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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.

enter image description here

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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.

enter image description here

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). enter image description here

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  • $\begingroup$ Oh god, can't understand it. Complicated.. but I'll try or you extend it if you think it can help $\endgroup$ – Sufyan Naeem Dec 21 '16 at 17:17
  • $\begingroup$ @SufyanNaeem Do the images and colors help? $\endgroup$ – A.K. Dec 21 '16 at 17:36
  • $\begingroup$ +1. Yes its lovely! But why tertiary carbon cation is more favourable? $\endgroup$ – Sufyan Naeem Dec 21 '16 at 18:30
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    $\begingroup$ @SufyanNaeem Because the more electron rich a carbon, the more stable a cation will be on it. since the methyl groups are more electron donating than the hydrogens a cation will be more stable on a more substituted group. $\endgroup$ – A.K. Dec 21 '16 at 19:35
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    $\begingroup$ @A.K. Note that tertiary anions are the least stable anions, not the most stable ones. Compare the reactivity of tBuLi with n-BuLi or sec-BuLi. But that’s probably not of too much relevance since your elimination would proceed via $\mathrm{E2}$ or $\mathrm{E1}$. $\endgroup$ – Jan Dec 23 '16 at 9:36
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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.

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  • $\begingroup$ But the book want me to give a conpound which gives 100% 2-methylbut-2-ene. So what should I do? $\endgroup$ – Sufyan Naeem Dec 21 '16 at 17:06

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