Question: Select the best conditions for the reactions.

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Expert Answer
Expert Answer

Solved Chegg Answers

General guidance

Concepts and reason

In general, primary alkyl halides undergo SN2{\rm{S}}{{\rm{N}}_{\rm{2}}}SN2​ reactions with strong nucleophiles and tertiary alkyl halides undergo SN1{\rm{S}}{{\rm{N}}_1}SN1​ reaction with weak nucleophiles. Fundamentals

Strong nucleophile is required for SN2{\rm{S}}{{\rm{N}}_{\rm{2}}}SN2​ reaction and these reactions are occurred in polar aprotic solvents (i.e. DMSO, DMF etc.).

Weak nucleophile is required for SN1{\rm{S}}{{\rm{N}}_1}SN1​ reaction. If a nucleophile is strong then there is a chance for the E2 product formation. Show less First Step | All Steps | Answer Only

Step-by-step

Step 1 of 2

Part a

The reaction is,

CH3
CH3
NaOCH3, DMSO
CH3
Br
H3C
НЫС
1-bromo-3-methylbutane
1-methoxy-3-methylbutane

Part a

Thus, the choice of reagent is NaOCH3,DMSO{\rm{NaOC}}{{\rm{H}}_{\rm{3}}}{\rm{,}}\;{\rm{DMSO}}NaOCH3​,DMSO .
Explanation

Here, sodium methoxide acts as a strong nucleophile and DMSO is polar aprotic solvent. 1-bromo-3-methylbutane reacts with NaOCH3,DMSO{\rm{NaOC}}{{\rm{H}}_{\rm{3}}}{\rm{,}}\;{\rm{DMSO}}NaOCH3​,DMSO to form 1-methoxy-3-methylbutane.

Step 2 of 2

Part b

The reaction is,

CH3
- CH3
ÇHz
|
Br
CH3OH
CH3
H₂C
H3C
2-bromo-2-methylbutane
2-methoxy-2-methylbutane

Part b

Thus, the choice of reagent is CH3OH{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}CH3​OH .
Explanation

Methanol is weak nucleophile and it reacts with 2-bromo-2-methylbutane to form 2-methoxy-2-methylbutane.

Answer

Part a

Thus, the choice of reagent is NaOCH3, DMSO

Part b

Thus, the choice of reagent is CH3OH

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