When the following compound is treated with sodium methoxide in methanol, two elimination products are possible. Explain why the deuterated product predominates by about a 7:1 ratio (refer to Problem 6-75).
Reference: Exercise 76:
Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the bond is slightly (5.0 kJ mol, or 1.2 kcal mol) stronger than the bond. Reaction rates tend to be slower if a bond (as opposed to a bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review Problem 4-56.)
(a) Propose a mechanism to explain each product in the following reaction.
(b) When the following deuterated compound reacts under the same conditions, the rate of formation of the substitution product is unchanged, while the rate of formation of the elimination product is slowed by a factor of 7.
Explain why the elimination rate is slower, but the substitution rate is unchanged.
(c) A similar reaction takes place on heating the alkyl halide in an acetone/water mixture.
Give a mechanism for the formation of each product under these conditions, and predict how the rate of formation of each product will change when the deuterated halide reacts. Explain your prediction.
Here mechanism of the reaction involves E2 reaction. It is known that the C-D cleavage can be up to 7 times slower than C-H cleavage, so the formed from the breaking of C-H should be 7 times fast and the ratio is 7:1.