Question

What Solvent is Best For\[E2\] Reaction?

Answer 1

Hint: Solvent plays a crucial role in reaction mechanism. It affects the rate of reactant towards a particular mechanism. Solvents are classified into polar and non-polar solvent. Polar solvents have charges due to electronegativity difference although they are not ionic compounds. For example, ${H_2}0$ is a polar molecule with hydrogen carrying a slight positive charge and oxygen carrying a slight negative charge due to greater electronegativity of oxygen than hydrogen. In contrast, non-polar solvents are perfectly covalent with no charges on them. Examples of non-polar molecules are \[CC{l_4}\], benzene, cyclohexane, etc.

Complete Step by Step Solution:

non-polar solvents are extremely rejected for \[E2\] elimination as non-polar molecules destabilise the transition state formed between the reactant and the product. They tend to decrease the rate of the reaction by increasing the activation energy of the transition state, hence are not used for \[E2\] elimination.

Polar solvents are used for \[E2\] elimination. Polar solvents are further of two types- polar protic and polar aprotic. Protic refers to protons that are hydrogen. Polar protic has hydrogen whereas polar aprotic does not have hydrogen. Examples of polar aprotic solvents are DMSO, DMF, Acetone. Examples of polar protic are $N{H_3}$and${H_2}0$.
Polar protic favours elimination over substitution. Whereas polar aprotic favours substitution over elimination. This is due to hydrogen bonding that the rate increases and elimination is favoured.

Note: Electronegativity of carbon is \[2.55\]and electronegativity of chlorine is \[3.16\]. Thus, the difference is about \[3.16 - 2.55{\text{ }} = 0.61\]. Thus, \[CC{l_4}\] is a polar molecule and there is an electronegativity difference between carbon and chlorine but it is not that large enough to bring polarity in the molecule. Electronegativity of oxygen is \[3.44\] and electronegativity of \[2.2\]. Thus, the difference \[ = 1.24\] which is greater than \[0.61\]. Hence the molecule is polar.