Question

Draw an alkyl chloride that would produce ONLY the following alkene in an E2 elimination. Use a dash or wedge bond to indicate relative stereochemistry on asymmetric centers, where applicable. Ignore inorganic byproducts, Draw the Alkyl chloride Reactant Strong Base

Answer 1

To produce a specific alkene in an E2 elimination, the alkyl chloride should have the leaving group and the proton on adjacent carbons, allowing for the concerted transfer of electrons and formation of a double bond.

Step-by-step explanation:

An alkyl chloride that would produce only a specific alkene in an E2 elimination reaction can be determined by considering the regioselectivity and stereochemistry of the reaction. The alkyl chloride should have the leaving group and the proton on adjacent carbons to allow for the concerted transfer of a set of electron pairs from the base to the more electronegative group. The reaction would proceed through a bimolecular mechanism, where the base captures the vicinal proton released during the elimination.

Answer 2

The image of the alkyl chloride that would only produce the given alkene in an E2 elimination is given below.

The task involves drawing a secondary alkyl chloride that would produce the given alkene in an E2 elimination reaction. The image shows an alkene that appears to be a cyclohexene with a methyl group at one of the carbon atoms adjacent to the double bond.

In an E2 elimination reaction, a hydrogen atom is removed from a carbon adjacent to the carbon bearing the leaving group (in this case, a chloride ion), and the electrons from the C-H bond are used to form the double bond.

Given the structure of the alkene, the corresponding secondary alkyl chloride would have a chlorine atom on the carbon adjacent to the carbon with the methyl group. The chlorine atom should be on the carbon that has a hydrogen atom in a position to be anti-periplanar to the chlorine, which is necessary for an E2 elimination to occur.

For a secondary alkyl chloride, the carbon bearing the chlorine atom must be bonded to two other carbon atoms. The chlorine atom will be located on the carbon adjacent to the one with the methyl group, and this carbon should have one hydrogen atom in the correct orientation for an E2 elimination.

Here's the step by step process to draw the starting alkyl chloride:

1. Draw the cyclohexane ring.

2. Add a methyl group to one of the carbons to make it a secondary carbon (which will be part of the alkene after elimination).

3. Identify the carbon adjacent to this methylated carbon where the double bond will form.

4. Place a chlorine atom on this adjacent carbon, ensuring that it's on the same side as a hydrogen on the methylated carbon to be anti-periplanar.

Let's create a drawing to illustrate this.

The generated image is intended to illustrate the structure of a secondary alkyl chloride before an E2 elimination reaction.

For an accurate depiction, imagine a cyclohexane ring with one of the carbons bearing a methyl group. Adjacent to this carbon with the methyl group, there should be a chlorine atom attached to the next carbon in the ring. Directly across this carbon (in the anti-periplanar position), there should be a hydrogen atom that will be abstracted by a strong base in the E2 elimination reaction. This arrangement allows the formation of the double bond in the cyclohexene after the reaction proceeds.

The diagram is given below: