JEE Chemistry Club

Gaurang Tandon

00:51

@AvnishKabaj haha ;P

@Jasmine there's two pi bonds... Explained by MOT.. Google for it...

Abcd

02:18

@AvnishKabaj please don't reply to trash messages. I send them to trash coz I have got their answer

Jasmine

02:33

So sigma bond is not always the first bond to be formed @GaurangTandon

Koolman

03:00

@Jasmine Does stability of carbocation depend on steric hindrance .

Gaurang Tandon

03:19

@Abcd please unpin the current pinned msg

@Jasmine actually it's by MOT; a lot of sigma bonds form already, pi bonds are the last two iirc

@Koolman depends on the carbocation; do you have an example?

Koolman

@GaurangTandon option a

Jasmine

@Koolman i think you are talking about the previous to the latest question i sent. I think not steric hinderence but the nature of the group attached has got sometihibg to do with stability. You go for steric hinderance while checking existance of molecule and rractivity

@GaurangTandon my concept in mot is poor i cant get what you mean to say

Koolman

@Jasmine then option a should be correct

Jasmine

@Koolman yes i think it should be correct. Today i will ask my teacher and confirm.

Koolman

okay

Jasmine

03:41

How are you saying @Koolmanthat A is correct? Whats your logic?

? @Koolman

I didn't mean that steric hindrance has got nothing to do with carbocation stability

Koolman

-I effect of alkyl group

Jasmine

No resonance possible in one to stabilise the positive charge so you check for alpha h. But its same

Tell me which is more electronegative sp2 hybridised or sp3 hybridized?

Sp2 of course as more s character closer to nucleus more force applied by nucleus

So in third the cation stabilised more or firs

@AvnishKabaj can you please help out

Abcd

03:59

Alkyls = +I effect not -I effect

@Jasmine Please share your problem

@GaurangTandon @AvnishKabaj $\ce{Ph-CH2-NH2 + EtO- -> ? }$

@Jasmine You won't understand C2 without knowing MOT. You should know how to write the electronic configuration using MOT at least. That much is in JEE syllabus. Around 2014, they had even asked the structure of the molecular orbitals.

Avnish Kabaj

@Jasmine @Abcd I'll get back to you two in a couple of hours

Abcd

@Jasmine Phenyl's +M effect dominates over its -I effect. Therefore it's an activating group in EAS

Gaurang Tandon

04:20

@Koolman the last structure has strong steric inhibition of resonance (SIR) effect, so the positive charge won't delocalise all throughout the ring; in the first structure, it will; more delocalisation of charge => more stability

@Abcd hehe what in the world is that o.O what's the answer? is this msc question? if so, page number?

Abcd

@GaurangTandon $\ce{PhCH2NH-}$

Koolman

04:40

@GaurangTandon so you mean SIR affect the stability of the cation

Abcd

(It's quite obvious it does)

@GaurangTandon anyway that question is controversial ... there's strong +I effect of alkyl stabilising the carbocation

Alkyl is stronger activating group than phenyl. So in all probability option A order is wrong

Jasmine

@Koolman yeah +I and not -I effect

@Abcd I know mot its just like I am very confused

Between c and c in c2 how many bonds exist?

I know to write the configuration as I have seen in jd lee

Is the theory of mot in accordance with all other theories?

Abcd

04:57

@Jasmine Huh?

@Jasmine please see the Wikipedia page on C2

Gaurang Tandon

@Koolman it does, that's what i said

@Abcd meh, it probably just abstracted the acidic hydrogen on N atom (ethoxide is strong base)

@Abcd but SIR plays a stronger role. There's a question on main site, let me share

7

Q: Which of these carbocations are more stable?

The line of text that follows this figure in my work-book reads: Cation (A) is more stable than cation (B), since the former is resonance stabilized but the latter isn't (due to steric effects). Don't both carbocations show resonance stabilization? More importantly, isn't cation (B) more ...

Avnish Kabaj

05:17

@Jasmine I agree with Gaurang Tandon it's most likely SIR

Is the answer (a, c)

?

@Jasmine what third?

@Abcd dunno

Abcd

06:00

@GaurangTandon How is that H even acidic

@AvnishKabaj she doesn't have the answer key

Gaurang Tandon

@Abcd i just made a guess based on the answer :/ is the question from msc?

Koolman

06:25

So according to answer should be c and d

Jasmine

Finally got the c2 thing as the 4 electrons enter pi bonding

@AvnishKabaj why not d?

And b is pakka more Inductive effect

So abcd should be the answer

I am not sure

But b I effect is very less affective oh so confusing. A is pakka

C is sure

Koolman

in b option the compound will have hyperconjugation

so it should be a,c,d

Jasmine

06:40

Yeah hyperconjugation dominates

A and c surely. But d is controversial

Koolman

yeah

Vikas Rv

@Koolman whats the given answer for that carbocation stability question ?

Koolman

@VikasRv I dont know the question was posted by Jasmine

Vikas Rv

Oh

Abcd

@GaurangTandon No. Its from my FIITJEE sheet

Avnish Kabaj

06:51

@Jasmine d is definitely not the answer primary carbocations are very unstable

Abcd

@AvnishKabaj Huh? There's no primary C+ in option D

You mis-saw probably

Gaurang Tandon

@Abcd misread xD

Abcd

@GaurangTandon but there's nothing to read that's why mis-saw

Vikas Rv

xD

But still if we have to choose between M and I effects, M effect is given priority so D cant be the answer na

Abcd

@VikasRv Huh?

@VikasRv Thats absolutely incorrect

@VikasRv SN1 with 3 degree is MUCH faster than that with primary allyl or benzyl

And you know whats formed in SN1

1

A: Why is allyl chloride more reactive towards substitution than alkyl chloride?

But how allyl halide is more reactive than alkyl halide? Reactivity towards $\ce{S_N1}$ The key to reactivity towards $\ce{S_N1}$ is the stability of the formed carbocation. Allyl system stabilises the carbocation through overlap with the vacant p orbital (@gsurfer999 has shown the resonanc...

Sid

07:04

That questions seems to be multiple-correct options..

Abcd

@Sid it is

Sid

which should mean, the answer is "All"?

I am not fully confident but fairly sure.

Abcd

@Sid d is controversial

I will ask a question about that tomorrow after returning from fiitjee

Sid

What's the controversy?

Abcd

@Sid Some say allylic one than benzylic is more stable due to "equivalent resonance structures"

Sid

07:08

@Abcd You mean, Allylic is more stable than Benzylic?

Abcd

@Sid what else have I written?

Vikas Rv

Abcd

@Sid I don't mean that, some people say that

@VikasRv What?

Vikas Rv

the rates given here are different :/

Abcd

@VikasRv Are you sure all of them are the same as the one in my table

9 mins ago, by Abcd

1

A: Why is allyl chloride more reactive towards substitution than alkyl chloride?

But how allyl halide is more reactive than alkyl halide? Reactivity towards $\ce{S_N1}$ The key to reactivity towards $\ce{S_N1}$ is the stability of the formed carbocation. Allyl system stabilises the carbocation through overlap with the vacant p orbital (@gsurfer999 has shown the resonanc...

@VikasRv Please tell me the mismatch one .

Please also mention the book name

Sid

07:12

I think we had a question about stability rates of carbocations in JEE-ADV 2017. I know from there that Tertiary, then Benzyl(Primary) and then, Allyl(Primary)

Abcd

@Sid Good.

That's the order I support too

Sid

There's probably some nice reason as to why that is true. I don't care enough right now to go searching for it.

Abcd

The reason is:
Phenyl is better at conjugation than allyl.

Vikas Rv

Sid

@Abcd that doesn't explain why tertiary is better than all of them.

But okay.

Abcd

07:15

@Sid Huh? Tertiary not also Has +I effect but also so much Hyperconjugation

16

Q: How is a free radical on allylic position more stable than at benzylic position?

Which is the most stable free radical among the given species? I was totally baffled as the book claims the answer to be option 1. How is a free radical on allylic position more stable? Moreover, the number of hyper conjugations is also way more in option 3 and also its a tertiary free...

organic-chemistry stability radicals

Allyl free radical appears to be more stable than benzylic one as per ron's answer

Vikas Rv

saplinglearning.com/media/loudon/loudon5ech17sec01.pdf

@Abcd this is where I found that table

Abcd

@VikasRv look we cant trust any random sites and pdf

Sid

In the end, you have to refer to those data and then choose which one satisfies the data. You can't compare Resonance effect and Hyper conjugation effects all on their own.

Abcd

Clayden is a great book

Vikas Rv

If it came in previous jee then I agree with it.

Abcd

07:17

@Sid .... It's quite intuitive to me

Vikas Rv

07:28

worldcat.org/title/modern-physical-organic-chemistry/oclc/…

even this book says the same

quora.com/Which-carbocation-is-more-stable-Benzyl-or-Tertiary

some are saying benzyl > tert and others are saying the opposite :(

anyway we should follow whatever was in previous jee questions

Carrick

hey, it's true that tertiary carbocation is more stable than the benzylic carbocation

Abcd

07:41

@VikasRv duh, dude that's ion affinity

Less stable one will have MORE ion affinity

Are you happy now?

@Carrick yeah...

Carrick

@Abcd I have a question btw...got some time to spare?

Abcd

@VikasRv quora is one of the worst site to refer to for chemistry

@VikasRv "follow"? What are you confused about in the order?

@Carrick I am doing Maths at the moment. You may post the question ... I'll see or someone else will answer

Carrick

@Abcd oh okay...

The question is: Is the nitrogen in this structure a chiral center?

Abcd

It is @Carrick

It's tetrahedral and 4 different groups are attached to it

Also, Nitrogenous inversion is prevented by constraining it to be the part of the ring

Carrick

@Abcd The book says there are only 4 stereogenic centres in this molecule(N isn't counted)

Abcd

07:50

Checking

Carrick

well I think the answer's wrong...N should be a stereogenic center right

Abcd

Maybe ...

Carrick

anyways thank you for your time

Jasmine

16:49

Sorry for the bad pic. Just see the second question ignore the rest please tell me what will be the order of reaction in 2nd question.

Koolman

Is the answer 3/2

Jasmine

Yes @Koolman i got something else show me how

Koolman

r=[A][B] and Keq = $[A]^2 / A_2$

substitute [A]

Jasmine

Can you send me handwritten pic?

Please,,

Li ion or mg ion whose size bigger?

Koolman

Jasmine

16:59

Ok @Koolman it was a liitlle tricky for me thanks though

Question 4 please send handwritten solution.

Number of chiral centres in resperine?

Abcd

18:46

@Jasmine I dont get why you have marked the sp2 carbons as chiral centers in your diagram.

@GaurangTandon : $\ce{CH3CH2MgCl ->[CH3CH2Cl] CH3C#CMgCl}$, how ?

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