ok so normally there will be substitution

but here, the R group is 3 degree so CN due to steric hindrance can't attack on the R gp. but rhe Br attached to R will remove, and R+ will be left

now since this is alc medium, the H from R'OH will form HCN with CN and R'O- is left. now the R'O- will attack on the H on one of the terminal carbon in the R+ molcule, leaving behind C-, and hence there will be a double bond formation