Question

Draw the structure of an alkyl halide that could be used in an E2 reaction to give the following alkene as the only alkene product:

Answer 1

Concepts and reason

In presence of strong bulky base (potassium tert-butoxide), secondary alkyl halides undergo ${{\rm{E}}_{\rm{2}}}$ elimination reaction. In ${{\rm{E}}_{\rm{2}}}$ elimination reaction, two adjacent substituents eliminated from an alkyl halide to form alkene. The mechanism of this reaction is concerted (i.e. elimination occur in one step).

Fundamentals

In ${{\rm{E}}_{\rm{2}}}$ elimination reaction, the eliminating groups must be anti to each other (i.e. $180{\;^{\rm{o}}}$ to each other). For example, in chair conformation of cyclohexane both eliminating groups (i.e. H and Br) must be in axial positions.

The given product is 4-methylcyclohex-1-ene.

Two alkyl halides are possible for the formation of given products. They are shown below.

The alkyl halide, 1-bromo-3-methylcyclohexane forms two alkene products when it reacts with strong base potassium tert-butoxide. On the other hand, 1-bromo-4-methylcyclohexane forms only single alkene product.

The reaction is,

Ans:

Thus, the major ${{\rm{E}}_{\rm{2}}}$ organic products is,