Would it not have been an aldehyde and a carboxylic acid since you get a carboxylic acid with excess potassium dichromate from a primary alcohol and an aldehyde from using less potassium dichromate and distilling as you go?

Reply 5

QS2Nabz

Q1 - 10.5
Q2 - -308 and 0.0118
6 marker - the alcohol, then aldehyde, then butane
last multiple choice question - D?
anyone get any similar answers ??

Reply 6

QS2Nabz

Original post by Ophelia.

but the second oh was attached to a secondary, so it must be a ketone ?? not sure, thats what i thought
and for the equations , was there 3 {o} and 2h20??

Reply 7

Ophelia.

Original post by QS2Nabz

but the second oh was attached to a secondary, so it must be a ketone ?? not sure, thats what i thought
and for the equations , was there 3 {o} and 2h20??

You're probably right, I was a little confused on that at first. I don't think I got the equation right either.What did you get for the thermometer and bumping crystals question?

Reply 8

Gswp

Original post by QS2Nabz

Q1 - 10.5
Q2 - -308 and 0.0118
6 marker - the alcohol, then aldehyde, then butane
last multiple choice question - D?
anyone get any similar answers ??

0.0118?

Joules?!

(edited 7 years ago)

Reply 9

Anna1909

I thought it was an aldehyde too but like it was primary and a secondary alcohol which I had never seen before 😁 It could have been a ketone and was the percentage uncertainty 1.3%?

Reply 10

Sam Webb

For the reflux question it had an alcohol that had two "OH" groups. IMO if you were to continuously oxidise that you could only oxidise it 3 times, once for the "OH" in the middle turning it into "O" with a double bond, and twice on the "OH" group at the end turning it into an "O" double bond and and "OH".

I reckon this means you would undergo the first stage of oxidation twice (turning "OH" into an "O" double bond). Therefore producing 2 H2Os)

CH2(OH)CH(OH)CH3 + 3[O] ----> CO(OH)COCH3 + 2H2O

What do you guys think?

Reply 11

Lola1244

Original post by QS2Nabz

Q1 - 10.5
Q2 - -308 and 0.0118
6 marker - the alcohol, then aldehyde, then butane
last multiple choice question - D?
anyone get any similar answers ??

I got D for the last one! Because it would increase pressure by 11.25X

Reply 12

Lola1244

Original post by Gswp

0.0118?

Joules?!

I got 11800 kj :redface:

oops

Reply 13

Sam Webb

Original post by Anna1909

I thought it was an aldehyde too but like it was primary and a secondary alcohol which I had never seen before 😁 It could have been a ketone and was the percentage uncertainty 1.3%?

Temperature change was 38 degrees C. Uncertainty was 0.5 degrees C, which they nicely told us instead of trying to trick us by saying 0.25 and expecting us to multiply it by 2 because there was 2 measurements taken.

This means:

0.5/38 x 100 = 1.32% :smile:

Reply 14

Applicant75235

http://www.thestudentroom.co.uk/showthread.php?t=4156405&p=65646977&posted=1#post65646977

Reply 15

Anna1909

Original post by Sam Webb

Do you reckon it would get the mark or not? I mean it was only one so it doesn't really matter I guess 😂 But thank you ☺️

Reply 16

Sam Webb

Original post by Anna1909

Do you reckon it would get the mark or not? I mean it was only one so it doesn't really matter I guess 😂 But thank you ☺️

You'd probs get the mark :smile:

Reply 17

Sam Webb

For the multiple choice question asking to calculate "C tripled bonded to O" bond enthalpy.

CH4 + H20 ----- > CO + 3H2 Enthalpy change +206kJ/mol

Each bond enthalpy:
[C-H] [H-O] [H-H] [CO]

There are 4 C-H bonds, 2 H-O bonds, and 3 H-H bonds. Only 1 CO bond. So the equation would be:

206 + (4x413) + (2x463) - (3x436) = 1476 kJ/mol

So the answer for this question was 1426 (C I think)

Reply 18

Kforkittyy

I got 11800kj as well!

Reply 19

Frazerc

For the mechanism question (7 marker) it said to name and 'outline' the mechanism. Were we meant to draw it or describe it? They gave us lines after all and very little space to draw it.