Q.15 - Organic Chemistry

Question

How does each of the following changes affect the rate of an E1 reaction?

a. doubling $\mathbf{\text{[RX]}}$ 

b. doubling $\mathbf{\text{[B:]}}$ 

c. changing the halide from ${\mathbf{\text{(CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{3}}}\mathbf{\text{CBr}}$ to ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\text{Br}}$

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

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

Step-by-Step Solution

Answer

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

 

a. Doubling the concentration of RX doubles the rate of an E2 reaction.

b. Doubling the concentration of B: doubles the rate of an E2 reaction.

c. Changing halide from ${\text{(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{CBr}$ to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ decreases the rate of an E2 reaction.

d. Changing the leaving group from ${\text{Cl}}^{\text{– }}{\text{to Br}}^{\text{–}}$ increases the rate of an E2 reaction

e. Changing the solvent from DMSO to ${\text{CH}}_{\text{3}}\text{OH}$ decreases the rate of an E2 reaction.

E1 reaction is only a substrate-dependent

E1 is a mode of elimination reaction where two-step eliminations take place.

The rate of a reaction is directly dependent upon the concentration of only the substrate and is hence known as the unimolecular elimination reaction.

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

Since the concentration of the substrate RX is doubled here, hence the rate of a reaction doubles according to the above relation.

When the concentration of base is doubled, the rate of the reaction remains the same according to the above relation.

Solvent dependent

E1 reactions are favorable in polar protic solvents such as water and methanol. So the rate of a reaction increases when the solvent is changed from a polar aprotic medium solvent such as DMSO to polar aprotic solvents such as methanol.

Type of leaving group present

When the leaving group is changed from ${\text{Cl}}^{\text{– }}{\text{to Br}}^{\text{–}}$ increases the rate of an E1 reaction since Bromide has a large size and strong conjugate basicity than Chloride.

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 ${\text{(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{CBr}$ (tertiary halide) to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ (primary halide), the rate of an E2 reaction decreases.

 

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

 

a. Doubling the concentration of RX doubles the rate of an E2 reaction.

b. Doubling the concentration of B: doubles the rate of an E2 reaction.

c. Changing halide from ${\text{(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{CBr}$ to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ decreases the rate of an E2 reaction.

d. Changing the leaving group from ${\text{Cl}}^{\text{– }}{\text{to Br}}^{\text{–}}$ increases the rate of an E2 reaction

e. Changing the solvent from DMSO to  ${\text{CH}}_{\text{3}}\text{OH}$ decreases the rate of an E2 reaction.