Compound A, C12H22O, undergoes reaction with dilute H2SO4 at 50°C to yield a mixture of two alkenes, B and C, C12H20. The major

Question

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Compound A, C12H22O, undergoes reaction with dilute H2SO4 at 50°C to yield a mixture of two alkenes, B and C, C12H20. The major

alkene product, B, gives only cyclohexanone after ozone treatment followed by reduction with zinc in acetic acid. Draw the structure of the minor alkene product, compound C. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. In cases where there is more than one answer, just draw one.

Chemistry

1 answer:

Nataly_w [17] · Answer 1 · 2021-12-29T12:34:40+03:00

Answer:

Compound C is an alkene containing two cyclohexane rings attached with each other through a single bond.

Explanation:

Only cyclohexanone is obtained after ozonolysis of major alkene product. This means major alkene product, B , should contain contain two cyclohexane rings attached with each other through a double bond.
Compound B and C are obtained after dehyration of compound A by treating it with dilute $H_{2}SO_{4}$ at $50^{0}\textrm{C}$ . So, compound A is an alcohol.
As compound B is obtained from compound A therefore compound A is an alcohol containing two cyclohexane rings attached with each other through a single bond.
Structures of all compounds are given below.