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Alecsey [184]
3 years ago
10

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

Chemistry
1 answer:
Ivan3 years ago
6 0

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

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Which measurement has the greatest number of significant figures?
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Answer:

Option B = 60,600 mg  (correct option)

Explanation:

First of all we will have an idea which numbers are consider as significant.

1 = All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

2= Leading zeros are not consider as a significant figures. e.g. 0.02 in this number only one significant figure present which is 2.

3= Zero between the non zero digits are consider significant like 105 consist of three significant figures.

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Option B have 5 significant figures.

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Calculate the volume occupied<br>at s.t.p by 6.89<br>gas [H = 10, N = 14​
Nadya [2.5K]

Answer:

9.07 L

Explanation:

<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0​].</em>

Step 1: Given and required data

  • Mass of NH₃ (m): 6.89 g
  • Molar mass of NH₃ (M): 17.0 g/mol

Step 2: Calculate the moles (n) of NH₃

We will use the following epxression.

n = m / M

n = 6.89 g / (17.0 g/mol) = 0.405 mol

Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP

At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).

0.405 mol × 22.4 L/1 mol = 9.07 L

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