Arrange the following compounds in the increasing order of their boiling points.
$C H_{3} C H_{2} C H_{2} C H O, C H_{3} C H_{2} C H_{2} C H_{2} O H, H_{5} C_{2} - O - C_{2} H_{5}, C_{2} H_{5} C H_{2} C_{2} H_{5}$

C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}

H_{5} C_{2} - O - C_{2} H_{5}

C H_{3} C H_{2} C H_{2} C H O

C H_{3} C H_{2} C H_{2} C H_{2} O H

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C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}

H_{5} C_{2} - O - C_{2} H_{5}

C H_{3} C H_{2} C H_{2} C H O

C H_{3} C H_{2} C H_{2} C H_{2} O H

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C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}

H_{5} C_{2} - O - C_{2} H_{5}

C H_{3} C H_{2} C H_{2} C H O

C H_{3} C H_{2} C H_{2} C H_{2} O H

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C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}

H_{5} C_{2} - O - C_{2} H_{5}

C H_{3} C H_{2} C H_{2} C H O

C H_{3} C H_{2} C H_{2} C H_{2} O H

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Solution

The correct option is A $C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}$ < $H_{5} C_{2} - O - C_{2} H_{5}$ < $C H_{3} C H_{2} C H_{2} C H O$ < $C H_{3} C H_{2} C H_{2} C H_{2} O H$
For boiling points, the increasing order is Hydrocarbons < Ethers < Aldehyde < Alcohol of similar molecular mass. Therefore the given order would be:
$C H_{3} C H_{2} C H_{2} C H_{2} C H_{3} < C_{2} H_{5} - O - C_{2} H_{5} < C H_{3} C H_{2} C H_{2} C H_{2} C H O < C H_{3} C H_{2} C H_{2} C H_{2} O H$

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