Chemistry: Post your doubts here!

[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.

 

[lolo lolo]

I really need chemistry paper 6 march 2019 paper please someonse send it

 

[lolo lolo]

I really need chemistry paper 6 march 2019 paper please someonse send it

chemistry 0620

 

[fourtris46]

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

thank you

 

[Omkar Kerkar]

https://imgur.com/a/S6hMoMx
The answer is D, can someone explain?
When they say "possible" oxidation products, does that mean that one OH (refer to the image) doesn't necessarily oxidise?
Thanks in advance

 

[Ebrahim12]

https://imgur.com/a/S6hMoMx
The answer is D, can someone explain?
When they say "possible" oxidation products, does that mean that one OH (refer to the image) doesn't necessarily oxidise?
Thanks in advance

Possible products refers to the different products that can be formed based on how oxidised each group gets. These are:
Alcohol Aldehyde
Alcohol Acid
Aldehyde Aldehyde
Aldehyde Acid
Acid Acid
4 Acids and 4 Aldehydes groups

 

[Omkar Kerkar]

Possible products refers to the different products that can be formed based on how oxidised each group gets. These are:
Alcohol Aldehyde
Alcohol Acid
Aldehyde Aldehyde
Aldehyde Acid
Acid Acid
4 Acids and 4 Aldehydes groups

Tysm!

 

[Ebrahim12]

s17/41
Can someone explain why the OH and CH groups are in their positions? couldn't their shift values be exchanged according to the booklet values?

Also I would've guessed the OH group has the higher shift value but the ms is showing the opposite

 

[Kanekii]

View attachment 64653

View attachment 64654

s17/41
Can someone explain why the OH and CH groups are in their positions? couldn't their shift values be exchanged according to the booklet values?

Also I would've guessed the OH group has the higher shift value but the ms is showing the opposite

CH is more electronegative than OH as CH is attached to benzene ring and secondary alcohol

 

[Ebrahim12]

CH is more electronegative than OH as CH is attached to benzene ring and secondary alcohol

But the proton in OH is directly BONDED to the oxygen, not just next to it. How are we supposed to know that its less shielded than the proton that's just next to it and a benzene ring?

 

[student3737]

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

So if there was no double bond nothing would be added there right?

 

[student3737]

Guys for the first image, in bi) how do you know what reacts with cold Hcl and cold naoh? Like I thought carboxylic acid reacts With cold Hcl but apparently only amine reacts with it, and for biii) I understand the functional groups for solubility in Hcl but how do you know which functional group is present based on solubility in NAOH?

In the second image, how do you find the mass of pesticide?

Thanks a lot in advance

 

[student3737]

But the proton in OH is directly BONDED to the oxygen, not just next to it. How are we supposed to know that its less shielded than the proton that's just next to it and a benzene ring?

The proton bonded to a benzene ring will always be most electronegative out of all the other atoms , even the ones which are bonded to OH are not as electronegative as the one bonded to a benzene ring

 

[Ebrahim12]

So if there was no double bond nothing would be added there right?

Yes

 

[Ebrahim12]

The proton bonded to a benzene ring will always be most electronegative out of all the other atoms , even the ones which are bonded to OH are not as electronegative as the one bonded to a benzene ring

Damn, really? do you know how other groups compare to each other in terms of shift value/electronegativity?

 

[Ebrahim12]

Guys for the first image, in bi) how do you know what reacts with cold Hcl and cold naoh? Like I thought carboxylic acid reacts With cold Hcl but apparently only amine reacts with it, and for biii) I understand the functional groups for solubility in Hcl but how do you know which functional group is present based on solubility in NAOH?

In the second image, how do you find the mass of pesticide?

Thanks a lot in advance

First question
Carboxylic acids don't react with HCL, cold or not. Are you thinking of amides or esters?
Did you answer b(ii)? the groups that react with cold NaOH will make the molecule soluble, from the list that's phenols and carboxylic acids.
(Both amide and ester alkaline hydrolysis need heat)

Sol in HCL: amine
Sol in NaOH: phenol, carboxylic acid

F is soluble in HCL, muct have amine
It's soluble in NaOH and has two O atoms left for the second group, must have carboxylic acid

H is insoluble in HCL, must have amide to count for the N atom.
It's soluble in NaOH and has one more O atom (one used in the amide group), must have phenol

Second question
6 = [in hex]/[in water]
Let m be the mass dissolved in hex, (0.0042-m) left in water
6 = [m/25]/[(0.0042-m)/25]

solve for m, m = 0.0036g

 

[mittul8yu98]

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.

amide with acid form ammonium ion not amine