Chapter 12, Problem 12.65P

Branching in an alkane chain decreases surface area. As a result, a branched-chain alkane tends to have a lower boiling point than a straight-chain alkane with the same number of carbons. Keeping this in mind, rank the alkane isomers in each group in order of increasing boiling point.

 a. ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}},{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}},{\left({\text{CH}}_{\text{3}}\right)}_{\text{4}}\text{C}$

 b. ${\left(\text{CH3}\right)}_{\text{2}}\text{CHCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}},{\text{ CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{CH}}_{\text{3}},{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$

Interpretation Introduction

(a)

Interpretation:

The compound in isomeric pair ( ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ ; ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ ; ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}\text{C}$

), the one that has higher boiling point should be determined.

Concept Introduction:

The boiling point of alkane increases with the increase in the number of carbon atom presents. With the increase in surface area of the molecule, the boiling point increases. Now, when branching increases then surface area decreases which results decrease in boiling point.

Answer to Problem 12.65P

Among this given isomeric pair, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ has higher boiling point.

The order of increasing boiling point is:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}{\text{C<(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{<CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$

Explanation of Solution

Considering these pair,

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ ; ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ ; ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}\text{C}$

In ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ there are no branching; but in ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ there is two methyl group is in branched chain; and in ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}\text{C}$ four methyl group is attached with central carbon atom. So branching increases. In a result, surface area decreases.

Thus, the increasing order of boiling point:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{4}}{\text{C<(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{<CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$

Interpretation Introduction

(b)

Interpretation:

The compound in isomeric pair( ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$ ; ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$ ; ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$ ), the one that has higher boiling point should be determined.

Concept Introduction:

The boiling point of alkane increases with the increase in the number of carbon atom presents. With the increase in surface area of the molecule, the boiling point increases. Now, when branching increases then surface area decreases which results decrease in boiling point.

Answer to Problem 12.65P

Among this given isomeric pair, ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$ has higher boiling point.

The order of increasing boiling point is:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{<CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{<CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$

Explanation of Solution

Considering these isomeric pair, ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$ ; ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$ ; ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$

In ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$ , there are no branching; but in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$ , there is two methyl group is attached in branched chain; and in ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH(CH}}_{\text{3}}{\text{)}}_{\text{2}}$ four methyl group is attached with two consequent carbon atom. So,branching increases. In a result, surface area decreases.

Thus, the increasing order of boiling point:

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{CHCH(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{<CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(CH}}_{\text{3}}{\text{)}}_{\text{2}}{\text{<CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{3}}$