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Q.43

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Organic Chemistry
Found in: Page 327

Short Answer

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.

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.

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Step by Step Solution

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  hydrogens

With more number of different types of hydrogens present, different products will be formed. As two different types of 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

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