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adelina 88 [10]
3 years ago
14

Show how you could prepare each of the following amines from benzaldehyde by reductive amination:

Chemistry
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

Check explanation.

Explanation:

Reductive amination is also known as reductive alkylation. Reductive amination is an good chemical reaction for the conversion of of carbonyl group into amine.

These are the steps to follow for the reaction;

(1). Step one: carbonyl group(either an aldehyde or a ketone) is used to convert amine to form an imine. The pH of the reaction should be slightly acidic.

(2). Step 2: the imine from step one is then converted into iminium ion under acidic condition too(for protonation).

(3). Step 3: the iminium ion is then reduced to the amine.

The equation of reaction are in the attached file.

Note: the attached picture is labelled (A) to (D).

Reaction A is the reaction of benzaldehyde with Benzylamine, B is the reaction of benzaldehyde with dibenzylamine, C is the reaction of benzaldehyde with N,N-Dimethylbenzylamine and D is the reaction of benzaldehyde with N-Benzylpiperidine.

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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma
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Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

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Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

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The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

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Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

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