Can somone help? It says I got 3 of the 20 blanks incorrect. dimethyl sulfoxideUnlike alcohols which are very unreactive owing to the inability of the hydroxyl to break away onhydroxylits own, alkyl bromides are very active because the halides are generally good leaving groups.Halides, therefore, are very useful in synthesis because they can be converted to otheraproticfunctional groups . The problem, however, is the difficulty in controlling their behaviors becausethey react in various pathways, such as SN1, SN2, E1 and E2.acceleratesteric hindrancepolarSN2 mechanism occurs when the leaving group is beaking away as the nucleophile attacks thealpha C. This mechanism, therefore, requires that the nucleophile attacks the alpha carbon at thebackside because frontside attack inhibits the interaction of the nucleophile and the alpha C becauseRate = k[RX]|nucleophile]primaryof steric hindrance due to the bulky halide leaving group. This backside attack also results intot - Buoconfiguration inversion of the product. SNI, on the other, is not limited to backside attack becausestrong baseby the time the nucleophile attacks, the leaving group is already gone. The problem, however,with SN1 is the possible proliferation of multiple products due to carbocation rearrangement.weak baseleaving groupTherefore, a tertiary alkyl halide reacts with a nucleophile via SN1, while a primary reacts viaalpha carbonSN2.E2SN2 reaction rates also vary with the strength of the nucleophile. This idea is validated by thegoodexperimental rate equation for SN2 which shows that Rate = k[RX][nucleophile] . Therefore, afunctional groupsprimary alkyl halide needs a nucleophile , but a strong base . If the nucleophile is a strongstrong nucleophilebase as well, such as t - Buo . primary alkyl halides undergo elimination instead via E2carbocationThe solvent affects the SN2 rate as well. Solvents in SN2 reactions of alkyl halides are generalinversionstrong base , but aprotic . Protic solvents turn bases into poor nucleophile. Therefore, proticRate = k[RX)solvents , such as alcohols and water slow down SN2 reactions. However, dimethyl sulfoxidenucleophileis a very good solvents for SN2 reactions.slow downtertiary

To fill the blanks with the appropriate choices.

Unlike alcohols which are very unreactive owing to the inability of the hydroxyl to break away on its own, alkyl bromides are very active because the halides are generally good leaving groups. Halides, therefore, are very useful in synthesis because they can be converted to other functional groups. The problem, however, is the difficulty in controlling their behaviors because they react in various pathways, such as SN1, SN2, El and E2. SN2 mechanism occurs when the leaving group is beaking away as the nucleophile attacks the alpha C. This mechanism, therefore, requires that the nucleophile attacks the alpha carbon at the backside because frontside attack inhibits the interaction of the nucleophile and the alpha C because of steric hindrance due to the bulky halide leaving group. This backside attack also results into configuration inversion of the product. SN1, on the other, is not limited to backside attack because by the time the nucleophile attacks, the leaving group is already gone. The problem, however, with SN1 is the possible proliferation of multiple products due to carbocation rearrangement. lide reacte clen SN1 roact

Unlike alcohols which are very unreactive owing to the inability of the hydroxyl to break away on its own, alkyl bromides are very active because the halides are generally good leaving groups. Halides, therefore, are very useful in synthesis because they can be converted to other functional groups. The problem, however, is the difficulty in controlling their behaviors because they react in various pathways, such as SN1, SN2, El and E2. SN2 mechanism occurs when the leaving group is beaking away as the nucleophile attacks the alpha C. This mechanism, therefore, requires that the nucleophile attacks the alpha carbon at the backside because frontside attack inhibits the interaction of the nucleophile and the alpha C because of steric hindrance due to the bulky halide leaving group. This backside attack also results into configuration inversion of the product. SN1, on the other, is not limited to backside attack because by the time the nucleophile attacks, the leaving group is already gone. The problem, however, with SN1 is the possible proliferation of multiple products due to carbocation rearrangement. lide reacte clen SN1 roact

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter15: Benzene And Aromaticity

Section15.SE: Something Extra

Problem 31AP

See similar textbooks