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A quite complicated and confusing Organic- question!
Ethanol reacts with 2- methylbutanoic acid to form an ester- the ester's formula is CH3 CH2 CH(CH3) CO2 CH2 CH3.
So when I want to write the balanded equation:
C5 H10 O2 + 3C2 H5 2OH -> 3CH3 CH2 CH(CH3) 2CO2 CH2 CH3, I know that this is wrong, but I don't know how to balance it correctly!!!
Could please somebody help me!
Also, how can I describe the reaction between one primary, one secondary and one tertiary alcohol of an alcohol with the molecular formula C4 H10 O with acidifiedaqueous potassium dichcromate (VI), naming the organic products of the different reactions.
Please could somebody try to explain that to me!!!!
Thanks a lot!!! Funny Mann
So when I want to write the balanded equation:
C5 H10 O2 + 3C2 H5 2OH -> 3CH3 CH2 CH(CH3) 2CO2 CH2 CH3, I know that this is wrong, but I don't know how to balance it correctly!!!
Could please somebody help me!
Also, how can I describe the reaction between one primary, one secondary and one tertiary alcohol of an alcohol with the molecular formula C4 H10 O with acidifiedaqueous potassium dichcromate (VI), naming the organic products of the different reactions.
Please could somebody try to explain that to me!!!!
Thanks a lot!!! Funny Mann
Reply 1
Reacting acidifyed potassium dichromate and sulphuric acid with a primary alcohol oxidation ocurrs to form aldehydes and carboxylic acids. To get an aldehyde, gently heat excess alcohol with a controlled amount of oxidising agent in distillation apparatus. Then the aldehyde is distilled off immediately. To get a carboxylic acid, vigourously oxidise the alcoholwith excess oxidising agent under reflux.
So C4 H10 O with oxidise to aldehyde butanal (C4 H8 O) - CH3 CH2 CH2 COH
and will oxidise to carboxylic acid butanoic acid (C4 H8 02) - CH3 CH2 CH2 COOH
Secondary alcohols will oxidise to form ketones with potassium dichromate.
So will oxidise to form butanone (C4 H8 O) - CH3 CO CH2 CH3
Tertiary alcohols will not react with potassium dichromate. They need to be burned to be oxidised.
So C4 H10 O with oxidise to aldehyde butanal (C4 H8 O) - CH3 CH2 CH2 COH
and will oxidise to carboxylic acid butanoic acid (C4 H8 02) - CH3 CH2 CH2 COOH
Secondary alcohols will oxidise to form ketones with potassium dichromate.
So will oxidise to form butanone (C4 H8 O) - CH3 CO CH2 CH3
Tertiary alcohols will not react with potassium dichromate. They need to be burned to be oxidised.
Reply 2
oops i forgot your equation lol.
CH3 CH (CH3) CH2 COOH + CH3 CH2 OH -> CH3 CH (CH3) CH2 CO2 CH2 CH3
CH3 CH (CH3) CH2 COOH + CH3 CH2 OH -> CH3 CH (CH3) CH2 CO2 CH2 CH3
Reply 3
Thank you so much for your answer, Iam really sorry, but you also explain this question to me?
A 9, 2 g sampel of ethanol and 20, 4g of 2- methylbutanoic acid were mixed with 2, 0g of concentrated sulphuric acid. The mixture was refluxed for four hours and then fractionally distilled to give 17, 4g of the crude ester. The ester was washed repeatedly with aqueous sodium carbonate until there was no more effervescence. After further washing with distilled water and drying, 15,6g of pure ester were obtained. a) Explain why crude ester was washed repetedly with aqueous sodium carbonate.
b) state which gas was responsible for the effervescence.
c) Calculate how many moles of each reactant were used (Mathanol, Mr:46? 2- methylbutanoic acid, Mr: 102)
d) Calculate the percentage yield of pure ester obtained in the above experiment.
I am sorry that I ask so many questions, but I really can't do that easily because I jumped one class and so I have to learn all that staff first.
Would be very grateful if you could help me agein. Thank you!!!!
A 9, 2 g sampel of ethanol and 20, 4g of 2- methylbutanoic acid were mixed with 2, 0g of concentrated sulphuric acid. The mixture was refluxed for four hours and then fractionally distilled to give 17, 4g of the crude ester. The ester was washed repeatedly with aqueous sodium carbonate until there was no more effervescence. After further washing with distilled water and drying, 15,6g of pure ester were obtained. a) Explain why crude ester was washed repetedly with aqueous sodium carbonate.
b) state which gas was responsible for the effervescence.
c) Calculate how many moles of each reactant were used (Mathanol, Mr:46? 2- methylbutanoic acid, Mr: 102)
d) Calculate the percentage yield of pure ester obtained in the above experiment.
I am sorry that I ask so many questions, but I really can't do that easily because I jumped one class and so I have to learn all that staff first.
Would be very grateful if you could help me agein. Thank you!!!!
Reply 4
Thank you so much for your answer, Iam really sorry, but you also explain this question to me?
A 9, 2 g sampel of ethanol and 20, 4g of 2- methylbutanoic acid were mixed with 2, 0g of concentrated sulphuric acid. The mixture was refluxed for four hours and then fractionally distilled to give 17, 4g of the crude ester. The ester was washed repeatedly with aqueous sodium carbonate until there was no more effervescence. After further washing with distilled water and drying, 15,6g of pure ester were obtained. a) Explain why crude ester was washed repetedly with aqueous sodium carbonate.
b) state which gas was responsible for the effervescence.
c) Calculate how many moles of each reactant were used (Mathanol, Mr:46? 2- methylbutanoic acid, Mr: 102)
d) Calculate the percentage yield of pure ester obtained in the above experiment.
I am sorry that I ask so many questions, but I really can't do that easily because I jumped one class and so I have to learn all that staff first.
Would be very grateful if you could help me again. Thank you!!!!
A 9, 2 g sampel of ethanol and 20, 4g of 2- methylbutanoic acid were mixed with 2, 0g of concentrated sulphuric acid. The mixture was refluxed for four hours and then fractionally distilled to give 17, 4g of the crude ester. The ester was washed repeatedly with aqueous sodium carbonate until there was no more effervescence. After further washing with distilled water and drying, 15,6g of pure ester were obtained. a) Explain why crude ester was washed repetedly with aqueous sodium carbonate.
b) state which gas was responsible for the effervescence.
c) Calculate how many moles of each reactant were used (Mathanol, Mr:46? 2- methylbutanoic acid, Mr: 102)
d) Calculate the percentage yield of pure ester obtained in the above experiment.
I am sorry that I ask so many questions, but I really can't do that easily because I jumped one class and so I have to learn all that staff first.
Would be very grateful if you could help me again. Thank you!!!!
Reply 5
no problem. I know what it is like to struggle in chem, I didn't have a teacher for 4 months!!!
a) I think the reason washing with water is to wash away the water soluble impurities to leave a pure ester. But washing with sodium carbonate causes carbon dioxide to form, in which after the washing the pH should be around 7. I don't really know any other purpose for this.
b) Carbon dioxide is responsible for the effervescence.
c) number of moles = mass/rmm. But i think you have to relate this to the equation in the mole ratios.
d) percentage yield = actual yield/theoretical yield x 100
the acid is just a catalyst so is not changed so you can ignore this
work out the ratio of moles from the equation. such as 1:1:1
work out the limiting reactant which is the smallest number of moles. This is the theoretical yield. Then convert the moles to grams.
Then divide the 17.4 g by the theoretical yield and x by 100 to get a percentage.
I cannot give you the actual answers without having to work it out myself. But if you still can't do it, I will work it out for you.
I hope this helps.
a) I think the reason washing with water is to wash away the water soluble impurities to leave a pure ester. But washing with sodium carbonate causes carbon dioxide to form, in which after the washing the pH should be around 7. I don't really know any other purpose for this.
b) Carbon dioxide is responsible for the effervescence.
c) number of moles = mass/rmm. But i think you have to relate this to the equation in the mole ratios.
d) percentage yield = actual yield/theoretical yield x 100
the acid is just a catalyst so is not changed so you can ignore this
work out the ratio of moles from the equation. such as 1:1:1
work out the limiting reactant which is the smallest number of moles. This is the theoretical yield. Then convert the moles to grams.
Then divide the 17.4 g by the theoretical yield and x by 100 to get a percentage.
I cannot give you the actual answers without having to work it out myself. But if you still can't do it, I will work it out for you.
I hope this helps.
Reply 6
First of all, thank you again.
I got for both, Ethanol and 2- methylbutanoic acid the amount of 0, 2 mol, therefore a ratio of 1:1. But what is the limiting reactant, is it just the 0, 2 mole??? This would not make much sense, because than the percentage yield would be 1, 15 % which seems very, very little.
I got for both, Ethanol and 2- methylbutanoic acid the amount of 0, 2 mol, therefore a ratio of 1:1. But what is the limiting reactant, is it just the 0, 2 mole??? This would not make much sense, because than the percentage yield would be 1, 15 % which seems very, very little.
Reply 7
That was rubbish what I did before. Is the percentage yield 45%?????
Reply 8
Yeah that sound more logical. Usually lab practicals do not get a yield above 50% which is expected due to poor measuring instruments and losses in filtration etc. Glad to help and thanks for the thankyou, i try to help people where i can but usually no one ever thanks me. Any other problems feel free to contact me, as i am doing a2 chem.
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