# Tag Info

28

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 ...

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

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 ...

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

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.

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

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 $\ce{... 1 I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group. 1 Consider the following assumptions to somehow force these substrates to proceed for$\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{...

1

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, ...

1

If hetro atom with lone pair is neighboring to carbocation, then that lone pair can be donated to empty orbital on carbocation (see figure below). These orbitals are said to be in conjugation. Conjugation leads to delocalization of electrons resulting in resonance structures. Quoting from "Organic Chemistry" by T.W. GRAHAM SOLOMONS , CRAIG B. FRYHLE .SCOTT ...

1

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-...

1

Those carbocations are termed as 'stable' as the O/N atom establishes a '2p-2p back bond' with the empty p orbital on the carbon atom carrying the positive charge. This helps the carbocation stabilize as the O/N atom shares its electron density to the starving cation. This effect is FAR more dominating than the pulling inductive effect of the O/N atom. C=...

1

The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize. However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, ...

1

The result is correct but the explanation is bad. The main reason is that when you compare CH3COOH and CH3ClCOOH , the main difference is that the alpha carbon in the former has more electron density than the latter one because of which when the conjugate base gets formed, it is more stabilised than ethanoic acid because of low electron density. You are ...

1

Yes, I once thought about this interesting phenomenon also. There is no good reason why hyperconjugative interactions cannot take place between the $\ce {N}$ lone pair orbital and the $\ce {C-H}$ $\sigma$. However, the interaction may be less favoured if you consider the geometry, compared to the the usual hyperconjugation to alkene $\pi$ systems or to ...

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