When secondary alcohol oxidized it produces alkanes. Draw the structure of 2,2 dimethyl 3-butanol + oxygen and give the names for it.

We can't draw structures on this forum. Here is a site that gives you the structure.

http://www.chemindustry.com/chemicals/0299156.html

To determine the products of the oxidation reaction of 2,2-dimethyl-3-butanol, we first need to understand the concept of oxidation of alcohols. Secondary alcohols, when oxidized, usually undergo an elimination reaction to produce ketones. However, if the oxidation is carried out under vigorous conditions, the ketone can further undergo oxidative cleavage to produce alkanes.

Here's the step-by-step explanation of how to draw the structure of the product by manually working through the oxidation process:

Step 1: Start by drawing the structure of 2,2-dimethyl-3-butanol. It has four carbon atoms, including the hydroxyl group, and the two methyl groups at the 2-position.

CH3 CH3
| |
CH3 - C - C - C - OH
|
CH3

Step 2: Recognize that the oxidation of this secondary alcohol will produce a ketone. The ketone formed would be 2,2-dimethyl-3-butanone.

CH3 CH3
| |
CH3 - C - C - C - O
|
CH3

Step 3: If under vigorous conditions, the ketone can further undergo oxidative cleavage, resulting in the production of alkanes. However, rather than drawing the structure explicitly, we can simply name the alkane product. In this case, the alkane generated from 2,2-dimethyl-3-butanone would be 2,2-dimethylbutane.

Therefore, the product resulting from the oxidation of 2,2-dimethyl-3-butanol is 2,2-dimethylbutane.

Remember, this process is based on the general observation for secondary alcohols under vigorous oxidation. It's always recommended to consult the specific reaction conditions in a laboratory setting for accurate predictions.