Show how you might synthesize this compound from an alkyl bromide and a nucleophile in an SN2 reaction. Use the wedge/hash bond tools to indicate stereochemistry where it exists. Only draw the reactants. Separate multiple reactants using the + sign from the drop-down menu. If there is more than one possible combination of alkyl bromide and nucleophile, draw only one combination. Do not include counter-ions, e.g., Na+, I-, in your answer.

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25-12-2019
Chemistry

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Show how you might synthesize this compound from an alkyl bromide and a nucleophile in an SN2 reaction. Use the wedge/hash bond tools to indicate stereochemistry where it exists. Only draw the reactants. Separate multiple reactants using the + sign from the drop-down menu. If there is more than one possible combination of alkyl bromide and nucleophile, draw only one combination. Do not include counter-ions, e.g., Na+, I-, in your answer.

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manuelacastro87 manuelacastro87 · Answer 1 · 2019-12-26T00:43:01+01:00

Answer:

See the image 1

Explanation:

If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. (see image 2)

The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. (see image 3)

What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product.

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-11-03T19:14:01+01:00

Show how you might synthesize this compound from an alkyl bromide and a nucleophile in an SN2 reaction. The 3-bromo-2-methylpentane is the alkyl bromide and azide is the nucleophile.

The mechanism of the reaction starts with the leaving group(bromine), which is followed by the azide(nucleophile) that attacks the electrophilic carbon at the bromine leaving the site.

The SN2 mechanism is used in this reaction, in which azide combines with alkyl bromide to generate the desired product. The reaction occurs by a coordinated process in which sodium azide, acting as a nucleophile, attacks the carbon-bearing bromine group while concurrently removing the bromine atom, generating an SN2 product.

Therefore, we can conclude that the 3-bromo-2-methylpentane is the alkyl bromide and azide is the nucleophile.

Learn more about nucleophiles here:

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