the reagents necessary to convert alcohol to ketone
which involves oxidation of alcohols.
<h3>
What is oxidation of alcohols?</h3>
- Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
- In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
- In order to cause primary alcohols to oxidize into aldehydes
(dichromate)
/pyridine (Collins reagent)- Chromium pyridinium compound (PCC)
- Dichromate of pyridinium (PDC, Cornforth reagent)
- Periodinane by Dess-Martin
- Oxalyl chloride with dimethylsulfoxide (DMSO) for Swern
- oxidation of secondary alcohols to ketones
(dichromate)
/pyridine (Collins reagent)- Chromium pyridinium compound (PCC)
- Dichromate of pyridinium (PDC, Cornforth reagent)
- Periodinane by Dess-Martin
- Oxalyl chloride and dimethyl sulfoxide (DMSO) (Swern oxidation)
/acetone (Jones oxidation)- Acetone with aluminum isopropoxide (Oppenauer oxidation)
To learn more about oxidation of alcohols with the given link
brainly.com/question/7207863
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<u>Question:</u>
Identify the reagents necessary to achieve each of the following transformations




The answer is Molecule. After a pair of shared electrons orbit around the nuclei of both atoms
Answer:
4.62 M
Explanation:
Molarity = moles/volumes (L), so you need to find the moles and the volumes in liters.
Finding the volume is easy because you just have to convert mL to L, so the volume is 0.45 L
Next, find the moles. You can do this by using the molar mass of aluminum to convert the grams to moles. The molar mass of aluminum is 26.98 g/mol.
56 g * (1 mol/26.98 g) = 2.08 mol
Now, divide the moles (2.08) by the volume (.45 L)
Molarity = 4.62 M
Answer:
<h2>15 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

But from the question
volume = final volume of water - initial volume of water
volume = 165 - 150 = 15 mL
We have

We have the final answer as
<h3>15 g/mL</h3>
Hope this helps you