Question: Suppose you have compounds A–D at your disposal. Using these compounds, devise two different ways to make E. Which one of these methods is preferred, and why?
The reaction of D with A is preferred due to the unhindered methyl halide.
An ion is said to be a nucleophile if it has a lone pair donation tendency. Therefore, nucleophile slowly attacks at the electron-rich crowded place.
The nucleophile is an electron-rich species. Therefore, it reacts at a very slow rate with an electron-crowded substrate. For example, tertiary alkyl halide reacts very slowly with the nucleophile phenoxide ion.
The nucleophiles react fast with the electrophilic compounds. For example, phenoxide reacts fast with bromoethane.
The given compound A is an alkyl halide. It has electrophilic carbon at C-I bond. Therefore, it cannot act as a nucleophile but can act as a substrate at which the nucleophile phenoxide ion (D) attacks quickly and forms the desired product (E).
The reaction is preferred because the nucleophile attacks at a fast rate at a less crowded alkyl halide.
The given compound B is a tertiary alkyl halide (electron crowded). It has electrophilic carbon at the C-I bond.
Therefore, it cannot act as a nucleophile but can act as a substrate at which nucleophile methoxide ion (C) attacks and forms the desired product.
But due to the crowded compound, the methoxide slowly reacts with the B and forms the desired product (E). Therefore, this reaction is not the preferred reaction.
Question: Quinapril (trade name Accupril) is used to treat high blood pressure and congestive heart failure. One step in the synthesis of quinapril involves the reaction of the racemic alkyl bromide A with a single enantiomer of the amino ester B. (a) What two products are formed in this reaction? (b) Given the structure of quinapril, which one of these two products is needed to synthesize the drug?
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