Q10P - Organic Chemistry (Mcmurry)

Question

Draw a line-bond structure for propene, ${\mathbf{CH}}_3\mathbf{CH}\mathbf{=}{\mathbf{CH}}_2$. Indicate the hybridization of the orbitals on each carbon, and predict the value of each bond angle.

Step-by-Step Solution

Answer

Line bond structure of propene

Carbons with their respective hybridization

                                     ${\mathrm{sp}}^3$Hybridized C atom bond angle

                                        $s{p}^2$Hybridized carbon atom bond angle

Step 1: Line bond structure

The line bond structure is a shorthand structure with covalent bonds represented by lines and atoms being shown by their respective symbols.

Step 2: Propene

The line bond structure of propene is represented as:

Line bond structure of propene

When an atom makes four covalent bonds. It is ${\mathrm{sp}}^3$ hybridized and makes bond angles of approximately ${109.5}^{°}$. The  hybridized carbon atom of propene is as shown:

                                       ${\mathrm{sp}}^3$  Hybridized C atom bond angle

When an atom makes two single bonds and one double bond, it is hybridized and makes bond angles of approximately. The hybridized carbon atoms of propene is as shown:

                                           ${\mathrm{sp}}^3$ Hybridized carbon atom bond angle

There are two bonds $C_1-C_2$. One is a sigma bond formed by overlap of an ${\mathrm{sp}}^2$ orbital of carbon 1 with  ${\mathrm{sp}}^2$ orbital of carbon 2. The other is a pi bond formed by overlap of a p orbital of carbon 1 with a p orbital of carbon 2. All four atoms connected to the carbon-carbon double bond lie in the same plane, and all bond angles between these atoms are ${120}^{°}$. The bond angle between hydrogen and the ${\mathrm{sp}}^3$ hybridized carbon is ${109}^{°}$.

                                      Carbons with their respective hybridization