Question

give any four points in favour of kekule's formula to show its resonance structure.​

Answer 1

Answer:

check the explanation and link below

Explanation:

Kekulé was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.

This diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing - and that must be attached to a hydrogen atom.

Answer 2

Answer:~

• In Kekule's structure, the carbon atoms in benzene are arranged in a hexagonal ring with each carbon atom is attached to two other carbon atoms.
• In Kekule's structure, the carbon atoms in benzene are arranged in a hexagonal ring with each carbon atom is attached to two other carbon atoms.The ring contains alternate single and double bond.
• In Kekule's structure, the carbon atoms in benzene are arranged in a hexagonal ring with each carbon atom is attached to two other carbon atoms.The ring contains alternate single and double bond.These double bonds are due to the π electrons present in benzene.
• In Kekule's structure, the carbon atoms in benzene are arranged in a hexagonal ring with each carbon atom is attached to two other carbon atoms.The ring contains alternate single and double bond.These double bonds are due to the π electrons present in benzene.These π electrons can delocalize and form two structures with different position of a double bond.
• In Kekule's structure, the carbon atoms in benzene are arranged in a hexagonal ring with each carbon atom is attached to two other carbon atoms.The ring contains alternate single and double bond.These double bonds are due to the π electrons present in benzene.These π electrons can delocalize and form two structures with different position of a double bond.These are the resonance contributor. Thus, they contribute equally to form a resonance hybrid.

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