Which of the following molecules will have hydrogen bonding in the pure liquid state?

A. H₃C-CH₃
B. Cl-CH₃
C. Cl-CH₂-CH₂-CH₃
D. CH₃-CH₂-CH₂-CH₃



Answer :

To determine which of the given molecules will exhibit hydrogen bonding in the pure liquid state, we need to remember the criteria for hydrogen bonding. Hydrogen bonding typically occurs in molecules where hydrogen atoms are directly bonded to highly electronegative atoms like fluorine (F), oxygen (O), or nitrogen (N). These atoms create a significant dipole moment, generating strong intermolecular attractions between hydrogen atoms and lone pairs of neighboring F, O, or N atoms.

Let’s analyze each option:

Option a.
[tex]\[ \text{H₃C-CH₃} \][/tex]
This is ethane. In ethane, there are no F, O, or N atoms bonded to hydrogen atoms. Therefore, this molecule does not meet the criteria for hydrogen bonding.

Option b.
[tex]\[ \text{Cl-CH₃} \][/tex]
This is chloromethane. Similarly, chloromethane does not have any hydrogen atoms bonded to F, O, or N. It does have Cl, which is electronegative, but not sufficient to form hydrogen bonds like F, O, or N.

Option c.
[tex]\[ \text{CH₃CH₂CH₂CH₃} \][/tex]
This is butane. Butane is a simple hydrocarbon with no electronegative atoms like F, O, or N bonded to hydrogen.

Option d.
[tex]\[ \text{CH₃Cl} \][/tex]
This molecule is also chloromethane. It was already analyzed in option b and does not fulfill the criteria for hydrogen bonding.

In all these molecules, there are no hydrogen atoms directly bonded to highly electronegative atoms such as fluorine, oxygen, or nitrogen. Therefore, none of the given options exhibit hydrogen bonding in the pure liquid state.

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