

CH -CH-CH,-CH -CH -CH3 I ~ 2 2 3

CH3d. Three of the carbon-carbon bonds have staggered conformers that are equally stable because

each is bonded to a carbon with three identical substituents.

il ilCH -CH-CH -CH -CH -CH~I 2 2 2 3

CH3

52. C and D are cis isomers. (In C, both substituents are downward pointing; in D, both substituentsare upward pointing.)

53. a. CH3cH2cH2cH2CH2Br 1 -bromopentane primary alkyl halide

CH,CH2CH2CHCH3 2 -bromopentane secondary alkyl halide. IBr

CH3CH,CHCH2CH3 3 -bromopentane secondary alkyl halide. ~ IBr

CH,I .CH3CHCH2CH2Br 1 -bromo-3-methylbutane primary alkyl halide

CH,I -CH3CH2CHCH2Br 1 -bromo-2-methylbutane primary alkyl halide

BrI

CH3CH2CCH3 2 -bromo-2-methylbutane tertiary alkyl halide- I -CH3

Chapter 3 33

BrI

CH3CHCHCH'lI .CH3

2 -bromo-3-methylbutane secondary alkyl halide

CH3I .CH3CCH2Br

ICH3

1 -bromo-2,2-dimethylpropane primary alkyl halide

b. Four isomers are primary alkyl halides.

c. Three isomers are secondary alkyl. halides.

d. One isomer is a tertiary alkyl halide.

54. a. butane d. 6-chloro-4-ethyl-3-methyloctane

e. 6-isobutyl-2,3-dimethyldecaneb. I-propanol

c. 5-isopropyl-2-methyloctane

55. a. ~CH2CH3 nH3C CH3 CH2CH3more stable

b. ~CH2CH3 ~H3hCH(CH3)2

more stable CH2CH3

c. ~CH2CH3

~CH3

more stable CH2CH3

CH3d. ~CH2CH3

H3C-9more stable CH2CH3

e. ~CH2CH3 n(CH3hCH (CH3hCH CH2CH3more stable

f. P-CH2CH3 (CH3)2C~

(CH3hCH more stable CH2CH3

34 Chapter 3

56. a. Each water molecule has two hydrogens that can form hydrogen bonds, whereas each alcoholmolecule has only one hydrogen that can form a hydrogen bond. Therefore, there are morehydrogen bonds between water molecules than between alcohol molecules.

:O-H----:O-HI IH HI

:O-CHlI -HIIII

:O-CHI 3

H

I

:O-HIH

b. Each water molecule has two hydrogens that can form hydrogen bonds, whereas eachammonia has three hydrogens that can form hydrogen bonds. However, oxygen is moreelectronegative than nitrogen, so the hydrogen bonds between water molecules are strongerthan the hydrogen bonds between ammonia molecules. Because the number of hydrogenbonds supports ammonia as having the higher boiling point but the strength of the hydrogenbonds supports water, we could not have predicted which would have the higher boiling point.However, being told that water has the higher boiling point we can conclude that the greaterelectronegativity of oxygen compared to nitrogen is more important than the number ofhydrogens that can form hydrogen bonds.

c. Each water molecule has two hydrogens that can form hydrogen bonds, whereas eachmolecule of hydrogen fluoride has only one hydrogen that can form a hydrogen bond.However, fluorine is more electronegative than oxygen. Again we cannot predict which willhave the higher boiling point, but we can conclude from the fact that water has the higherboiling point that in this case the greater number of hydrogens that can form hydrogen bondsis more important than the greater electronegativity of fluorine compared to oxygen.

57. Six ethers have molecular formula C5HI20.

CH30CH2CH2CH2CH3

/o~

butyl methyl etherI-methoxybutane

ethyl propyl etherl-ethox ypropane

see-butyl methyl ether2-methox ybu tane

CH3I -CH3CH20CHCH3

/"--oAethyl isopropyl ether2-ethoxypropane isobutyl methyl ether

I-methoxy-2-methylpropanetert-butyl methyl ether2-methoxy-2-methylpropane

Chapter 3 35

58. The most stable conformer has two CH3 groups in equatorial positions and one in an axialposition. (The other conformer would have two CH3 groups in axial positions and one in anequatorial position.) .

CH1

H3C~

H3C

59. a. 3-ethyl-2,5-dimethylheptane b. 1,4-dichloro-5-methylheptane

60. Dibromomethane does not have constitutional isomers.

If carbon were flat, the two structures shown below would be constitutional isomers, because theBr's would be 90° apart in one compound and 180° apart in the other compound, so they wouldbe different compounds. However, since carbon is tetrahedral the two structures are identical, sodibromomethane does not have constitutional Isomers.

HI

H-C-BrIBr

BrI

H-C-HIBr

61. CH20HHO~O\

HO~OHOH

62. #1 CH3CH2CH2CH2CH2CH2CH3heptane

CHI 3#4 CH3CCH2CH2CH3

I. CH3

2,2-dimethylpentane

CH1I .#2 CH3CHCH2CH2CH2CH3

2-methylhexane

CH1I .#5 CH3CH2CCH2CH1I .

CH3

3,3-dimethylpentane

CH3 CH3I . I#6 CH3CH-CHCH2CH3

2,3-dimethylpentane

CH1I .#3 CH3CH2CHCH2CH2CH3

3-methylhexane

CH-~ CH-~I - 1-

#7 CH3CHCH2CHCH3

2,4-dimelhylpentane

#9 CH1CH2CHCH2CH-'l- I -CH2CH3

3-ethylpentane

36 Chapter 3

CH3 CH3I 1-

#8 CH3C-CHCH1I -CH3

2,2,3-trimethylbutane

63. The most stable conformer has all the substituents in equatorial positions.

CICI~CICI~Cl

CI

ICH3

64. 5 4 CH H$CH2CH3 HCH3a. CH31li32 H H H ~CH2CH3

H H H H HH H

A B C

Hx!XH H~:CH1CH3C*H

H H H HCH2CH3 CH3CH2 HD E F

b.A A

rc E

Pontentialenergy

IB F

D

0 60 120 180 240 300 360Degrees of Rotation

65. a.

Hx!XCH3b. H~H

H H H CH3CH3CH2 CH2CH3

Chapter 3 37

66. a. one equatorial and one axialb. both equatorial and both axialc. both equatorial and both axial

d. one equatorial and one axiale. one equatorial and one axialf. both equatorial and both axial

67. Both trans-l ,4-dimethylcyc1ohexane and cis-l-tert-butyl- 3-methylcyc1ohexane have a conformerwith two substituents in the equatorial position and a conformer with two substituents in the axialposition. cis-l-tert-Butyl-3-methyIcyc1ohexane will have a higher percentage of the conformerwith two substituents in the equatorial position, because the bulky tert-butyl substituent will havea greater preference for the equatorial position.