Answer:

Answer:

**(a)** The authoritative knots of the asterisked carbon in twain the starting symbolical and issue are **lactam (cyclic amide)** and **secondary amine**, respectively.

The oxidation equalizes of the asterisked carbon in twain the starting symbolical and issue are +2 and -2, respectively.

Explanation: x + 1 – 2 – 1 = 0, i.e. x = +2 and x + 1 + (2*1) – 1 = 0, i.e. x = -2

i.e. There is a **enmesh diminution ( [H] )**.

**(b)** The authoritative knots of the asterisked carbon in twain the starting symbolical and issue are **ketal (bis ether strong to ketonic carbon)** and **ketone**, respectively.

The oxidation equalizes of the asterisked carbon in twain the starting symbolical and issue are 0 and 0, respectively.

Explanation: x + 1 – 1 – 1 + 1 = 0, i.e. x = 0 and x + 1 – 2 + 1 = 0, i.e. x = 0

i.e. There is a **no** **enmesh modify**.

**(c)** The authoritative knots of the asterisked carbon in twain the starting symbolical and issue are **acetal (bis ether strong to aldehydic carbon)** and **ester**, respectively.

The oxidation equalizes of the asterisked carbon in twain the starting symbolical and issue are 0 and 0, respectively.

Explanation: x + 1 – 1 – 1 + 1 = 0, i.e. x = 0 and x + 1 – 2 – 1 = 0, i.e. x = +2

i.e. There is a **enmesh oxidation ( [O] )**