It is not equal to alkali. Alkalinity is evaluated from a thermodynamic perspective.

A strong nucleophile will give a relatively fast SN2 reaction

π bonds can also function as nucleophiles But the nucleophilicity is not very strong

lone electron pair

induction effect

The greater the steric hindrance, the slower the reaction rate, and the tertiary halogenated hydrocarbons will not react.

The greater the steric hindrance, the higher the activation energy

The stronger the nucleophilicity, the faster the reaction rate

strong or weak

Protic solvents

polar aprotic solvents

Why attack from behind

Multiple substituted places produce olefins

Bases with small steric hindrance are easy to form

Produce olefins at less substituted sites

Bases with large steric hindrance are easy to form, such as:

At this time, there is only one β-H, and there is only one configuration, so one product is obtained

Compared

coplanar: The C-C bond can rotate. The formation of π bond requires parallel overlapping of p orbitals and requires the four groups to be coplanar.

The steric hindrance is smaller when the substituent is located in the axial direction

Breaking of C-D bonds requires more ability, higher activation energy, and slower reaction rate

Protic solvents are more suitable as they stabilize ionic intermediates and transition states, resulting in smaller Ea

Due to the steric inversion structure of ions, more

Bases with large steric hindrance are mainly eliminated

strong Nu moderate base

CN negative ion is not a strong base, and the product is mainly SN2

sulfonate ions (RSO3−) is an excellent leaving group

Prepared from the corresponding alcohol

Similar to halogenated hydrocarbons, substitution and elimination will occur

Hydrooxide (HO−) is a bad leaving group and cannot be directly substituted like halogenated hydrocarbons.

Use a strong acid to convert HO- into a good leaving group H2O, and then substitution occurs

Elimination can also occur under strong acid