Q27 E - Organic Chemistry (Mcmurry)

Question

Predict the product(s) for each elimination reaction below. In each case show the mechanism. What do the mechanisms have in common? Why?

Step-by-Step Solution

Answer

1. 
   Formation of the desired product
2. 
   Formation of the desired product
3. 
   Formation of the desired product

Step 1: Elimination reaction

Two types of reaction take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at neighboring hydrogen

Step 2: Formation of the following products

a.

Treatment of 2-methyl butan-2-ol with a strong acid gives 2-methyl-1-ene and 2-methyl but-2-ene. Here, 2-methyl but-2-ene is the major product. Thus, the reaction is as follows:

Formation of the desired product

The mechanism is as follows:

Protonation of the alcoholic group forms a good leaving group in the product. The more stable tertiary carbocation is formed by expelling out the leaving group. The water molecule abstracts the proton present on the carbon, which is adjacent to the carbocation and forms the less substituted alkene. The water molecule abstracts the proton from another carbon, which is adjacent to the carbocation and forms the more substituted alkene.

Mechanism for the reaction

b.

Treatment of 3-chlorobut-1-ene with sodium acetate in the presence of acetic acid gives buta-1,3-diene. Thus, the reaction is as follows:

Formation of the desired product

 

The mechanism is as follows:

Heterolytic cleavage of Cl-Cl bod gives an allylic carbocation. Now, the acetate ion abstracts the hydrogen and forms an alkene and acetic acid.

Mechanism for the reaction

c.

Treatment of (1S,2S)-1,2-dimethylcyclohexanol with strong acid,  gives 1,2-dimethylcyclohex-1-ene as the major product. Thus, the reaction is as follows:

Formation of the desired product

The mechanism is as follows:

Protonation of the alcoholic group forms a good leaving group in the product. The more stable tertiary carbocation is formed by expelling out the leaving group. The water molecule abstracts the proton from another carbon, which is adjacent to the carbocation and forms the more substituted alkene.

Mechanism for the reaction