Organic Chemistry : Help with Rearrangement Reactions

Study concepts, example questions & explanations for Organic Chemistry

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Example Questions

Example Question #41 : Reactions Types

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Which of the following is the correct major product of the above reaction?

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Possible Answers:

Correct answer:

Explanation:

Here we see an  reaction with rearrangement. The bromine, an excellent leaving group, leaves the carbon chain and a carbo-cation (positively charged carbon) is formed on that carbon. A positive charge is more stable on a more substituted carbon, and so the positive charge rearranges itself onto the branched carbon. Essentially, the positive charge and a hydrogen on the branched carbon switched positions. The methanol was then free to attack the branched carbon to form the major product shown.

Example Question #50 : Reactions Types

What is the major product of the reaction shown?

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Possible Answers:

IV

I

V

III

II

Correct answer:

IV

Explanation:

This reaction adds  and  (eliminate II). The reaction is Markovnikov (Eliminate I). A hydride shift occurs putting the carbocation on the more substituted carbon before addition of (eliminate III and V).

Example Question #51 : Reactions Types

What is the major product of the reaction shown?

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Possible Answers:

II

IV

I

None of these

III

Correct answer:

II

Explanation:

After carbocation is formed, a rearrangement reaction stabilizes positive charge by putting it on a tertiary carbon. This is done by a methyl shift. Recall that tertiary carbocations are the most stable due to the inductive effect of alkyl groups on the electron-deficient carbocation.

Example Question #52 : Reactions Types

What is the major product of the reaction shown?

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Possible Answers:

Correct answer:

Explanation:

The first step of this reaction will be protonation of the hydroxyl oxygen to create a good leaving group. When the leaving group leaves, what's left is a secondary carbocation that is vicinal to (next to) a quaternary carbon. A methyl shift is thermodynamically favored in this case, as the rearrangement will leave a tertiary carbocation. Following the rearrangement the nucleophile (bromide) will attack the tertiary carbocation, forming a sigma bond with the carbon. The answer is thus .

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