I guess this alkylhalide, in picture below, would be the answer but this alkyl halide may also give 2-methyl-1-butene because there are 2 beta-carbons and therefore 2 beta-positioned hydrogens I.E.if Hydrogen on the leftside of alpha-carbon leaves with chloride then formation of 2-methyl-1-butene would take place in stead of 2-methyl-2-butene but I need a alkylhalide which would always give 2-methyl-2-butene upon dehydrohalogenation.
You should switch the positions of the chlorine and hydrogen to be eliminated.
For your given molecule a secondary carbo-anion intermediate (at the 3/green carbon) will have to compete with the formation of a primary carbo-anion (at the 1/red carbon) intermediate to provide the desired product.
If you use 2-chloro-3-methyl-butane your desired product will form a much more favorable ternary carbo-anion intermediate (at the 3/green carbon) which will compete again with a primary carbo-anion intermediate(at the 1/red carbon).
2-Chloro-2-methylbutane is correct molecule only You will get upon dehydrohalogenation the major product as 2-methylbut-2-ene always; however, it won't be exclusive product. The other one i.e. 2-methylbut-1-ene will also be there albeit in smaller amounts
The % of the major product however could be improved using high polarity solvents, which favour SN1. Free carbocation will preferably give more substituted alkene.