# Which has a greater dipole moment: methylamine or methanol? [closed]

Which has a greater dipole moment: methylamine $$(\ce{CH3NH2})$$ or methanol $$(\ce{CH3OH})?$$

• It's difficult to compare the dipole moment from the structure of both the molecules. I guess you should go for experimental values. – Sameer nilkhan Dec 26 '19 at 4:26
• As per experimental value dipole moment of methylamine is 1.31D and of methanol is 1.69D. – Sameer nilkhan Dec 26 '19 at 4:30

The bond dipole moment of a $$\ce{H-C}$$ bond is $$0.3 D$$, that of a $$\ce{C-N}$$ bond, $$0.4 D$$, that of a $$\ce{H-N}$$ bond, $$1.3 D$$, that of a $$\ce{C-O}$$ bond, $$0.7 D$$, and that of a $$\ce{H-O}$$ bond, $$1.5 D$$, each in the direction of the second atom designated.

Locate the carbon atom of each molecule at the origin of a three-dimensional ($$r$$, $$\theta$$, $$z$$) coordinate system with the nitrogen atom of the $$\ce{CH3NH2}$$ molecule and the oxygen atom of the $$\ce{CH3OH}$$ molecule on the $$z$$-axis of each coordinate system.

To find the dipole moment of the $$\ce{CH3NH2}$$ molecule, we add the vector dipole moment of a $$\ce{H-C}$$ bond (with the carbon atom above the hydrogen atom),

$$0.3 D \cdot u_z$$

the vector dipole moment of the $$\ce{C-N}$$ bond (with the nitrogen atom above the carbon atom),

$$0.4 D \cdot u_z$$

and the vector dipole moment of a $$\ce{H-N}$$ bond (with the nitrogen atom below the hydrogen atom),

$$1.3 D \cdot (-u_z)$$

obtaining as a net vector sum

$$(0.3 D + 0.4 D - 1.3 D) \cdot u_z = -0.6 D \cdot u_z$$,

the magnitude of which is $$0.6 D$$.

To find the dipole moment of the $$\ce{CH3OH}$$ molecule, we add the vector dipole moment of the $$\ce{H-C}$$ bond (with the carbon atom above the hydrogen atom),

$$0.3 D \cdot u_z$$,

the vector dipole moment of the $$\ce{C-O}$$ bond (with the oxygen atom above the carbon atom),

$$0.7 D \cdot u_z$$

and the vector dipole moment of the $$\ce{H-O}$$ bond (with the oxygen atom below the hydrogen atom and the $$\ce{C-O-H}$$ bond angle approximately $$104.5^\circ$$),

$$1.5 D \cdot [\cos(180^\circ-104.5^\circ)\cdot u_z + \sin(180^\circ-104.5^\circ)\cdot u_r]$$,

the net vector sum of which is

$$0.3 D \cdot u_z + 0.7 D \cdot u_z + 1.5 D \cdot [\cos(180^\circ-104.5^\circ)\cdot u_z + \sin(180^\circ-104.5^\circ)\cdot u_r]$$

= $$[1.0 D + 1.5 D \cos(75.5^\circ)]\cdot u_z + \sin(75.5^\circ)\cdot u_r$$

= $$(1.0 D + 1.5 D \cdot 0.25038000405)\cdot u_z + 0.96814764037\cdot u_r$$

= $$1.37557000608 D \cdot u_z + 1.45222146057 \cdot u_r$$

the magnitude of which is $$2.00 D$$ (rounded).

One might thus conclude that methanol's dipole moment (experimental value being $$1.69 D$$) is greater than that of methylamine (experimental value being $$1.31 D$$).