Question

Please help with drawing they are generally the same type of questions.

Solvolysis of 1-bromo-1-methylcyclopentane in methanol yields an initial intermediate, a second intermediate, and then the product (1-methoxy-1-methylcyclopentane) shown. Draw the structure of the second intermediate. Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars, including charges where needed. The single bond is active by default. Elimination and Substitution: Reaction of 2-bromo-2-methylbutane and EtONa Predict the major organic product formed when the compound shown below undergoes a reaction with EtONa. Interactive 3D display mode Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Draw the major product obtained when the following alkyl halide undergoes an E^2 reaction. If you expect no reaction to occur, submit the starting material as your answer. Interactive 3D display mode Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. Draw the major product obtained when the following alkyl halide undergoes an E_2 reaction. If you expect no reaction to occur, submit the starting material as your answer. Interactive 3D display mode Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. 2-Bromobutane is treated with sodium methoxide in methanol at 323 K. Draw the major product of the reaction. Make sure to consider the stereochemistry of the reaction. 100 degree C Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.

Answer 1

Ans-1)

Step-l) Formation of a very stable tertiary carbocaion As a solvent methanol is polar solvent the starting bromo compound will ionize on it's own accord and form a stabel 3 carbocation Br also a much good leaving group And hence first intermediate is expected to be this 3° carbocation only Br, CH CH3 H3COH Polar solvent 1-bromo-1 methylcyclopentane 30 carbocation Intermediate 1 Step-Il) This carbocation then get attacked by weakly nucleophilic Methanol molecules i.e. solvent moleule Intermediate -2 is stable enough as Methanol is very weakly basic. They don readily abstract this proton CH3 CH3 но 30 carbocation Intermediate 1 Intermediate-2 Step-llI) Deprotonation of intermediate-2 H2C он H3C CH3 CH3 1-methoxy-1-methylcyclopentane Intermediate-2

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