# Homework Solution: Consider the following chemical reactions (NONE of which you have to know at… Consider the following chemical reactions (NONE of which you have to know at this point). In each the functional group of the asterisked (*) carbon in both the starting material and product (it's the same C in both). Then determine that carbon's oxidation level, and indicate on the right whether there has been a net oxidation (|O| . reduction ([H]). or no change (circle your choice). Understand that the equations ignore certain other species, and so appeal to be unbalanced. Func. grp._______ func. grp._____________ axid. Level______ oxid. Level____ Consider the subjoined chemical reactions (NONE of which you feel to perceive at this aim). In each the authoritative knot of the asterisked (*) carbon in twain the starting symbolical and issue (it’s the identical C in twain). Then state that carbon’s oxidation equalize, and betoken on the equitable whether there has been a enmesh oxidation (|O| . diminution ([H]). or no modify (foe your dainty). Understand that the equations disown actual other figure, and so address to be unbalanced. Func. grp._______ func. grp._____________ axid. Equalize______ oxid. Equalize____

(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] )