Answer to Question #225411 in General Chemistry for nik

Question

Consider THREE (3) compounds; dimethyl ether, ethanol, and propane having boiling points, NOT necessarily in the correct order of ─42.1°C, ─24.8°C, and 78.4°C

Predict with explanation the melting and boiling points for methylamine based on the given compounds.

Answer 1

Both the melting point and the boiling point of organic compounds of each homological series are higher, the greater the number of carbon atoms in the members of this series. As a rule, isomers with a branched chain have a higher volatility (i.e., they have a lower boiling point) than their corresponding isomers with an unbranched chain.

Alcohols and carboxylic acids, i.e. compounds containing the —OH group, have abnormally high boiling points: ethanol.

dimethyl ether: C₂H₆O, H₃С-О-СН₃ : ─24.8°C

The boiling point and density of esters are lower than the corresponding alcohols, this is due to the inability to form solvates and hydrogen bonds, since esters are non-polar compounds

ethanol: С₂H₆O, С₂H₅OH : 78.4°C

propane: C₃H₈, CH₃CH₂CH₃: ─42.1°C Alkane molecules interact weakly with each other, so alkanes melt and boil at significantly lower temperatures than substances with polar molecules that are close to them in mass

methylamine: CH₃NH₂ In comparison with alcohols, aliphatic amines have lower boiling points . This indicates that amines are associated to a lesser extent than alcohols, since the strength of hydrogen bonds with the nitrogen atom is less than with the participation of more electronegative oxygen.

Therefore, the boiling point of methylamine is less than that of ethanol. The bond N-H is polar, so primary and secondary amines form hydrogen bonds, so their boiling point is higher than that of alkanes and esters. Boiling point from -24.8 °C to 78.4°C. (-6°C).