CH3CH2CHCH3 --H2CrO4--> ?
l
OH
it came out wrong the OH is attached to the 3rd carbon, not the first
Secondary alcohols are oxidized to ketones.
CH3CH=CHCH3 Is the product cis and/or trans. This looks like a dehydration reaction producing an alkene.
http://www.chemguide.co.uk/organicprops/alcohols/oxidation.html
H2CrO4 produces oxidation and H2SO4 produces elimination. I apologize.
The given reaction involves the oxidation of a compound CH3CH2CHCH3 using H2CrO4. The product formed will depend on the specific reaction conditions and the functional groups present in the starting compound.
To determine the product, we need to consider the possible oxidation reactions that can occur with the functional groups present in CH3CH2CHCH3.
In this case, CH3CH2CHCH3 is a alkane (specifically, butane). Alkanes are generally unreactive under normal conditions and are not easily oxidized. H2CrO4, a strong oxidizing agent, is commonly used to convert alcohols to carbonyl compounds.
However, there is an OH group mentioned in the reaction. This indicates the presence of an alcohol functional group. If the OH group is attached to any of the carbon atoms in CH3CH2CHCH3, it means we have an alcohol.
When an alcohol is reacted with H2CrO4, it undergoes oxidation to form a carbonyl compound. The specific carbonyl compound formed depends on the position of the OH group in the starting molecule.
If the OH group is attached to the terminal carbon atom (end of the carbon chain), for example, CH3CH2CH(OH)CH3, it will be oxidized to form a ketone. In this case, the product would be CH3CH2C(O)CH3, also known as 2-butanone or ethyl methyl ketone.
If the OH group is attached to any of the internal carbon atoms, the resulting compound will form an aldehyde. For example, if the OH group is attached to the second carbon atom (from the left) in CH3CH2CH(OH)CH3, the product would be CH3C(O)CH2CH3, also known as propanal.
In summary:
CH3CH2CHCH3 (butane) + H2CrO4 (strong oxidizing agent) → CH3CH2C(O)CH3 (2-butanone) or CH3C(O)CH2CH3 (propanal)