Chapter 8.2, Problem 4E

Interpretation Introduction

Interpretation:Thewave number for C-C stretching vibration needs to be determined, given that vibration for C=C is 1640 cm^-1.The wave number for carbon -carbon triple bond needs to be determined wherein the k is appropriate multiple of carbon-carbon double bond.

Concept introduction:

There is large variance in force constant with small differences in bond length. The wavenumber is inversely proportional to square root of reduced mass. Hence, when there is increase in reduced mass, there is decrease in wave numbers.

Answer to Problem 4E

The wave number for C-C bond is 1159.6 cm^-1 and stretching frequency of C-C triple bond is 2009 cm^-1.

Explanation of Solution

The relation between wave number and reduced mass is depicted in equation

  $\overline{\nu }=\frac{1}{2\pi c}\sqrt{\frac{k}{m^{*}}}\mathrm{...............}(1)$

Where

  $\begin{array}{l}\overline{\nu }\text{ = wave number}\\ \text{k = force constant}\\ m^{*}\text{ = reduced mass}\\ \text{c = speed of light}\end{array}$

Computing for reduced mass, the equation is

  $m^{*}\text{ = }\frac{{\text{m}}_{\text{c}}^{}{\text{ x m}}_c}{{\text{m}}_{\text{c}}^{}{\text{ + m}}_c}\mathrm{................}(2)$

Where

  ${\text{m}}_{\text{c}}^{}\text{ = atomic mass of carbon}$

Substituting atomic mass of carbon as 12 and

  $m^{*}\text{ = }\frac{\text{12g x 12g}}{12\text{g + 12 g}}=\frac{144g}{24g}=6g$

From equation (1), all the factors are constant except reduced mass and the reduced mass is inversely proportional to square root of reduced mass.

  $\frac{{\overline{\nu }}_{C-C}}{{\overline{\nu }}_{C=C}}=\sqrt{\frac{k_{(C-C)}{m}^{*}{}_{(C-C)}}{k_{(C=C)}{m}^{*}{}_{(C-C)}}}\mathrm{.....................}(3)$

Where,

  $\begin{array}{l}{\overline{\nu }}_{C-C}\text{ stands for wave number for single bonds}\\ {\overline{\nu }}_{C=C}\text{ stands for wave number for double bonds}\end{array}$

But given that $k_{(C=C)}=2k_{(C-C)}$ and ${\overline{\nu }}_{C=C}=1640{\text{ cm}}^{-1}$

  $\begin{array}{l}\frac{{\overline{\nu }}_{C-C}}{{\overline{\nu }}_{C=C}}=\sqrt{\frac{k_{(C-C)}{m}^{*}{}_{(C-C)}}{k_{(C=C)}{m}^{*}{}_{(C-C)}}}\\ \frac{{\overline{\nu }}_{C-C}}{1640c{m}^{-1}}=\sqrt{\frac{k_{(C-C)}{m}^{*}{}_{(C-C)}}{2k_{(C-C)}{m}^{*}{}_{(C-C)}}}\\ \frac{{\overline{\nu }}_{C-C}}{1640c{m}^{-1}}\text{= 0}\text{.707}c{m}^{-1}\\ {\overline{\nu }}_{C-C}=1159.65c{m}^{-1}\end{array}$

It is known that force constant for C-C triple bond is three times that of C-C single bond.

  $\begin{array}{l}{k}_{(C=C)}=2k_{(C-C)}\\ k_{(C-C)}=\frac{k_{(C=C)}}{2}\\ \end{array}$

Hence, computing for C-C triple bonds.

  $k_{(C\equiv C)}=3k_{(C-C)}=\frac{3k_{(C=C)}}{2}$

  $k_{(C\equiv C)}=\frac{3k_{(C=C)}}{2}$

For triple bond, the equation is

  $\begin{array}{l}\frac{{\overline{\nu }}_{C\equiv C}}{{\overline{\nu }}_{C=C}}=\sqrt{\frac{k_{(C\equiv C)}{m}^{*}{}_{(C-C)}}{k_{(C=C)}{m}^{*}{}_{(C-C)}}}=\sqrt{\frac{3k_{(C=C)}{m}^{*}{}_{(C-C)}}{2k_{(C=C)}{m}^{*}{}_{(C-C)}}}\\ \frac{{\overline{\nu }}_{C-C}}{1640c{m}^{-1}}\text{= }\sqrt{\frac{3}{2}}=1.224\\ {\overline{\nu }}_{C-C}\text{ = 2008}{\text{.6cm}}^{-1}\\ \end{array}$

Hence, the stretching frequency of C-C triple bond is 2009cm^-1.

Conclusion

Thus, the wave number for C-C bond is 1159.6 cm^-1 and stretching frequency of C-C triple bond is 2009 cm^-1.