# Tag Info

35

Yes, it has a lot to do with mass. Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (...

17

This sort of reaction isn't investigated seriously anymore. But 80 years ago, Kharasch et al. (J. Am. Chem. Soc. 1939, 61 (6), 1559–1564) wrote that Both results agreed within the limit of error (3%) and indicated an equimolar mixture of 2- and 3-bromopentanes [...] So, for all intents and purposes, there is no major product. It is not always possible to ...

8

Some of the conformers of salicylic acid are given below, all of which are planar in configuration.$\mathrm{^{[1]}}$ Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) ...

7

I will show you how I understand this effect. I like to think that partial positive charge on $\ce{C1}$ attracts the electrons from $\ce{C2}$ because of the electrostatic force. With this model your questions can be answered this way: Yes, $\ce{C1}$ will pull electrons from every adjacent atom, including hydrogen. It may pull them, however, to less extent ...

6

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB). $\ce{Me-\overset{+}{N}(^iPr)3}$ We only need to look at ...

5

Your argument of inductive effect and hyperconjugation, made me to tell you a story about theory behind these two effects. I think after reading that you may able to understand how far we have come from them: Wagner and Saytzeff (Ref.1; in German) prepared 2-pentene by dehydrohalogenation of 3-iodopentane. Addition of hydrogen iodide to this pentene yielded ...

4

A $\ce{C-D}$ bond has a shorter distance, thus can be in a more favorable position to donate electron density towards an electron deficient center. The $\ce{D}$ has a greater mass, and would show a shorter bond distance. This is reflected in the $+ I$ effects noted in the literature.

4

When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant. Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be ...

4

For $b$, you have equivalent resonating structures as the structures are identical. The structures $a'$ and $b'$ are not correct as nitrogen cannot expand it's octet. For $a''$ and $b''$, the first structure $(a'')$ is more stable as there is negative charge localised on electronegative atom as compared to $b''$ where it is localised on carbon. For the ...

4

The other answer has given the major conformer of the protonated form of the o-hydroxybenzoic acid, but I believe there is a part of your question that still remains to be answered. You asked why the hydrogen bond makes the hydrogen bond more acidic than the para isomer, when the hydrogen bond is already present in the protonated form of the ortho-isomer. ...

4

According to the IUPAC goldbook, the definition of the mesomeric effect is as follows, The effect (on reaction rates, ionization equilibria, etc.) attributed to a substituent due to overlap of its p- or π-orbitals with the p- or π-orbitals of the rest of the molecular entity. Delocalization is thereby introduced or extended, and electronic charge may flow ...

4

Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is $$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$ $$\begin{array}{lc} \hline \text{compound} & K_\mathrm{eq} \\ \hline \text{acetaldehyde} & 6 \times 10^{-7} \\ \text{acetone} & 5 \times 10^{-9} \\ \hline \end{array}$$ ...

3

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions. A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S. This means that the number of destabilizing ...

3

We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. ...

3

Alkanes or $\mathrm{sp^3}$ hybridized carbons show +I effect. Alkenes contain $\mathrm{sp^2}$ hybridized carbons. Due to greater $\mathrm{s}$ character they have higher electron withdrawing tendency. So they show -I effect. Coming to the first question: In the given figures we have $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized carbons in the given ...

2

You are correct as resonance is most powerful factor for stabilisation of carbonation So C is most stable and after that due to hyperconjugation B would be more stable than A. Order for governing stability is $\text{resonance}>\text{hyperconjugation}>\text{inductive}$

2

The inductive effect is the polarization of σ (sigma) bonds within a chemical species caused usually by differences in electronegativity between the neighboring atoms. The higher the polarization it causes, the higher the inductive effect. In this case, the decreasing order of electronegativity is: $\ce{-F > -OR > -NH2}$ So, for practical purposes, ...

2

The explanation can[1] be given by analysing their respective resonance[2] structures which is as follows: (Credits: ChemWriter) Now, inductive effects would work. In case of first (c's resonance), positive charge would be stabilised by $\ce{+I}$ effect of methyl group; whereas in the second (d's resonance) one, $\ce{+I}$ effect would become dead upto ...

2

How to decide whether +M effect or -I effect will operate in this case? The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$. In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ ...

2

There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group. Some particular examples: \begin{align} \ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH} \end{align} About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-...

2

I'm not sure if G is an activating or deactivating compound. You may want to look at the relative order of activation/deactivation of some substituents, (source: www.chem.ucalgary.ca) As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{... 2 Yes, both your answers and reasoning are correct. For the second compound a slight ambiguity might arise, because the$\mathrm{sp^2}$-hybridized carbons show$\text{-I}$effect which would destabilize the carbocation, however considering the fact that there are not only$\ce{2 Me}$— groups attached which show$\text{+I}$, they are also closer than one of ... 2 Some cheminformatic descriptors for the puzzling compounds can be found on the PubChem website: SMILES N+(=O)O ; Nitroacetic acid; 625-75-2; alfa-Nitro acetic acid; 2-nitroacetic acid; ALFA-NITROACETICACID; CID: 43581; MF: C2H3NO4; MW: 105.05g/mol; InChIKey: RGHXWDVNBYKJQH-UHFFFAOYSA-N ; IUPAC Name: 2-nitroacetic acid ; and SMILES N+(C)(C)C; ... 2 The acidity of benzoic acid$(\ce{C6H5-COOH})$and its substituted derivatives,$\ce{R-C6H4-COOH}(e.g., \ \ce{4-NO2-C6H4-COOH})$have been studied extensively using the Hammett Equation. Similarly, acetic acid$(\ce{CH3-COOH})$and its substituted derivatives,$\ce{R-CH2-COOH}(e.g., \ \ce{Cl-CH2-COOH})$have been studied extensively as well using ... 2 What you should be looking at is the stability of the carbocation intermediate formed when the strong acid$\ce{HBr}$protonates the alkene. This stability is more sensitive to alkyl substitution than the stability of the uncharged alkene, so the net effect is that alkyl substitution lowers the activation energy versus having no substitution. Ergo a faster ... 1 For alkyl groups attached to the phenol ring, there are two main points that your reasoning does not take account of - Inductive effect is distance dependent. As the distance increases, its effect becomes very small. Even if the group were near, the general trend is that hyperconjugation has stronger effect than inductive effect. You can refer the ... 1 Since electron-withdrawing groups (EWGs) lower orbital energy (we can rationalise this in terms of reduced electron density leading to reduced shielding), they will stabilise multiple bonds - you see this in the Diels-Alder reaction, where EWGs on the dienophile reduce the energy of its LUMO and so accelerate the reaction by reducing the gap between the ... 1 In the benzyne mechanism, mesomeric effects are not involved as when protons are removed, the electrons are situated in orbitals perpendicular to the pi-system, in non-bonding orbitals (which makes sense since the electrons come from a sigma bond). Therefore the electrons left over are not delocalised, which is why such a strong base is required. This also ... 1 I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group. 1 Consider the following assumptions to somehow force these substrates to proceed for$\ce{S_N1}$reaction mechanism: Low temperature Polar protic solvent Low concentration of nucleophile w.r.t. reactant Now, in an$\ce{S_N1}$reaction, the formation of carbocation is must. The primary factor that determines the reactivity of organic substrates in an$\ce{...

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