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defon
3 years ago
6

Compound A has molecular formula C7H15Br. Treatment of compound

Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
3 0

Answer:

See images below.

Explanation:

The sodium ethoxide is a strong nucleophile and a strong base. When it acts as a strong nucleophile (a substance that has unpaired electrons and needs to share or donate it to be stable), an elimination reaction occurs. When it acts as a strong base, a substitution reaction occurs.

For a halide, the elimination takes place when the halogen is no steric hindrance, so the halogen must be in position 1. Then the compound A must be the must steric hindrance compound with formula C₇H₁₅Br, which is the compound showed below (3-Bromo-3-ethyl-pentane), to promote only elimination product. It happens because it's a tertiary halide.

When it reacts with sodium ethoxide (H₃C-H₂C-O-Na), the Br⁻ will leave the compound with close hydrogen, and a double bond will be formed between the carbons, so compound B must be the one showed below (3-Ethyl-2-pentene).

When the compound B is treated with dilute sulfuric acid, an addition reaction will occur, so the double bond will be broken and a hydrogen and a hydroxyl must be added to the compound. For the Markov-Nikov rule, the hydrogen will be in the most hydrated carbon. So, compound C will be the 3-Ethyl-3-pentanol showed below.

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Explanation:

1.

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4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

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Hopefully this was clear and you understood!

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