Q10.77CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the following compounds, the C atoms form a single ring. Draw a Lewis structure for each compound, identify cases for which resonance exists, and determine the carbon-carbon bond order(s):

1.  ${\mathbf{C}}_3{\mathbf{H}}_4$
2. ${\mathbf{C}}_3{\mathbf{H}}_6$
3. ${\mathbf{C}}_4{\mathbf{H}}_6$
4. ${\mathbf{C}}_4{\mathbf{H}}_4$
5. ${\mathbf{C}}_6{\mathbf{H}}_6$
   

Step-by-Step Solution

Answer

a.

Here, the resonance does not exist. The C-C bond order for ${\mathbf{C}}_3{\mathbf{H}}_4$   is 1.5.

b.

Here, the resonance does not exist. The C-C bond order for ${\mathbf{C}}_3{\mathbf{H}}_6$   is 1.

c.

Here, the resonance does not exist. The C-C bond order for ${\mathbf{C}}_4{\mathbf{H}}_6$  is 1.

d.

Here, the resonance exists. The C-C bond order for  ${\mathbf{C}}_4{\mathbf{H}}_4$ is 1.5.

e.

Here, the resonance exists. The C-C bond order for ${\mathbf{C}}_6{\mathbf{H}}_6$  is 1.5.

Step 1: Definition

Lewis structures are those diagrams that show the bonding between atoms of a molecule, as well as lone pair of electrons that may exist in the molecule.

For resonance to exist in a molecule, there should be presence of conjugation (two alternate double bond or one single bond with some charge on the next atom).

Example; pi-pi conjugation.

Step 2: Subpart (a) Lewis’s structure of C 3 H 4

In the given molecule, there are 3 carbon atoms and 4 hydrogen atoms, so the ring will be three-membered having a double bond in the ring. The Lewis structure of  ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$ is as follows.

Since it has one double bond and 2 single bonds, the order is more than 1 between two atoms. Here, the bond order is 1.3.

 

 $=\frac{2+1+1}{3}=\frac{4}{3}=1.333$

 

In the given structure, there is no pi-pi conjugation, therefore here resonance does not exist. 

Step 3: Lewis’s structure of C 3 H 6

In the given molecule, there are 3 carbon atoms and 6 hydrogen atoms so the ring will be three-membered with saturation. The Lewis structure of   ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}$ is as follows.

As all are single bonds present in the ring, so bond order is 1.

In the given structure, there is no pi-pi conjugation, therefore here resonance does not exist. 

 

Step 4: Lewis’s structure of C 4 H 6

In the given molecule, there are 4 carbon atoms and 6 hydrogen atoms so the ring will be four-membered with saturation. The Lewis’s structure of  ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is as follows.

As all are single bonds present in the ring, so bond order is 1.

In the given structure, there is no pi-pi conjugation, therefore here resonance does not exist. 

Step 5: Lewis’s structure of C 4 H 4

In the given molecule, there are 4 carbon atoms and 4 hydrogen atoms so the ring will be four-membered having two double bonds.

Hence,  ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}$ has 2 double bond (bond order 2) and 2 single bonds (bond order1), resonance occurs in ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}$ , due to of which all bonds have partial double bond character and partial single bond character. Therefore, bond order is 1.5.

 $=\frac{2+2+1+1}{4}=\frac{6}{4}=1.5$

In the given structure, there is pi-pi conjugation, therefore here resonance exists. 

Step 6: resonance structure of C 6 H 6

In the given molecule, there are 6 carbons and 6 hydrogens so the ring will be six-membered with three double bonds. The Lewis structure is as given below.

The delocalization (resonance) is present in ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}$ . It has partial double bond and partial single bond character. Therefore, the bond order is 1.5.

 

 $=\frac{2+2+2+1+1+1}{6}=\frac{9}{6}=1.5$

In the given structure, there is pi-pi conjugation, therefore here resonance exists.