Q.35 - Organic Chemistry

Question

Consider the following E2 reaction.

a. Draw the by-products of the reaction and use curved arrows to show the movement of

electrons.

b. What happens to the reaction rate with each of the following changes? [1] The solvent is changed to DMF. [2] The concentration of ${\mathbf{\text{-OC(CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{3}}}$ is decreased. [3] The base is changed to ${}^{\mathbf{-}}\mathbf{\text{OH}}$. [4] The halide is changed to ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH(Br)CH}}}_{\mathbf{\text{3}}}$ [5] The leaving group is changed to ${\mathbf{\text{I}}}^{\mathbf{\text{–}}}$.

Step-by-Step Solution

Answer

a. The by-products of the reaction and curved arrows to denote the movement of electrons are shown below:-  

b.

[1] The rate of a reaction increases when the solvent is changed to DMF.

[2]  The rate of a reaction decreases when the concentration of  ${\text{-OC(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ decreases.

[3]  The rate of a reaction increases when the base is changed to  ${}^{-}\text{OH}$.

[4] The rate of a reaction increases when the halide is changed to  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(Br)CH}}_{\text{3}}$.

[5] The rate of a reaction increases when the leaving group is changed to  ${\text{I}}^{\text{–}}$.

E2 reaction is both substrate and base-dependent

E2 is a mode of elimination reaction where eliminations take place in a single concerted step.

The rate of a reaction is directly dependent upon the concentration of both the substrate and the base is hence known as the bimolecular elimination reaction.

 $\text{r ∝ }\left[\text{Substrate}\right]\left[\text{Base}\right]$

According to the above relation, when the concentration of the tert-butoxide base ( ${\text{-OC(CH}}_{\text{3}}{\text{)}}_{\text{3}}$) is decreased, the rate of a reaction decreases simultaneously.

Solvent dependent

E2 reactions are favorable in polar aprotic solvents such as DMSO and DMF. So the rate of a reaction increases when the solvent is changed from a polar protic medium. An example would be the change of alcohol to polar aprotic solvents, such as DMF.

Type of leaving group present

When the leaving group present in the structure has a larger size and strong conjugate basicity, then the rate of an E2 reaction increases.

When the leaving group is changed from  ${\text{Br}}^{-}$ to ${\text{I}}^{-}$ increases the rate of an E2 reaction since Iodide has a large size and strong conjugate basicity than bromide.

Type of the substrate present

Tertiary halides are the most reactive towards an elimination reaction, followed by secondary and primary halides. This is because tertiary halides give a more substituted product than the secondary and primary halides. 

An increase in substitution of a double bond increases the molecule's stability due to the +I effect provided by the alkyl group.

So when changed from primary halide to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(Br)CH}}_{\text{3}}$ (secondary halide) the rate of an E2 reaction increases.

Use of Base

When a strong base like hydroxides is used, then the rate of E2 reaction always increases. 

Hence keeping in mind all the points, the following changes that affect the rate of an E2 reaction are as follows:

a. The by-products of the reaction and use curved arrows to show the movement of electrons are shown below:-               

Representation of reaction scheme using curved arrows

b.

[1] The rate of a reaction increases when the solvent is changed to DMF.

[2] The rate of a reaction decreases when the concentration of  ${\text{-OC(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ decreases.

[3] The rate of a reaction increases when the base is changed to  ${}^{-}\text{OH}$.

[4] The rate of a reaction increases when the halide is changed to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH(Br)CH}}_{\text{3}}$.

[5] The rate of a reaction increases when the leaving group is changed to  ${\text{I}}^{\text{–}}$.