CHEMISTRY COLLEGE

The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism.

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Answer 1

Answer:

Answer:

See the attached file for the structure

Explanation:

See the attached file

Related Questions

The quantity of mass of an object contained within its volume is a measure of

Answers

the answer is density

the quantity of mass of an object contained within its volume is a measure of density

In basic enzyme kinetics, a large Km indicates that the substrate binds A. Permanently B. Transiently C. Covalently D. with low specificity E. Weakly

Answers

Answer: D

Explanation:

Km value is a signature of the enzyme. It is the characteristic feature of a particular enzymes for a specific substrate. Km denotes the affinity of the enzyme for substrate. The lesser the numerical value of Km, the affinity of the enzyme for the substrate is more.

In the velocity x substrate graph in a fixed quantity if enzyme. As substrate concentration is increase, the velocity is also increasing at the initial phase but the curve fatten afterwards. This is because as more substrate is added, all enzymes molecules become saturated. Further increase in substrate cannot make any effect in the reaction velocity.

The maximum velocity is called Vmax. Km is the concentration of substrate that Vmax is half.

The larger the numerical value of Km, the lesser the enzyme binds the substrate

A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 358C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

Answers

Answer:

(a) $\Delta S_(ref)=3.876(kJ)/(K)$

(b) $S_(heat\ source)=-1.678(kJ)/(K)$

Explanation:

Hello,

(a) In this case, such refrigerant, we can notice that at the given conditions, the initial entropy from property tables (Cengel 7th ed) is:

$s_(initial)=s_f+xs_(fg)=0.15457+0.4*0.78316=0.4678(kJ)/(kg*K)$

Now, for the final condition, we first need to compute the initial specific volume as it remains the same (rigid tank) after the thermodynamic process:

$v_(initial)=v_f+xv_(fg)=0.0007533+0.4*(0.099867-0.0007533)=0.0404(m^3)/(kg)$

Then, at 400 kPa we evaluate the given volume that is also between the liquid and vapor specific volume, thus, we calculate the quality at the end of the process:

$x_f=(0.0404-0.0007907)/(0.051201-0.0007907) =0.786$

With it, we compute the final entropy:

$s_(final)=0.24761+0.785*0.67929=0.781(kJ)/(kg*K)$

Finally, entropy change for the refrigerant turns out:

$m_(ref)=(0.5m^3)/(0.0404(m^3)/(kg) )=12.4kg \n\n\Delta S_(ref)=12.4kg *(0.781(kJ)/(kg*K)-0.4678(kJ)/(kg*K) )\n\n\Delta S_(ref)=3.876(kJ)/(K)$

(b) In this case, by using the first law of thermodynamics we compute the acquired heat by the refrigerant from the heat source by computing the initial and final internal energy respectively (no work is done):

$Q=\Delta U$

$u_(initial)=38.28+0.4*186.21=112.764(kJ)/(kg)\n \nu_(final)=63.62+0.786*171.45=198.40(kJ)/(kg)$

Hence:

$Q=12.4kg*(198.40-112.764)(kJ)/(kg) =1059.1kJ$

Finally, the entropy change of the heat source (which release the heat, therefore it is negative):

$S_(heat\ source)=(1059.1kJ )/((358+273.15)K) \n\nS_(heat\ source)=-1.678(kJ)/(K)$

$\Delta S _(tot)=3.876(kJ)/(K)-1.678(kJ)/(K)\n\n\Delta S _(tot)=2.198(kJ)/(K)$

Which has thermodynamic agreement as it is positive

Regards.

Final answer:

The entropy changes in this process can be partially calculated using principles from thermodynamics. However, without the exact heat transfer, not all values can be determined.

Explanation:

The calculation of the entropy change in this thermodynamic process involves principles from thermodynamics and requires steps to determine the initial and final states of the refrigerant. First, we would need to find the entropy at the initial and final states using the refrigerant properties table for refrigerant-134a and the provided information (200 kPa and 40% quality initially, 400 kPa finally). The entropy change of the refrigerant is the difference between the final and initial entropy.

Next, the entropy change of the heat source is calculated as the heat transfer divided by the absolute temperature of the source. However, the problem does not provide the amount of heat transferred from the source, making it impossible to determine this value directly.

Finally, in an isolated system, the total entropy change of the process is the sum of the entropy changes of the refrigerant and the heat source. Here, the precise values cannot be calculated due to a lack of specific data including exact heat transfer.

Learn more about Entropy Change here:

brainly.com/question/35154911

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A sample of phosgene gas at an initial concentration of 0.500 m is heated at 527 °c in a reaction vessel. at equilibrium, the concentration of co (g) was found to be 0.046 m. calculate the equilibrium constant for the reaction at 527 °c.

Answers

Answer : The equilibrium constant will be - 0.454

Explanation : The reaction is given below;

$COCl_(2) \ \textless \ ----\ \textgreater \ CO + Cl_(2)$

We need to find Equilibrium constant - $K_(c)$ ;

$K_(c)$ = [CO] [ $Cl_(2)$ ] / $[COCl_(2) ]$

So, $K_(c)$ = [0.046] X [0.046] / [0.5 - 0.046]

Therefore, $K_(c)$ = 0.454

1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-137 decays over 60 years, how many grams will remain?

Answers

Answer:

1.25 g

Explanation:

Now we have to use the formula;

N/No = (1/2)^t/t1/2

N= mass of cesium-137 left after a time t (the unknown)

No= mass of cesium-137 present at the beginning = 5.0 g

t= time taken for 5.0 g of cesium-137 to decay =60 years

t1/2= half life of cesium-137= 30 years

Substituting values;

N/5= (1/2)^60/30

N/5= (1/2)^2

N/5= 1/4

4N= 5

N= 5/4

N= 1.25 g

Therefore, 1.25 g of cesium-137 will remain after 60 years.

On average, about ____________ of incoming solar radiation is reflected back to space.A 50%
B 30%
C 20%
D 10%

Answers

I believe it B 30% hope that helps

I think it’s B I’m not sure sorry if it’s wrong

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