Explain how sustitution and elimination reactions compete in the same reaction ?

Explanation

The substitution reactions and elimination reacts always take place in a competition. The path of the reaction and the product formed depends on the following factors : (i) Nature of substrate (ii) Strength of the nucleophiles (iii) Strength of the base (iv) Nature of solvents (v) Temperature of the reaction

High temperature favours elimination reaction whereas low temperature favours substitution reaction.

In case of `3^(@)` - alkyl halides, the `S_(N)1` is major product when substitution reaction and elimination reactions take place in competition in the presence of weak base.
The tertiary but oxide is a strong base but bulky nucleophile. So, it will prefer to abstract a proton from tertiary halide and thus cause the elimination reaction to form alkene as a major product. However, if alkyl halide is primary, the `S_(N)2` reaction takes place. The ethoxide ion is a strong nucleophile and also a stronge base. With tertiary halide it causes both elimination and substitution `(S_(N)1)` reaction, however, elimination product (alkene) will be major due to strong basic character of ethoxide ion. If alkyl halide is primary, the ethoxide ion cause `S_(N)2` reaction.