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Chemistry: Post your doubts here!

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5 (c) http://freeexampapers.com/A-Level/Chemistry/CIE/2012-Jun/9701_s12_ms_23.pdf

May/June/2012

Variant 23

I've attached the mark scheme! Please help guys, this question alone carries 4 marks :/
I'll start from begining If you will.
Only COOH reacts with carbonates so it has COOH group
SO by the information we have we can deduce structure of F
it has COOH, rest of it is C2H5O. if its has OH it will be CH3CH(OH)COOH. That fits all the information.
The type of reaction has to be oxidation because KMnO4 is an oxidising agent.
When H is treated with conc. H2SO4 it will be a dehydration reaction. So it will become CH2=CHCOOH. when oxidised with cold KMnO4 it becomes CH2(OH)CH(OH)COOH.

SO we have two groups in F that will react with Na. OH and COOH.
so one mole of F will react with 2 mol of Na to produce 2 H which is actually 2/2 H2 or 1 H2
That means that n(F)=n(H2).
n(F)=0.6/Mr and Mr=3(12)+6+3(16)=90
n(F)=0.006667 mol = n(H2)
volume = n*24 dm3 = 0.006667*24=0.16 dm3=160cm3


If anything is not clear feel free to ask without any hesitation.
 
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If a carbon has 4 different groups attached to,would it exhibit cis-trans isomerism?

For e.g. 2 bromo,2 hydroxy,1 methyl ethene ? Would this exhibit cis trans ??
Metanoia Please answer this query..
 
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I'll start from begining If you will.
Only COOH reacts with carbonates so it has COOH group
SO by the information we have we can deduce structure of F
it has COOH, rest of it is C2H5O. if its has OH it will be CH3CH(OH)COOH. That fits all the information.
The type of reaction has to be oxidation because KMnO4 is an oxidising agent.
When H is treated with conc. H2SO4 it will be a dehydration reaction. So it will become CH2=CHCOOH. when oxidised with cold KMnO4 it becomes CH2(OH)CH(OH)COOH.

SO we have two groups in F that will react with Na. OH and COOH.
so one mole of F will react with 2 mol of Na to produce 2 H which is actually 2/2 H2 or 1 H2
That means that n(F)=n(H2).
n(F)=0.6/Mr and Mr=3(12)+6+3(16)=90
n(F)=0.006667 mol = n(H2)
volume = n*24 dm3 = 0.006667*24=0.16 dm3=160cm3


If anything is not clear feel free to ask without any hesitation.

THANKS A MILLION DOLLARS FOR THIS!

It's just that I wanna clarify something in this statement:
"one mole of F will react with 2 mol of Na to produce 2 H which is actually 2/2 H2 or 1 H2"

so one mole of F will react with 2 mol of Na because F has two functional groups? OH and COOH? Also what do you mean by 2/2 H2? :s I get the 1H2 part because 2H is equal to H2.

Again, I appreciate the time you guys take out to help us, confused students. :)
 
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THANKS A MILLION DOLLARS FOR THIS!

It's just that I wanna clarify something in this statement:
"one mole of F will react with 2 mol of Na to produce 2 H which is actually 2/2 H2 or 1 H2"

so one mole of F will react with 2 mol of Na because F has two functional groups? OH and COOH? Also what do you mean by 2/2 H2? :s I get the 1H2 part because 2H is equal to H2.

Again, I appreciate the time you guys take out to help us, confused students. :)
Its both that will react,hence 2 H Producing groups in one molecule ,

Since one will give 2H it can be simplified as just an H2 molecule.
 
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THANKS A MILLION DOLLARS FOR THIS!

It's just that I wanna clarify something in this statement:
"one mole of F will react with 2 mol of Na to produce 2 H which is actually 2/2 H2 or 1 H2"

so one mole of F will react with 2 mol of Na because F has two functional groups? OH and COOH? Also what do you mean by 2/2 H2? :s I get the 1H2 part because 2H is equal to H2.

Again, I appreciate the time you guys take out to help us, confused students. :)
2/2 is 1. I thought it might clearify the process but ir caused confusion. Sorry. Skip that bit.
 
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Which part in the syllabus does it have the bit where AgCl or AgBr react with NH3 to give Ag(NH3)2Br?? I have found it twice in the past papers (old though like 2004 and 2006)

In excess NH3, a complex cation is formed [Ag(NH3)2]+
 
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(e) The food additive E330 is another organic compound which occurs naturally in fruit. E330 has the following composition by mass: C, 37.5%; H, 4.17%; O, 58.3%. Calculate the empirical formula of E330.

I know how to solve it till here:

C H O

37.5/12 4.17/1 58.3/16

3.125/3.125 4.17/3.125 3.64/3.125

1 1.33 1.16

What am I supposed to do after this??
 
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(e) The food additive E330 is another organic compound which occurs naturally in fruit. E330 has the following composition by mass: C, 37.5%; H, 4.17%; O, 58.3%. Calculate the empirical formula of E330.

I know how to solve it till here:

C H O

37.5/12 4.17/1 58.3/16

3.125/3.125 4.17/3.125 3.64/3.125

1 1.33 1.16

What am I supposed to do after this??
Multiply 1.33 and 1.16 with numbers and see which is closest to whole number. (Trial and error method)
Hence trying when u reach to number 6 you'll see you get C6H8O7
 
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Here you go:
Those lone pairs tho
How are compounds A and B formed? :s PLEASE HELP!!
You always seem to be panicking in your questions :p
A is with an alcohol so the malic acids Carboxy groups will react with them to form a diester
B is with a carboxylic acid so it'll react with the single OH present to form an ester.


Just keep these 2 things in mind and draw the structural formula like your simply making an ester from an acid and an alcohol.

Some of these questions arent that much of a problem,so i think you should at least revise the syllabus of organic a few times,and then attempt the past papers,no use just doing them without a clear concept.
 
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Those lone pairs tho

You always seem to be panicking in your questions :p
A is with an alcohol so the malic acids Carboxy groups will react with them to form a diester
B is with a carboxylic acid so it'll react with the single OH present to form an ester.


Just keep these 2 things in mind and draw the structural formula like your simply making an ester from an acid and an alcohol.

Some of these questions arent that much of a problem,so i think you should at least revise the syllabus of organic a few times,and then attempt the past papers,no use just doing them without a clear concept.
What do u mean by "those lone pairs tho"?
 
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