Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine

Question

Unsymmetrical dimethylhydrazine ($\ce{H2NN(CH3)2}$) has a melting point of −57 °C, but its isomer, 1,2-dimethylhydrazine ($\ce{(CH3NH)2}$) has a melting point of −9 °C. That looks to me like a significant difference, so there must be some mechanism which makes these two compounds extremely different in this respect. However, I'm unable to figure out why that difference exists; could you please explain that?

These are the structures of the compounds (from wikimedia commons):

Unsymmetrical dimethylhydrazine:

1,2-dimethylhydrazine:

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Here're some of the arguments which I considered:

1. The packing in crystal structures could be 'different' in a significant way. This could potentially be supported by the fact that the difference between their boiling points is much lower: unsymmetrical dimethylhydrazine has a boiling point of 64 °C while 1,2-dimethylhydrazine's is 87 °C. However, this is an extremely vague argument; it could be strengthened by some details regarding how the different packings contribute to the melting points.

2. Hydrogen bonding should be a pretty significant factor to consider while analyzing these compounds. I suspected that steric hindrance may allow only one $\ce{N}$ atom for hydrogen bonding in unsymmetrical dimethylhydrazine. However, this would suggest that 1,2-dimethylhydrazine should have a lower melting point, which is clearly not the case.

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_{We had a discussion about these two compounds' melting points in the Physics SE chat a while back, so I thought I'd pop over here and ask.}

Answer 1

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:

$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$

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

Answer 2

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.