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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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What are the compounds empirical and molecular formulas

Liono4ka [1.6K]

Answer:

The empirical formula is the simplest ratio of atoms while molecular is the actual formula.

Explanation:

Empirical Formula is the simplest ratio of atoms present in the compound.

The molecular formula shows the actual number of atoms present in the compound.

For example, CH is the empirical formula and C6H6 is the molecular formula of benzene.

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2 years ago

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anzhelika [568]

Answer:Not so good, but I'll manage. What about you?

Explanation:

4 0

3 years ago

Read 2 more answers

At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2

saul85 [17]

2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>

4 0

3 years ago

Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun

Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂= $\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}$ =0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

$\frac{0.92678}{0.0309}$ =Si³⁰

Si³⁰=29.99288

8 0

3 years ago

Consider the following balanced equation:

algol [13]

Moles of PF₃ : 4

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

$\tt P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)

Limiting reactant : the smallest ratio (mol divide by coefficient)

P₄ : F₂ =

$\tt \dfrac{1.25}{1}\div \dfrac{6}{6}=1.25\div 1\rightarrow F_2~limiting~reactant(smallest~ratio)$

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

$\tt \dfrac{4}{6}\times 6~moles=4~moles$

8 0

3 years ago