Show the two (neutral) organic starting materials that would be used to synthesize the following ketone by a hydroxide mediated crossed‑aldol reaction the product formed is alpha-beta unsaturated carbonyl compound.
<h3>What is the Aldol-condensation ?</h3>
The condensation response among distinctive molecules of an aldehyde or ketone in a protic solvent which includes water or alcohol constitutes the crossed aldol response. When condensation is among distinctive carbonyl compounds, it's far referred to as crossed aldol condensation.
Applying the retrosynthetic evaluation technique, the systems of the beginning substances are decided via way of means of the usage of the bottom catalyzed for the crossed aldol mechanism. The beginning substances are aldehyde and ketone compounds that react with every different to present the goal molecule.
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Divide each wight by the relative atomic mass
C = 216 / 12 = 18
H = 36 / 1 = 36
O = 288/16 = 18
Ratio of C:h:O = 1:2:1
Empirical formula is CH2O Could be formaldehyde HCHO.
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
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In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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Answer: Ba(CN)2 + (NH4)2SO4 => BaSO4 + 2 NH4CN
Explanation: The reaction will produce BaSO4 Barium sulfate and NH4CN Ammonium cyanide.
