Organic Chemistry Lab - Zyban Synthesis B) How do I go about solving this proble

Question

Organic Chemistry Lab - Zyban Synthesis
B) How do I go about solving this problem? We started with 0.403 g of propiophenone, I don't know how to calculate the theoretical yield with so many steps like this. I've provided a picture of the reaction and materials used below the question.
Thank you!!

6. Identify the reaction components and solvents as more soluble in the aq organic (o) layer during separation of substitution and protonation steps. 3 pts ts as more soluble in the aqueous (a) or H N HN CI CI NH Br

Explanation / Answer

A) The solubility that you have assigned for the compounds is correct for almost all the compounds, except the last one. The last compound is dichloromethane, DCM, which is not very soluble in water. DCM is more soluble in organic solvents.

While determining solubility of organic compounds, a very useful rule of thumb is “Like dissolves like.”

The first and the fourth compounds in the diagram you uploaded are ion-pair adducts and ion-pair compounds are more soluble in water because of ion-dipole and dipole-dipole interactions.

The second, third and fifth compounds contain cyclic hydrocarbon (second) or aromatic (third and fifth) rings. Now, hydrocarbons and aromatic compounds are in general less polar than ion-pairs or acids or ketones and hence are more soluble in organic solvents. In these cases, it is mainly due to the less polar nature of the compounds that they are more soluble in organic solvents. The final compound, DCM, is soluble in organic solvents. Although it is electron-withdrawing chlorine atoms and one would expect the molecule to have a considerable degree of polarity, in reality, the dipole-dipole interactions of the C-Cl bond(s) is nullified by the dipole-dipole intraction of the H-C bond(s) making DCM non-polar and hence, soluble in organic solvents.

B) Now, for the second part of the question, we only need to consider the initial and the final products. Assuming that the reaction goes to 100 percent completion, we know that 1 mole of propiophenone should yield 1 mole of Zyban analog (as per the reaction sequence). This problem then reduces to a simple molar ratio problem.

We start with 0.403 gm of propiophenone and the M.W. of propiophenone is 134.2 gm/mole.

Hence, moles of propiophenone used in the reaction is

(0.403 gm)/ (134.2 gm/mole) = 0.003 mole.

Now, 1 mole of priophiophenone gives 1 mole of Zyban analog (3a) in your question.

Hence, 0.003 mole of prophiophenone will yield 0.003 mole of Zyban analog. Now, molecular weight of Zyban analog (3a) is given as 205.3 gm/mole

Therefore, theoretical yield of Zyban analog considering 100% reaction completion will be

(0.003 mole) * (205.3 gm/mole) = 0.6159 gm

Please note that molecular weights are dimensionless quantities and I have used gram molecular weight for my calculations. The last two digits may be slightly off from the supplied answer due to different rounding patterns.

Ans: The theoretical yield of Zyban analog is 0.6159 gm.

5) Now, again from the reaction sequence supplied, it is clear that 1 mole of HCl protonates 1 mole of the Zyban analog.

Now, the reaction sequence afforded 0.003 mole of Zyban analog and hence, 0.003 mole of HCl will be required to protonate the same.

The molarity of the supplied HCl solution is 2M.

Let us assume that x L of 2M HCl is required to protonate 0.003 mole of Zyban analog. From the previous step, we know that this amount must equal 0.003 mole of HCl.

Hence, we have

(x L) * (2 mole/ 1 L) = 0.003 mole

or, x = 0.003/2 = 0.0015

Hence, the millilitres of 2M HCl required is

(0.0015 L) * (1000 mL/ 1 L) = 1.5 mL.

Ans: The volume of 2M HCl required is 1.5 mL.

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