Chemistry: Post your doubts here!

[Ebrahim12]

may june 2016 42

Can someone explain why in the ms they take the concentration of ch3nh2 as y instead of y - 3.966 x10^-4 ?

 

[Kanekii]

Need help with (ii)

 

[Ebrahim12]

View attachment 64621 Need help with (ii)

Atomic mass is the weighted average of isotopic masses:
10.8 = 19.78%(10.0129) + 80.22%(m)
m = 10.9941 (6sf)

 

[Kanekii]

Atomic mass is the weighted average of isotopic masses:
10.8 = 19.78%(10.0129) + 80.22%(m)
m = 10.9941 (6sf)

How did you ended up with 10.8?

 

[MShaheerUddin]

Please explain ii

 

[Ebrahim12]

How did you ended up with 10.8?

Atomic weight from periodic table.

 

[Ebrahim12]

Please explain iiView attachment 64625

Few things to note:
All structural isomers have the same molecular weight of 266.5
Only the chloride ions not bonded to to Cr react (the ones outside the brackets)
n(AgCl) = n(chloride ions)

So, the number of moles of chloride ions outside the brackets is the same as the number of moles of AgCl

For A:
1g of isomer reacts with 3.75x10^-3 mol of AgCl
1g of isomer has 3.75x10^-3 mol of Cl^-1
(1/266.5) mol of isomer has 3.75x10^-3 mol of Cl^-1
3.75x10^-3 mol of isomer has 3.75x10^-3 mol of Cl^-1

This gives us a mole ratio of 1:1 between the isomer and the chloride ions outside the brackets
Therefore, 3-n = 1
n = 2

A is [Cr(H2O)4Cl2]Cl.2H2O

Same reasoning gives n value and molecular formula for B and C.

 

[Khitan]

Having problem/confusion in deducing molecule thorugh NMR pattern.
Taking alot alot of time and it is of only 1-2 or maximum 3 marks.
what should i do not getting better in it?How you people doing?

 

[Kanekii]

Having problem/confusion in deducing molecule thorugh NMR pattern.
Taking alot alot of time and it is of only 1-2 or maximum 3 marks.
what should i do not getting better in it?How you people doing?

There are some qs where its of 6 marks the best thing you could for it is Practice, Practice actually helps alot.

 

[fourtris46]

does anyone have the 2019 march paper4?

 

[student3737]

For step 4: why is the reagent aqueous Hcl and heat? We are turning amide to amine so shouldn’t it be reduction with LiAlH4?

 

[student3737]

Also for this one , why is N zero order, why not first? Thanks a lot in advance

 

[Ebrahim12]

Also for this one , why is N zero order, why not first? Thanks a lot in advance

The rate equation depends on the slow step in the mechanism. Here step one is the slow and it needs no Y atoms and two X atoms.

 

[Ebrahim12]

For step 4: why is the reagent aqueous Hcl and heat? We are turning amide to amine so shouldn’t it be reduction with LiAlH4?

There is three ways to turn an amide to an amine

1) Acid hydrolysis, using HCL(aq) and heat
This breaks the amide linkage completely and you end up with two molecules, the amine and the acid.
The amine formed will react with the acid to form a salt.

2) Alkaline hydrolysis, using NaOH(aq) and heat
Same as acid, except the amine doesn't form salt, but the acid does.

3) Reduction, using LiAlH4
Here you end up with one molecule, an amine, the carbons in the carboxylic acid group remain attached while the oxygen is removed.

 

[Ebrahim12]

does anyone have the 2019 march paper4?

https://www.facebook.com/463818014075291/posts/671605593296531/?sfnsn=mo

 

[Kanekii]

There is two ways to turn an amide to an amine
1) Hydrolysis, using HCL(aq) and heat
This breaks the amide linkage completely and you end up with two molecules, the amine and the acid
2) Reduction, using LiAlH4
Here you end up with one molecule, an amine, the carbons in the carboxylic acid group remain attached while the oxygen is removed

But the amine will react with the acid and form ammonium ion in acidic hydrolysis

 

[Kanekii]

For step 4: why is the reagent aqueous Hcl and heat? We are turning amide to amine so shouldn’t it be reduction with LiAlH4?

I think its supposed to be NaOH(aq) + Heat

 

[Ebrahim12]

But the amine will react with the acid and form ammonium ion in acidic hydrolysis

Fair enough, edited.

 

[student3737]

For this compound Q, if we add excess Bromine , why does bromine get added to the right end side? I thought it was only on the benzene ring but in the marking scheme they said bromine is added on benzene and on the right end side

 

[Ebrahim12]

For this compound Q, if we add excess Bromine , why does bromine get added to the right end side? I thought it was only on the benzene ring but in the marking scheme they said bromine is added on benzene and on the right end side

It's electrophilic addition of bromine across the double bond.