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Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?
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Lakshmi Priya
Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?
Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:
For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.
Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.
Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.
Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.
Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:
For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.
Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.
Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.
Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.
Yes, you are correct. And I even checked the values for confirmation, but the book states the reason that u11>u12 to be resonance. They state that in u11 due to resonating structures, there will be charge involved in dipole moment, whereas not that effective in u12More
The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.
The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.
Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:
For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.
Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.
Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.
Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.
Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:
For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.
Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.
Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.
Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.
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cis-1,2-Dichloroethene has the larger dipole.
cis-1,2-Dichloroethene, 1.9 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 1.9 D.
1,1-dichloroethylene, 1.3 D
The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.
trans-1,2-Dichloroethene, 0 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 0 D.
cis-1,2-Dichloroethene has the larger dipole.
cis-1,2-Dichloroethene, 1.9 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 1.9 D.
1,1-dichloroethylene, 1.3 D
The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.
trans-1,2-Dichloroethene, 0 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 0 D.
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