Question

Substitution of hydrogen by deuterium is a convenient way to label organic molecules at specific sites for the purpose of investigating reaction mechanism. At what frequency would the$C-D$stretch of a phenyl ring appear if the frequency of the$C-H$stretch is $\mathbf{3062}{\mathbf{cm}}^{-\mathbf{1}}$?

Short answer

The stretching frequency of $C-D$is$2246c{m}^{-1}$.

Step-by-step solution

Definition of Organic Molecule

Organic molecules are often made up of carbon atoms arranged in rings or long chains to which other atoms of elements such as hydrogen, oxygen, and nitrogen are linked.

Given Information

The frequency of the$\mathbf{C}-\mathbf{H}$ stretch i.e.${\nu }_{C-H}=3062c{m}^{-1}$

Relation between stretch frequency of C−H  and C−D

Write the relation between stretch frequency of $C-H$and $C-D$.

$\frac{{\nu }_{C-D}}{{\nu }_{C-H}}=\sqrt{\frac{{\mu }_{C-H}}{{\mu }_{C-D}}}$Where, μ is reduced mass.

Find values use in the relation and simplify.

$\begin{array}{l}{\mu }_{C-H}=\frac{12}{13}\\ {\mu }_{C-H}=0.923amu\\ {\mu }_{C-D}=\frac{24}{14}\\ {\mu }_{C-D}=1.714amu\end{array}$

Use above values in relation

$\begin{array}{c}\frac{{\nu }_{C-D}}{{\nu }_{C-H}}=\sqrt{\frac{{\mu }_{C-H}}{{\mu }_{C-D}}}\\ \sqrt{\frac{0.923}{1.714}}=0.733\end{array}$

Find the stretching frequency

Calculate the stretching frequency.

$\begin{array}{c}{\nu }_{C-D}=0.733\times {\nu }_{C-H}\\ =0.733\times 3023\\ =2245.93\\ \approx 2246c{m}^{-1}\end{array}$

Stretching frequency of $C-D$is$2246c{m}^{-1}$