Question

On being heated with a solution of sodium ethoxide in ethanol, compound A (C7H15Br) yielded a mixture of two alkenes B and C, each having molecular formula C7H14. Catalytic hydrogenation of the major isomer B or the minor isomer C gave only 3−ethylpentane. Draw structures for compounds A, B, and C consistent with these observations.

Answer 1

Compound A is 2-bromo-3-ethylpentane. Compound B is 3-ethylpent-2-ene. Compound C is 3-ethylpent-1-ene. Structures are given below.

Step-by-step explanation:

Sodium ethoxide in ethanol gives E2 ellimination by removing HBr when it is treated with 2-bromo-3-ethylpentane.

2-bromo-3-ethylpentane (
`C_(7)H_(15)Br` has two possible positions to give ellimination products.

3-ethylpent-2-ene is the major product as it contains more substituted double bond and 3-ethylpent-1-ene is the minor product as it contains less substituted double bond.

Both 3-ethylpent-2-ene and 3-ethylpent-1-ene gives only one product due to catalytic hydrogenation through addition of
`H_(2)` in double bond.

Structures are given below.