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Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine
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Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine
According to the Carnelley’s Rule:
That of two or more isomeric compounds, those whose
atoms are the more symmetrically and the more compactly
arranged melt higher than those in which the atomic
arrangement is asymmetrical or in the form of long chains
If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
That of two or more isomeric compounds, those whose atoms are the more symmetrically and the more compactly arranged melt higher than those in which the atomic arrangement is asymmetrical or in the form of long chains
If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.
According to the Carnelley’s Rule:
If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
References and further reading:
Effect of molecular symmetry on melting temperature and solubility
Melting Point and Molecular Symmetry
Carnelley’s Rule and the Prediction of Melting Point
According to the Carnelley’s Rule:
If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
References and further reading:
Effect of molecular symmetry on melting temperature and solubility
Melting Point and Molecular Symmetry
Carnelley’s Rule and the Prediction of Melting Point
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Melting points of related compounds: hydrazine, 2 °C; monomethylhydrazine, −52 °C (Wikipedia).
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.
Melting points of related compounds: hydrazine, 2 °C; monomethylhydrazine, −52 °C (Wikipedia).
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.
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