The Wolff-Kishner reaction involves the reaction of an aldehyde/ketone with hydrazine in the presence of KOH. The process is use
ful for converting an aldehyde or ketone into an alkane. The reaction involves formation of a hydrazone, followed by base-catalyzed double-bond migration, loss of N2 gas to give a carbanion, and protonation to give the alkane. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions
1 answer:
Answer:
in the attached image is the reaction mechanism.
Explanation:
The first reaction (reaction 1) shown in the attached image is the Wolff-Kishner reduction, which is characterized when the carbonyl is reduced to an alkane in the presence of a hydrazine and a base. In reaction 1, the aldehyde reacts with hydrazine to produce oxime. This mechanism begins with the attack of the amine on the carbonyl group. Proton exchange happens and the water leaves the molecule.
In reaction 2, the KOH is deprotoned in nitrogen and organized to form the bond between the nitrogen molecule. this deprotonation releases the nitrogen gas
You might be interested in
Answer:
600 g K₂SO₄
Explanation:
First write down the complete, balanced chemical equation for such question. In this case:
Cr₂(SO₄)₃ + 2K₃PO₄ → 3K₂SO₄ + 2CrPO₄
Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.
- Molar mass(MM) of Cr₂(SO₄)₃:
= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)
= 2*(52) + 3*(32) + 12*(16)
= 104 + 96 + 192
= 392 g
- Molar mass(MM) of K₂SO₄:
= 2*(MM of K) + 1*( MM of S) + 4*( MM of O)
= 2*(39) + 32 + 4*(16)
= 78 + 32 + 64
= 174 g
Here comes the concept of Limiting reagent:
The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.
Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.
Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?
Yeild of K₂SO₄ :
That is 599.3 g.
Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.
Therefore, 600g of K₂SO₄ is produced.