JEE Chemistry Club

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Anonymous

02:25

@Jasmine The simple solution to this is more the +I groups attached to a carbocation, more the stability. Hence, (D) is correct as well.

Gaurang Tandon

@Abcd in just five minutes, so idk if that counts

Anonymous

Anonymous

I'm just learning the basics of using curly arrows, but I'm confused if the product will be cis or trans, how will I know? I know trans is more stable, but still, I don't know many things.

Gaurang Tandon

@Jasmine possibly when a grignard reagent has a long chain (six carbons), and there's an acyl group at the end, then an intramolecular attack can occur

@IceInkberry which named rxn is it?

Anonymous

It is not named, it is in the exercises. The reactants and products are given and you have to properly use curly arrows to show the mechanism.

Anonymous

02:30

$\ce PhCHBrCHBrCO2H$ + $\ce NaHCO3$ --> $\ce PhCH=CHBr$ + $\ce NaHCO3$

Anonymous

So, that is the second step in this

Gaurang Tandon

hmm, I've no clue :/

@AvnishKabaj any idea above ---^

Anonymous

Gaurang Tandon

nope, still no clue

looking at it seems like cis should be the product instead :-O

Anonymous

I thought the same, but I don't trust myself :P

Gaurang Tandon

02:36

you can post this question on main though, seems good enough

Anonymous

Also, @AvnishKabaj and @Abcd, thanks for recommending Clayden, it is absolutely fantastic(though I'm still on the basics).

Anonymous

@GaurangTandon Okay, I think there might be a simple thing which I might not have learnt or we might be missing out. If I get no answer here, I'll post it.

Gaurang Tandon

Avnish Kabaj

03:10

@IceInkberry anti periplanar

Look for it

In clayden there's a diagram and everything

Anonymous

@AvnishKabaj What page?

Avnish Kabaj

04:40

@IceInkberry ¯\_(ツ)_/¯

Anonymous

@AvnishKabaj Ah, okay, fine. Nvm.

Avnish Kabaj

@IceInkberry there's an index

395

2 hours later…

Abcd

06:25

@J_B892 I was talking about information from March.If you have the pdf, go to page 158

or do you want me to send pictures?

Anonymous

06:48

@AvnishKabaj Ah yeah, got it. Thanks! :D

Anonymous

Forgot that there was an index

Koolman

07:35

@Jasmine Does option A is correct or not ?

J_B892

08:08

@Abcd Possible for you to send the link for the pdf?

2 hours later…

Abcd

09:52

@J_B892 rushim.ru/books/mechanizms/march6ed.pdf

@GaurangTandon In 3 out of 4 options OH- is above NH2- ...looks like MSC is asserting that he is sure of the order.

2 hours later…

Abcd

12:14

@GaurangTandon I am listening

Gaurang Tandon

my msg didn't get sent

Abcd

@GaurangTandon what was it?

Gaurang Tandon

i was saying that there's also a possibility that he has made a typo in one question, and then copied it throughout the other questions

Abcd

@GaurangTandon Huh?

Gaurang Tandon

Abcd

12:18

@GaurangTandon What do you mean

Gaurang Tandon

i mean that those three options that say that OH- is better than NH2- may all be wrong

Abcd

@GaurangTandon You are well aware with the way he asks question., He expects you to exclude one and then choose others one by one. That's how he sets his questions.

So he expects you to first identify OH->NH2- and then solve the rest of the puzzle

Gaurang Tandon

i'll still say it may be wrong

Abcd

@GaurangTandon can you ask Ron, or Zhe

Gaurang Tandon

well you may go ahead in the periodic table room...

Abcd

12:22

@GaurangTandon This is that old JEE room: chat.stackexchange.com/rooms/48921/the-jee-launchpad

Gaurang Tandon

nice

Abcd

I was surprised to see Zhe actively participate in chemistry there

Gaurang Tandon

me too

1 hour later…

Abcd

13:43

@DC (aka @IceInkberry) did you check my comment on your answer

Anonymous

13:58

@Abcd Yes, I did. I forgot to mention that polarizability depends on charge as well. I mean, just look at defination of polarizability :P

Abcd

@IceInkberry NH2- and OH- have same charge

Anonymous

@Abcd Open chemical bonding from class 11. Covalent character. Polarizibility.

Abcd

@IceInkberry Huh?

Anonymous

@Abcd You are misunderstanding the term.

Abcd

@IceInkberry Huh?

Anonymous

14:01

It's not charge on the whole molecule but rather on the individual atoms. Not totally ionic, but due to electronegativity differences, charges are generated.

Anonymous

And it represents the covalent character of the molecule

Anonymous

Polarizability is directly proportional to covalent character

Abcd

... NH2 - and OH- have comparable polarisibility of charge, I insist.

Polarisability is just how spread out that charge is

@IceInkberry Do you have any source to verify that order?

Anonymous

Order in the question?

Abcd

@IceInkberry But $\text{polarisability} \ne \text{covalent character}$

@IceInkberry Yes

Anonymous

14:05

chemistry.umeche.maine.edu/CHY251/Ch7-Overhead4.html

Abcd

@IceInkberry Good find (cc: @GaurangTandon), but I don't think your reasoning for that is correct.

@IceInkberry Anymore sources?

Anonymous

Abcd

@IceInkberry I know this.

Anonymous

For polarizability thing

Abcd

14:11

@IceInkberry I know this

Anonymous

You said no

Abcd

I dont have problem with this

@IceInkberry When did i?

Anonymous

6 mins ago, by Abcd

@IceInkberry But $\text{polarisability} \ne \text{covalent character}$

Abcd

@IceInkberry Which book btw?

Anonymous

JD Lee

Abcd

14:12

$\ne$ means "is not equal to"

Anonymous

I thought you meant they are not related

Anonymous

Why did you say that then? :P

Abcd

@IceInkberry Can you convince stubborn me that your answer is correct

Anonymous

I have more sources. Let me search my history :P

Anonymous

But, I kmow for sure that nucleophilicity is directly proportional to polarizability. I have read it

Anonymous

14:16

know*

Abcd

I know that too

Anonymous

Meaningless? o.o

Anonymous

It determines how it will interact with solvent.

Abcd

Never mind.

Anonymous

14:35

Heh, maybe I might be wrong, but you couldn't prove it.

Gaurang Tandon

I read that, but I really didn't understand why we need to bring in polarizability in this case. We've had other cases of comparison, never in my life I needed to use polarizability :/

Abcd

14:52

@GaurangTandon In the case of S- and O- how will you compare the nucleophilicity?

The reason $\ce{S-}$ is a better nucleophile is: it is more polarisable

@GaurangTandon Never in your life? I am sure your reso teacher told you this :) ^

Avnish Kabaj

15:15

S- is a better nucleophile because it's homo is high in energy

Polarizability / surface charge density are just substitute concepts

Gaurang Tandon

15:31

@Abcd I was told that's because S- is bigger in size facepalm

@AvnishKabaj cries

Abcd

@GaurangTandon No I have seen this polarisability explanation in March as well. There's nothing wrong with it.

@GaurangTandon thats wrong though

@GaurangTandon I think it's more related to HSAB

HSAB theory

HSAB concept is an initialism for "hard and soft (Lewis) acids and bases". Also known as the Pearson acid base concept, HSAB is widely used in chemistry for explaining stability of compounds, reaction mechanisms and pathways. It assigns the terms 'hard' or 'soft', and 'acid' or 'base' to chemical species. 'Hard' applies to species which are small, have high charge states (the charge criterion applies mainly to acids, to a lesser extent to bases), and are weakly polarizable. 'Soft' applies to species which are big, have low charge states and are strongly polarizable. The concept is a way of applying...

Anonymous

16:03

@AvnishKabaj YESSS, I had read that.

Anonymous

Just today in clayden

Anonymous

@Abcd Nucleophilicity does not necessarily depend on basicity. It depends on solvent. I think we need to make that clear, because we often get confused with it.

Abcd

@IceInkberry I never said Nucleophilicity depends on basicity

Anonymous

Pardon me

Anonymous

I read wrong things and do wrong things at times

Anonymous

16:11

@Abcd Sulphur seems like between hard and soft according to the image.

Anonymous

Ah, ignore my previous message. I haven't learnt enough to be a part of this discussion.

Abcd

room topic changed to JEE/High School Chemistry Problems: Problem Solving Strategies of Chemistry.SE For Mathjax see: chemistry.meta.stackexchange.com/questions/89/… JEE Problem Solving Rooms : Physics: chat.stackexchange.com/rooms/54160/problem-solving-strategies Maths: chat.stackexchange.com/rooms/76340/jee-maths-problems [inorganic-chemistry] [jee] [organic-chemistry] [physical-chemistry] [problem-solving] [questions]

Anonymous

@Abcd Wait a minute, I might be wrong, but is $\ce{O^-}$ more basic than $\ce{S^-}$. How?

Abcd

@IceInkberry Indeed

Anonymous

Explain how ^^"

Anonymous

16:20

Sulphur has more electron electronegativity?

Abcd

@IceInkberry HUH?

Anonymous

Mistakekekekekek

Abcd

@IceInkberry Ink in berry

Anonymous

What I mean is sulphur is less electronegative and hence will donate electrons easily than oxygen?

Anonymous

@Abcd Answer quick. I have to sleep. I can't sleep without knowing this. lol.

Anonymous

16:26

@Abcd Ink coloured frozen berry

Abcd

@IceInkberry Proton is a harder acid.

A carbon atom is a softer acid

Oxygen is a harder base

Sulphur is a softer base

@DC got it?

HSAB Theory in action

Anonymous

I didn't know it until now. Mind=Blown. Thanks.

Abcd

Wc @DC

Vikas Rv

17:11

Is it only for hydrogen that different subshells ( s,p,d etc) of same shell have same energy or does it apply to all hydrogen like atoms ?

Jasmine

17:36

@Abcd @Koolman i dont have the key but i think A should be correct.

@VikasRv degenerate orbitals for monoatomic ions or atoms have same order but rest the energy level differs. In short yes for hydrogen like atoms it should be same.

Can anyone tell me the number of bonds present in c2. Actually what is c2? How even it exists?

1 hour later…

J_B892

18:44

@Abcd Thanks!

4 hours later…

Avnish Kabaj

23:09

in trash, 7 hours ago, by Abcd

@AvnishKabaj How would you explain the greater basicity of O- than S- using the homo lumo concept?

C-H antibonding orbital is lower in energy than C-X due to similar orbital size. So you need a low energy homo for better interaction.

@GaurangTandon no need to man there's no point in knowing homo lumo at highschool level . I toh get confused while doing problems. So I end up using orbital size only : P

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