Explanation: This is a reaction of oxidation of in the presence of acidified . Acidified is a strong oxidizing agent.
To balance out the on the reactant side, we write on the product side.
Balancing out the following reaction gives us:
Answer:
a. Hydrocarbons have low boiling points compared to compounds of similar molar mass.
b. Hydrocarbons are hydrophobic.
d. Hydrocarbons are insoluble in water.
Explanation:
As we know that the hydrocarbons is a mix of carbon and hydrogen. In this the availability of the electronegative atom is not there that shows there is no bonding of the hydrogen plus it is dissolved. Also, the hydrocarbons is considered to be a non-polar but as compared to the water, water is a polar
In addition to this, the strong bond is no existed that shows the lower boiling points
Therefore option A, B and D are right
Answer is: <span>C. propanal.
1-propanol is primary alcohol. With weak oxidizing agent primary alcohol gives aldehyde and with strong oxidizing agent primary alcohol gives carbonic acid.
</span>Oxidation of 1-propanol with <span>Na</span>₂<span>Cr</span>₂<span>O</span>₇<span> and sulfuric acid</span><span> gives propanal and o</span><span>xidation with </span>chromic acid<span> gives </span>propionic acid<span>.</span>
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
To view more about rational reaction, refer to:
brainly.com/question/20308523
#SPJ4