What are the Bond Angles in Cyclopropane?

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Cyclopropane, a cycloalkane with three carbon atoms, features a triangular structure where each carbon is bonded to two hydrogens and the other two carbons. The confusion arises regarding the bond angles; while tetrahedral geometry suggests angles of 109.5 degrees due to sp3 hybridization, the actual C-C-C bond angles in cyclopropane are 60 degrees. This discrepancy is attributed to the ring strain present in cyclopropane, making it more reactive and resulting in weakened C-C bonds compared to other cycloalkanes. The study of such angle strains falls under the branch of organic chemistry.
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Cyclopropane is a cycloalkane with three carbon atoms. Each carbon is bonded to two hydrogens as well as the other two carbons, forming a triangular bond between the three carbons.

Each carbon has a tetrahedral molecular geometry since it has sp3 hybrization. However, I am rather confused about the bond angles that the carbon atoms make with another. I am talking about the C-C-C angles.

Because the structure is tetrahedral, the bond angles should be 109.5 degrees. But because the cyclopropane is an equilateral triangle, they must be 60 degrees. Which is correct and why not the other?

Thanks!

BiP
 
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I believe you should be able to find a reasonable answer to your questions here:

Cyclopropane Ring Strain.

In short, the angles between the carbons are 60°. Cyclopropane is notoriously reactive (and its C-C bonds are weakened relative to other cycloalkanes) as a result.
 
Thank you so much it all makes sense now. Just curious, under what branch of chemistry would you study these sorts of things, such as the angle strains of cycloalkanes? Organic?

Thanks.

BiP
 
Bipolarity said:
Thank you so much it all makes sense now. Just curious, under what branch of chemistry would you study these sorts of things, such as the angle strains of cycloalkanes? Organic?

Thanks.

BiP

Yep!
 
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