Q30P - Organic Chemistry (Mcmurry)

Question

Predict the product of each reaction below and indicate if the mechanism is likely to be ${\mathbf{SN}}_1\mathbf{,}{\mathbf{SN}}_2\mathbf{,}{\mathbf{E}}_1\mathbf{,}{\mathbf{E}}_2{\mathbf{orE}}_{1\mathrm{cB}}$

Step-by-Step Solution

Answer

a.

Formation of the desired product

b.

Formation of the desired product

c.

Formation of the desired product

d.

Formation of the desired product

Step 1: Elimination reaction

In the ${\mathbf{E}}_2$ reaction, C-H and C-X bond-breaking occur simultaneously when a base abstract ${\mathbf{H}}^{\mathbf{+}}$ from one carbon while the leaving group departs from the neighboring carbon. In the ${\mathbf{E}}_1$ reaction, C-X bond-breaking occurs first, the substrate dissociates to yield a carbocation in the slow rate-limiting step before losing ${\mathbf{H}}^{\mathbf{+}}$ from an adjacent carbon in a second step. In the ${\mathbf{E}}_{1\mathrm{cB}}$ reaction, C-H bond-breaking occurs first, when a base abstract proton gives a carbanion, followed by loss of the leaving group from the adjacent carbon in a second step.

Step 2: (a) Formation of the following product

Treatment of 2-hydroxy hexa-1-one with NaOH in the presence of water as a solvent form hex-1-ene-1-one as the major product. This reaction follows ${\mathrm{E}}_{1\mathrm{cB}}$ mechanism, in this elimination of -OH group occurs.

Formation of the desired product

Step 3: (b) Formation of the following product

Treatment of sodium phenoxide with an alkyl halide in the presence of DMF forms ethoxy benzene as the major product. This reaction follows ${\mathrm{SN}}_2$ mechanism. In this Na is replaced by I. 

Formation of the desired product

Step 4: (c) Formation of the following product

Treatment of alkyl halide with methanol forms demethylated product and an alkene. This reaction follows ${\mathrm{E}}_1$ mechanism. In this elimination of iodine takes place. 

Formation of the desired product

Step 5: (d) Formation of the following product

Treatment of alkyl halide with sodium ethoxide in the presence of ethanol to form an alkene and ethoxy alkane. This reaction follows ${\mathrm{E}}_1$  and ${\mathrm{SN}}_1$ mechanism. In this elimination of Br takes place. 

Formation of the desired product