Q72E - Organic Chemistry (Mcmurry)

Question

Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of deuterated and nondeuterated phenylethylenes in an approximately 7;1 ratio. Explain

Step-by-Step Solution

Answer

The C-D bond is about 5 kJ/mol stronger than a C-H bond. The removal of the hydrogen or deuterium is the rate-limiting step for ${\text{E}}_{\text{2}}$  reaction. Thus, this reaction must be ${\text{E}}_{\text{2}}$ elimination. The reason that we obtain 7 times more deuterated product is because of the added stability of the deuterium carbon bond that makes the overall reaction rate slow. The strength of the deuterium carbon bond is somewhat higher than that of a carbon-hydrogen bond thus making its rate slower. This leads to a higher abundance of the deuterated product. 

Step 1: Evidence for bimolecular elimination mechanism

The evidence that supports the ${\mathbf{E}}_{\mathbf{2}}$ mechanism is provided by a phenomenon known as the deuterium isotope effect. For reasons that we won’t go into, a carbon-hydrogen bond is weaker by about 5 kJ/mol than the corresponding carbon–deuterium bond. Thus, a C-H bond is more easily broken than an equivalent C-D bond, and the rate of C-H bond cleavage is faster.

Step 2: Explanation

The C-D bond is about 5 kJ/mol stronger than a C-H bond. The removal of the hydrogen or deuterium is the rate-limiting step for ${\text{E}}_{\text{2}}$  reaction. Thus, this reaction must be ${\text{E}}_{\text{2}}$ elimination. The reason that we obtain 7 times more deuterated product is because of the added stability of the deuterium carbon bond that makes the overall reaction rate slow. The strength of the deuterium carbon bond is somewhat higher than that of a carbon-hydrogen bond thus making its rate slower. This leads to a higher abundance of the deuterated product.