@AvatarShiny Don't edit questions put on hold if you don't make them worth reopening.

Ok

Is there a list of reagents and their mechanisms somewhere in the corner of the world wide web because I can't seem to find one

I feel that the answer should be $a$ and $d$.

Because: For option a, each molecule of $I$ has its enantiomer in $IV$

For option d, $II'$s 1st Molecule has its enantiomer in $IV$ while the second molecule has its homomer, (identical molecule) in $IV$ so this pair is also valid as per the question.

But the answer key only states $a$. Why is option d not included?

It's wrong.

@AvatarShiny why

The answer key is wrong

Sometimes ms chouhan

Can be wrong too

well, we can't be very certain until a chemist approves...

Maybe there's some logic we are missing out on.

Could be but idts

@Abcd Salt: $\ce{[A+][B-]}$ (i) $\ce{$S$-A+ + $S$-B-}$; (ii) $\ce{$S$-A+ + $R$-B-}$; (iii) $\ce{$R$-A+ + $S$-B-}$; (iv) $\ce{$R$-A+ + $R$-B-}$. Therefore (i) and (iv), ergo a.

I cant read anything.

$\require{mchem}\ce{A}$

there you go...

$\require{\mhchem}{\ce{[A+][B-]}$ (i) $\ce{$S$-A+ + $S$-B-}$; (ii) $\ce{$S$-A+ + $R$-B-}$; (iii) $\ce{$R$-A+ + $S$-B-}$; (iv) $\ce{$R$-A+ + $R$-B-}}$

@Abcd mhchem!

not mchem

i edited it for you

\require{mhchem}

@Martin-マーチン I don't see the edit.

I still see Dollar signs

and ces

well, how do you load MathJax?

well....

I have a tab. (bookmarked tab)

That is labelled "start chatjax"

I click that, and mathjax loads.

umkay... did you get that on math?

@Martin-マーチン Yes, I got it on Math.SE.

from the chat room there.

This one also loads mhchem

anyway... the stereo centres are left-right: (i) S-S; (ii) S-R; (iii) R-S; (iv) R-R; therefore (i) and (iv) are enantiomeric salts, the others are not

Unable to start chatjax

@Martin-マーチン But look 2 and 4 have one identical.

I mean ii) and iv)

yes, that exactly is the problem

only one

Well, obviously you have to treat the pair as one

Yeah, realised that. Thanks.

I didn't get that

also II and III should be the same, but that was no option

@AvatarShiny There's a negative radical and a positive radical.

That makes one full compound.

ion, not radical

So we actually have only 2 compounds too consider and we find that compound 1 and compound 4 are enantiomers @AvatarShiny

@Martin-マーチン oh, i thought they mean the same.

no, radical has unpaired electrons

there can be radicals that are ions

can you give some examples?

Why not 2 and 4

@AvatarShiny because they are not enantiomers.

Look at +ve + -ve as a complete unit.

Don't examine them separately.

Ok I'll draw and see once

Got it thanks

How does r-s determine the solubilty though?

@Rick Don't you know that enantiomers have identical chemical properties in achiral solvents?

and solubility is a chemical property.

sorry, it's a physical property.

But in spite of that, enantiomers have identical physical properties too.

ok..yes..got it

@Abcd What?

In Stork enamine reactions, why does the reaction proceed to give us the enamine (C=C) and not the imine (C=N)? I mean, the two can tautomerize back and forth, so why do we say only the enamine is formed?

I thought imines are more stable than enamines?

@Rick You are confused about the nomenclature

In that context, the tautomers are an enamine and an iminium ion

You cannot form an imine with the with a secondary amine

You can form an iminium ion, which may prefer the enamine tautomer

@Zhe ok yes..

why is the enamine preferred in this case?

Because the other case is an iminium?

There's also a good amount of strain there

But the C=N locks the conformation

oh..yes got it now: so basically in Stork enamine reactions, if a primary amine is used, we get the imine and if we use a secondary amine, we get the enamine (because in the iminuium ion that gets formed, the only H nearby to get deprotonated is from the CH3 away (N doesn't have one on it)

That actually reminds me of when I came full circle by taking a class with Rick Danheiser and then having Gilbert Stork's grandson in my class...