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Arlecino [84]
2 years ago
10

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.

to learn more about aldol addition

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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)

To know more about products

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3 years ago
A 10.0cm3 volume of alcohol has a mass of 7.05g. What the density
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A 10.0cm3 volume of alcohol has a mass of 7.05g. 0.705 \frac{g}{c m^{3}} is the density

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<u>Explanation:</u>

The ratio between the mass of a substance and its unit volume is referred as density.

\text {Density}=\frac{\text { mass }}{\text { volume }}=\frac{7.05 g}{10.0 \mathrm{cm}^{3}}=0.705 \frac{g}{\mathrm{cm}^{3}} is the Answer

<u>Note:</u>

1000 g = 1 kg

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When we used to transform the given values into S.I units, then

\text {Density}\left(\frac{k g}{m^{3}}\right)=\frac{\text {Mass}(k g)}{\text {Volume}\left(m^{3}\right)}

Alcohol’s volume can be written as, 10.0 c m^{3} = 10 \times 10^{-2} m^{3}

Alcohol’s mass can denote as, 7.05 g = 7.05 \times 10^{-3} \mathrm{kg}

Then, \text { Density }\left(\frac{k g}{m^{8}}\right)=\frac{, 00705(k g)}{0.1\left(m^{3}\right)} = .0705 \frac{\mathrm{kg}}{\mathrm{m}^{\mathrm{s}}}

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