Question

What affects $SN2$ reaction rate?

Answer 1

$SN2$ is a bimolecular nucleophilic substitution reaction. These are single-step reactions. Several factors affect the rate of $SN2$ reaction. Such as solvents, steric hindrance, nucleophile, etc. Steric hindrance is the hindrance offered by bulky groups in a reaction and because of this primary alkyl are said to be more reactive towards $SN2$reactions.

Complete answer:
$SN2$reaction is a bimolecular nucleophilic substitution reaction. It is a reaction that depends upon the concentration of both alkyl halide and nucleophile. It is a second-order reaction because the rate-determining step depends on the nucleophile concentration as well as the concentration of substrate.
Many factors affect the rate of $SN2$ reaction. Some of them are stated below:
Strength of the nucleophile: species with negative charge are stronger than an analogous neutral species (e.g.$- OH > {H_2}O > - N{H_2} > N{H_3}$).
Steric effect: when bulky groups interfere with the reaction, because of their size, this is called steric hindrance because of their size. Steric hindrance affects nucleophilicity but not basicity. Primary alkyl halides are more reactive toward $SN2$ reaction as they are less hindered
Leaving group: the substrate should have a good leaving group. A good leaving group should be electron-withdrawing, relatively stable, and polarizable. They are weak bases and neutral molecules (e.g.$C{l^ - },B{r^ - },{H_2}O$). Strong bases are not good leaving groups.
Solvent: polar protic solvents decrease nucleophilicity (e.g. acetonitrile, DMSO, and acetone), and polar aprotic solvents increase nucleophilicity.

Note:
In the $SN2$ reaction, the nucleophile approaches the carbon atom to which the leaving group is attached. As the nucleophile forms a bond with this carbon atom, the bond between the carbon atom and the leaving group breaks. The bond-making and bond-breaking actions occur simultaneously. Eventually, the nucleophile has formed a complete bond to the carbon atom, and the bond between the carbon atom and the leaving group is completely broken.