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Daniel [21]
3 years ago
14

Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists the

se compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane c. 2-methylpentane < 2,3-dimethylbutane d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Chemistry
1 answer:
Softa [21]3 years ago
5 0

Answer:

a. 2,3-dimethylbutane < 2-methylpentane < n-hexane

Explanation:

The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.

The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.

n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.

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4 0
3 years ago
Read 2 more answers
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