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kogti [31]
3 years ago
12

Draw the curved arrow mechanism for the addition of HCN in water with NaOH to 3,4-dimethylcyclopentan-1-one to give the correspo

nding cyanohydrin in the fewest steps. Draw the arrows that lead to the resonance structures with full octets around each atom other than hydrogen. Draw all electrons and charges if necessary on all structures; do not show any inorganic side products or counterions. Reagents needed for each step are provided in the boxes.
Chemistry
1 answer:
Crank3 years ago
6 0

Answer:

See picture and explanation below

Explanation:

In the attached picture you have the answer for this.

In the first step, we have an acid base reaction between HCN and NaOH. When this happens, its formed the following:

HCN + NaOH --------> CN⁻ + H₂O + Na⁺

As a second step, The CN⁻ attacks the carbonile group in the cyclopentane. This causes to open the double bond, and then, the Cyano enters the molecule.

The final step is another acid base reaction, where the oxygen substracts a hydrogen atom of the water in the medium, and then, the alcohol group is formed. With this step, the cyanohydrin is finally formed. See the mechanism below in the picture.

Hope this helps

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I hope this helps. let me know in the comments if anything is unclear. 
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Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

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Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

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We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

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Answer:

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