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OCR Chemistry A 2017 Exam Thread (New A Level)

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Original post by epsilon9012
Did everyone get C2H4O for the combustion substance in the multiple choice question?


Yep
Original post by epsilon9012
An the number if water molecules was 6.02x10^22


Wasn't it 3.01 x 10^22? Because there were 2 moles of water?
Original post by Callum200299
incorrect , you need an extra carbon on your carboxylic acid part


What carboxylic acid apert its an ester
Original post by epsilon9012
Did everyone get C2H4O for the combustion substance in the multiple choice question?


Yeah that was the right answer
Original post by epsilon9012
What carboxylic acid apert its an ester


the part of the ester made from the carboxylic acid ?
Original post by Clintbarton
Wasn't it 3.01 x 10^22? Because there were 2 moles of water?


It was 0.1 mol of NaOH and an excess of acid, so it formed 0.1 mol of water
Original post by Clintbarton
Wasn't it 3.01 x 10^22? Because there were 2 moles of water?


No, it was a trick as there were two carboxylic acid groups but only enough NaOH to react with one of them generating 1 mole of water per mole of Dioic acid.
Sneaky question
MCQ which I thought would catch a lot of people out, the most acidic in aqueous solution question between 4 different functional grouped compounds.

Answer was actually the acyl chloride because it reacts with the water to form HCl which is a strong acid, but it's a sneaky question and I imagine people put the carboxylic acid.
Original post by epsilon9012
You have them the complete wrong way round there


How many marks would you lose if you got it the wrong way round
Did anyone get ?
C40H58 + 11h2 -----> C40H80
How did you do for the mlutiple choices question and asking about which one is the most acidic in water? I got D (acly chlorid+ water= carboxlic acid +HCl)
Original post by AnnaRainbows
Butan-1-ol + [O] --> Butanoic acid and Water, using acidified potassium dichromate and under reflux

Butan-1-ol + [O] --> Butanal and water, using acidified potassium dichromate and distillation technique

that's what I put and then I drew the structures in displayed formula


you'll need 2 [o] for the formation of carboxylic acid
Original post by epsilon9012
You have them the complete wrong way round there


How many marks would you lose for putting them the wrong way round
What were the names of the molecules which had to be named?
Original post by jadeemma
What were the names of the molecules which had to be named?


I put:
hex-3enol (apparently it has to be hex-3-en-1-ol)
3-methylphenol
Original post by RihWHYB
I put:
hex-3enol (apparently it has to be hex-3-en-1-ol)
3-methylphenol


If i recall correctly, it wasn't a benzene ring, it was just a normal cyclohexane ring. I put 3-methylcyclohexanol
Original post by chrisantonio123
How many marks would you lose for putting them the wrong way round


Ok so according to past papers if your explanation is perfect then you lose 1 mark i your explanantion is averge but you got the compound wrong then you lose 2 marks so its not really that much an i think you'll be fine
Original post by YAREYOUSOSMELLY
If i recall correctly, it wasn't a benzene ring, it was just a normal cyclohexane ring. I put 3-methylcyclohexanol


3 methyl cyclohexan-1-ol and no there was no phenol there
Reply 1998
I have a question to do with the way grades will be given this year. As there are no course work this year, will the total grade boundaries be much lower. This given to the fact that the course work system was broken, in the sense that most people had on there course work A-B. So students who got an U overall, had the similair grade in there course work as a student who got an A overall. Just wondering what everyone else thinks
OCR are gettin more cunning every year.

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