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Redrawn structures with chemfig (uniform font size, geometry and scaling)
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andselisk ♦
andselisk ♦
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questionReaction between 1,7‐dichloroheptan‐4‐one and alcoholic KOH

The given answer is 1,1-dicyclopropylmethanone:

1,1-dicyclopropylmethanone1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is whyWhy not make a more stable five membered-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one:?

2-ethenylcyclopentan-1-one2-ethenylcyclopentan-1-one

question

The given answer is 1,1-dicyclopropylmethanone:

1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is why not make a more stable five membered ring in the first step and then again perform E2 to get 2-ethenylcyclopentan-1-one:

2-ethenylcyclopentan-1-one

Reaction between 1,7‐dichloroheptan‐4‐one and alcoholic KOH

The given answer is 1,1-dicyclopropylmethanone:

1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?

2-ethenylcyclopentan-1-one

specific title; add alt text; \text(mathjax)=>plain text; tags
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Gaurang Tandon
Gaurang Tandon
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In the given reaction, why Why are two cyclopropylthree-membered rings preferred to a single five membered-membered ring in the reaction of a dihalide with alcoholic KOH?

enter image description here

question

The given answer is 1,1-dicyclopropylmethanone:

enter image description here1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is why not make a more stable five membered ring in the first step and then again perform $\text{E2}$ toE2 to get 2-ethenylcyclopentan-1-one:

enter image description here2-ethenylcyclopentan-1-one

In the given reaction, why are two cyclopropyl rings preferred to a five membered ring?

enter image description here

The given answer is:

enter image description here

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is why not make a more stable five membered ring in the first step and then again perform $\text{E2}$ to get:

enter image description here

Why are two three-membered rings preferred to a single five-membered ring in the reaction of a dihalide with alcoholic KOH?

question

The given answer is 1,1-dicyclopropylmethanone:

1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is why not make a more stable five membered ring in the first step and then again perform E2 to get 2-ethenylcyclopentan-1-one:

2-ethenylcyclopentan-1-one

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Archer
Archer
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In the given reaction, why are two cyclopropyl rings preferred to a five membered ring?

enter image description here

The given answer is:

enter image description here

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

My question is why not make a more stable five membered ring in the first step and then again perform $\text{E2}$ to get:

enter image description here