Q75. - Organic Chemistry

Question

Question: The principles of this chapter can be applied to organic molecules of any size. Answer the following questions about amoxicillin, an antibiotic from the penicillin family: 

                                               

a. Predict the hybridization and geometry around each highlighted atom.

 b. Label five polar bonds using the symbols  ${\mathbf{\delta }}^{\mathbf{+}}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{ }{\mathbf{\delta }}^{\mathbf{-}}$ . 

c. How many π bonds does amoxicillin have? Label them.  

d. Find a C- H bond containing a carbon atom having a hybrid orbital with 33% s character.

Step-by-Step Solution

Answer

Answer

  a.                    

b.                                       

 

 c. Six

 

d.  The part of amoxicillin that has -hybridized C-H bond is shown below:

                                                

   

Step-by-Step Solution Step 1: Determination of hybridization and geometry

The hybridization and geometry are determined as follows:

• The single-bonded carbon has ${\mathrm{sp}}^3$ hybridization and tetrahedral geometry.
• The double-bonded carbon has${\mathrm{sp}}^2$ hybridization and trigonal planar geometry.

• The triple-bonded carbon has $\mathrm{sp}$ hybridization and linear geometry

Step 2: Existence of dipole

The high electronegative atoms attract the shared electrons and distort the electron cloud. 

 

A dipole is created by the partial positive charge$\left({\delta }^{+}\right)$ (acquired by the less electronegative atom) and the partial negative charge $\left({\delta }^{-}\right)$(acquired by the highest electronegative atom).   

Step 3: Information about amoxicillin

a. The double-bonded carbon atom and nitrogen atom have ${\mathrm{sp}}^2$hybridization and trigonal planar geometry. The single-bonded carbon atom and nitrogen have ${\mathrm{sp}}^{3^{}}$hybridization and tetrahedral geometry.

 

b. The electronegative value is ranked in the order given below:

  $\mathrm{H}\simeq \mathrm{C}<<\mathrm{N}<\mathrm{O}$       

Therefore, the dipole is created between the atoms in the following manner:

                               

                                                                               Dipole of amoxicillin

 

 

c. The double bonds represent the pi bonds. Therefore, six $\mathrm{\pi }$bonds are present in the structure.

                                       

                                                                            Six pi bonds in amoxicillin

 

d. Each ${\mathrm{sp}}^2$hybrid orbital has 33% s character. Therefore, the hydrogen bonded with ${\mathrm{sp}}^2$ hybridized carbon has 33% s character

                                                                           

                                                        ${\mathrm{sp}}^2$   hybridized C-H bond that contains 33% s character