Question

a) Show the full balanced equation of Benzaldehyde + Sodium Cyanide -> Benzoin. Stoichiometry ratio should be 2:1. Ratio of -CN to benzoic is 1:1 if that even helps. Rely on Sodium Cyanide being 0.54M in 95% Ethanol. b) Use the density of benzaldehyde to calculate its mass. 400 microliters were used in my experiment, and the MW is 106.13. C) Thus, Find the Limiting Reagent, Theoretical yield, and % Yield of Benzoin. Obtained yield in experiment was .0246g benzoin crystals, and again, 400microliters of benzaldehyde used AND 4mL of Sodium Cyanide (0.54M in95% ethanol) was used. PLEASE SHOW NICE LEGIBLE WORK SO I CAN UNDERSTAND WHAT IS GOING ON.

Answer 1

a.   Benzoin condensation of benzaldehyde under cyanide catalysis to aromatic α - hydroxy ketones ( benzoin )

2 Benzaldehyde + 2 Sodium Cyanide (0.54 M in ethanol) 1 Benzoin + 1 cyanide ion

  

the cyanide ion reacts with benzaldehyde to form cyanohydrin,this reaction will start from cyanohydrins which is obtained from the addition of cyanide ions to aldehydes.

b. banazladehyde density =1.04

1 millilitere =1000 microlieters

400 microliters equal to 0.4 milli liters

=0.4*1.04

=0.416 g or 416 mg.

C. limiting reagent .

benzaldehyde 416 mg, but sodium cyanide 105.84 mg (0.54*49=26.46, it meams 1 ml conatins 26.46 mg so 26.46* 4 =105.84 mg)

as per stachimetry reuired wt 96 .03 mg (416/106.13*49=192.0, 192/2=96.03 (half mole required with respect to benzaldehyde so dividedby 2)

so benzaldehyde is limiting reagent.

theoritical yield:

=416/106.13*212.26

=831.99/2 (two molecules will combine as one benzoin molecule it should be equal to weight as 100%)

=41599 mg or 0.4159 g

so theoretical yield 0.4159 g

% yield of benzoin

= obtained yield/theoretical yield

=0.0246/0.4159

= 5.91%

% yield of benzoin =5.91%