write an equation to represent the oxidation of an alcohol.
identify the reagents that may be used to oxidize a given alcohol.
identify the specific reagent that is used to oxidize primary alcohols to aldehydes rather than to carboxylic acids.
identify the product formed from the oxidation of a given alcohol with a specified oxidizing agent.
identify the alcohol needed to prepare a given aldehyde, ketone or carboxylic acid by simple oxidation.
write a mechanism for the oxidation of an alcohol using a chromium(VI) reagent.
The reading mentions that pyridinium chlorochromate (PCC) is a milder version of chromic acid that is suitable for converting a primary alcohol into an aldehyde without oxidizing it all the way to a carboxylic acid. This reagent is being replaced in laboratories by Dess‑Martin periodinane (DMP), which has several practical advantages over PCC, such as producing higher yields and requiring less rigorous reaction conditions. DMP is named after Daniel Dess and James Martin, who developed it in 1983.
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
Cr2O2−7+14H++6e−→2Cr3++7H2O
Reacting 1-chloro-2-ethylcyclohexene with hydrogen gas using a platinum catalyst would give a product of 1-chloro-2-ethylcyclohexane.
Hydrogen gas is a reducing agent, which in this reaction, simply mean that the alkene double bond in the cyclohexene will disappear because one of the two bonds forming the double bond (in the alkene) will be connected to a hydrogen atom. The platinum catalyst is necessary to allow the reaction to proceed at a much lower (activation) energy than would have been required.
Answer:
Lead to Sulfur = 2 : 1
Explanation:
Given
Represent lead with L and Sulfur with S
L1 = 6.46g for S1 = 1 g
L2= 3.23g for S2 = 1 g
Required
Determine the simple whole number ratio of L to S
Divide L1 by L2
L = L1/L2
L = 6.46g/3.23g
L = 2
Divide S1 by S2
S = 1g/1g
S = 1
Represent L and S as a ratio:
L : S = 2 : 1
Hence, the required ratio of Lead to Sulfur is 2 to 1
Answer:
C = 107.97 mol/L
Explanation:
Given data:
Volume of solution = 1.38 mL (1.38 mL× 1 L /1000 mL = 0.00138 L)
Mass of ammonium sulfite = 17.36 g
Concentration of solution =?
Solution:
We will calculate the number of moles of ammonium sulfite.
Number of moles = mass/molar mass
Number of moles = 17.36 g / 116.15 g/mol
Number of moles = 0.149 mol
Concentration:
C = n/V
C = concentration
n = number of moles of solute
v = volume in L
C = 0.149 mol / 0.00138 L
C = 107.97 mol/L
