Structures of molecules of two compounds are given below : (a) Which of the two compounds will have intennolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding. (b) The melting point of a cornpound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two cornpounds will show higher melting point. (c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it.

Structures of molecules of two compounds are given below : (a) Whic

1.	Structures of molecules of two compounds are given below : (a) Which of the two compounds will have intennolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding. (b) The melting point of a cornpound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two cornpounds will show higher melting point. (c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it.
Answer» Solution :(a) In compound (I) Intramolecular H-bonding is formed Here H-atom, in between the two HIGHLY ELECTRONEGATIVE atoms, is PRESENT within the same molecule. In ortho-nitrophenol (compound n. H-atom is in between the two oxygen atoms. Compound (II) forms intermolecular H- bonding. In para-nitrophenol cm there is a gap between `NO_(2)` and OH group. So, H-bond exists between H-atom of one molecule and O -atom of another molecule as shown below. (b) Compound (ID Possess higher melting point because large number of molecules are JOINED together by H-bonds. (c) Due to intramolecular H bonding, compound (I) is not able to form H-bond with water, so it is less soluble in water. While molecules of compound II form H-bonding with H20 easily, so it is soluble in water.