What is the oxidation state of the carbon atom bonded to the hydroxyl group in isoborneol? What is the oxidation state of the same carbon atom in camphor?

If you are looking for something specific you should post a specific questions. In general, the oxidations state of ANY atom is almost anything you want to make it. If I am to balance an equation involving camphor or isoborneol, I would look at the average oxidation state of the C atoms. That said, the molecular formula is C10H18O; therefore, to be zero for the molecule, I would assign +1 to H atoms making +18, -2 to the oxygen atom, making +16 overall and that leaves -16 for 10 C atoms or -16/10 = -8/5 for each of the 10 C atoms. Same thing for camphor. If this answer is not satisfactory, please clarify and repost. Please remember that oxidations states is just a book keeping system we use to help us from time to time.

JYD, that is word for word the question in my lab manual for the lab due today. Any chance you are in Orgo Lab at UNC Chapel Hill?

To clarify the question DrBob222, is there a difference between the oxidation states of the other carbons in isoborneol and the carbon bonded to the hydroxyl group? And after oxidation, does a difference exist in camphor as well? Does saying the oxidation state goes from -8/5 to -7/5 adequately sum up the situation?

Well, aren't you asking some chemistry questions to get my funny molecules jumping!

In isoborneol, the carbon atom bonded to the hydroxyl group has an oxidation state of +1. Think of it as a positive carbon trying to make everyone happy by sharing its electrons.

Now, when it comes to camphor, that carbon atom loses a little bit of its positive attitude and ends up with an oxidation state of +2. It's like going from a happy-go-lucky carbon to a slightly more serious one. Life can be tough, even for carbon atoms.

So, in a nutshell, in isoborneol the carbon atom is a +1 carbon, while in camphor it's a +2 carbon. Hope that answers your question, and remember, chemistry can be as whimsical as a circus, sometimes!

To determine the oxidation state of the carbon atom bonded to the hydroxyl group in isoborneol and camphor, we need to consider the electronegativities of the atoms and the overall charge of the molecule.

In both isoborneol and camphor, the carbon atom bonded to the hydroxyl group is part of an alcohol functional group (-OH). In most cases, oxygen has an oxidation state of -2. Since there is only one oxygen atom in the hydroxyl group, we can assign it an oxidation state of -2.

For isoborneol, we need to consider the other atoms and their oxidation states. The sum of the oxidation states of all the atoms in a neutral molecule should equal zero. Since hydrogen usually has an oxidation state of +1, we assign it a +1 oxidation state. Carbon usually has an oxidation state of +4 in organic compounds. Oxygen, as mentioned earlier, has an oxidation state of -2. The overall charge of the molecule is zero.

Let's assume the oxidation state of the carbon atom bonded to the hydroxyl group in isoborneol is x. So, the equation will be as follows:

+1 (hydrogen) + x (carbon) -2 (oxygen) = 0

By rearranging the equation, we have:

x + 1 - 2 = 0

x - 1 = 0

x = 1

Therefore, the oxidation state of the carbon atom bonded to the hydroxyl group in isoborneol is +1.

Now, let's consider camphor. Like isoborneol, we assign +1 to hydrogen and -2 to oxygen. However, since camphor is a ketone, the carbon atom bonded to the oxygen in the carbonyl group (C=O) has an oxidation state of +2.

Using the same approach, we assign the oxidation state of the carbon atom bonded to the hydroxyl group in camphor as y. The equation will be:

+1 (hydrogen) + y (carbon) -2 (oxygen) + 2 (carbon in the carbonyl group) = 0

By rearranging the equation, we have:

y + 1 - 2 + 2 = 0

y + 1 = 0

y = -1

Therefore, the oxidation state of the carbon atom bonded to the hydroxyl group in camphor is -1.

In summary, the oxidation state of the carbon atom bonded to the hydroxyl group in isoborneol is +1, while in camphor, it is -1.