@Abcd well, Idk. I was planning to write an answer myself, as a sort of a post it note.

@Abcd ??

@Abcd in most of the shifts

Anti periplanarity is needed since maximum overlap between the orbitals takes place

@AvnishKabaj look carbocation is sp2 and has vacant p orbital and rotation can take place about single bond so why not attack from any side?

@GaurangTandon they taught that?

hhmmm

@Abcd they didn't; I studied it from MSC via examples

@Abcd still waiting

@GaurangTandon uploading ... till then you tell me how you'd solve this question...

@Abcd configuration is retained/inverted of those species which are being attacked by others; here the species shown is itself attacking; attacking species don't undergo any change

@GaurangTandon what do you mean? What's your answer?

B is the answer

@GaurangTandon NGP right?

re-read what I said; there's no logic of NGP in this question

@GaurangTandon ,,,, Oxygen's lone pair attacks dude

In option B

that still doesn't invalidate what I'm saying

I think I'm unable to convey properly what I mean

@AvnishKabaj can you rephrase?

@GaurangTandon and that is called NGP

@GaurangTandon See in the first example Anti periplanar is not possible so epoxide is formed

WIthout any pinacol rearrangement

@Abcd no, am sorry, you didn't get what I'm saying; let's wait for Avnish

@Abcd checking

@GT in the second page there are many other examples for you

@GaurangTandon Oh you are saying The ionic compound attacks CH3I

@Abcd yes...

@GaurangTandon why can't there be NGP instead?

By lone pair of oxygen?

@Abcd why would there be? what would be the product?

@Abcd in eg 14, why did the ring contract?

@GaurangTandon After carbocation formation there's alkyl shift (that alkyl shifts which is anti periplanar to leaivng group , here OH)in pincol rearrangement ...

@Abcd yes there is, but why did the ring contract? There was an alkyl shift in eg 12 and 13 as well, but there the ring remained cyclohexane only

@GaurangTandon which bond is anti to OH ?

@Abcd oh i see your point, this is interesting...

@GaurangTandon there was no such question in MSC?

@Abcd nope, i'm sure there wasn't...

@GaurangTandon oh, he missed some important questions then...

right, could be...

i'll ask my teacher about this topic, and let you know what I come to know

@GaurangTandon hmm you are right

@GaurangTandon What about option C

there can be NGP there

@Abcd sorry, i think sn1 => racemic mixture is the safest bet

@GaurangTandon Are you going to say , it's too hindered for NGP by Br?

i was going to say we don't need to insert ngp in every second question, unless there's some conditions (like an NGP product given in multi-choice options) that would require us to do so

idk

ask avnish, or on main

@AvnishKabaj There can't be any NGP in option C because of steric hindrance right?

@Abcd i'm still doubtful of this though; in e2 mech, the LG leaves and Nu attacks all in one step, so it makes sense for them to be anti-peri-planar. however, in sn1 mech (like pinacol-colone), there are two separate steps. the first one results in formation of a planar carbocation. once that is formed, attack can be from either direction, not necessarily only from that alkyl group which was originally anti to the -OH

@GaurangTandon I know ... quite confused about that... But this is even given in Peter Skyes

@Abcd hmm, page number?

@GaurangTandon How to write strikethrough language in chat

@GaurangTandon 5.19 of Indian Peter skyes

@Abcd idk

@GaurangTandon I think the reason is:

sorry, I don't have peter sykes, can you please send me the relevant screenshots (use snipping tool)? that way I can also show my teacher

@GaurangTandon I have hard copy

@Abcd well, then send a camera photo? _/\_

OK

@Abcd found sth interesting; from - "An alkyl group which is situated trans- to the leaving –OH group alone may migrate. If otherwise, ring expansion occurs, i.e. the ring carbon itself migrates to the carbocation centre. This reveals another interesting feature of the reaction, viz. that it is largely concerted. There appears to be a connection between the migration origin and migration terminus throughout the reaction."

@GaurangTandon thats the question you asked me about..

@Abcd yes, though let's read peter sykes first...

@GaurangTandon Whats your favourite PCM subject atm?

@Abcd M (doesn't matter though, balance is best)

okay, please wait till evening till I get back with a proper response

and thanks for the pics

Isn't C the most stable one over here?

OH group involves in H-bonding and t-butyl group and Me group are in equatorial groups for enhanced stability (as compared to equatorial-axial)

Haha...C is right though, ain't it?

@J_B892 No

Me has to be at the same side of isopropyl

@GaurangTandon Can H2O as a base abstract deuterium as well? Why or why not?

Soo...A?

@J_B892 No , OH has to be on oppsoite side of IsoPr

Is the answer d?

Its given A

@J_B892 nitpick: should be "stable chair" in question

@Abcd nitpick: iPr

@GaurangTandon Wrong answer in his textbook dude

@GaurangTandon wrong answer in his textbook dude

@J_B892 shouldn't be A; C is most reasonable

Yeahh...@GaurangTandon

@GaurangTandon wrongggggggggg

in C, -OH is opposite to iPr (axial vs equatorial)

Dudes

@GaurangTandon That is not what opposite meanssss

can you get any more stable than what is given in option C @Abcd?

@GaurangTandon its not about stability ... Look at the question please

@J_B892 fwiw, Stop. Solving. Himanshu. Pandey. Buy MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed (not an advertisement)

its about cis trans

@J_B892 fwiw, Stop. Solving. Himanshu. Pandey. Buy MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed (an advertisement)

@Abcd it's actually asking most stable conformation

@GaurangTandon I do have that book too. But i wanted more practice on problems to improve my speed and accuracy

@GaurangTandon ... just give me a min

@J_B892 your choice, i found msc good enough

@J_B892 MSC has ample of problems too

Solve it twice

@Abcd @GaurangTandon Hmm...Okay sounds reasonable

guys i'm logging out; have hw to finish; cya later!

@GaurangTandon Here's that D abstraction thing

i'll check and tell

@J_B892 @GaurangTandon ^

@J_B892 The problem was with his diagram...

Option A is correct. See my diagram^

I got multiple pings what's up

@Abcd tell me in a new room

hello

In thermodynamics , during an adiabatic process when work is done on the surroundings/on the system there occurs a temperature change

I was wondering how this doesn't affect the heat energy

for change in heat energy must be zero during an adiabatic process

Isn't temperature a measure of change in heat energy?

@susanJ no

Temperature is a measure of internal energy

kk

@GaurangTandon Did you see it? @AvnishKabaj Please see that question.

That alkene one

@AvnishKabaj you probably found it upsetting, but know that that's what happens when you challenge people on the internet to decode your name via an anagram...

@J_B892 good job @Abcd that's correct; i messed up big time

@GaurangTandon that alkene question?

@Abcd this is from MSC? chapter name?

five seems plausible, though they've probably given answer 3

@GaurangTandon No its from my FIITJEE worksheet

@GaurangTandon No 4

@GaurangTandon are you sure H2O can extract D

@Abcd what's the soln? i see no way to get 4 as the answer?

thats my main question

@Abcd why not?

@GaurangTandon fiitjee and other worksheets dont give solution they just give answer

@GaurangTandon C-D bond is stronger

:(

well you may ask on main; that's really all I had :/

@Abcd Got it yet?

Oh sorry sending tomorrow

@GaurangTandon @AvnishKabaj Is this a mistake in MSC? In question he asks for Z 2 butene and in his solution he has shown E-2 Butene.

Is my answer correct if it was Z-2 Butene

@GaurangTandon I know its NGP but why is enantiomer formed in reaction 2 ?