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
Br
Se
As
Ge
Ga
Rb
Elements in the top right corner of the periodic table have the highest electronegativity. Elements on the right side have a higher electronegativity than those on the left, same with the ones on the top in comparison to those on the bottom.
Answer:
See explanation and image attached for details
Explanation:
The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.
Answer:
0.8749 grams of hydrogen gas was formed from the reaction.
Explanation:
P = Pressure of hydrogen gad= 744 Torr = 0.98 atm
(1 atm = 760 Torr)
V = Volume of hydrogen gas= 11 L
n = number of moles of hydrogen gas= ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of vapor = 27.0 °C = 300.15 K
Putting values in above equation, we get:
Using an ideal gas equation:
n = 0.4374 moles
Mass of 0.4374 moles of hydrogen gas:
0.4374 mol × 2 g/mol = 0.8749 g
0.8749 grams of hydrogen gas was formed from the reaction.
First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
