• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

Messages
1,983
Reaction score
3,044
Points
273
thanks!
no just (b)...also for (c) no matter what reaction with give, the CH2 and no. of C and H of (CH2CH) will stay the same right? as in no. of C and H on the right hand sight of the formula like HO(CH2CH)O and H2N(CH2C)OO(H) or is there no rule like that?
image.jpg image.jpg
I hope it answers your question. You have to keep in mind that aminoacid was not the only product but the only organic product and examiner has shown only organic product coz we are interested in it. The rest of them will be impurities and would be distilled off.
 
Messages
93
Reaction score
51
Points
28
When is which Reducing and Oxidation agent used?
I have noticed in Organic questions especially that we have to use exact Agent for reactions.
 
Messages
1,983
Reaction score
3,044
Points
273
When is which Reducing and Oxidation agent used?
I have noticed in Organic questions especially that we have to use exact Agent for reactions.
CIE's syllabus is wrong in many ways. if you will do BS in chemistry you will learn this. MnO4 and NaBh4 are not used in organic reaction are they are very reactive and can oxidise them to oxides and reduce the compound into their elements. but you will have to learn this:
potassium dichromate is not a strong oxidizing agent. we usually use it to oxidize C-OH to C=O.
it can't oxidise every group.
potassium permanganate is although a very strong oxidising agent. if concentrated it can oxidise all of the organic compound into CO2 and H2. it is used to oxidise C=C into diol if very dilute and cold. if concentrated and hot it can even rupture the double bond or even oxidise CH2= into CO2 and H2
and for reducing agent use any of them. it doesn't matter. although NaBH4 is stronger i guess.
 
Messages
192
Reaction score
107
Points
53
upload_2014-4-18_18-48-9.png

Whats order with respect to { H+}
Marking scheme Answer is First order .. .. ..
Question is from November 2011 , type 41
 
Messages
675
Reaction score
862
Points
103
Could you link me to the paper or send me pics of the entire question?

If a new concept is tested, it's done by exemplifying a reaction and the question follows the example.
I've looked in all 3 variants twice but couldn't find the question.
 
Messages
675
Reaction score
862
Points
103
View attachment 39319

Whats order with respect to { H+}
Marking scheme Answer is First order .. .. ..
Question is from November 2011 , type 41

Look, seeing 1 & 2, we know that the CH3CHO is order 1

There are two ways to do this. Both are the same. You can either make CH3CHO 0.25 or make CH3OH 0.10

You need a changing H+ conc but constant conc's for the other two. Lets suppose you don't have that. You can find that the orders of CH3CHO and CH3OH are 1 (that's baby stuff). With this is mind, you'll need to consider 3 experiments.

Lets take 1 3 and 4.

Comparing 3 and 4, if you had 0.25 of CH3CHO, you could directly compare the two. So what if you dont? You know the order and it's relative rate so lets make CH3CHO 0.25

Now since it's [CH3CHO] order 1,

0.25/0.2 = relative rate /3.20
Relative rate with CH3CHO 0.25 CH3OH 0.16 and H+ 0.1 = 4

comparing this new experiment to 3 gives you order 1.
 
Messages
675
Reaction score
862
Points
103
q7a(iv) / q5b(iv)
AbbbbY

Dude. Read under the figure in the mark scheme.


"(allow the 2-, 3- or 4- isomer) [6]"

The figure they've shown is for examiners, not students. This is what the CIE maintains in each of their teacher-examiner meetings. The line shown is just a representative of Cl being on 2 3 or 4.
 
Messages
192
Reaction score
107
Points
53
Look, seeing 1 & 2, we know that the CH3CHO is order 1

There are two ways to do this. Both are the same. You can either make CH3CHO 0.25 or make CH3OH 0.10

You need a changing H+ conc but constant conc's for the other two. Lets suppose you don't have that. You can find that the orders of CH3CHO and CH3OH are 1 (that's baby stuff). With this is mind, you'll need to consider 3 experiments.

Lets take 1 3 and 4.

Comparing 3 and 4, if you had 0.25 of CH3CHO, you could directly compare the two. So what if you dont? You know the order and it's relative rate so lets make CH3CHO 0.25

Now since it's [CH3CHO] order 1,

0.25/0.2 = relative rate /3.20
Relative rate with CH3CHO 0.25 CH3OH 0.16 and H+ 0.1 = 4

comparing this new experiment to 3 gives you order 1.
Sorry , but i am still not able to understand :(
 
Messages
675
Reaction score
862
Points
103
Sorry , but i am still not able to understand :(

Okay I'll do it again and this time in full:

Finding order wrt CH3CHO
Considering experiments 1 and 2, if 0.25/0.2 = 1.25/1.0 then it's order 1. So, order 1.

Like wise, wrt CH3OH,
seeing exp 2 and 3,
0.16/0.10 = 2/1.25 so order 1.

wrt to H+ is tricky because all experiments have varying concentrations. This will change the relative rate. So what if it does? Simply find the new relative rate if it were a certain concentration. You already know CH3CHO and CH3OH are order 1.

A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

If my second experiment (of the 2) had 0.25, I could compare them. If my first had 0.20, I could also compare them. So what if it doesn't. We'll make it that.
Method 1:
Both 0.25:

converting 0.20 to relative rate with 0.25:

B: 0.20 0.16 0.10 = 3.20
B': 0.25 0.16 0.10 = X

Since it's order 1, 0.25/0.20 = X/3.20
Solving for X will give you 4.

You can use these to solve for order of H+. (i.e, 0.10/0.05 = 4/2)

Method 2:
Both 0.20:


A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

I need to remove 0.05. Again, relative rate will be linked directly since it's order 1.

A: 0.25 0.16 0.05 = 2.00
A': 0.20 0.16 0.05 = Y

0.20/0.25 = Y/2.00
Solving for Y will get you 1.60

Solving for H+, 0.1/0.05 = 3.2/1.6 so again order 1.

If you don't get it, read and re read what I wrote. Then write it down. Once you start writing you'll understand what's going on. It's really simple once you get the hang of it. These questions used to give me nightmares some time ago. Practice is the key!

Oh, also, another tip, the A Level syllabus is limited to Order 2. So, if the relative rate isn't changing, it's order 0, if it's changing but the ratio I showed you above is the same for both, it's order 1 and anything else (i.e different ratio) is order 2. That's it.


Also, it's looking oh so awfully long for a few marks right now. Once you get the hang of it, you'll be doing it all on your calculator only just like I did when I first solved it.
 
Last edited:
Messages
192
Reaction score
107
Points
53
Okay I'll do it again and this time in full:

Finding order wrt CH3CHO
Considering experiments 1 and 2, if 0.25/0.2 = 1.25/1.0 then it's order 1. So, order 1.

Like wise, wrt CH3OH,
seeing exp 2 and 3,
0.16/0.10 = 2/1.25 so order 1.

wrt to H+ is tricky because all experiments have varying concentrations. This will change the relative rate. So what if it does? Simply find the new relative rate if it were a certain concentration. You already know CH3CHO and CH3OH are order 1.

A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

If my second experiment (of the 2) had 0.25, I could compare them. If my first had 0.20, I could also compare them. So what if it doesn't. We'll make it that.
Method 1:
Both 0.25:

converting 0.20 to relative rate with 0.25:

B: 0.20 0.16 0.10 = 3.20
B': 0.25 0.16 0.10 = X

Since it's order 1, 0.25/0.20 = X/3.20
Solving for X will give you 4.

You can use these to solve for order of H+. (i.e, 0.10/0.05 = 4/2)

Method 2:
Both 0.20:


A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

I need to remove 0.05. Again, relative rate will be linked directly since it's order 1.

A: 0.25 0.16 0.05 = 2.00
A': 0.20 0.16 0.05 = Y

0.20/0.25 = Y/2.00
Solving for Y will get you 1.60

Solving for H+, 0.1/0.05 = 3.2/1.6 so again order 1.

If you don't get it, read and re read what I wrote. Then write it down. Once you start writing you'll understand what's going on. It's really simpel once you get the hang of it. These questions used to give me nightmares some time ago. Practice is the key!

Oh, also, another tip, the A Level syllabus is limited to Order 2. So, if the relative rate isn't changing, it's order 0, if it's changing but the ratio I showed you above is the same for both, it's order 1 and anything else (i.e different ratio) is order 2. That's it.


Also, it's looking oh so awfully long for a few marks right now. Once you get the hang of it, you'll be doing it all on your calculator only just like I did when I first solved it.
Thanks man for your effort :);)
Got it :)
 
Messages
65
Reaction score
90
Points
28
Messages
65
Reaction score
90
Points
28
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s13_qp_51.pdf

Q1 c (iv) and (vi)? please help :(
thanks.
 
Top