Cl- is good leaving group so I believe the reaction would be SNAE.

Oh okay...but if you look at something like vinyl chloride it doesn't undergo SN coz it has a partial double bond character right? In this case too -Cl is a good leaving group. So what is the difference in these two cases? what causes a difference in reaction even though they're quite similar

@J_B892 the mechanism is basically diffrent the the case of vinyl chloride SN2 and SN1 are not possible , where as in case of reduction of acid halide the mechanism is $SN^{\text{AE}}$ which starts by attack of H- on the carbonyl grp. and then oxygen revert the lone pair to eliminate Cl- which is a nice leving grp and thus is easily removed

oh okay cool...Thanks!

okay welcome :-)

@J_B892 As far as any conjugation is considered... You are thinking of a 3p lone pair with a carbonyl group made of 2p orbitals... And in your supposed canonical structure,chlorine is developing a positive charge... That itself should convince you that there is going to be no major contribution from ur canonical strucuture

@YUSUFHASAN @AdvilSell Hi!

yo

Hi!

I've seen in many reactions with Bromine water, if phenol group is present with other groups like COOH, SO3H, the latter ones are replaced by Bromine.

yes

Which class of reactions are these? Are they there in the syllabus?

I read somewhere it's called IPSO substitution.

Yeah.. That is the name

yeah they are called ipso

though they are not in syllabus

I don't find it anywhere in NCERt/Solomons

but still asked in KVPY

@AdvilSell It was there in adv 2013

@Dante Hmmm...

Why do they give such random reactions..

Thats because IPSO substitution is in syllabus.. Tho not explicity mentioned.. and we use it in reactions,but it is not named so by foreign authors

@YUSUFHASAN oh is it so..

It is an indian thing,"ipso substitution".. Other people just call it the ipso position for all purposes

@YUSUFHASAN I see.

@YUSUFHASAN oh i c

@Dante though they are not that difficult to understand ........

Not difficult to understand but why the hell are they not there in even the most famous JEE reference books? Drives me mad

@Dante I personally take "JEE"" reference books with a grain of salt.. If we are talking standard literature then it's sthg :)

Hmm. Anyways it's my mistake that I wasted more than a year of JEE preparation and now I don't have time to read any standard books...

There's a good reservation for PWD candidates in IIT, does that mean every class will have at least one physically handicapped student?

@AdvilSell ?

yeah

I'm just curious

I don't know seriouslly

ok

I had a slight doubt. In acidic buffer solution of ethanoic acid and sodium acetate, the ethanoate ions from sodium acetate react with hydtogen ions in the solution to make more ethanoic acid... Wouldn't this change PH

It will.

Henderson's equation takes care of all that.

Ohkay

It makes few approximations also though.

@Zerix Buffer solutions are said to be buffer when the reaction mixture is in equilibrium.. So the relative concentrations of the reactants and products will not change unless some other substance is added to the mixture... Then the buffer solution reacts with this new substance to reach a new equilibrium,which will have the same concentration of H+ ions

Yes. That's the way how it works. But I guess the pH slightly changes despite the equilibrium because more acid is being formed

Or the acid gets slightly depleted.....

@dante I think you must ping someone in the periordiv table grp.

They might get pissed off, all are PhD scientists, I'm scared.

they might understand....

Not taking the risk, orthocresol will probably reply

ok XD

@Zerix pH changes till +-1 can be accomodated by the buffer solution... Under the buffer capacity,consider that you have as much conjugate base and acid available,so that the pH change can be negated

What does dehydrohalegenation of 1 acetyl 2 chloride cyclohexane yield?

Will the double bond be formed where the acetyl group is attached or on the other side of chlorine?

@YUSUFHASAN @AdvilSell

@Dante Towards acetyl?

Yeah

I don't get why. $\beta$ elimination has hydrogen removal as first step.

Shouldn't it be difficult since it's sterically hindered?

Thermodynamically stable product will be formed... The usual assumption... E2 products eliminate via TS..partial negative due to C-H bond breaking is stabilised by acetyl

@Dante Acetyl is sp2 hybridised(central atom).. It is mostly planar.. And since a very small anion OH- is removing the H, no steric hindrance operates(basicity is a thermodynamic concept, only gibbs free energy/stability matters)

@YUSUFHASAN Oh, so you mean transition state will be more stable?

@YUSUFHASAN @Dante I have a doubt here isn't dehydrohalogenation have a base extracting H+ ???

This means that the OH- anion has all the time in the world to lasily approach the H from above the plane,below the plane or wherever it wants to... It is not like a nucleophile which is in a hurry to eliminate and therefore has to look for the easiest place ro approach(sorry for the anthromorphisation,but it is one explanation which workd for me)

@AdvilSell Yes. Thats what i am stressing.. It is an acid base interaction.. Not a nucleophilic one.. So don't see steric here

Yes, I've heard that explanation before. Thanks!

@AdvilSell That's no starter right?

@Dante isnt this the starter

@AdvilSell That H abstraction will only lead to a enol form which is unstable.. Goes back to keto... And u cant close a 4 memebered ring.. No starter

@Dante Yeah, u r correct!

@YUSUFHASAN I agree but as far as acid base interaction go the H+ will be abstrated from the position circled

Can someone link a source which explains why meta substituted products are thermodynamically more stable?

Yep.. Abstract that H.. Nthg will happen for 100 molecules(suppose).. For the 101st molecule,the other H goes and the reaction goes forward.. Other molecules be like"Hey,this path looks funky, Chatlier also told us that if we go down in concentration somewhere,we gotta go for backup!".. So ultimately.. All the molecules get in the new fashion and the unfavorable path gets favored

@Dante U talking about the xylenes?

@YUSUFHASAN haha :-) okay

Not just Xylenes, I just read that meta substituted Toulene Sulphonic acid is also more stable than other isomers

I assumed all meta stuff is more stable, is that true?

Yeah.. Reason is similar.. Look up on SE for the xylenes... I cant find the link right now.. It is simply hyperconjugaion and stuff... Somebody had given a nice answer on this

???

Ye wala?

Haan... U got it,i think?

No, had just opened the link

I'll read the answer now.

Got it.

@YUSUFHASAN I have the same doubt you asked in the comments.

Is that any special kind of hyper conjugation?

@Dante It is basically a way of representing the same thing.. Instead of making the complete resonance structure of C-H bond breaking and the π bond shifting,he has simply broken the π bond heterolytically first and shown the intermediates.. An unconventional way,sure

U break the C-H also heterolytically,combine plus and minus signs to make pi bonds.. And u get the same thing

right

@YUSUFHASAN @AdvilSell

yup

Which Cl- is abstracted by AlCl3 and why?

@dante is answer a

No

Acylation was preferred idk why

still why on meta amide is a strongly activating grp must be on o/p

Para itself, answer is given on next page

Which I didn't send

@Dante what i didn't get you ???

Answer is given same as options 'a' but acylation instead. Option C is the answer which is not there in the picture, it was on next page]

oh ok

then let me think

Resonance structures of alkyl cation seem more stable to me than acyl cation

Btw, what do you think about my argument on yesterday's hydrolysis question in periodic table group?

" highly unstable methyl carbocation" must be correct according to me

I see.

@Dante Hmm.. U say that there are more options for this question?

Above one?

Yep.. It has only two options?

@YUSUFHASAN Out of 4, 2 are obviously not correct cuz they're meta substitutions. Out of the remaining two, one does acylation on para position while the other does alkylation on the para position.

Hmm.. So the answer must be acylation on the para position

?

I'm confused between them

@YUSUFHASAN Yeah

How?

Yep... I told you told this already.. Backbonding by oxygen in the carbonyl cation.. No stabilisation in the alkyl carbocation

Oh yeah....... but alkyl carbocation is also stabilized by carbonyl's pi bond. I tried drawing resonance structures of both, alkyl cation's resonance structures 6 covalent bonds each while that of acyl's have 5 and 6 respectively.

I remember reading in solomons that number of covalent bonds affects the stability of resonating structures to a great extent, that's why I got confused.

And btw, what do you think about my explanation regarding hydrolysis question I sent yesterday.

Again.. If u think u have got some major contributing structure for the alkyl carbocation.. Think about electron source and electron sinks in accordance with electronegativity... The carbonyl O can only backbond at its own carbon.. And there is no means for the carbocation to move here

@Dante I didnt see that I think.. Which explanation u talking bout?

Most recent one in periodic table.

@Dante Yeah of course, ethers are very unreactive and probably HI (conc.) is one of the few acids that can cleave it.. So no chance for dil.HCl to go along that path

Thats why they r used as solvents

Okay, then what were you and advil trying to figure out in that yesterday?

What we were trying to figure out was the position of the unsaturation.. Whether it should be alpha-beta or gamma-delta.. Cuz the mech. seems to give the wrong answer.. Nobody was touching the ether

en.m.wikipedia.org/wiki/File:Birch_Anisol.png You sent this yesterday.

Shifting of unsaturation is last step in the mechanism.

Yeah.. But i feel that it shouldnt shift.. Mechanistically it seems that way to me

@YUSUFHASAN Protonation of double bond after keeping in acidic medium for long---> Hydride shift so that carbocation enjoys resonance with pi bond---> deprotonation

Does that make sense?

Which pi bond do you want to resonate with? The carbonyl is not going to do that...

Why?

Carbonyl.. Electron sink.. Can't send electrons against electronegativity

Hmmm..

@PolarBear This is what I think :

the amide grp will be directly reduced to a amine as it happens usually as nitrogen donate the lone pair to kick of OLiH3 complex and then H- break the double bond to restore the amine , then for the aldehyde thus formed the nitrogen will again kick off the OLiH3 complex and again H- attack as a nucleophile to reform the amine.

@PolarBear this is what I propose

but just a minor correction please perform the ketone reduction first as its nuclephilic

Oh yeah, seems feasible!

@PolarBear yeah

@YUSUFHASAN 's a little different eh?

@PolarBear I messed a bit here

the reduction of ketone will be preffered first as its more nucleophilic

Yep.. Plus,to be honest,i know that lactam's as a cyclic compound exist.. I have never seen this ring you have made as an intermediate

I would prefer not to make unknown compounds here

@YUSUFHASAN yeah I told that i messed up a bit

@YUSUFHASAN I got the thought of lactum a bit latter

Ah, I see. Thanks both!

I couldn't think of that

@Dante you here

@dante ???????

@AdvilSell Sorry, I was asleep. Good morning!