Q.43 - Organic Chemistry

Question

Elimination of HBr from 2-bromobutane affords a mixture of but-1-ene and but-2-ene. With sodium ethoxide as base, but-2-ene constitutes 81% of the alkene products, but with potassium tert-butoxide, but-2-ene constitutes only 67% of the alkene products. Offer an explanation for this difference.

Step-by-Step Solution

Answer

Potassium tert-butoxide is a bulky base than sodium ethoxide, so it tends to take protons from primary carbon to avoid getting suffered from steric repulsion. So, the concentration of but-2-ene decreases.

Sodium ethoxide abstracts a proton from secondary carbon to give products where the concentration of but-2-ene is much higher.

Types of base used

Sodium ethoxide is not a bulky base, so it takes up protons from the secondary carbon to give a more substituted stable but-2-ene product. Hence the concentration of but-2-ene is much higher in this case. 

But when potassium tert-butoxide is used, it is a bulky base that abstracts a proton from the less hindered side, i.e., from the primary carbon to give but-1-ene in some more amount. As a result, the concentration of but-2-ene gets decreased in this case.

Presence of the β hydrogens

With more number of different types of  $\mathbf{\text{β}}$ hydrogens present, different products will be formed. As two different types of $\text{β}$  hydrogens are present, so two different products are formed. 

Based on the explanations, potassium tert-butoxide is a bulky base than sodium ethoxide, so it tends to take protons from primary carbon to avoid suffering from steric repulsion. 

So, the concentration of but-2-ene decreases in this case, although it is more stable due to more substitution (Zaitsev rule). 

While sodium ethoxide is not a bulky base abstracts proton from secondary carbon to give products where the concentration of but-2-ene is much higher.

Representation of the elimination and the substituted products