Q73. - Organic Chemistry

Question

Question: Answer the following questions about acetonitrile $\left({\mathrm{CH}}_{3}C\equiv N\right)$ :

a. Determine the hybridization of both C and N atoms. 

b. Label all bonds as $\mathit{\sigma }$ or$\mathbf{\pi }$  .

c. In what type of orbital does the lone pair on N reside? 

d. Label all bonds as polar or nonpolar.

Step-by-Step Solution

Answer

Answer

 

 a.  ${\mathrm{sp}}^3,\mathrm{sp},\mathrm{sp}$ respectively

 b.

 

 c.       $\mathrm{sp}$

d. 

Step-by-Step Solution Step 1: Determination of hybridization on carbon and nitrogen

The multiplicity of the bonds determines the hybridization around carbon and nitrogen. 

 

If the single bond is around carbon and nitrogen, then there is ${\mathrm{sp}}^3$  hybridization. If it is a double bond, then there is ${\mathrm{sp}}^2$  hybridization, and for the triple bond, there is $\mathrm{sp}$  hybridization.

Step 2: Determination of sigma and pi bonds

A single bond denotes a sigma bond, and the overlapping of the s-s orbital forms it.

 

The double bond denotes the pi bond, and it is formed by the overlapping of the p-p orbital. The triple bond contains two pi bonds and one sigma bond.

Step 3: Determination of polar and nonpolar bonds

The polar bonds are determined by the large electronegativity difference between the two atoms, which form the corresponding bond. 

 

For example, a bond between carbon and oxygen (C-O) is polar, and the bond between carbon and hydrogen (C-H) is nonpolar. 

Step 4: Information about acetonitrile

a. As discussed earlier, the single-bonded carbon is shown by ${\mathrm{sp}}^3$ hybridization, and triple-bonded carbon and nitrogen are shown by $\mathrm{sp}$  hybridization.

                                     

                                     Structure of acetonitrile 

 

b. Sigma bond is shown by a single bond, and a double bond shows pi bond.

                                                 

                                                                     Sigma and pi bonds

 

 

c. In the given structure, the nitrogen has   hybridization. Therefore, the lone pair of electrons reside in  sp -hybridized orbital.   

 

 

d. The bond between carbon and nitrogen is only the polar bond due to the greater electronegativity difference between nitrogen and carbon.

                                 

                                      Polar and nonpolar bonds