Question

When 2-methyl butyl bromide is treated with sodium ethoxide in ethanol, what will be the major product?
A) 2-methyl but-2-ene
B) 3-methyl but-1-ene
C) 2-methyl but-1-ene
D) 2-methyl sodium-butoxide

Answer 1

In${{E}_{2}}$elimination reaction, the base abstracts the acidic proton from alkyl halide followed by the removal of a halide to give the alkene. The major product is obtained by following Saytzeff's rule. The highly substituted alkene across the double bond is the major product.

Complete step by step answer:
Elimination reactions are a type of reaction in which the substituents are removed from the molecule. It can be either a one-step or a two-step reaction. The one-step reaction is called the E2 elimination. It follows the concentrating mechanism. It is a bimolecular reaction. E2 mechanism results in the formation of a pi bond.
The E2 elimination reaction is carried out in presence of a strong base. Such that the base must have the ability to remove an acidic proton from the molecule.
Here, the 2-methyl butyl bromide which is alkyl halide undergoes the reaction with sodium ethoxide which acts as a base in presence of ethanol. This is an example of E2 elimination. The reaction for the conversion of the alkyl halide to an alkene is given as:

& \begin{matrix} {} & {} & {} & \text{Br} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} \\\ {} & {} & {} & \text{ }\\!\\!|\\!\\!\text{ } & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} \\\ \text{C}{{\text{H}}_{\text{3}}} & -\text{CH} & -\text{CH} & -\text{CH} & \xrightarrow[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}]{\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{ONa}} & \text{C}{{\text{H}}_{\text{3}}} & -\text{C} & \text{=CH} & -\text{C}{{\text{H}}_{\text{2}}} & \text{+} & \text{C}{{\text{H}}_{\text{3}}} & -\text{CH} & -\text{CH} & \text{=C}{{\text{H}}_{\text{2}}} & {} \\\ {} & \text{ }\\!\\!|\\!\\!\text{ } & {} & \text{ }\\!\\!|\\!\\!\text{ } & {} & {} & \text{ }\\!\\!|\\!\\!\text{ } & {} & {} & {} & {} & \text{ }\\!\\!|\\!\\!\text{ } & {} & {} & {} \\\ {} & \text{C}{{\text{H}}_{\text{3}}} & {} & \text{H} & {} & {} & \text{C}{{\text{H}}_{\text{3}}} & {} & {} & {} & {} & \text{C}{{\text{H}}_{\text{3}}} & {} & {} & {} \\\ {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} & {} \\\ \end{matrix} \\\ & \begin{matrix} \text{(2-Methyl butyl bromide)} & \text{ (2-Methyl but-2-ene)} & \text{ (} \\\ \end{matrix}\text{3-Methyl but -1-ene)} \\\ \end{aligned}$$ Here, the base sodium ethoxide abstracts the proton from the $\beta $ position followed by the removal of $$\text{Br}$$the ion $$\text{HBr}$$. In 2-methyl butyl bromide there are two $\beta $ positions possible that correspond to the halogen. Each β can lose its hydrogen in the form of $$\text{HBr}$$ to form a double bond between the alpha and beta position. Since there are two $\beta $ positions. It is more likely to form two alkene products as shown in the reaction. Now, let's apply a saytzeff's rule for the determination of a major product. From the alkenes obtained, the (I) alkene which is 2-methyl but-2-ene has the minimum number of hydrogen across the double bond. Therefore it is the major product of the reaction. Thus the major product obtained is 2-methyl but-2-ene. **Hence, (A) is the correct option.** **Additional information:** During the elimination reaction of alkyl halide and alcohol different alkenes are formed. Saytzeff’s or Zaitsev's rule predicts the major product. The Saytzeff’s or Zaitsev's rule states that out of the alkenes formed the one with less number of hydrogen atoms bonded across the double bond or the most substituted double bond is the major product of the reaction. The empirical rule to predict the major alkene product in an elimination reaction. According to rule, the major product of alkene is the one that corresponds to the removal of hydrogen from the alpha position carbon which is having less number of hydrogen. **Note:** Here use the saytzeff's rule for the determination of major products. The highly substituted alkene is one which has less number of hydrogen across the double bond. Try to follow the path of mechanism. The hydrogen can be removed from the two adjacent positions.