Explain how substitution and elimination reactions compete in the same reaction.

(N/A) Substitution and elimination reactions always take place in competition. The reaction pathway and the product formed depend on the following factors:
$(i)$ Nature of the substrate
$(ii)$ Strength of the nucleophile
$(iii)$ Strength of the base
$(iv)$ Nature of the solvent
$(v)$ Temperature of the reaction

Nature of substrate and solvents	Nature of Nucleophile or Base and Reaction preferred
$3^{\circ}$ alkyl halide,Polar protic solvent	Strong nucleophile but weak base: $S_{N}1$
$3^{\circ}$ alkyl halide,Polar aprotic solvent	Strong base but weak nucleophile (Heat): Elimination $(E2)$
$3^{\circ}$ alkyl halide,Polar protic solvent	Weak base: Elimination $(E1)$
$1^{\circ}$ alkyl halide or methyl,Polar aprotic solvent	Strong nucleophile but weak base: $S_{N}2$

High temperature favours elimination reactions,whereas low temperature favours substitution reactions.
In the case of $3^{\circ}$ alkyl halides,$S_{N}1$ is the major product when substitution and elimination reactions compete in the presence of a weak base.
The tertiary butoxide ion is a strong base but a bulky nucleophile. Therefore,it prefers to abstract a proton from a tertiary halide,causing an elimination reaction to form an alkene as the major product. However,if the alkyl halide is primary,the $S_{N}2$ reaction takes place. The ethoxide ion is a strong nucleophile and also a strong base. With a tertiary halide,it causes both elimination and substitution $(S_{N}1)$ reactions; however,the elimination product (alkene) will be major due to the strong basic character of the ethoxide ion. If the alkyl halide is primary,the ethoxide ion causes an $S_{N}2$ reaction.