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2 years ago

How many different products can be formed from a poorly planned crossed aldol addition?

Chemistry

2 answers:

Lerok [7]2 years ago

6 0

There are four different products can be formed from a poorly planned crossed aldol addition.

So the correct option is D

Two distinct ketone and aldehyde reactants are used in a crossed condensation reaction. Due to presence of distinct ketone and aldehyde reactants there are possibility of formation of several enolate ion nucleophiles as well as multiple carbonyl electrophile .

Example : Ethanal and propanone undergoes cross-aldol condensation reaction and give results in two products.

First case: ethanal acts as enolate ion

Second case: propanone acts as enolate ion

Third case: when self condensation is also a product.

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zheka24 [161]2 years ago

3 0

There are 4 different products will be formed from a poorly planned crossed aldol addition.

In a crossed aldol condensation, two distinct ketone and aldehyde reactants are used. Due to the possibility of several enolate nucleophiles as well as multiple carbonyl electrophiles, similar reactions typically result in a variety of distinct condensation products.

Ethanol and propanol undergo cross-aldol condensation, which results in two products:

one where ethanol serves as an enolate ion and the other where propanol serves as an enolate ion.
2-Methylbut-2-enal and Pent-2-enal are just the two cross-aldol derivatives.

Therefore, the correct answer will be option (d)

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Draw a structure for an alcohol that exhibits a molecular ion at M+ = 74 and that produces fragments at m/z = 59, m/z = 56 and m

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Answer:

Please refer to the attachment below for answer and explanation.

Explanation:

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3 years ago

4.How element gold is different from oxygen?

V125BC [204]

Answer:

Gold is a metal, more specifically a transition metal, whereas Oxygen is a nonmetal, more specifically a reactive nonmetal. Using this information, you can compare and contrast metals, nonmetals, and metalloids.

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Reccomend this site for questions llike these: https://ptable.com/#Properties

7 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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4 years ago