Question

write down the oxidation state of each of the labeled atoms in this molecule: h ch3 h ch3 h h - c - c - c - c - c - nh2 h ch3 h h h atom oxidation state a b c d e

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
b	- 2
c	- 2
d	0
e	- 2