Compound A has the formula C9H19Cl. B is a C9H19Br compound. A and B undergo base-promoted E2 elimination to give the same alken
e C as the major product as well as different minor products. C reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form 2,6-dimethylheptane. Addition of HCl to C yields A as the major product. Propose structures for A and B.
1 answer:
You may find in the attached picture the structure for compound (A) and (B).
Explanation:
The compound (A) and (B), which are haloalkanes, will undergo base-promoted E2 elimination and it will form the alkene (C) with the double bond in the second position. Zaitev's rule, which is an empirical rule, tell us that the halogen will leave with the hydrogen from the neighbor carbon which have the lowest amount of hydrogen, in the picture this specific hydrogen have a blue color.
Further compound (C) it will form compound (A), as a major product, by addition of hydrobromic acid. The Markovnikov's rule, which is also empirical, tell us that the halogen from the acid will add the carbon, involved in the double bond, that have the lowest amount of hydrogen.
Compound (C) will participate to a hydrogenation reaction to form 2,6-dimethylheptane.
Learn more about:
Zaitev's rule
Markovnikov's rule
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