May 31, 2024, 11:00:42 AM
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Hard to say, but my intuition strongly suggests that distance loses to electronegativity. You are looking at the change in electronegative effects over the distance (approximately) of a C-C bond length (~1.5 Å), and electrostatic effects decay quite fast with distance. 
2
I agree this is dubious, two competing factors changing at the same time. The only way to say for sure is to compare pKa data (no, I don't have a source at hand).
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I dont know how relevant this is, but aniline in its protonated form is much less reactive towards electrophiles than it is as free base. Also, phenol in its anionic form is much more reactive towards electrophiles than it is in its non-charged form
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Organic Chemistry Forum / Re: Negative Induction effect with respect to Electron Deficiency
« Last post by einsbuzzing on Today at 12:17:23 AM »
The inductive I and also mesomer M effect is used in organic covalent bonded molecules, not in ionic bonded compounds like salts.

https://en.m.wikipedia.org/wiki/Inductive_effect
Atoms that have an electron-withdrawing effect have the −I effect.  This is usually caused by high electronegativity or a positive charge.  Strongly electronegative particles attract electrons particularly strongly.

Particles that have an electron-pushing effect have a +I effect.  This happens e.g.  B. if the particle is negatively charged or has a low electronegativity.  The +I effect can also be observed in the formation of hybrid orbitals, e.g.  B. the methyl group CH3 donates electrons, even if this is not obvious due to the C–C single bond.

So in case of ionic bonds its difficult to say. A anion has normaly a high electronegativity, but also a negative charge. For a cathion its opposit.
[/quote]

What is the impact of I on Cations? A professor informed us that cations exhibit a greater propensity for attraction and the -I effect due to their electron deficiency.
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I’m tutoring ochem this summer, and my students had a question from the professor about ranking the acidity of various compounds. Most were easy enough, but they were confused about whether 2-chloropentanoic acid or 3-fluoropentanoic acid was more acidic. Their professor said 3-fluoro was more acidic, and they asked me why, and I didn’t really have a good answer. I told them it’s because fluorine is so electronegative that even 2 positions away, the inductive effect still makes it a stronger acid than chlorine. But the truth is, if I had been answering the question, I probably would have said that 2-chloropentanoic acid is the most acidic, because I was always taught that distance is the most important factor with the inductive effect. Is the professor right? I was trying to find pKa values online to answer for myself but couldn’t find any. And what’s the best way to approach these ranking questions where the acids differ in both EWG and position of EWG?
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Analytical Chemistry Forum / Re: Acid chloride TLC
« Last post by Tom_Boomer on Yesterday at 12:20:04 PM »
Thank you for your response wildfyr! In regards to:
the product on the baseline is the starting material, the one that runs is the ester.
I assumed the same at first, however, I started to wonder as to why at the sample at T0 there would be no dot in that case, even though the starting product dissolves in the DMF. Hence, me wondering what it could be.
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Basic physics. Work/energy and force are different fundamental physical concepts. Knowing the force at _some_ point of the bond breaking process is not enough to say how much work needs to be done to achieve separation.
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Oh, you've excelled yourself this time!
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High School Chemistry Forum / Re: CH4 combustion exercise
« Last post by mjc123 on May 29, 2024, 06:00:47 PM »
Please read the forum rules; you have to show an attempt of your own before receiving help.

In any case, I don't know what a lambda factor is, I don't know what the selected equilibrium reaction is, and I don't have the basic thermodynamic information to hand. Is this information that is provided in the question paper but you are withholding from us? How do you expect anyone to solve this problem?
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Why isn't Bond Enthalpy/Bond disassociation energy measured as a force?

I understand every bond chemically has a length and energy to break, and energy is Newton*meters, but force is Newtons.
Is the Bond enthalpy/Bond disassociation energy equivalent to the force needed to break the bond * the bond length?

Why don't we say, to break the bond from O to H we need to put magnets on left of the O and right of the H and apply some pulling force of XYZ?
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