5

2Alkanes

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2.1 (b) If 2n + 2 = 30, then n = 14; that is, there are fourteen carbons.

2.2 (b) Following the logic of part (a), add 20° to the boiling point of tridecane to obtain 256° as the boilingpoint of tetradecane. The condensed structural formula is CH3(CH2)12CH3.

2.4 Newman projections for conformations about the C1–C2 bond of butane.

CH2CH3

HHHH

HH

FED

CBA

CH3CH2

H

HH

H H H

CH2CH3H

CH3CH2 H

HH

H

H

CH2CH3

HH

H HH

H

H

H

CH2CH3

HH

H H

(b) The curve of potential energy vs. angle of rotation (shown in Fig. IS2.1 on the following page) is muchlike that for ethane, because all staggered forms are identical, and all eclipsed forms are identical. Thestaggered forms A, C, and E are at the energy minima, and the eclipsed forms B, D, and F are at themaxima.

(c) Butane exists entirely in the staggered form.

2.5 (b) The compound is named as a nonane; both possible numbering schemes for substituents are (4,5,5,6),but one gives the ethyl group the lowest number, and the other gives the methyl group the lowestnumber. The numbering is determined by the group that is cited first in the name—the ethyl group.Thus, the name is 4-ethyl-5,5,6-trimethylnonane.

6 Chapter 2 Alkanes

Angle of rotation, θ

PO

TE

NT

IAL

EN

ER

GY

eclipsed forms

staggered forms

Figure IS2.1 Potential-energy curve for the solution to Problem 2.4b.

2.7 (b)

(CH3)2CHCH2CH2CHCHCH2CH2CH3

CH2CH2CH3

C(CH3)3

6-tert-butyl-2-methyl-5-propylnonane

2.9 There are four ethyl groups and six methyl groups in the structure of 4-sec-butyl-5-ethyl-3-methyloctane.

2.10 (b) (d)

2.11 (b) The name is 1,1,2-trimethylcyclobutane.

2.15 (b) C, 83.63 mass %; H, 16.37 mass %.

2.16 (b) The empirical formula of a compound with molecular formula C6H14 is C3H7.

2.17 (b) Combustion of a hydrocarbon C8H14 follows the equation

+ 7 H2O8 CO2O2

232

C8H14 +

According to this equation, 11.5 moles of oxygen are consumed per mole of hydrocarbon. Because 10.00mg (0.000091 mole) of hydrocarbon is used in the analysis, then (11.5)(0.000091) mole = 0.00105 mole =0.0336 g = 33.6 mg of O2 is consumed.

2.18 (b) Any structure with five carbons, an —OH group, and either a ring or a double bond is correct. Two ofmany examples are the following:

OH H2C CH CH2CH2CH2 OH

Instructor’s Solutions to Additional Problems Chapter 2 7

(d) The elemental analysis corresponds to an empirical formula C2H3; the smallest possible molecularformula is C4H6. An alkyne with this formula is H3C—C———C—CH3 (dimethylacetylene or 2-butyne).

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2.20 (b) Use the rule of thumb that an additional carbon adds about 20–30° to the boiling point. Because thesecond compound has one fewer carbon than the first, its boiling point should be about 20–30° less. Theestimated boiling point is 122–132°.(d) The estimated boiling point is 50–60°.

2.21 (b) The isomers of octane with six carbons in their principal chains:

3-ethylhexane3,3-dimethylhexane2,2-dimethylhexane

CH3CH2CHCH2CH2CH3

CH3CH2

CH3CH2CCH2CH2CH3

CH3

CH3

CH3CCH2CH2CH2CH3

CH3

CH3

2,5-dimethylhexane2,4-dimethylhexane2,3-dimethylhexane

CH3CHCH2CH2CHCH3

CH3

CH3

CH3CHCH2CHCH2CH3

CH3

CH3

CH3CHCHCH2CH2CH3

CH3

CH3

CH3CH2CHCHCH2CH3

CH3

CH3

3,4-dimethylhexane

2.22 (b) The label abbreviations are given in part (a) in the Solutions Manual.

p

p

p p

p

s

tq

2.23 (b) Cyclopentane has five carbons and only secondary hydrogens:

cyclopentane

(d) An alkane that has a molecular mass of 84.2 can only have the formula C6H12. The reason is that analkane of molecular mass = 84 with five carbons would have 24 hydrogens, and this is not possible (why?);and an alkane with seven carbons would have no hydrogens. The only alkanes with this formula arecyclohexane and the other constitutionally isomeric cycloalkanes. (There are nine other possiblestructures.)

CH3cyclohexane methylcyclopentane

2.24 (b) 4-ethyl-5-methyloctane(d) 4-isopropyloctane(f) 1,1,2,2-tetraethyl-3,4-dimethylcyclobutane

8 Chapter 2 Alkanes

2.25 (b) (d)

CH3CCH2CH2CH3

CH3

CH3

CH3CHCHCHCHCH2CH3

CH3

CH3

CH2CH2CH3

CH3

2.26 (b) This compound should be named 2,4,6,6-tetramethyloctane:

CH3CCH2CHCH2CHCH3

CH3

CH2CH3

CH3

CH3

2,4,6,6-tetramethyloctane

(d) This compound should be named 4-tert-butyloctane:

CH3C CHCH2CH2CH3

CH3

CH3

CH2CH2CH2CH3

4-tert-butyloctane

2.29 Both curves would have the same general appearance, that is, three barriers of equal height. (See diagrambelow.) However, the curve for 2,2,3,3-tetramethylbutane should have greater potential-energy barriersbecause each eclipsed form has three severe methyl-methyl interactions.

Angle of rotation, θ

PO

TE

NT

IAL

EN

ER

GY

curve for ethane

curve for 2,2,3,3-tetramethylbutane

2.34 The C—O bond is shorter than the C—C bond. Consequently, when the methyl groups are gauche, thedistance between them is smaller in the ether [compound (a)] than in butane [compound (b)].Consequently, van der Waals repulsions in the gauche form of the ether should be larger than those of thegauche form in butane, and the energy of the gauche form of the ether should be higher relative to the antiform, in which van der Waals repulsions are absent. In other words, the ether should have the greaterenergy difference between the gauche and anti forms.

2.37 (b) The molecular formula of decalin is C10H18. Use the method described in part (a) in the SolutionsManual. Decalin contains 86.88% carbon.

2.38 (b) Two of several structures consistent with the analysis in part (a) that have two tertiary carbons and allothers secondary:

Instructor’s Solutions to Additional Problems Chapter 2 9

2.40 (b) Any two of the following structures are correct.

C

O

NH CH2CH3CH3 NHC

O

CH3CH2 CH3

C

O

NCH3

CH3

CH3

C

O

NH

CH3

CH2CH3

The relationship between X and Y: they are constitutional isomers, because they have the same atomiccomposition, but different connectivities.

2.42 (a) ketone(b) nitrile(c) alcohol(d) ether

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