Explain why the product distribution is observed for the following reactions: **Title:** Understanding Product Distribution in Elimination Reactions**Introduction:**In organic chemistry, elimination reactions often lead to a mixture of products. The distribution of these products can be influenced by various factors such as the structure of the substrate, the type of base used, and reaction conditions. Here, we explore two reactions involving potassium tert-butoxide (KOtBu) to understand the observed product distribution.**Reaction Details:**1. **First Reaction:** - **Substrate:** 2-bromobutane (a secondary alkyl halide). - **Base:** KOtBu. - **Products and Distribution:** - Major product: 2-butene (70%). - Minor product: 1-butene (30%). **Explanation:** - The more substituted alkene, 2-butene, is favored due to Zaitsev's rule, which states that the more substituted alkene is more stable and therefore, typically the major product.2. **Second Reaction:** - **Substrate:** 2-bromo-2-methylbutane (a tertiary alkyl halide). - **Base:** KOtBu. - **Products and Distribution:** - Major product: 2-methyl-2-butene (99%). - Minor product: 2-methyl-1-butene (1%). **Explanation:** - This reaction highly favors the formation of the more substituted alkene, 2-methyl-2-butene. The bulky base KOtBu also plays a critical role in selectively forming the more stable alkene.**Conclusion:**In both reactions, the distribution of products aligns with Zaitsev's rule. The steric effect of the KOtBu base favors the formation of more substituted, stable alkenes. Understanding these principles is crucial for predicting product outcomes in elimination reactions.

This can be explain easily by the mechanism of the reaction. In the first reaction when E or trans product form the transition sate (I) of the reaction is very stable as compare to the transition sate of the Z or cis product. In transition sate I the bulky CH3 group are situated far away from each other and Hydrogen and Bromine are anti periplanar to each other, so reaction is favorable and the transition energy is low. But the transition sate II is unfavorable because the two bulky group are in the same side create steric hindrance and also Br and H in the same side, so the transition energy is very high and it formed the minor product as 30% . Same in case of reaction 2 here trans or E product form more predominantly (90%) , because here the two group is more bulky compare to first reaction. so the transition sate of Z or cis product is less favorable so also product form less amount.