Do you think you can put NaOH with an acid catalyst for the hydrolysis of the Estee???Also does anyone remember the moles we were given when we had to work out KC that got an answer of 9.09x10-3 I know one was 0.58 another was 3.76 I think???? What were the other two?
i put naoh aq with acid catalyst too, the h2o and acid catalyst would break the ester up and then the nah would form the sodium ethanoate or whatever it was
Step 3 was a substitution question to go from Br to a CN group. So I said KCN aqueous and alcohol.
The final step was reduction, so Hydrogen & Nickel Catalyst.
I was going to put KCN aqueous and alcoholic, but in the question it said "one condition" so I went with alcoholic. Not sure what kind of answer they were looking for there, did they want aqueous and alcoholic, or aqueous or alcoholic?
i put naoh aq with acid catalyst too, the h2o and acid catalyst would break the ester up and then the nah would form the sodium ethanoate or whatever it was
Yeah that's what I thought but now I'm thinking an acid catalyst will form the carboxylic acid not the carboxylic salt because when the Salt is formed the acid catalyst will turn it to a CA (
Yeah that's what I thought but now I'm thinking an acid catalyst will form the carboxylic acid not the carboxylic salt because when the Salt is formed the acid catalyst will turn it to a CA (
Conditions are a dilute strong base, heat under reflux and conc h2so4 catalyst and the base had to be NaOH also question said conditionS hence I think the answer should've been:
Dilute NaOH, HUR and conc H2SO4 catalyst but in the past they've been generous with putting dilute/conc
Your reasoning is right but using what I said/you said you will produce the alcohol and sodium salt which is what we needed. Only if you used a strong acid would you not et the sodium salt but get a carboxylic acid instead
I was going to put KCN aqueous and alcoholic, but in the question it said "one condition" so I went with alcoholic. Not sure what kind of answer they were looking for there, did they want aqueous and alcoholic, or aqueous or alcoholic?
Markschemes in past have 'aqueous or alcoholic' but I put aqueous ethanol because it covers both answers.
Conditions are a dilute strong base, heat under reflux and conc h2so4 catalyst and the base had to be NaOH also question said conditionS hence I think the answer should've been:
Dilute NaOH, HUR and conc H2SO4 catalyst but in the past they've been generous with putting dilute/conc
Your reasoning is right but using what I said/you said you will produce the alcohol and sodium salt which is what we needed. Only if you used a strong acid would you not et the sodium salt but get a carboxylic acid instead
Markschemes in past have 'aqueous or alcoholic' but I put aqueous ethanol because it covers both answers.
Hi thank you for doing this! if possible could you add in the marks for the questions so i can estimate how badly i did.. also i dont remember a question about how to distinguish between enantiomers?? Did i miss this??
Hi, thanks for this:
I disagree with the answer to Q1 - I think they would have looked for something using the idea of Bronsted-Lowry bases/acids, because that's what we studied in the topic. Otherwise the reaction you've written is not really a reaction? You have CH3COOH on both sides.
Also for question 7, I disagree again. The rate equation contains the reagents that are used in the rate determining step in the proportions used in the rate determining step. If you increase the concentration of one, it doesn't change the mechanism, nor the proportions used, so the rate equation is still the same, I believe.
DEFINITELY no question on distinguishing between enantiomers
I disagree with the answer to Q1 - I think they would have looked for something using the idea of Bronsted-Lowry bases/acids, because that's what we studied in the topic. Otherwise the reaction you've written is not really a reaction? You have CH3COOH on both sides.
Also for question 7, I disagree again. The rate equation contains the reagents that are used in the rate determining step in the proportions used in the rate determining step. If you increase the concentration of one, it doesn't change the mechanism, nor the proportions used, so the rate equation is still the same, I believe.
DEFINITELY no question on distinguishing between enantiomers
IDK if these are right, just what I thought
I said the rate equation was the same and stated that iodine was not in it or whatever hence rate remains the same but I rushed it - any marks do you think?