Question

Which of the following statements about regioselectivity of elimination reactions is not correct?
A.More substituted and more stable alkenes are formed preferentially by both \[{E_1}\] and \[{E_2}\] mechanisms
B.Substrates with a poorer nucleophile tend to give the less substituted alkenes
C.Sterically hindered, bulky bases give the less substituted alkenes.
D.Reactions by the \[{E_1}\] mechanism are generally less regioselective than those by the \[{E_2}\] mechanism.

Answer 1

Hint: The elimination reaction leads to the formation of a double bond in more than one place and we know that alkene stability is influenced by degree of substitution of the double bond. Stability depends on the number of C atoms attached to the double bond, so greater the number of carbons attached to the double bond, the more stable it is. And not only alkene stability is the factor that plays a role in elimination but steric hindrance also plays a role. In case of very bulky bases especially, the least substituted alkene forms, even though it is less stable.

Complete step by step answer:
Regioselectivity means that the reaction selectively produces one Regio isomer as the major product and an elimination reaction leads to the formation of a double bond in more than one place. There are two mechanisms of elimination i.e. \[{E_2}\] and \[{E_1}\] . The elimination \[{E_1}\] mechanism is unimolecular elimination and \[{E_2}\] mechanism is bimolecular
 \[{E_1}\] reactions are regioselective and follow Zaitsev rule. It forms more than one alkene product depending on the position of double bond and base; they form major and minor products. The major product is called the saytzeff (Zaitsev) product which is a more stable product and a more substituted alkene and the minor product is Hofmann product it is less substituted alkene.
 \[{E_2}\] reactions are stereoselective and result in the formation of trans-double bonds most preferably and favour the formation of Zaitsev products. Also, the increase in the \[{E_2}\] reaction rate with increasing alkyl substitution can be rationalized with transition state stability.
However, not only alkene stability is the factor that plays a role in elimination but steric hindrance can also play a role. In case of very bulky bases especially, the least substituted alkene forms, even though it is less stable.

Therefore, the correct answer is option (D).

Note: An elimination reaction can lead to formation of a double bond in more than one place. If the halide is on one carbon and there are protons that can be removed on either side, then by taking one proton or the other may lead to two different products. This reaction could have different regiochemical outcomes which means it could happen at two different places in the molecule.