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

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

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