JEE Chemistry Club

1:02 PM

@DivyankaChaudhari Is it option (b) ?

Anonymous

@Tanuj Yes

@DivyankaChaudhari well the reason why -OH is more activating than -OR groups on benzene is kinda experimental ( the lone pair of oxygen in -OH overlaps with the delocalised ring electron system of benzene )

Abcd

11

Q: Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?

Why is phenol more reactive towards electrophilic substitution than methoxybenzene? Isn't the lone pair on the methoxy group oxygen more available for participating in resonance into the ring than the lone pair on hydroxyl oxygen due to the +I effect of the methyl group? Or is it because of the s...

cc: @DivyankaChaudhari

You'll understand the 2nd answer if you know basics of MOT @DivyankaChaudhari

Tanuj

@Abcd have you studied ideal gases ? (Gas laws etc and deviation from real gas) ?

Abcd

@Tanuj Who has not studied them

Tanuj

1:17 PM

@Abcd Okay , try to answer this question then.

Anonymous

I don't really get that

Anonymous

But, is it due to hyperconjugation as well

Anonymous

I mean the methoxy one has more alpha H

Abcd

@DivyankaChaudhari :/ See. There's an Antibonding molecular orbital of CH bond

Agreed?

Tanuj

Abcd

1:18 PM

@Tanuj let me explain DC first

Tanuj

@Abcd sure man. Have a look at it when you're free

Anonymous

Yes @Abcd

Abcd

@DivyankaChaudhari And oxygen has lone pairs, which can be donated to that AntiBonding MO

So that disturbs the resonance, making it less effective

Got it @DC?

Anonymous

Ah, yeah

Anonymous

I'm so weak at MOT

Anonymous

1:24 PM

I was wondering how it has ABMO

Anonymous

But got it

Anonymous

@Abcd Thanks, you are my saviour for MOT answers :P

Abcd

Wc

@Tanuj Option A is correct

Solving others..

C is correct

:44208867 Why is D correct?

:44208867 Its hard to comment on B actually

But since dependence of Volume on Z is quadratic

I think we can assume that P doesnt depend linearly on it

@DC why is D correct?

Anonymous

For high pressure, Z= 1 + Pb/Rt

Anonymous

RT*

Abcd

1:37 PM

@DivyankaChaudhari How?

Anonymous

P Increases, Z increases

Gaurang Tandon

@Abcd B is simple actually; b=0 is given, so Z=1-a/RTV or Z=1-kP (assuming constt temperature, and k is a constant)

Abcd

@GaurangTandon why b= 0?????

Anonymous

Gaurang Tandon

@Abcd duh. graph of Z vs P is less than 1 at all pressures, so b has to be zero

Abcd

1:41 PM

@GaurangTandon hmm

@Tanuj You can read the d discussion and get your answer

Anonymous

@GaurangTandon didn't quite get how you converted 1/RTV to P

Gaurang Tandon

@DivyankaChaudhari V is inversely propn to P (assuming constt T)

Anonymous

Oh, constant T, missed it

Anonymous

Ahh

Anonymous

But wait, can we assume that?

Anonymous

1:45 PM

@GaurangTandon ^

Gaurang Tandon

what else do we do :/

Anonymous

Thinking

Tanuj

How did you figure out option (A) ?

Gaurang Tandon

Hi @Tanuj :)

Tanuj

@GaurangTandon Hey man ! :)

Gaurang Tandon

1:54 PM

figure out A the same way as we did B; because Z>1 at all pressures implies a=0 and so Z=1+Pb/RT

Tanuj

@GaurangTandon why would Z>1 mean a=0 ?

Gaurang Tandon

because it's the only way to get Z>1. if a and b were both non zero, you'd get graph C; if only b was zero, you'd get graph B; it's factual

Anonymous

@Tanuj Put a=0 in Vander waals eqn and derive Z

Tanuj

Z=1+Pb/RT @DivyankaChaudhari

But how do you know that a=0 indeed for gas (A) ?

Anonymous

@Tanuj For any pressure, from the eqn Z>1. The graph also shows the same.

Tanuj

2:06 PM

@DivyankaChaudhari Okay , I don't think I got the question.

So all the given graphs are correct , but one of the option is not , right ?

Anonymous

Yes

Anonymous

I have asked this question to my teacher. If I get an answer, will let know.

Tanuj

@DivyankaChaudhari hmm , what question ?

Anonymous

Option B seems correct, but one statement has to be wrong

Tanuj

Also , when I put b=0 and solve for Z , I get Z=1-a/RTV

Anonymous

2:11 PM

Should probably get to work than wasting time on one question.

Anonymous

@Tanuj Yes

Abcd

→ 1 message moved to trash

Tanuj

@Abcd for option (B) , I get Z=1-a/RTV , using PV=RT , I get $Z=1-a/PV^2$

But that doesn't match with the graph , I mean , it's not even linear

Abcd

@Tanuj Whats the answer

@Tanuj Write V= ZRT/P

So you get :

$$Z= 1- \dfrac {aP}{ZR^2T^2}$$

which is not linear

So option B is incorrect.

Anonymous

3:15 PM

@Abcd How is it not linear?

Abcd

3:34 PM

@DivyankaChaudhari mutliply both sides by Z

Anonymous

3:46 PM

Got it

Anonymous

Couldn't properly read the eqn due to (not enabled)mathjax

Anonymous

Smart

Anonymous

But what's wrong with GT's method

Abcd

@DivyankaChaudhari I didnt see it.

what did GT do

Anonymous

2 hours ago, by Gaurang Tandon

@Abcd B is simple actually; b=0 is given, so Z=1-a/RTV or Z=1-kP (assuming constt temperature, and k is a constant)

Abcd

3:55 PM

@DivyankaChaudhari How did he take - a/RTV = kP

@DivyankaChaudhari we cant assume constant temperature unless its mentioned

Anonymous

He put V=RT/P, which is probably little incorrect

Anonymous

2 hours ago, by Divyanka Chaudhari

Oh, constant T, missed it

Anonymous

2 hours ago, by Divyanka Chaudhari

But wait, can we assume that?

Anonymous

2 hours ago, by Gaurang Tandon

what else do we do :/

JavaScriptCoder

yeah

Abcd

3:57 PM

@DivyankaChaudhari yes thats y its wrong

JavaScriptCoder

at first glance, isn't there a n?

as in

> pv = nrt

so V=RT/P only happens when n = 1?

Abcd

@JavaScriptCoder : $(p+ an^2/v^2)(v- nb)= nRT$

We are talking about Real gases @JavaScriptCoder

Maybe you ll learn about them later

JavaScriptCoder

ok with the stuff about particles interactions

right?

with van der waals stuff?

Abcd

hmm

JavaScriptCoder

yeah see this

> $p(V-b) = RT$

replace V with V-B to account for V-b

that was not a good explanation

@Abcd then from this you can derive the equation you mentioned

Abcd

4:03 PM

@JavaScriptCoder not following...

we are already done with the question :P

do you see any discrepancy

Let me know if thats the case

JavaScriptCoder

no i dont

see this

Van der Waals equation

The van der Waals equation (or van der Waals equation of state; named after Johannes Diderik van der Waals) is based on plausible reasons that real gases do not follow the ideal gas law. The ideal gas law treats gas molecules as point particles that do not interact except in elastic collisions. In other words, they do not take up any space, and are not attracted or repelled by other gas molecules. To account for the volume that a real gas molecule takes up, the van der Waals equation replaces V in the ideal gas law with (V-b), where b is the volume per mole that is occupied by the molecules. This...

it has both the equation you @mentioned and I mentioned

:P just had to say that I know about real gases

Abcd

we discussed p(V-b)= RT too

@JavaScriptCoder haha okay

JavaScriptCoder

its all good

Gaurang Tandon

4:16 PM

I'm afraid i was wrong. I just checked 2006 paper, it's there, answer is B. Sorry for the confusion!

Anonymous

Yay, we got the right answer :D

Anonymous

Confusions create discussions

Anonymous

For the answer!

JavaScriptCoder

For truth!

FOR GLORY! (and rep)

Anonymous

I'm laughing

JavaScriptCoder

4:25 PM

If it was a sin to want money, i am not a sinner

if it were a sin TO WANT GLORY (and rep) then by god i am the greatest sinner alive

(paraphrased from Henry IV's speech before Agincourt in Shakespeare's Henry IV)

Wait its Henry V lol

Anonymous

Thee shouldst wend and doth some productive w'rk than recalling what thing which henry hath said.

JavaScriptCoder

4:41 PM

lalalalaicanthearyouium

Anonymous

6:05 PM

Anonymous

Please help. I know, these questions I'm asking are too easy, but in classes we didn't have any doubt session of GOC yet and I'm stuck. I have my exam on sunday :( Thanks

Abcd

@DivyankaChaudhari Is it:

d)1

c) 2

b)3

a)4

Anonymous

d-1 and c-2

Anonymous

Could you tell me how?

Abcd

@DivyankaChaudhari Mistyped

@DivyankaChaudhari Sure

Anonymous

6:10 PM

Also, @Abcd, do you have Ms chouhan solution pdf ?

Abcd

@DivyankaChaudhari No I have hard copy

Anonymous

I have to solve from pdfs until I get hardcopy

Anonymous

Oh, nvm :(

Abcd

@DivyankaChaudhari its their in market stores better buy from there

Anonymous

I have ordered it, yet to get.

Abcd

6:11 PM

@DivyankaChaudhari we decide free radical stability just as we decide Carbocation stability

d) 3 degree + conjugation

Anonymous

Never heard of it.

Abcd

the free electron can be delocalised

@DivyankaChaudhari Now you have :)

c) conjugation + 2degree

b) 3 degree only

a) 2 degree only

Anonymous

Oh, I now get it!

Anonymous

Could you tell me why 3 degree are more stable

Anonymous

Sorry for such basic qsn again but you know what situation I'm in

Abcd

6:13 PM

@DivyankaChaudhari most number of alpha hydrogens

aka hyperconjugation

Anonymous

Oh yeah

Abcd

which leads to electron delocalisation

Anonymous

I feel so dumb

Abcd

delocalisation = happiness = stability

Anonymous

Thanks

Anonymous

6:14 PM

happiness+ stability?

Anonymous

=*

Abcd

typo

Anonymous

One more

Anonymous

Abcd

@DivyankaChaudhari Find the most basic N

Anonymous

6:16 PM

Upper one?

Abcd

@DivyankaChaudhari edited

@DivyankaChaudhari Is the answer b?

Anonymous

yes

Abcd

@DivyankaChaudhari okay see:

Uppermost Nitrogen's lone pair is involved in resonance with ring

while lower nitrogen's lone pair is free

Anonymous

Yes

Abcd

so It can take H+ and work like a base

Anonymous

6:19 PM

Oh yeah

Abcd

so thats why H+ goes there

Anonymous

Got it

Anonymous

Lewis base

Abcd

Okay nice

Anonymous

Thanks again

Anonymous

6:19 PM

Also, could you please tell me what should i do now that my GOC is so bad and I've my exam

Anonymous

I've never experienced this situation before

Anonymous

I'm totally tensed

Abcd

@DivyankaChaudhari solve more and more questions.

You like organic @DC ?

Anonymous

I am fascinated

Anonymous

But

Anonymous

6:21 PM

We are rushing through it

Anonymous

And so

Abcd

How many hours are you studying per day @DC?

Anonymous

Depends, I have my school everyday as well as coaching, 3-4hrs.

Anonymous

That's too less

Anonymous

You must be studying like 6-7 hours :P

Abcd

6:25 PM

secret

@DivyankaChaudhari dontask

Avatar Shiny is topper guy

I am not

Anonymous

Who is that?

Abcd

@DivyankaChaudhari That was Avnish Cabbage's original name on SE

@DivyankaChaudhari yeah exactly

Anonymous

I got topper's marks

Abcd

In 4 days there will be a topper

people will be dying for him

Good night

.

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