
Which of the following isomers is more volatile: o-nitrophenol or p-nitrophenol?
Answer
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Hint: To solve this question we need to know that in case of o-nitrophenol there is intramolecular hydrogen bonding and in case of p-nitrophenol there is intermolecular hydrogen bonding.
Complete step by step answer:
In this question, by volatile compound we mean that compound which can vaporize readily that is which can boil easily. Also, intramolecular forces are between atoms within a molecule and intermolecular forces are between different molecules.
The o-nitrophenol can also be written as $2 - $nitrophenol in which the nitro group is attached at ortho position (adjacent carbon of the hydroxyl group). In o-nitrophenol, there is intramolecular hydrogen bonding between the hydrogen of $ - OH$ group and oxygen of $ - N{O_2}$ group which is present at ortho position. Because of this, it is not associated with other molecules or with water and thus it is slightly free to vaporize. So, it gets vaporized easily because there are non intermolecular forces. Its structure can be shown as:
The p-nitrophenol can also be written as $4 - $nitrophenol in which the nitro group is attached at para position (at fourth carbon starting from the carbon of hydroxyl group). In p-nitrophenol, there is intermolecular hydrogen bonding between the hydrogen of $ - OH$ group of one p-nitrophenol and the oxygen of $ - N{O_2}$ group of other molecules of p-nitrophenol. Due to intermolecular hydrogen bonding, several molecules of p-nitrophenol are associated with each other which results in high boiling point and makes it less volatile than ortho-nitrophenol. The structure of para-nitrophenol is:
Hence, we can say that o-nitrophenol is more volatile than p-nitrophenol because of intermolecular hydrogen bonding in p-nitrophenol.
Note:
We need to remember in this question that intramolecular hydrogen bonding occurs in compounds in which hydrogen is near to the other atom as we saw in case of o-nitrophenol in which hydrogen was near to the oxygen of hydroxyl group.
Complete step by step answer:
In this question, by volatile compound we mean that compound which can vaporize readily that is which can boil easily. Also, intramolecular forces are between atoms within a molecule and intermolecular forces are between different molecules.
The o-nitrophenol can also be written as $2 - $nitrophenol in which the nitro group is attached at ortho position (adjacent carbon of the hydroxyl group). In o-nitrophenol, there is intramolecular hydrogen bonding between the hydrogen of $ - OH$ group and oxygen of $ - N{O_2}$ group which is present at ortho position. Because of this, it is not associated with other molecules or with water and thus it is slightly free to vaporize. So, it gets vaporized easily because there are non intermolecular forces. Its structure can be shown as:
The p-nitrophenol can also be written as $4 - $nitrophenol in which the nitro group is attached at para position (at fourth carbon starting from the carbon of hydroxyl group). In p-nitrophenol, there is intermolecular hydrogen bonding between the hydrogen of $ - OH$ group of one p-nitrophenol and the oxygen of $ - N{O_2}$ group of other molecules of p-nitrophenol. Due to intermolecular hydrogen bonding, several molecules of p-nitrophenol are associated with each other which results in high boiling point and makes it less volatile than ortho-nitrophenol. The structure of para-nitrophenol is:
Hence, we can say that o-nitrophenol is more volatile than p-nitrophenol because of intermolecular hydrogen bonding in p-nitrophenol.
Note:
We need to remember in this question that intramolecular hydrogen bonding occurs in compounds in which hydrogen is near to the other atom as we saw in case of o-nitrophenol in which hydrogen was near to the oxygen of hydroxyl group.
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