Aqa chem4 15th june 2011 (resit) thread

hey im stuck on chapter 11 (hate it!)
the exam style question at the back, question 2)a)i) and ii)

Q: the reaction of but-1-ene with chlorine produces 1,2-dichlorobutane, C4H8CL2
i) given tht chlorine exists as a mixture of 2 isotopes, 35CL and 37CL, predict the number of molecular ion peaks and their m/z values in the mass spectrum of C4H8CL2.
ii) the mass spec of 1,2-dichlorobutane contains peaks at m/z 77 and 79, draw the structure of the fragment ion which produces the peak at m/z=77 and write an equation showing its formation from the molecular ion.

can any1 explain how to work this out (NOT JUST THE ANSWER! BUT THE EXPLANATION PLEASE!!)

ALSO! how exactly do integration ratio's work!? :s-smilie:

im getting annoyed at this chapter!

(edited 12 years ago)

Reply 42

foianorsx205

What I am massively thankful for is that they don't make the proton n.m.r content TOO hard, as they had done in the past on the previous syllabus in AQA, so fingers crossed they keep up to the trend! I am not going to be happy with synthetic routes however, they are a bitch, no matter how often I try to get it in my head, there will always be ONE reagent I will not be able to remember!

Reply 43

Original post by Arcanine

Ta-da!

Thanks

Reply 44

Nas479

Does anyone have the january 2011 cheistry unit 4 exam paper and markscheme that would be a great help to me thanks. This exam is realy difficult im doing this for the second time

Reply 45

http://www.a-levelchemistry.co.uk/AQA%20Chemistry/AQA%20A2%20Chemistry/Unit%204/Unit%204%20Past%20Papers%20and%20Revision%20Aids/Unit%204%20summary%20home.htm

Original post by Nas479

Does anyone have the january 2011 cheistry unit 4 exam paper and markscheme that would be a great help to me thanks. This exam is realy difficult im doing this for the second time

Reply 46

FristyKino

Original post by Mobs25

hey im stuck on chapter 11 (hate it!)
the exam style question at the back, question 2)a)i) and ii)

Q: the reaction of but-1-ene with chlorine produces 1,2-dichlorobutane, C4H8CL2
i) given tht chlorine exists as a mixture of 2 isotopes, 35CL and 37CL, predict the number of molecular ion peaks and their m/z values in the mass spectrum of C4H8CL2.
ii) the mass spec of 1,2-dichlorobutane contains peaks at m/z 77 and 79, draw the structure of the fragment ion which produces the peak at m/z=77 and write an equation showing its formation from the molecular ion.

can any1 explain how to work this out (NOT JUST THE ANSWER! BUT THE EXPLANATION PLEASE!!)

ALSO! how exactly do integration ratio's work!? :s-smilie:

im getting annoyed at this chapter!

For part i:
2 peaks. Because there are 2 chlorine isotopes.
The two peaks will be: 126 (for Cl-35) and 128(for Cl-37)

For part ii:
Peak at 77 = CH3CH2CHCl (note its not the HCL acid) (Cl 35 is used here not 37)
Equation: CH3CH2CHClCH2Cl ---> CH3CH2CHCl(radical sign) + CH2Cl+

I hope i've got my radican sign and + sign on the right ones. Other than that I believe it is all correct.

Reply 47

dolmio

Original post by FristyKino

For part i:
2 peaks. Because there are 2 chlorine isotopes.
The two peaks will be: 126 (for Cl-35) and 128(for Cl-37)

For part ii:
Peak at 77 = CH3CH2CHCl (note its not the HCL acid) (Cl 35 is used here not 37)
Equation: CH3CH2CHClCH2Cl ---> CH3CH2CHCl(radical sign) + CH2Cl+

I hope i've got my radican sign and + sign on the right ones. Other than that I believe it is all correct.

remember for fragmentation equations that you show a + and a radical dot on the compound on the LHS (the 'reactant' I guess), examiners are quite picky about that! :smile:

Reply 48

FristyKino

Original post by dolmio

remember for fragmentation equations that you show a + and a radical dot on the compound on the LHS (the 'reactant' I guess), examiners are quite picky about that! :smile:

Ahh yes you are quite right abou that - that totally slipped my mind

Reply 49

Original post by JK471993

Need to get 76? in A2 Chemistry for an A. Definitely need an A in this unless I get an A in History (not gonna happen).

Chem 4 is okay compared to Chem 5 but I hope they don't ask us stuff from AS.

I find remembering reagents the hardest thing.

Hey this is something i'm realy confused about. Do you mean 76% of the actual paper will get an A? So i get 76% of the paper i will 100ish/120 ums points?

Reply 50

al_habib

1

guys i've got questions for you..

an ester contains a benzene ring. The mass spectrum of this ester shows a molecular ion peak at m/z=136.
Deduce the molecular formula for this ester.

How do you go about this question , its taking me ages to get my head around

Reply 51

Original post by al_habib

guys i've got questions for you..

an ester contains a benzene ring. The mass spectrum of this ester shows a molecular ion peak at m/z=136.
Deduce the molecular formula for this ester.

How do you go about this question , its taking me ages to get my head around

Ok i'm not sure if this is right but i think its close if not right. First work out molecular formula of benzene C6H6 = 78 then as its an ester you know there are 2 oxygen atoms so thats another 32. 78+32=110.
you also know that there are 2 carbons attached to the oxygens (structure of an ester -cooc-) so thats another 24.
110+24 = 134. then the carbon attached to single bonded oxygen must have 4 bonds so thats add another 3 (for the 3 hydrogens)
134+3=137
Now when i worked out the molecular formula of benzene i forgot to minus a hydrogen (as there is an ester group attached) so minus 1 and you get 136
So the molecular formula is C6H5COOCH3

(edited 12 years ago)

Reply 52

jimmy303

1

Original post by al_habib

x

Tiger_Lilly92

x

Tiger Lilly you're right, but there is another option as far as I'm aware.

By your method you've joined methanol (CH3OH) with benzoic acid (C6H5COOH) to get Methyl benzoate (C6H5COOCH3).

However, it is also possible for one of the carbons in the ester bond to also be part of the benzene ring itself.

Ethanoic Acid (CH3COOH) with phenol (C6H5OH) would give Phenyl Ethanoate (CH3COOC6H5).

Both have the relative molecular mass of 136. Both are right, I just wanted to make you aware of the fact that the carbons of an ester bond are not necessarily seperate from other functional groups should a question like this come up in the exam but with only the one option (i.e. HCOOC6H5, R.M.M. = 122).

Your method is perfect, though it may be better to remove the hydrogen as you add the benzene ring as I often forget if I leave it to the end :tongue:

Reply 53

Original post by jimmy303

Tiger Lilly you're right, but there is another option as far as I'm aware.

By your method you've joined methanol (CH3OH) with benzoic acid (C6H5COOH) to get Methyl benzoate (C6H5COOCH3).

However, it is also possible for one of the carbons in the ester bond to also be part of the benzene ring itself.

Ethanoic Acid (CH3COOH) with phenol (C6H5OH) would give Phenyl Ethanoate (CH3COOC6H5).

Both have the relative molecular mass of 136. Both are right, I just wanted to make you aware of the fact that the carbons of an ester bond are not necessarily seperate from other functional groups should a question like this come up in the exam but with only the one option (i.e. HCOOC6H5, R.M.M. = 122).

Your method is perfect, though it may be better to remove the hydrogen as you add the benzene ring as I often forget if I leave it to the end :tongue:

Thank you i never thought of that and yeah i did forget, i got 137 and then realised :P

Reply 54

EavieQu

Heya

Can anyone explain Question 5 a (ii) on this paper??

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-CHM4-W-MS-JUN07.PDF

It's an old spec paper, and I don't know if this stuff still comes up, but I'm just confused about which part of the specification this question comes under?!

Thankyou!!

Reply 55

Mobs25

Original post by FristyKino

For part i:
2 peaks. Because there are 2 chlorine isotopes.
The two peaks will be: 126 (for Cl-35) and 128(for Cl-37)

For part ii:
Peak at 77 = CH3CH2CHCl (note its not the HCL acid) (Cl 35 is used here not 37)
Equation: CH3CH2CHClCH2Cl ---> CH3CH2CHCl(radical sign) + CH2Cl+

I hope i've got my radican sign and + sign on the right ones. Other than that I believe it is all correct.

the answer for 2ai) is 3 peaks with 126, 128 and 130. I got the two peaks but mine had m/z values, 126 and 130. the third peak i assume is from the Cl35.5 (in periodic table). thus this one will have a m/z value of 127. But the answer states otherwise. :s-smilie:

btw for the second part, your radical sign and plus sign should be the other way round thus,
CH3CH2CHClCH2Cl ---> CH3CH2CHCl+ + CH2C*

(edited 12 years ago)

Reply 56

jimmy303

1

Original post by EavieQu

Heya

Can anyone explain Question 5 a (ii) on this paper??

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-CHM4-W-MS-JUN07.PDF

It's an old spec paper, and I don't know if this stuff still comes up, but I'm just confused about which part of the specification this question comes under?!

Thankyou!!

In Esterification, you learn that with conc. H2SO4 alcohols and carboxylic acids bond to form an ester.

On the specification: understand that amino acids have both acidic and basic
properties, including the formation of zwitterions. Thus when the amino acid is mixed with NaCO3 (a base) the carboxylic acid group reacts with it to produce CO2 and water (as is learnt when you study carboxylic acids - they act as weak acids, liberating CO2 from carbonates).

For the last, it is a synoptic question based on unit 2 but it is touched upon in unit 4 if I remember with successive substitution. Chloromethane undergoes nucleophilic substitution by the amine group on the amino acid, but successive substitution doesn't occur as neither is in excess (a 1:1 ratio).

Overall the question's pulled bits from everywhere in the spec. Amines, amino acids, carboxylic acids and esterification.

Hope that clears it up!

Mobs25

the answer for 2ai) is 3 peaks with 126, 128 and 130. I got the two peaks but mine had m/z values, 126 and 130. the third peak i assume is from the Cl35.5 (in periodic table). thus this one will have a m/z value of 127. But the answer states otherwise.

Dichlorobutane contains two chlorine atoms.

4 carbons, 8 hydrogens give (4x12) + 8 = 56.

The chlorine atoms are a mix of the two isotopes.

With two Cl35 isotopes: 56 + 35 + 35 = 126.
With two Cl37 isotopes: 56 + 37 + 37 = 130
With one each: 56 + 35 + 37 = 128

:smile:

Reply 57

Mobs25