Tutor profile: Mausam A.

Which of the following molecules will show a vibrational infrared spectrum and why? : H2, HCl, NH3, CO2, and H2O.

CONCEPT - Vibrational spectra are shown by a molecule when the vibrational motion is accompanied by a change in dipole moment i.e dµ/dr = 0 where µ = dipole moment and r = bond length. ANSWER - All the molecules except H2 will be infrared-active. EXPLANATION - H2 due to its symmetric structure does not have a dipole moment nor is the vibrational change is accompanied by a change in dipole moment. Again, although CO2 is a symmetric molecule with no dipole moment, the asymmetric stretching motion and bending mode of vibration do result in a change in dipole moment.

Arrange the following compounds in order of increasing Sn1 reactivity : Cyclopropyl chloride, Cyclobutyl chloride, and Cyclopentyl chloride

An Sn1 reaction proceeds through the rate-determining formation of planer carbonation in which the angle between any two bonds is 120. The normal bond angles of three, four, and five-membered rings are of the order of 60, 90, and 108 respectively. These represent deviations of 60, 30, and 12 respectively from the trigonal angle. The bond angle strain in cycloalkyl carbocations thus increases in the order cyclopentyl cation < cyclobutyl cation < cyclopropyl cation. Therefore, the stability of the carbonation increases in the order of cyclopropyl cation < cyclobutyl cation < cyclopentyl cation. Since Sn1 reactivity depends on the stability of the carbonation, the reactivity order of these cycloalkanes is as follows : Cyclopropyl chloride < Cyclobutyl chloride < Cyclopentyl chloride

H-F has a dipole moment of 1.82D; its boiling point 19.34-degrees celsius. Ethyl fluoride has an almost identical dipole moment and has a larger molar mass, yet its boiling point is -37.7 degrees celsius. Explain

Since the dipole moments of both hydrogen fluoride and ethyl fluoride are almost the same, the strengths of dipole-dipole interactions involved in these compounds are almost the same and therefore, the compounds are expected to boil at nearly the same temperature. However, because of larger molecular weight C2H5F is expected to boil at a higher temperature than H-F. The difference in the boiling point can be explained in the light of the hydrogen bonding force. Because of the presence of a hydrogen atom bonded to the highest electronegative fluorine atom, the molecules of hydrogen fluoride remain associated through strong intermolecular hydrogen bonding. On the other hand, H-bonding is not possible in the case of C2H5F. Since it requires extra energy to break hydrogen bonds, HF boil at a much higher temperature than the C2H5F.