Chemistry: Post your doubts here!

[abruzzi]

Read the details given in the question for the compound.
It says it gives white ppt with Br2 aq means it will have a phenol.gives yellow ppt with tri idoform test means there will be COCH3 group present and it is mentioned that it is 1,4 di substituted.so phenol at 1 and COCH3 at 4.got it!

so it was that simple huh. thanks a tonne

 

[hope4thebest]

We need to look at the bonding in the -CONH2 group.

Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond.

A pi bond is made by sideways overlap between p orbitals on the carbon and the oxygen.

In an amide, the lone pair on the nitrogen atom ends up almost parallel to these p orbitals, and overlaps with them as they form the pi bond.

image: http://www.chemguide.co.uk/organicprops/amides/amidedeloc.gif

The result of this is that the nitrogen lone pair becomes delocalised - in other words it is no longer found located on the nitrogen atom, but the electrons from it are spread out over the whole of that part of the molecule.

This has two effects which prevent the lone pair accepting hydrogen ions and acting as a base:

• Because the lone pair is no longer located on a single atom as an intensely negative region of space, it isn't anything like as attractive for a nearby hydrogen ion.
• Delocalisation makes molecules more stable. For the nitrogen to reclaim its lone pair and join to a hydrogen ion, the delocalisation would have to be broken, and that will cost energy.

Compare E values of the metal getting reduced at the cathode with that of H2 (which is zero). So if it's positive, the metal gets discharged. If negative, hydrogen gets discharged.

Ag+ + e– ⇌ Ag E = +0.80 so Ag discharged
Fe2+ + 2e– ⇌ Fe E = –0.44 so H2 discharged
Mg2+ + 2e– ⇌ Mg E = –2.38 so H2 discharged

Compare E values of the anion with that of
O2 + 4H+ + 4e– ⇌ 2H2O E = +1.23

F2 + 2e– ⇌ 2F– E = +2.87 more positive so O2 discharged
S2O82– + 2e– ⇌ 2SO42– E = +2.01 more positive so O2 discharged
Br2 + 2e– ⇌ 2Br– E = +1.07 less positive so Br2 discharged

Thank You so much brother but dont we consider this equation= O2+2H2O+4e- ⇌ 4OH- (E=+0.40 V) for the anode reactions?

 

[SOTZAIDI]

I just want to know one answer ... for now ;D ..... thst in question paper they ask for structural formulae but in mark scheme they draw skeletal, do we have to draw skeletal or we can simply write structural formulae in draw structures question?? this is for Paper 4 a2 structured que

 

[hope4thebest]

What type of question is this ? Help needed people! Thank You

 

[hope4thebest]

I just want to know one answer ... for now ;D ..... thst in question paper they ask for structural formulae but in mark scheme they draw skeletal, do we have to draw skeletal or we can simply write structural formulae in draw structures question??

Its perfectly fine with structural formula but not displayed of course.

 

[itallion stallion]

View attachment 42087
What type of question is this ? Help needed people! Thank You

I also need help in this!

 

[thats_me]

We need to look at the bonding in the -CONH2 group.

Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond.

A pi bond is made by sideways overlap between p orbitals on the carbon and the oxygen.

In an amide, the lone pair on the nitrogen atom ends up almost parallel to these p orbitals, and overlaps with them as they form the pi bond.

image: http://www.chemguide.co.uk/organicprops/amides/amidedeloc.gif

The result of this is that the nitrogen lone pair becomes delocalised - in other words it is no longer found located on the nitrogen atom, but the electrons from it are spread out over the whole of that part of the molecule.

This has two effects which prevent the lone pair accepting hydrogen ions and acting as a base:

• Because the lone pair is no longer located on a single atom as an intensely negative region of space, it isn't anything like as attractive for a nearby hydrogen ion.
• Delocalisation makes molecules more stable. For the nitrogen to reclaim its lone pair and join to a hydrogen ion, the delocalisation would have to be broken, and that will cost energy.

Compare E values of the metal getting reduced at the cathode with that of H2 (which is zero). So if it's positive, the metal gets discharged. If negative, hydrogen gets discharged.

Ag+ + e– ⇌ Ag E = +0.80 so Ag discharged
Fe2+ + 2e– ⇌ Fe E = –0.44 so H2 discharged
Mg2+ + 2e– ⇌ Mg E = –2.38 so H2 discharged

Compare E values of the anion with that of
O2 + 4H+ + 4e– ⇌ 2H2O E = +1.23

F2 + 2e– ⇌ 2F– E = +2.87 more positive so O2 discharged
S2O82– + 2e– ⇌ 2SO42– E = +2.01 more positive so O2 discharged
Br2 + 2e– ⇌ 2Br– E = +1.07 less positive so Br2 discharged

For this question why didn't you use the equation with the electrode potential +0.40 for oxygen ?

 

[Farez]

I just want to know one answer ... for now ;D ..... thst in question paper they ask for structural formulae but in mark scheme they draw skeletal, do we have to draw skeletal or we can simply write structural formulae in draw structures question?? this is for Paper 4 a2 structured que

They'll accept any appropriate representation of the structure (structural, displayed, skeletal) as long as it is right.

View attachment 42087
What type of question is this ? Help needed people! Thank You

This has to do with optical isomerism. As you can see it has two chiral centress, so you'll have rotate them around, and see which matches with F. It's impossible to explain in words. Try to visualize it.

There's no point in spending hours behind this question because it's only worth 2 marks, and it may never come again.

 

[xXGTZXx]

View attachment 42087
What type of question is this ? Help needed people! Thank You

The anw is H
The anw is in the Question, they say free rotation of the C--C bond can occur.
So just rotate the C--C bond in b/w both sides until one of them mathces

Q like these require some common sense and read the Q carefully have the Anw is already there, if something like this pops stay cool and read the Q again. If u dont get skip and get back to it later

 

[Spongebobfan]

5bi
production of O2
given cathode equation from question: 2H2O(l) + 2e– -> H2(g) + 2OH–(aq), find E value in data booklet = -0.83V
now we find anode equation, we know the reaction was neutral at the start so reactant must be ph7 (H2O) so must be O2 + 4H+ + 4e– ⇌ 2H2O = +1.23V
E = (1.23 – (–0.83)) = 2.06V

5bii
cathode reaction is still the same 2H2O(l) + 2e– H2(g) + 2OH–(aq) -0.83V
anode for production of Cl2 Cl 2 + 2e– -> 2Cl – +1.36V
Eo = (1.36 – (–0.83)) = 2.19V

5ci
Cl2:O2 ratio as [Cl-] increases
we look at the the equation from production of O2 and Cl2
in production of O2, Cl- concentration is not related so there will be no change
in production of cl2, if you increase Cl- concentration equilibrium of Cl 2 + 2e– -> 2Cl – shifts back to the reactants (left) so E value decrease (less positive)

5cii
same as part i, because [Cl-] increase, equilibrium shifts to reactants , E value decrease so more Cl2 is formed
O2 does not change
so Cl2(g) : O2(g) increases

for 5b in calculating the E0, can you please explain why the E0 is +2.06V and +2.19V and not negative instead?
because the cathode is -0.83V and in each of the anode, the O2- and Cl- are being oxidised to O2 and Cl2.
so the E0= -0.83-(-1.23) ??
pleasee explain my A2 paper is tomorrow help!

 

[Jinosupreme]

GUYS NEED HELP, FOR PAPER4 chemistry

For NMR,

One year mark scheme says

"Drawing suitable diagrams showing the origin of two energy states"

How to draw?

 

[A star]

GUYS NEED HELP, FOR PAPER4 chemistry

For NMR,

One year mark scheme says

"Drawing suitable diagrams showing the origin of two energy states"

How to draw?

they just want you to draw to show one atom rotating with and one against the spin

 

[Jinosupreme]

they just want you to draw to show one atom rotating with and one against the spin

so I can just draw a circle then arrow spin to left, then another circle arrow spin to right?

 

[Farez]

so I can just draw a circle then arrow spin to left, then another circle arrow spin to right?

 

[AbbbbY]

isnt hte hydrogen suppose to be directly attached to NOF ?

Amino Acid chains. C=O And N-H. Hydrogen bonds between O and N. And guess what! The O isn't directly attached to H!

Yep, it is in the water. You need to have a hydrogen attached to O,N or F which comes into close proximity to another electronegative atom with a lone pair, as is the case of CH3F and H2O. Am i right? Anyone knows if there is H bond there? And why is it insoluble in H2O?

You're very right here. Any other day, I'd have said you were right and Hydrogen Bonds exist b/w CH3F and H2O. BUT, not today. Not while you're still an A Level student. Not till you're done with P4 tomorrow. The A Level syllabus is limited to, rather stupidly, NOF bonded DIRECTLY to H with one exception of aa's in Applications. So, as far as questions go, you're not gonna use any logic. You're not gonna use your brain. You're going to see if N O or F is directly attached to H or not.

 

[andreiaafonso14]

NEED HELP PAPER 4 CHEMISTRY

Question says:
Calculate the pH of the buffer formed when 10.0cm3 of 0.100 mol dm-3 NaOH is added to 10.0cm3 of 0.250 mol dm-3 CH3CO2H, whose pKa=4.76.

I keep getting the wrong concentrations for the salt and acid according to the mark scheme... HELP!

 

[hope4thebest]

NEED HELP PAPER 4 CHEMISTRY

Question says:
Calculate the pH of the buffer formed when 10.0cm3 of 0.100 mol dm-3 NaOH is added to 10.0cm3 of 0.250 mol dm-3 CH3CO2H, whose pKa=4.76.

I keep getting the wrong concentrations for the salt and acid according to the mark scheme... HELP!

NaOH + CH3CO2H ----------- CH3CO2Na + H2O
moles of NaOH= C*V = 10/1000 * 0.100 = 0.001 moles of acid= C*V = 10/1000 * 0.250 = 0.0025
Hence acid is excess. Both will react with 1 ratio 1 so acid remaining would be 0.0025-0.001= 0.0015 moles.
The moles of salt would be 0.001 as moles of CH3CO2Na are 0.001.
Then you will calculate concentrations of both salt and acid. Total volume is 20 cm3. Use the formula C=n/V to calculate concentrations.
Then use pH= pKa + log(salt/acid)

 

[hope4thebest]

Farez brother
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_42.pdf
Marking scheme: http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_ms_42.pdf
Question 2 part (b)
I have calculated the ratio 5.66 but cant understand the next step. (100–A) / A = 5.66
Why do we use volumes instead of concentrations in this ratio.
Thank You!

 

[andreiaafonso14]

NaOH + CH3CO2H ----------- CH3CO2Na + H2O
moles of NaOH= C*V = 10/1000 * 0.100 = 0.001 moles of acid= C*V = 10/1000 * 0.250 = 0.0025
Hence acid is excess. Both will react with 1 ratio 1 so acid remaining would be 0.0025-0.001= 0.0015 moles.
The moles of salt would be 0.001 as moles of CH3CO2Na are 0.001.
Then you will calculate concentrations of both salt and acid. Total volume is 20 cm3. Use the formula C=n/V to calculate concentrations.
Then use pH= pKa + log(salt/acid)

Thanks!

 

[SYED ALI ARIZ]

Any important topics for tommorows p42???