Question

Question: Draw the structure of compounds A–E in the following reaction scheme:

Short answer

Answer:

Step-by-step solution

Step-by-Step SolutionStep 1: E2 elimination reaction and conversion of A to B

In this reaction, two substituents are removed from the substrate in a single-step mechanism in the presence of a base.

2-Chloro-2-cyclohexylethane undergoes E2 elimination in the presence of the strong base potassium tert-butoxide to give 2-Chloroethene (Hoffmann product) as shown below:

Conversion of A to B

Bromination (conversion of B to C)

Alkenes react with a cold solution of bromine molecule with or without tetrachloromethane to undergo an electrophilic addition reaction.

2-Chloroethene undergoes bromination to give 1,2-Dibromo-2-cyclohexylethane, as shown below:

Conversion of B to C

E2 elimination reaction (conversion of C to D)

A type of organic reaction in which a pair of atoms are removed from the molecule in the presence of an acid, base, or under high temperature in a concerted step is known as the elimination reaction.

1,2-Dibromo-2-cyclohexylethane undergoes an elimination reaction in the presence of potassium tert-butoxide to form an alkyne. Two equivalents of H-Br are removed from the vicinal halide by successive E2 elimination reaction, as shown below:

Conversion of C to D

Deprotonation (conversion of D to E)

Sodium amide is a string base that acts as a nucleophile to deprotonate the alkyne, as shown below:

Conversion of D to E

Methylation reaction (conversion of E to desired product)

The organic reaction involving the substituting of an atom or any group in a molecule by a methyl group is known as the methylation reaction.

The deprotonated alkyne undergoes methylation in the presence of methyl iodide to give 3-cyclohexyl-prop-2-yne, as shown below:

Conversion of E to the desired product