The Foundations of Chemistry

we remove a second carbon from the unbranched isomer and, if possible, form additional
branching, and so on. Finally, we add hydrogen atoms to each skeleton to complete the
structure.

Solution

The three compounds are:

The compound with the most branching would be expected to be the one with the lowest
boiling point, and the straight-chain isomer would be expected to have the highest boiling
point.

CH 3 C(CH 3 ) 3 , bp9.5°C; CH 3 CH 2 CH(CH 3 ) 2 , bp27.9°C;

and CH 3 CH 2 CH 2 CH 2 CH 3 , bp36.1°C.

(Models of these compounds are shown in the margin.)

You should now work Exercise 15.

H

HH

HH

C C

HH

H

C C H same as H

HH

H

C C

HH

HH

C C H

H

HHC

H

HHC

H

HH

HH

C C

HH

HH

C C

H

H

C H, H

H

H

C

H

H

C

H

HCH

HH

H

C H, and

C

CH 3 CH 2 CH 2 CH 2 CH 3 , CH 3 CH 2 CH(CH 3 ) 2 ,

and CH 3 C(CH 3 ) 3.

TABLE 27-3 Isomeric C 6 H 14 Alkanes

Normal

IUPAC Name Formula bp (°C)

hexane CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 68.7

2-methylpentane 60.3

3-methylpentane 63.3

2,2-dimethylbutane 49.7

2,3-dimethylbutane 58.0

CH 3 CH 3

CH 3 CH CHCH 3

CH 3

CH 3

CH 3 CH 2 CCH 3

CH 3

CH 3 CH 2 CHCH 2 CH 3

CH 3

CH 3 CH 2 CH 2 CHCH 3

TABLE 27-4 Numbers of

Possible

Constitutional

Isomers of

Alkanes

Formula Isomers

C 7 H 16 9

C 8 H 18 18

C 9 H 20 35

C 10 H 22 75

C 11 H 24 159

C 12 H 26 355

C 13 H 28 802

C 14 H 30 1,858

C 15 H 32 4,347

C 20 H 42 366,319

C 25 H 52 36,797,588

C 30 H 62 4,111,846,763