Compound a has molecular formula c7h15br. treatment of compound
a with sodium ethoxide yields only one elimination product (compound
b) and no substitution products. when compound b is treated with dilute sulfuric acid, compound
c is obtained, which has molecular formula c7h16o.
draw the structures of compounds a, b, and c.

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.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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