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Does 1-bromo-4-chlorobenzene have a permanent dipole moment?
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Kasim Muflahi
Does 1-bromo-4-chlorobenzene have a permanent dipole moment?
There are two aspects to dipole moments:
The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference.
The second is the bond length.
So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.
Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2
There are two aspects to dipole moments:The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference.The second is the bond length.So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.
Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2
As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places.
Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.
\begin{array}{lr}\hline
\text{Method} & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline
\text{BP86/def2-SVP} & 0.07\\
\text{BP86/def2-TZVPP} & 0.11\\
\text{B3LYP/def2-TZVPP} & 0.09\\
\text{B3LYP+D3(BJ)/def2-TZVPP} & 0.09\\
\text{PBE0/def2-TZVPP} & 0.10\\
\text{M11/def2-TZVPP} & 0.10\\
\text{B2PLYP/def2-TZVPP} & 0.09\\
\text{MP2/def2-TZVPP} & 0.06\\
%\text{CCD//MP2/def2-TZVPP} & ----\\
\hline\end{array}
In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:
We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.
However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine.
This causes a larger charge difference than the polarisation of the bonding structure.
Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.
As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places. Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.\begin{array}{lr}\hline\text{Method} & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline\text{BP86/def2-SVP} & 0.07\\\text{BP86/def2-TZVPP} & 0.11\\\text{B3LYP/def2-TZVPP} & 0.09\\\text{B3LYP+D3(BJ)/def2-TZVPP} & 0.09\\\text{PBE0/def2-TZVPP} & 0.10\\\text{M11/def2-TZVPP} & 0.10\\\text{B2PLYP/def2-TZVPP} & 0.09\\\text{MP2/def2-TZVPP} & 0.06\\%\text{CCD//MP2/def2-TZVPP} & ----\\\hline\end{array}
In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:
We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences. Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.
However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine. This causes a larger charge difference than the polarisation of the bonding structure.
Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.
There are two aspects to dipole moments: The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference. The second is the bond length. So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.
Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2
There are two aspects to dipole moments:The first is the amount of charge transfer between the linked atoms - this is determined by the electronegativity difference.The second is the bond length.So, the greater difference in electronegativity between Cl and C vz Br and C is partially compensated by the longer Br-C bond vz Cl-C.
Many dipole moment are determined in solution and between different solvents they may differ by more than 0.2 units. I.e. 0 < μ < 0.2
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As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places. Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$. \begin{array}{lr}\hline \text{Method} & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline \text{BP86/def2-SVP} & 0.07\\ \text{BP86/def2-TZVPP} & 0.11\\ \text{B3LYP/def2-TZVPP} & 0.09\\ \text{B3LYP+D3(BJ)/def2-TZVPP} & 0.09\\ \text{PBE0/def2-TZVPP} & 0.10\\ \text{M11/def2-TZVPP} & 0.10\\ \text{B2PLYP/def2-TZVPP} & 0.09\\ \text{MP2/def2-TZVPP} & 0.06\\ %\text{CCD//MP2/def2-TZVPP} & ----\\ \hline\end{array}
In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:
We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.
However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine. This causes a larger charge difference than the polarisation of the bonding structure.
Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.
As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places. Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.\begin{array}{lr}\hline\text{Method} & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline\text{BP86/def2-SVP} & 0.07\\\text{BP86/def2-TZVPP} & 0.11\\\text{B3LYP/def2-TZVPP} & 0.09\\\text{B3LYP+D3(BJ)/def2-TZVPP} & 0.09\\\text{PBE0/def2-TZVPP} & 0.10\\\text{M11/def2-TZVPP} & 0.10\\\text{B2PLYP/def2-TZVPP} & 0.09\\\text{MP2/def2-TZVPP} & 0.06\\%\text{CCD//MP2/def2-TZVPP} & ----\\\hline\end{array}
In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:
We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.
However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine. This causes a larger charge difference than the polarisation of the bonding structure.
Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.
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