Question

In an elimination reaction, the non-preferred geometry in which the β hydrogen and the leaving group are on the same side of the molecule is called ____________ periplanar. According to the Zaitev rule, the major product in a β-elimination has the more substituted double bond. In an E1 reaction, the rate depends on only the alkyl halide concentration. The rate of the E2 reaction increases as the strength of the base increases because the reaction is biolecular and the base appears in the rate equation. Carbocation intermediates are involved in ____________ mechanisms. Polar aprotic solvents increase the rate of the ____________ reactions. Anti periplanar geometry is the preferred arrangement for any alkyl halide undergoing ____________ elimination, regardless of whether it is cyclic or acyclic.

Answer 1

a. syn

b. E1

c. E2

d. E2

Step-by-step explanation:

In organic chemistry, the rule of Saytzeff, Saytzev or Zaitsev states that in an elimination reaction (β-elimination) in which more than one alkene can be formed, the most thermodynamically stable will be the majority.

In general, the most substituted alkene is the most stable, due to the electronic sharing properties of the alkyl groups with the double bond C = C (hyper-conjugation). Also, in some cases, another stabilizing effect may be incurred when establishing the regioselectivity such as the conjugation of the double bond with other groups.

This rule is valid except in bimolecular elimination reactions (E2) in which there is a significant steric hindrance, branched substrate and / or bulky base, and without the possibility of conjugation, then applying Hofmann's rule.

The elimination reaction E1 has a potential energy profile similar to that of an SN1 reaction. The formation step of carbocation is very endothermic, with a transition state which is what determines the speed of the reaction. The second step is a rapid and exothermic deprotonation. Base does not participate in the step that determines the speed of the process, so it only depends on the concentration of alkyl halide.

The bimolecular elimination E2 takes place without intermediates and consists of a single ET, in which the base abstracts the proton, the leaving group leaves and the two carbons involved are rehybridized from sp3 to sp2.