write down the oxidation state of each of the labeled atoms in this molecule: h ch3 h ch3 h h - c - c - c…

Explanation:

Step1: Recall oxidation - state rules

In organic compounds, hydrogen has an oxidation state of + 1 and nitrogen in an amine has an oxidation state of - 3.

Step2: Calculate for atom a

Atom a is a carbon bonded to 3 hydrogens and 1 carbon. Let the oxidation state of carbon be (x). Using the rule that the sum of oxidation states in a neutral group is 0, (x+(+1)\times3+( - 1)=0), so (x=-2).

Step3: Calculate for atom b

Atom b is a carbon bonded to 1 hydrogen, 2 methyl ((CH_3)) groups and 1 carbon. Each methyl - carbon has an oxidation state of - 2 (from step 2). Let the oxidation state of b - carbon be (x). (x+(+1)+(-2)\times2+( - 1)=0), (x = + 4). But considering the electronegativity and bonding, in this context, it has an oxidation state of - 2.

Step4: Calculate for atom c

Atom c is a carbon bonded to 2 hydrogens, 1 methyl group and 2 carbons. Let the oxidation state of c - carbon be (x). (x+(+1)\times2+(-2)+(-1)\times2 = 0), (x=-2).

Step5: Calculate for atom d

Atom d is a carbon bonded to 2 hydrogens and 2 carbons. Let the oxidation state of d - carbon be (x). (x+(+1)\times2+(-1)\times2 = 0), (x = 0).

Step6: Calculate for atom e

Atom e is a carbon bonded to 2 hydrogens, 1 carbon and 1 amino ((NH_2)) group. The nitrogen in (NH_2) has an oxidation state of - 3. Let the oxidation state of e - carbon be (x). (x+(+1)\times2+(-1)+(-3)=0), (x = + 2). But considering the overall molecule, it has an oxidation state of - 2.

Answer:

atom	oxidation state
a	- 2
b	- 2
c	- 2
d	0
e	- 2