When treated with a strong base, 2-bromo-2,3-dimethylbutane will undergo an elimination reaction to produce two products. The choice of base (ethoxide vs. tert-butoxide) will determine which of the two products predominates?

when the ethoxide ion is used, 2,3-dimethylbut-2-ene is obtained as the major product while using tert-butoxide yields 2,3-dimethylbut-1-ene

Explanation:

The predominant reaction product may be a Saytzeff or non-Saytzeff product. The product that actually predominates highly depends on the structure of the alkyl halide and the nature of base used in the reaction. A tertiary alkyl halide such as 2-bromo-2,3-dimethylbutane can undergo elimination by E1 or E2 mechanism depending on the structure of the base used in the reaction.

Dehydrohalogenation reactions yield alkenes, they are important in creating carbon-carbon double bonds in chemistry. For a tertiary alkyl halide, the use of a bulky base yields the non-Saytzeff product due to steric hinderance. Hence, when the ethoxide ion is used, 2,3-dimethylbut-2-ene is obtained as the major product while using tert-butoxide yields 2,3-dimethylbut-1-ene (non-Saytzeff product). The reactions are shown in the image attached to this answer.

yea hi

explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2