How to Draw the Organic Product Formed in the Following Reaction

How to Draw the Organic Product Formed in the Following Reaction

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Introduction

In organic chemistry, we often must draw the organic products formed in a reaction. These organic products can be either single molecules or polyatomic ions. The type of reaction will give us a clue as to what types of products we might expect to see. In this guide, we will go over how to draw the organic product(s) formed in the following reaction:

The Reaction: The addition of HBr to an alkene

Product Type: A single molecule – an alkane

The Reaction

The reaction is a reversible one and the organic product will be the reactant that is in excess. In this case, that would be the alcohol. To draw the organic product, start with the carbon chain that has the OH group on it. Then, add any other groups that were on the reactant side of the equation. In this case, there would be a methyl group (CH3) and a halogen (X).

The Product

The product of the following reaction is an organic compound with the molecular formula C6H12O6. This product is a carbohydrate, specifically a hexose sugar.

Drawing the Product

In order to draw the organic product formed in the following reaction, you will need to know the structure of the reactants and the products. The reactants are a ketone and an alcohol, and the product is an ester.

The first step is to draw the structure of the ketone. The ketone has a carbon atom bonded to two oxygen atoms, and a carbon atom bonded to two hydrogen atoms. The second step is to draw the structure of the alcohol. The alcohol has a carbon atom bonded to three hydrogen atoms, and a hydroxyl group (-OH).

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The third step is to draw the product, which is an ester. The ester has a carbon atom bonded to two oxygen atoms, and a carbon atom bonded to two hydrogen atoms. The fourth step is to add any necessary lone pairs of electrons. In this case, there are no lone pairs of electrons on any of the atoms in the product.

The final step is to show how the reactants are arranged in space in order to form the product. In this case,the ketone and alcoholreact through a nucleophilic displacement reaction, in whichthe oxygen atomofthe alcohol attackedthe carbonylcarbonofketone toproducean ester.

Conclusion

The mechanism for the formation of the organic product in the following reaction can be summarized as follows:

1) The nucleophile, A, attacks the electrophilic carbon atom of the carbonyl group to form a new bond.

2) The electrons in the bonds between A and the carbonyl carbon are shifted to create a new double bond between them. This creates a resonance-stabilized carbanion intermediate.

3) The new double bond is attacked by the nucleophile, B, which displaces A from the carbonyl carbon.

4) The electrons in the bond between B and the carbonyl carbon are shifted to create a new double bond between them. This creates a resonance-stabilized carbanion intermediate.

5) The proton on the oxygen atom is lost to form a water molecule, which leaves behind a negative charge on the oxygen atom.

6) The negative charge on the oxygen atom is used to attack the positively-charged hydrogen atom on A, which displaces B from the carbonyl carbon.
7) The electrons in the bonds between A and the carbonyl carbon are shifted to create a new double bond between them. This creates a resonance-stabilized carbanion intermediate.
8) The proton on the oxygen atom is lost to form a water molecule, which leaves behind a negative charge on the oxygen atom. 9) The negative charge on the oxygenatom is used to attack 10)The 11)The alkyne group 12 13

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