(A) Towards SN2 reaction, order of reactivity is ${\mathrm{CH}}_3\mathrm{Br}>{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}>{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHBr}>{\left({\mathrm{CH}}_3\right)}_3\mathrm{CBr}$
(R) Greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of S_N1 reaction.

Explanation:

Statement (A):

The order of reactivity for the SN2 reaction is:

CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr.

This is correct because the SN2 mechanism relies on a direct backside attack by the nucleophile. The smaller and less hindered the alkyl halide, the easier the attack. Methyl bromide (CH₃Br) has no steric hindrance, making it the most reactive, while tertiary bromide (CH₃₃CBr) is the least reactive due to significant steric hindrance.

Statement (R):

"Greater the stability of the carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of SN1 reaction."

This statement is correct, but it applies to the SN1 reaction. In SN1, the reaction rate depends on the formation of a stable carbocation intermediate, which is faster for tertiary alkyl halides due to their greater carbocation stability.

Relation Between (A) and (R):

While both statements are correct, (R) does not explain (A) because the SN2 reaction mechanism does not involve carbocation formation. Instead, the reactivity in SN2 is determined by steric factors, not carbocation stability.

Final Answer

Both (A) and (R) are correct, but (R) is not the correct explanation of (A).