Question

# The correct order of decreasing the boiling point of the following alcohols is:

A

$\mathrm{I}>\mathrm{III}>\mathrm{II}$

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B

$\mathrm{II}>\mathrm{III}>\mathrm{I}$

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C

$\mathrm{I}>\mathrm{II}>\mathrm{III}$

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D

$\mathrm{III}>\mathrm{II}>\mathrm{I}$

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Solution

## The correct option is A $\mathrm{I}>\mathrm{III}>\mathrm{II}$Boiling Point: It is defined as the point at which the pressure exerted by the surroundings is equal to the pressure exerted by the vapours of the liquid.In the case of alcohol, boiling point can be determined by the ability of the compound to form H-bonding among the alcohol molecules. More will be the extent of H-bonding, more will be its boiling point.Hydrogen bonding is highly affected by the hindrance. A compound with hindrance will have less tendency to form H-bond hence its boiling point will be less.Among $Pen\mathrm{tan}-1-ol$, $2-methylbu\mathrm{tan}-2-ol$ and $3-methylbu\mathrm{tan}-2-ol$;$Pen\mathrm{tan}-1-ol$ have least hindrance so it will have maximum extent of H-bonding. Similarly, $3-methylbu\mathrm{tan}-2-ol$ will have less hindrance as compared to $2-methylbu\mathrm{tan}-2-ol$. Hence boiling point of $3-methylbu\mathrm{tan}-2-ol$ will be greater than $2-methylbu\mathrm{tan}-2-ol$.Hence, Maximum boiling point will be of $Pen\mathrm{tan}-1-ol$ and minimum boiling point will be of $2-methylbu\mathrm{tan}-2-ol$.Therefore the correct option is A. $\mathrm{I}>\mathrm{III}>\mathrm{II}$

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