Question

Assuming all the four valencies of carbon atom in propane pointing towards the corners of a regular tetrahedron; the distance (in $\overset{{}^\circ }{\mathop{A}}\,$) between the terminal carbon atoms in propane will be: (write the value of the nearest integer) Given: $C-C$ single bond length is 1.54 $\overset{{}^\circ }{\mathop{A}}\,$.

Answer 1

Hint: The 3 carbon chains of propane can have a cyclic structure. Due to this property of propane in this problem, the propane is assumed to be a cyclopropane, then by using trigonometric operations, the unknown values can be determined.

Complete step-by-step answer: We have been given a propane molecule, with $C-C$ single bond length is 1.54 $\overset{{}^\circ }{\mathop{A}}\,$. We have to find the distance between the terminal carbon atoms. So, let's assume that the propane molecule is in the form of a triangle. With A and B as terminal carbon atoms, and Z as the middle carbon atom. So, the figure will be as follows,

As we know that C-C bond angle is $109{}^\circ 28'$, so the value of $\theta =109{}^\circ 28'$, and the lengths of ZB = ZA = 1.54 $\overset{{}^\circ }{\mathop{A}}\,$.
Now, $\dfrac{AO}{AZ}=\sin \left( \dfrac{\theta }{2} \right)$ = $\left( \dfrac{109{}^\circ 28'}{2} \right)$ = $\sin \,54{}^\circ 44'$ = $\sin \,54.73{}^\circ $ ,
Therefore, AO = 0.816 $\times $ AZ
AO = 0.816$\times $1.54
AO = 1.257$\overset{{}^\circ }{\mathop{A}}\,$
Now, as we know AB = 2AO, therefore,
AB = 2$\times $1.257$\overset{{}^\circ }{\mathop{A}}\,$
AB = 2.514$\overset{{}^\circ }{\mathop{A}}\,$
Hence, the distance between terminal carbon atoms is calculated to be 2.514$\overset{{}^\circ }{\mathop{A}}\,$.

Note: The value of the atomic distances is measured in angstrom$\overset{{}^\circ }{\mathop{A}}\,$, which is $1\overset{{}^\circ }{\mathop{A}}\,={{10}^{-8}}cm\,={{10}^{-10}}m$. The value of the sin of angle $54.73{}^\circ $is found to be 0.816.