Identify A and B respectively in the following conversion
Ethene $\xrightarrow{A}$ Bromoethane $\xrightarrow{B}$ Ethyl propionate
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In organic reactions, always check the reactivity of the reagents with the functional groups in the starting material and product. In this case, HBr and silver acetate are the appropriate reagents.
Step 1: Understanding the reaction mechanism.
The reaction starts with ethene (C$_2$H$_4$), which reacts with an electrophile to form bromoethane (C$_2$H$_5$Br) in the first step. The appropriate reagent to add a bromine atom to ethene is HBr. In the second step, the bromoethane reacts with a nucleophile to form ethyl propionate. The nucleophile in this case is C$_2$H$_5$COOAg (silver acetate), which displaces the bromine atom to form the ester. Step 2: Analyzing the options. (A) Br$_2$/AlBr$_3$, C$_2$H$_5$ONa: This is not correct, as the reaction does not involve a bromine addition via AlBr$_3$ in the first step. (B) Br$_2$/AlBr$_3$, C$_2$H$_5$COOAg: This reagent combination is also incorrect, as the addition of Br$_2$ does not occur in this mechanism. (C) HBr, C$_2$H$_5$ONa: This is not correct because sodium ethoxide (C$_2$H$_5$ONa) would not be the nucleophile in the second step for ester formation. (D) HBr, C$_2$H$_5$COOAg: This is the correct combination, as HBr adds across the double bond of ethene to form bromoethane, and then silver acetate (C$_2$H$_5$COOAg) displaces the bromine atom to form ethyl propionate. Step 3: Conclusion.
The correct answer is (D) HBr, C$_2$H$_5$COOAg, as it correctly describes the reagents for the two steps in the reaction mechanism.
