Chemistry: Post your doubts here!

[AbbbbY]

Which varient will you be doing this yr ?

12 22 34 42 52

 

[sitooon]

Meaning of Cl being inside ?

 

[NinjaInPyjamas]

q4 butene will convert into 1,2-dihydroxybutane(v). the secondary -OH will oxidise to =O and primary -OH will oxidise to -COOH
q5 do you want full question answered?

thanks!
no just (b)...also for (c) no matter what reaction with give, the CH2 and no. of C and H of (CH2CH) will stay the same right? as in no. of C and H on the right hand sight of the formula like HO(CH2CH)O and H2N(CH2C)OO(H) or is there no rule like that?

 

[sitooon]

Ans should be C(CH2)4 (OH)2
but its C(CH2OH)4

 

[AbbbbY]

so do


A!

This is a pretty simple question which idk why a lot of students have trouble answering.

Tackle it like this.

If the compound was an alkane and had no rings or double bonds, number of H atoms present = 2n+2 = 2(20)+2 = 42

Now, every time there're 2H less, you'll either have a double bond or a ring. The question states that there's one aldehyde (C=O) and 1 ring structure which has 1 C=C so this accounts for 6 missing Hydrogens, thereby leaving you with 8 missing Hydrogens.

8/2 = 4 Hydrogens in the aliphatic side chain.

However, total C=C are 5 because you have to include the C=C from the cyclohexene + cis formation hence A

so do we always take the compound to be an alkane and then calculate from there?

and can you please tell me how there are 5 C=C bonds? ^_^ I can count only 4[/quote]


Umm I'm not sure but that's the easiest way, if not the only covered by A Level syllabus.

__

How can you count just 4? I just made you count 5!
4 in the aliphatic side chain, and since it is asking for the total C=C's, 1 from the cyclohexene = 5.

 

[AbbbbY]

View attachment 39292
Meaning of Cl being inside ?

Never seen anything like this before. Where did you find this?

 

[sitooon]

Never seen anything like this before. Where did you find this?

Mj2013

 

[Suchal Riaz]

View attachment 39292
Meaning of Cl being inside ?

i am in as level but my understanding is that Cl is maybe pi bonded to delocalised pi electrons. i mean Cl's pi electrons are also delocalised. but it is very strange. i am not sure what is happening.

 

[Suchal Riaz]

thanks!
no just (b)...also for (c) no matter what reaction with give, the CH2 and no. of C and H of (CH2CH) will stay the same right? as in no. of C and H on the right hand sight of the formula like HO(CH2CH)O and H2N(CH2C)OO(H) or is there no rule like that?

I hope it answers your question. You have to keep in mind that aminoacid was not the only product but the only organic product and examiner has shown only organic product coz we are interested in it. The rest of them will be impurities and would be distilled off.

 

[M Haseeb Javed]

When is which Reducing and Oxidation agent used?
I have noticed in Organic questions especially that we have to use exact Agent for reactions.

 

[Suchal Riaz]

When is which Reducing and Oxidation agent used?
I have noticed in Organic questions especially that we have to use exact Agent for reactions.

CIE's syllabus is wrong in many ways. if you will do BS in chemistry you will learn this. MnO4 and NaBh4 are not used in organic reaction are they are very reactive and can oxidise them to oxides and reduce the compound into their elements. but you will have to learn this:
potassium dichromate is not a strong oxidizing agent. we usually use it to oxidize C-OH to C=O.
it can't oxidise every group.
potassium permanganate is although a very strong oxidising agent. if concentrated it can oxidise all of the organic compound into CO2 and H2. it is used to oxidise C=C into diol if very dilute and cold. if concentrated and hot it can even rupture the double bond or even oxidise CH2= into CO2 and H2
and for reducing agent use any of them. it doesn't matter. although NaBH4 is stronger i guess.

 

[yousef]

Whats order with respect to { H+}
Marking scheme Answer is First order .. .. ..
Question is from November 2011 , type 41

 

[AbbbbY]

Mj2013

Could you link me to the paper or send me pics of the entire question?

If a new concept is tested, it's done by exemplifying a reaction and the question follows the example.
I've looked in all 3 variants twice but couldn't find the question.

 

[sitooon]

q7a(iv) / q5b(iv)
AbbbbY

 

[AbbbbY]

View attachment 39319

Whats order with respect to { H+}
Marking scheme Answer is First order .. .. ..
Question is from November 2011 , type 41

Look, seeing 1 & 2, we know that the CH3CHO is order 1

There are two ways to do this. Both are the same. You can either make CH3CHO 0.25 or make CH3OH 0.10

You need a changing H+ conc but constant conc's for the other two. Lets suppose you don't have that. You can find that the orders of CH3CHO and CH3OH are 1 (that's baby stuff). With this is mind, you'll need to consider 3 experiments.

Lets take 1 3 and 4.

Comparing 3 and 4, if you had 0.25 of CH3CHO, you could directly compare the two. So what if you dont? You know the order and it's relative rate so lets make CH3CHO 0.25

Now since it's [CH3CHO] order 1,

0.25/0.2 = relative rate /3.20
Relative rate with CH3CHO 0.25 CH3OH 0.16 and H+ 0.1 = 4

comparing this new experiment to 3 gives you order 1.

 

[AbbbbY]

q7a(iv) / q5b(iv)
AbbbbY

Dude. Read under the figure in the mark scheme.


"(allow the 2-, 3- or 4- isomer) [6]"

The figure they've shown is for examiners, not students. This is what the CIE maintains in each of their teacher-examiner meetings. The line shown is just a representative of Cl being on 2 3 or 4.

 

[yousef]

Look, seeing 1 & 2, we know that the CH3CHO is order 1

There are two ways to do this. Both are the same. You can either make CH3CHO 0.25 or make CH3OH 0.10

You need a changing H+ conc but constant conc's for the other two. Lets suppose you don't have that. You can find that the orders of CH3CHO and CH3OH are 1 (that's baby stuff). With this is mind, you'll need to consider 3 experiments.

Lets take 1 3 and 4.

Comparing 3 and 4, if you had 0.25 of CH3CHO, you could directly compare the two. So what if you dont? You know the order and it's relative rate so lets make CH3CHO 0.25

Now since it's [CH3CHO] order 1,

0.25/0.2 = relative rate /3.20
Relative rate with CH3CHO 0.25 CH3OH 0.16 and H+ 0.1 = 4

comparing this new experiment to 3 gives you order 1.

Sorry , but i am still not able to understand

 

[AbbbbY]

Sorry , but i am still not able to understand

Okay I'll do it again and this time in full:

Finding order wrt CH3CHO
Considering experiments 1 and 2, if 0.25/0.2 = 1.25/1.0 then it's order 1. So, order 1.

Like wise, wrt CH3OH,
seeing exp 2 and 3,
0.16/0.10 = 2/1.25 so order 1.

wrt to H+ is tricky because all experiments have varying concentrations. This will change the relative rate. So what if it does? Simply find the new relative rate if it were a certain concentration. You already know CH3CHO and CH3OH are order 1.

A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

If my second experiment (of the 2) had 0.25, I could compare them. If my first had 0.20, I could also compare them. So what if it doesn't. We'll make it that.
Method 1:
Both 0.25:

converting 0.20 to relative rate with 0.25:

B: 0.20 0.16 0.10 = 3.20
B': 0.25 0.16 0.10 = X

Since it's order 1, 0.25/0.20 = X/3.20
Solving for X will give you 4.

You can use these to solve for order of H+. (i.e, 0.10/0.05 = 4/2)

Method 2:
Both 0.20:


A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

I need to remove 0.05. Again, relative rate will be linked directly since it's order 1.

A: 0.25 0.16 0.05 = 2.00
A': 0.20 0.16 0.05 = Y

0.20/0.25 = Y/2.00
Solving for Y will get you 1.60

Solving for H+, 0.1/0.05 = 3.2/1.6 so again order 1.

If you don't get it, read and re read what I wrote. Then write it down. Once you start writing you'll understand what's going on. It's really simple once you get the hang of it. These questions used to give me nightmares some time ago. Practice is the key!

Oh, also, another tip, the A Level syllabus is limited to Order 2. So, if the relative rate isn't changing, it's order 0, if it's changing but the ratio I showed you above is the same for both, it's order 1 and anything else (i.e different ratio) is order 2. That's it.


Also, it's looking oh so awfully long for a few marks right now. Once you get the hang of it, you'll be doing it all on your calculator only just like I did when I first solved it.

 

[Grenade]

ote="Hassan Ali Abid, post: 744106, member: 31550"]Which variant ???[/quote]
CIE

 

[yousef]

Okay I'll do it again and this time in full:

Finding order wrt CH3CHO
Considering experiments 1 and 2, if 0.25/0.2 = 1.25/1.0 then it's order 1. So, order 1.

Like wise, wrt CH3OH,
seeing exp 2 and 3,
0.16/0.10 = 2/1.25 so order 1.

wrt to H+ is tricky because all experiments have varying concentrations. This will change the relative rate. So what if it does? Simply find the new relative rate if it were a certain concentration. You already know CH3CHO and CH3OH are order 1.

A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

If my second experiment (of the 2) had 0.25, I could compare them. If my first had 0.20, I could also compare them. So what if it doesn't. We'll make it that.
Method 1:
Both 0.25:

converting 0.20 to relative rate with 0.25:

B: 0.20 0.16 0.10 = 3.20
B': 0.25 0.16 0.10 = X

Since it's order 1, 0.25/0.20 = X/3.20
Solving for X will give you 4.

You can use these to solve for order of H+. (i.e, 0.10/0.05 = 4/2)

Method 2:
Both 0.20:


A: 0.25 0.16 0.05 = 2.00
B: 0.20 0.16 0.10 = 3.20

I need to remove 0.05. Again, relative rate will be linked directly since it's order 1.

A: 0.25 0.16 0.05 = 2.00
A': 0.20 0.16 0.05 = Y

0.20/0.25 = Y/2.00
Solving for Y will get you 1.60

Solving for H+, 0.1/0.05 = 3.2/1.6 so again order 1.

If you don't get it, read and re read what I wrote. Then write it down. Once you start writing you'll understand what's going on. It's really simpel once you get the hang of it. These questions used to give me nightmares some time ago. Practice is the key!

Oh, also, another tip, the A Level syllabus is limited to Order 2. So, if the relative rate isn't changing, it's order 0, if it's changing but the ratio I showed you above is the same for both, it's order 1 and anything else (i.e different ratio) is order 2. That's it.


Also, it's looking oh so awfully long for a few marks right now. Once you get the hang of it, you'll be doing it all on your calculator only just like I did when I first solved it.

Thanks man for your effort
Got it