Isomers which produce tertiary alkyl halides when reacted with hydrogen halide
Answers
Explanation:
In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 106 times more basic than methanethiolate. As a result, we see a clear-cut difference in the reaction products, which reflects nucleophilicity (bonding to an electrophilic carbon) versus basicity (bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in R–Br and first order in Nu:(–)), suggesting a bimolecular mechanism for each. The substitution reaction is clearly SN2. The corresponding designation for the elimination reaction is E2. An energy diagram for the single-step bimolecular E2 mechanism is shown on the right. We should be aware that the E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the R–groups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of C–H may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of C–H breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions.
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane