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Original post by severus.snape
what exactly do you mean by synoptic by the way? :P I am new to A2 :/
:colondollar:


Like they'll do a bit of a back ground check on you :tongue:

Anything from previous units can come up... from what I've seen mostly unit 4 organic :smile:
Original post by posthumus
Like they'll do a bit of a back ground check on you :tongue:

Anything from previous units can come up... from what I've seen mostly unit 4 organic :smile:


Which is thankfully the nice part of Unit 4. Can't beat a nice acyl chloride, those reactive little monkeys!
I have attached qsn and answer, How is the answer that? Why does product B show different structure, as if the CL is joined onto CH3? and Why cant chlorine join onto all 5 positions including the CH3 group?
Original post by Gnome :)
So if you have a burette that measures to 0.1cm3, you uncertainty is +/- 0.05, so 0.05*2=0.1 :smile:


Thank you :smile: do you know what the percentage error for a pipette or a thermometer would be ? How you would calculate it ?
Reply 1124
any body have link to A2 Bio and Phy threads ?
Original post by Knoyle quiah
I have attached qsn and answer, How is the answer that? Why does product B show different structure, as if the CL is joined onto CH3? and Why cant chlorine join onto all 5 positions including the CH3 group?


Sorry, What is the question ?
Reply 1126
Original post by Knoyle quiah
I have attached qsn and answer, How is the answer that? Why does product B show different structure, as if the CL is joined onto CH3? and Why cant chlorine join onto all 5 positions including the CH3 group?


Couldn't find ur Question (or its nt clear), try this fundementals

1. To Add Cl to benzene u need AlCl3 with Cl2, (electrophilic substitution)
2. Adding Cl to CH3 is a free radical substitution (from AS), for tht u need UV light, so this -CH3 can be anywhere in benzene (from AS - CH3 no need to be attached in it too if CH4+Clwith UV can give CH3Cl)
(edited 10 years ago)
Reply 1127
Original post by severus.snape
what exactly do you mean by synoptic by the way? :P I am new to A2 :/
:colondollar:


Synoptic means questions can come out from your AS level together with your A2. :smile: like for organic chemistry they can ask u from alkane right up to aldehyde. :wink:


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Reply 1128
Could someone please explain this question from Unit 4? :smile: I know the answer cannot be C or D...
Sorry, the Unit 4 thread is not so active atm

Q) The mass spectrum of CH3CH2Br will have:

A a peak at m/e = 109
B peaks at m/e = 108 and 110
C peaks at m/e = 93, 94 and 95
D peaks at m/e = 79, 80 and 81
Original post by SKK94
Could someone please explain this question from Unit 4? :smile: I know the answer cannot be C or D...
Sorry, the Unit 4 thread is not so active atm

Q) The mass spectrum of CH3CH2Br will have:

A a peak at m/e = 109
B peaks at m/e = 108 and 110
C peaks at m/e = 93, 94 and 95
D peaks at m/e = 79, 80 and 81


My gut feeling is A. Could be wrong though.....
Original post by SKK94
Could someone please explain this question from Unit 4? :smile: I know the answer cannot be C or D...
Sorry, the Unit 4 thread is not so active atm

Q) The mass spectrum of CH3CH2Br will have:

A a peak at m/e = 109
B peaks at m/e = 108 and 110
C peaks at m/e = 93, 94 and 95
D peaks at m/e = 79, 80 and 81

I think the answer is B (for the parent ion peak at least) because the RAM of bromine is 80, this however is an average resulting from the mass of the two most common stable isotopes of Bromine; one being Br-79, and the other being Br-81.
Br-80 is much less common and only has a half life of about 18 minutes so it won't appear in the mass spectrum.
Therefore there will be one peak for the bromoethane with Br-79 in it (29 + 79 = 108) and one peak for the bromoethane with Br-81 in it (29 + 81 = 110).
(edited 10 years ago)
Hi, could anybody explain why the answer here is D and not the others?

The acid used in a standard Hydrogen electrode to provide a 1 mol dm^-3 solution of Hydrogen ions is
A - Ethanoic acid
B - Phosphoric(V) acid
C - Sulfuric acid
D - Hydrochloric acid
Original post by GeorgeL3
Hi, could anybody explain why the answer here is D and not the others?

The acid used in a standard Hydrogen electrode to provide a 1 mol dm^-3 solution of Hydrogen ions is
A - Ethanoic acid
B - Phosphoric(V) acid
C - Sulfuric acid
D - Hydrochloric acid

I can't find my notes on this but it's because sulfuric acid gives 2mol of hydrogen ions so you would need to use 0.5moldm-3 of sulfuric acid in order to give 1moldm-3 of hydrogen ions. The electrode needs to be under standard conditions with all concentrations at 1moldm-3. Phosphoric acid gives 3mol so you'd need 0.33moldm-3. Then ethanoic acid is too weak.

I doubt this was helpful but I can't find my notes on this lol.
Original post by LeaX
I can't find my notes on this but it's because sulfuric acid gives 2mol of hydrogen ions so you would need to use 0.5moldm-3 of sulfuric acid in order to give 1moldm-3 of hydrogen ions. The electrode needs to be under standard conditions with all concentrations at 1moldm-3. Phosphoric acid gives 3mol so you'd need 0.33moldm-3. Then ethanoic acid is too weak.

I doubt this was helpful but I can't find my notes on this lol.

Thanks, yeah I see what you mean, the reason it confused me though is that surely using the correct concentrations and volumes, any of them could give a 1 moldm^-3 solution of H+ ions.
I guess it must just be that for standard conditions it all has to be 1 moldm^-3.
Original post by GeorgeL3
I think the answer is B (for the parent ion peak at least) because the RAM of bromine is 80, this however is an average resulting from the mass of the two most common stable isotopes of Bromine; one being Br-79, and the other being Br-81.
Br-80 is much less common and only has a half life of about 18 minutes so it won't appear in the mass spectrum.
Therefore there will be one peak for the bromoethane with Br-79 in it (29 + 79 = 108) and one peak for the bromoethane with Br-81 in it (29 + 81 = 110).


Wow we need to know all this ??? :redface:
Original post by posthumus
Wow we need to know all this ??? :redface:

Yeah you need to know from unit 1 or 2 that RAM is just the average mass of the isotopes rather than the actual mass of the atom, that's why they're all decimals. You also need to be able to calculate RAM when given values.
Then you need to know that Bromine and Chlorine in particular have a different pattern since Bromine's main isotopes are 79 and 81 which average at 80 and chlorines main ones are 35 and 37 which average at 36.
If you want more information then look over here:
http://www.chemguide.co.uk/analysis/masspec/mplus2.html
Original post by GeorgeL3
Yeah you need to know from unit 1 or 2 that RAM is just the average mass of the isotopes rather than the actual mass of the atom, that's why they're all decimals. You also need to be able to calculate RAM when given values.
Then you need to know that Bromine and Chlorine in particular have a different pattern since Bromine's main isotopes are 79 and 81 which average at 80 and chlorines main ones are 35 and 37 which average at 36.
If you want more information then look over here:
http://www.chemguide.co.uk/analysis/masspec/mplus2.html


Oh okay that's not too bad I guess :smile: For the RAM of the Halogens usually anyway... it +-1 neutrons from the RAM always

Thank you !!
Reply 1137
Original post by GeorgeL3
Thanks, yeah I see what you mean, the reason it confused me though is that surely using the correct concentrations and volumes, any of them could give a 1 moldm^-3 solution of H+ ions.
I guess it must just be that for standard conditions it all has to be 1 moldm^-3.


Yup, A standard H electrode should always have 1moldm-3 H+ solution at 298K with 1atm Hydrogen gas.

Any standard solution means its should have 1M at 25C (298K) with 1atm.
This is bit of a rookie question but how do you know which group attached to the ring is the starting number (when naming whole compound), if I was going to use the Benzene prefix rather than Phenol, btw the bonds dont show in the picture , would 1,3 dimethyl,2-chloro,5-hydroxybenzene be correct?
Original post by Knoyle quiah
This is bit of a rookie question but how do you know which group attached to the ring is the starting number (when naming whole compound), if I was going to use the Benzene prefix rather than Phenol, btw the bonds dont show in the picture , would 1,3 dimethyl,2-chloro,5-hydroxybenzene be correct?


You will never get a question which asks you to chose a starting point on the 6 carbon atoms on the benzene ring.

However when naming a compound that involves a benzene ring only and groups attached, you always start with the group on top as position 1.

However in your question, they would accept all answers, doesn't matter where you start tbh. Check the ms.

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