Question

The C-H stretching vibration in chloroform (CHCl3) occurs at 3000 cm-1
. Calculate the C–D stretching frequency in
CDCl3

Answer 1

Let's assume that the C-H and C-D bonds can be regarded as isolated from the rest of the molecule. Then, the vibrational stretching frequency $\nu$ is related to the reduced mass of the system of two atoms $\mu$ :

$\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$ ,

where $k$ is the force constant of the chemical bond. Assuming that the force constant is the same for CHCl₃ and CDCl₃:

$\frac{\nu_H}{\nu_D} = \sqrt\frac{\mu_D}{\mu_H}$

The reduced masses of the systems are:

$\mu_H = \frac{m_Cm_H}{m_C+m_H} = \frac{12·1}{12 + 1} = 0.923$ a.u.

$\mu_D = \frac{m_Cm_D}{m_C+m_D} = \frac{12·2}{12 + 2} = 1.71$ a.u.

Therefore, the C-D stretching frequency is:

$\nu_D = 3000·\sqrt{\frac{\mu_H}{\mu_D}} = 2201 \text{ cm}^{-1}$

Answer: the C–D stretching frequency in CDCl₃ is 2201 cm^-1.