Q.11 - Organic Chemistry

Question

How does each of the following changes affect the rate of an E2 reaction?

a. tripling $\mathbf{\text{[RX]}}$

b. halving $\mathbf{\text{[B:]}}$

c. changing the solvent from ${\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\text{OH}}$ to DMSO 

d. changing the leaving group from $\mathbf{\text{I–}}$to  $\mathbf{\text{Br–}}$

e. changing the base from ${}^{\mathbf{-}}\mathbf{\text{OH}}$ to ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}$

f. changing the alkyl halide from ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\text{Br}}$ to ${\mathbf{\text{(CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}\mathbf{\text{CHBr}}$

Step-by-Step Solution

Answer

The following changes that affect the rate of an E2 reaction are as follows:

 

a. Tripling the concentration of RX triples the rate of an E2 reaction.

b. Halving the concentration of B: halves the rate of an E2 reaction.

c. Changing the solvent from  ${\text{CH}}_3\text{OH}$ to DMSO increases the rate of an E2 reaction.

d. Changing the leaving group from ${\text{I}}^{-}{\text{ to Br}}^{-}$ decreases the rate of an E2 reaction

e. Changing the base from ${}^{-}\text{OH}$ to ${\text{H}}_2\text{O}$ decreases the rate of an E2 reaction.

f. Changing the alkyl halide from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ to ${\text{(CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{CHBr}$ increases the rate of an E2 reaction.

E2 reaction is a base and substrate-dependent

E2 is a mode of elimination reaction where a single-step elimination takes place.

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

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

Solvent and base dependent

The rate of a reaction increases when the solvent is changed from a polar protic medium solvent such as methanol to polar aprotic solvents like DMSO. E2 reactions are favorable in polar aprotic solvents.

 

E2 is favorable for strong bases like hydroxides and unfavorable for weak bases like water. So, when the base is changed from a strong base like hydroxides to a weak base like water, the rate of an E2 reaction decreases.

Type of leaving group present

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

When the leaving group is changed from  decreases the rate of an E2 reaction since Bromide is has a small size and weak conjugate basicity than Iodide.

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 stability of the molecule due to the +I effect provided by the alkyl group. So, when changed from (primary halide) to  (secondary halide), the rate of an E2 reaction increases.

 

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

a. Tripling the concentration of RX triples the rate of an E2 reaction.

b. Halving the concentration of B: halves the rate of an E2 reaction.

c. Changing the solvent from  ${\text{CH}}_3\text{OH}$ to DMSO increases the rate of an E2 reaction.

d. Changing the leaving group ${\text{I}}^{-}{\text{ to Br}}^{-}$ decreases the rate of an E2 reaction

e. Changing the base from ${}^{-}\text{OH}$ to ${\text{H}}_2\text{O}$ decreases the rate of an E2 reaction.

f. Changing the alkyl halide from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ to ${\text{(CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{CHBr}$ increases the rate of an E2 reaction.