Q3 P. - Organic Chemistry (Mcmurry)

Question

Use the electronegativity values shown in Figure 2-2 to rank the following bonds from least polar to most polar: H₃C–Li, H₃C–K, H₃C–F, H₃C–MgBr, H₃C–OH

Step-by-Step Solution

Answer

                                                             Table 2-2

Using above table we can calculate the ΔEN and rank the answer in the increasing order of ΔEN. The most polar bond has the largest ΔEN. Thus, in order of increasing bond polarity:

                               ${\mathrm{H}}_3\mathrm{C}-\mathrm{OH}<{\mathrm{H}}_3\mathrm{C}-\mathrm{MgBr}<{\mathrm{H}}_3\mathrm{C}-\mathrm{Li}={\mathrm{H}}_3\mathrm{C}-\mathrm{F}<{\mathrm{H}}_3\mathrm{C}-\mathrm{K}$

Electronegativity

Electronegativity is a chemical property that describes the tendency of an atom or functional group to attract electron toward itself.it is dimensionless property because it is only a tendency.

Calculate &Delta;EN using figure 2-2

The polarity of bond is given by the difference in electronegativity between the two atoms that form said bond.The value of electronegativity for each element is given in table 2-2.

So, ΔEN value can be calculate as

Carbon:    EN = 2.5               Carbon:        EN = 2.5                   Fluorine:    EN = 4.0

Lithium:    EN = 1.0               Potassium:   EN = 0.8                   Carbon:      EN= 2.5

________________              __________________                  _________________

               ΔEN = 2.5                                 ΔEN = 1.7                     ΔEN = 1.5    

 

 

 

Carbon:    EN = 2.5                                    Oxygen:    EN = 3.5               

Magnesium:     EN = 1.2Carbon:   EN = 2.5   

_____________________                                 ___________________

                       ΔEN = 1.3                                                     ΔEN = 1.0

Calculate the rank of following bond from least polar to more polar

Using above table we can calculate the ΔEN and rank the answer in the increasing order of ΔEN. The most polar bond has the largest ΔEN. Thus, in order of increasing bond polarity:

    

                  ${\mathrm{H}}_3\mathrm{C}-\mathrm{OH}<{\mathrm{H}}_3\mathrm{C}-\mathrm{MgBr}<{\mathrm{H}}_3\mathrm{C}-\mathrm{Li}={\mathrm{H}}_3\mathrm{C}-\mathrm{F}<{\mathrm{H}}_3\mathrm{C}-\mathrm{K}$