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Aqa chem 4/ chem 5 june 2016 thread

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Original post by Alevels16
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
Reply 1641
Original post by High Stakes
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?
Original post by thomaarrss
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 :frown:(
Can anyone confirm?

I put heat under reflux to separate alcohol is this right?

Was nucle add elim the answer to any questions?
Original post by Alevels16
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 :frown:(


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

Original post by Fibsy
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.
(edited 7 years ago)
Original post by MuaazAnsari
the electroplie produced was ClCH2CO


Thank you finally someone who agrees!
Original post by Suits101
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.


Sounds right :smile: but I didn't put heat under reflux :frown:
Grade boundaries anyone? For an A* and an A?
Original post by Alevels16
Sounds right :smile: but I didn't put heat under reflux :frown:


Question was 3 marks -

Reagent: NaOH (1)
Conditions: dilute (1), conc H2SO4 OR heat under reflux
Separation technique: - (1)

That would be my guess. Dilute in my opinion is essential and then either of the other two conditions. :smile:
Original post by 🙈🙈🙉🙊
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

IDK if these are right, just what I thought :smile:
Original post by DexterMabel
I had applied for veterinary and now I'm going to have to get an Ace chem 5 if I want to get in 😭😭


I feel you :s-smilie: - I'm a medicine applicant :/
Original post by Capsicle102
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

IDK if these are right, just what I thought :smile:


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?
There is a news article on the daily mail about chem 4 :smile:
Original post by Suits101
Question was 3 marks -

Reagent: NaOH (1)
Conditions: dilute (1), conc H2SO4 OR heat under reflux
Separation technique: - (1)

That would be my guess. Dilute in my opinion is essential and then either of the other two conditions. :smile:


I put dilute as the condition,do you think I'll still get the mark or should I have put dilute NaOH for the reagent?
I remember getting 8.33...x10^-something as one of the answers for something? :ahhhhh:
Original post by Saldudsjen123
I put dilute as the condition,do you think I'll still get the mark or should I have put dilute NaOH for the reagent?


I did originally!

Dilute should be the condition though so you're fine :smile:

Even if you did put reagent as dilute NaOH then most likely they'd understand it should've been a condition and marked it right anyway
Did you use water in your equilibrium constant or not?

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