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How do I change the old-style numerals in the Georgia font?

I am planning to use Georgia Font in Xelatex for a formal report.

The font looks great except the numbers appears informal to me.

Informal here is some numbers like 4, and 5 are below the line. How can I fix this ?

• See en.wikipedia.org/wiki/Text_figures for understanding of the "informal" appaerance of the numbers. – Kyle_the_hacker Jun 24 at 14:41
• Note that most people will consider lining figures in running text to be more informal than text figures. Many common fonts still default to lining figures, which means only properly, often professionally, typeset text uses text figures. What everyone uses by default in their Word documents is of course the informal variety; what you see in professionally typeset books is considered more formal. I would advise against ‘fixing’ this, because it doesn’t need fixing – at least not in running text. Tables and titles are a different matter. – Janus Bahs Jacquet Jun 24 at 23:40

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# Which Stack Exchange sites use MathJax?

As the title says, I could use an exhaustive list of all Stack Exchange sites that use MathJax. If the list included any site-specific customizations made to the MathJax configuration on each site, that would be even better.

-

These sites support MathJax:

2. Astronomy
3. Aviation
4. Biology (with mhchem extension)
5. Chemistry (with mhchem extension)
6. Code Review (activated 2014-04-07, with \$ delimiters) 7. Cognitive Sciences 8. Computational Science 9. Computer Science 10. Cross Validated 11. Cryptography 12. Data Science 13. Earth Science (with mhchem extension) 14. Economics 15. Electrical Engineering (with \$ delimiters)
16. Engineering
17. History of Science and Mathematics
18. Mathematica
19. Mathematics
20. Mathematics Educators
21. MathOverflow
22. Physics
23. Puzzling
24. Quantitative Finance
25. Robotics
26. Signal Processing
27. Space Exploration
28. Theoretical Computer Science
29. Worldbuilding

This list was complete as of 2014-01-17 and has been updated by the community since then. If you notice that this list has become outdated, please update it. Thanks!

Run the Stack Snippet to get the live list based on /sites from the Stack API:

$.get('https://api.stackexchange.com/2.2/sites?pagesize=500&filter=!*L6SiaRiUSk*Z2zr', function(data) { var i,$result = $('#result'), sites = []; for(i=0; i < data.items.length; i++) { if (data.items[i].site_type === 'main_site' && data.items[i].markdown_extensions && data.items[i].markdown_extensions.indexOf('MathJax')>-1) { sites.push(data.items[i].name); } } sites.sort(); for(i = 0; i < sites.length; i = i + 1) {$result.append(
\$('

').html(sites[i])
);
}
});

-
As of the time I'm posting this comment, this is the complete set. – Kasra Rahjerdi Jan 17 '14 at 20:12
Just as a remark: It is intentionally deactivated on TeX.SE, cf. Can we turn off math-tex please?, and to clear up a common misunderstanding: It is also off-topic there to ask MathJax related questions. – Speravir Jan 18 '14 at 5:10
@Speravir: apparently, the topicality of MathJax questions on TeX.SE is "marginal". – Ilmari Karonen Jan 19 '14 at 20:14
That is sad, I want to use MathJax at superuser. – Pedro77 Sep 15 at 13:10


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Activity Series

## Illustrates how to predict the outcome of reactions

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Activity Series

Credit: Sodium: User:Ajhalls/Wikimedia Commons; Silver: User:Daderot/Wikimedia Commons
Source: Sodium: http://commons.wikimedia.org/wiki/File:Large_Sodium_Explosion.jpg; Silver: http://commons.wikimedia.org/wiki/File:Camp_cup_and_tumbler,_1795-1800,_Paul_Revere_silver_collection,_Worcester_Art_Museum_-_IMG_7624.JPG

#### What’s the difference between the two pictures above?

We see above two metals that can be exposed to water. The picture on the left is of sodium, which gives a violent reaction when it comes in contact with water. The picture on the right is of silver, a metal so unreactive with water that it can be made into drinking vessels. Both metals have a single s electron in their outer shell, so you would predict similar reactivities. However, we have a better tool that allows us to make better prediction about what will react with what.

### The Activity Series

Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivities. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series. Table below is an activity series of most common metals and of the halogens.

 Activity of Metals Activity of Halogens Li K       React with cold water, replacing Ba     hydrogen. Sr Ca Na F2 Cl2 Br2 I2 Mg Al      React with steam, but not cold Zn     water, replacing hydrogen. Cr Fe Cd Co Ni     Do not react with water. React Sn    with acids, replacing hydrogen. Pb H2 Cu Hg    Unreactive with water or acids. Ag Pt Au

For a single-replacement reaction, a given element is capable of replacing an element that is below it in the activity series. This can be used to predict if a reaction will occur. Suppose that small pieces of the metal nickel were placed into two separate aqueous solutions: one of iron(III) nitrate and one of lead(II) nitrate. Looking at the activity series, we see that nickel is below iron, but above lead. Therefore, the nickel metal will be capable of replacing the lead in a reaction, but will not be capable of replacing iron.

\begin{align*}& \text{Ni}(s)+\text{Pb(NO}_3)_2(aq) \rightarrow \text{Ni(NO}_3)_2(aq)+\text{Pb}(s)\\ & \text{Ni}(s)+\text{Fe(NO}_3)_3(aq) \rightarrow \text{NR (no reaction)}\end{align*}

In the descriptions that accompany the activity series of metals, a given metal is also capable of undergoing the reactions described below that section. For example, lithium will react with cold water, replacing hydrogen. It will also react with steam and with acids, since that requires a lower degree of reactivity.

#### Sample Problem: Single-Replacement Reactions

Use the activity series to predict if the following reactions will occur. If not, write NR. If the reaction does occur, write the products of the reaction and balance the equation.

A. \begin{align*}\text{Al}(s)+\text{Zn(NO}_3)_2 (aq) \rightarrow\end{align*}

B. \begin{align*}\text{Ag}(s)+\text{HCl}(aq) \rightarrow\end{align*}

Step 1: Plan the problem.

For A, compare the placements of aluminum and zinc on the activity series. For B, compare the placements of silver and hydrogen.

Step 2: Solve.

Since aluminum is above zinc, it is capable of replacing it and a reaction will occur. The products of the reaction will be aqueous aluminum nitrate and solid zinc. Take care to write the correct formulas for the products before balancing the equation. Aluminum adopts a 3+ charge in an ionic compound, so the formula for aluminum nitrate is Al(NO3)3. The balanced equation is:

\begin{align*}2\text{Al}(s)+3\text{Zn(NO}_3)_2(aq) \rightarrow 2\text{Al(NO}_3)_3(aq)+3\text{Zn}(s)\end{align*}

Since silver is below hydrogen, it is not capable of replacing hydrogen in a reaction with an acid.

\begin{align*}\text{Ag}(s)+\text{HCl}(aq) \rightarrow \text{NR}\end{align*}

### Summary

• Metals and halogens are ranked according to their ability to displacement other metals or halogens below them in the series.

### Review

1. What does the activity series tell us?
2. Can a metal undergo any of the reactions listed below it in the series?
3. List two metals that cobalt will displace and two that will displace it.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes


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## Introduction

Chemo- and regioselective modification of proteins is a chemical challenge because polypeptide chains contain numerous similarly reactive functional groups1, 2. Often the polymer, fluorophore, or small-molecule drug is rendered electrophilic for conjugation chemistry at nucleophilic protein side-chains. Many undesired side reactions occur and purification often fails. Major advances to overcome these chemistry challenges have been made, including the use of non-natural clickable amino acids within proteins for bio-orthogonal chemistry3, 4, engineered enzymatic systems5 such as sortase A6, and non-enzymatic routes that leverage selective chemistry between a matched reaction pair7,8,9,10,11 such as the reaction of a N-terminal cysteine with a thioester11.

Recently we reported an enzyme-free, one-step, regio- and chemoselective protein-modification method based on cysteine arylation12. We discovered that the π-clamp tetrapeptide (Phe-Cys-Pro-Phe) possessed unique reactivity to achieve self-labeling with a perfluoroaryl (PFA) electrophile (Fig. 1a). The reaction even proceeded selectively in complex glycosylated IgG proteins. Each residue within the π-clamp is important as mutations often diminished the reaction efficiency. Further, the π-clamp-mediated arylation reaction rate is tunable by over four orders of magnitude with different salts13. Ammonium sulfate and other structure-stabilizing salts accelerated the reaction, while the addition of a denaturing salt such as guanidinium chloride impeded the reaction. Taken together, these results prompted us to carry out a systematic investigation of the π-clamp-promoted reaction. Here we employ kinetic studies, solution and solid-state NMR, and density functional theory (DFT) calculations to determine the structure-reactivity relationships that underscore this π-clamp mediated reaction. Our findings suggest several structural and chemical features contribute to the unique reactivity of π-clamp including the trans prolyl amide bond, lowered reaction ΔH, reduced cysteine pK a and side chain-perfluoroaryl electrophile interactions.

## Results

### Trans prolyl amide accelerates π-clamp-mediated arylation reaction

We first investigated if the prolyl amide conformation being either cis or trans in the π-clamp was important. Proline is conformationally unique among amino acids as the backbone phi (ϕ) torsion angle is restricted by the side chain ring structure. The Xaa-Pro amide bond can undergo cis/trans isomerization while other amide bonds are found in the trans conformation for most proteins14, 15. Pro is thought to be a key player in the π-clamp, as mutation to D-Pro abrogates the reaction rate12. To understand if the conformation of the Cys-Pro amide bond perturbs the reaction we used mutation studies. Proline analogues containing certain functional groups can constrain the Cys-Pro conformation; analogues α-methylproline (αMePro)16, 17 and 5,5-dimethylproline (5,5-dmP)18, 19 strongly promote the trans and cis conformations, respectively. Three peptides incorporating proline (the π-clamp, peptide 1A), α-methylproline (peptide 1B) and 5,5-dimethylproline (peptide 1C) at residue 3 were synthesized. The rate constants of their reactions with the PFA probe 2 were measured (Figs S1S3). We found the α-methylproline mutation (1B) increased the rate constant by a factor of 2.5 relative to π-clamp 1A, whereas the 5,5-dimethylproline mutation (1C) led to a significant loss of reactivity (Fig. 1b), indicating that trans Pro promoted the π-clamp-mediated conjugation.

Computational tools were employed to calculate the Gibbs free energy change (∆G) of π-clamp-mediated arylation reaction, to further study whether a trans proline analogue provided any energetic advantage. We previously investigated the reaction energy pathway for cis Pro π-clamp12. To calculate ∆G of the arylation reaction between peptide 1B and PFA probe using density functional theory (DFT), we first extracted three snapshots from molecular dynamics (MD) simulations of the trans proline π-clamp and then manually added a methyl group to form αMePro. Similarly, for starting structures of peptide 1 C, we extracted four snapshots from MD simulations of the cis proline π-clamp and manually added two methyl groups to form 5,5-dmP. For the product’s starting structure, we manually connected the PFA group to the peptide cysteine. After performing geometry optimizations on the starting structures of the peptide and product (Fig. S4), ∆G of the arylation reaction was calculated using DFT (see Section 3 in supporting information for details). We found that trans proline provided a thermodynamic advantage (lower Gibbs free energy of reaction) over cis proline by 11.8 kJ/mol for π-clamp-mediated arylation (Fig. 1c and Table S3). Taken together, both mutation and computational studies show a trans prolyl amide bond is preferred in the π-clamp-mediated arylation reaction.

### Enthalpy of activation is decreased in π-clamp-mediated arylation

The reactivity of π-clamp cysteine was higher than that of other cysteines, leading to the site-selectivity of π-clamp. Compared to the reaction between PFA probe 2 and the π-clamp double-glycine mutant peptide 3 (GCPG, secondary rate constant k = 0.00065 ± 0.00005 M−1 s−1, Fig. S5), the rate constant for π-clamp peptide 1 A (0.63 ± 0.02 M−1s−1) was enhanced 1000-fold. Ammonium sulfate further accelerated π-clamp mediated arylation without affecting the regioselectivity13.

We investigated whether and how the π-clamp cysteine arylation reaction activation parameters were different from other cysteine using Eyring equation. We chose to compare π-clamp peptide 1 A, double glycine mutant 3, π-clamp proline analogue mutants 1B1 C and π-clamp mutants with phenylalanine mutated to leucine (1D), cyclohexylalanine (1E) and pyrenylalanine (1 F). PFA probe 2 was reacted with these peptides and the reaction rate constants (k) were measured at different temperatures (T) (Table S4, Figs S6S36) to calculate the enthalpy of activation (ΔH), the entropy of activation (ΔS), and the Gibbs energy of activation (ΔG) (see Methods Section for details). As summarized in Table S5, the ∆H for π-clamp and mutants (1A1 F) were lowered by 37–84 kJ/mol compared to that of the double glycine mutant 3, contributing to the significantly enhanced reaction rate. Similarly, to shed light on whether ammonium sulfate changed activation parameters to accelerate the π-clamp-mediated arylation, the rate constant for the reaction between 1 A and 2 in the presence of two molar (2 M) ammonium sulfate was measured at different temperatures (Figs S38S42). We found both ΔH and ΔS were changed by ammonium sulfate (Table 1), but the change of ΔH is more significant and is responsible for the reaction acceleration.

### π-clamp cysteine pKa is reduced even for Phe mutants

We have shown that the cysteine pK a of π-clamp is lowered by 0.6 pH unit compared to that of the double glycine mutant12. To test whether the cysteine pK a values in π-clamp mutants are also perturbed and how important the cysteine pK a value is for enhancing its reactivity, we measured the cysteine pK a in 1D, 1E, 1 F and 3.

Several methods have been developed to measure the cysteine pK a in biomolecules. Spectrophotometric titration20,21,22, where UV absorbance of the cysteine containing molecule is plotted against different pH values, is a convenient method. This method is based on the fact that ionization of cysteine thiol to thiolate results in a large increase in its molar absorption coefficient at 240 nm20. To calculate the cysteine pK a of 1D, 1E and 3, we measured the absorbance at 240 nm (A240) of the peptides dissolved in buffers at different pHs. The absorbance of the buffer was used as a blank. Then the pK a was determined (Fig. 2a–c) by fitting the UV absorbance and pH values with the following equation:

$$y=\frac{A\times {10}^{pH}+B\times {10}^{p{K}_{a}}}{{10}^{p{K}_{a}}+{10}^{pH}}$$

where y is the absorbance of the peptide at 240 nm, A is the upper plateau of absorbance at high pH, and B is the lower plateau at low pH. The cysteine pK a in the double glycine mutant 3 was measured to be 7.95, while the cysteine pK a of both 1D and 1E were 0.6 pH unit lower than that of 3.

The A240 method could not be applied to 1 F because of the strong absorbance of the pyrenyl group at 240 nm. Instead we measured pK a based on the reactivity of cysteine at different pHs. The pK a values of cysteine in proteins have been measured by determining their reaction rates with a reagent that is more reactive with thiolate than thiol22, 23. Here we used the rate constant of the SNAr reaction between 1 F and 2 to fit for pK a, since thiolate is more reactive than thiol in the perfluoroarylation reaction. The rate constants were plotted against pH and fit with the same equation as in the A240 method, where y is the reaction rate constant, and A and B are the upper and lower plateaus for the reaction rates (Fig. 2d). The cysteine pK a of 1 F was determined to be 0.4 pH unit lower than 3. In Fig. 2e, we summarize the cysteine pK a values and the rate constants for the peptides studied. We observed a more than 10-fold rate constant difference between 1D and 1E, even though their pK a values were similar. In addition, peptide 1 F was more reactive than peptides 1D and 1E, but the cysteine pK a of 1 F was higher. Although all three π-clamp mutants had a lower cysteine pKa compared to that of the double glycine mutant 3 which facilitated ionization of the thiol to the more reactive thiolate, this pK a effect was not the only factor to account for the enhanced reactivity of the π-clamp.

### ssNMR indicates π-clamp adopts 1:1 cis/trans Pro3 while the perfluoroarylated product adopts only trans

To characterize the backbone conformation of the π-clamp and the S-perfluoroarylated product (PFA-tagged π-clamp), 1D and 2D 13C magic-angle-spinning (MAS) NMR spectra were measured on π-clamp peptide 5 (Fig. S43) and π-clamp PFA-tagged product 6 (Fig. S44), in which the π-clamp motif Phe-Cys-Pro-Phe were uniformly 13C and 15N labeled. The 13C cross-polarization (CP) spectrum of peptide 6 (Fig. 3a) shows sharp resonances, but with more peaks than expected for four labeled residues, indicating the presence of multiple resolvable peptide conformations. The Pro3 conformational heterogeneity is manifested as two resolved Cβ peaks around 30 ppm, and two partially resolved Cδ peaks around 48 ppm. In the carbonyl region, structural polymorphism is also readily observed, especially among the multiple Cys2 CO peaks, ~169 ppm, which are resolved from the carbonyl signals of the other labeled residues. The 15N CP spectrum (Fig. 3b) further highlights this conformational polymorphism: the N-terminal Phe1 amino group shows four resolved 15N signals at 22–35 ppm, indicating four distinct conformations. Likewise, Pro3 exhibits four resolved 15N peaks at 130–135 ppm.

To further resolve and assign the 13C and 15N chemical shifts, we measured 2D 13C-13C correlation spectra (Fig. 4) of peptides 5 and 6, and 2D 15N-13C correlation spectra (Fig. S45) of peptide 6. The 2D 13C-13C correlation spectrum of peptide 5 (Fig. 4a) shows larger 13C line widths than the PFA-tagged peptide 6 (Fig. 4b) due to the conformational distribution and the dry state of the peptide. Measurement of the dried, unreacted peptide 5 by solid-state NMR was necessary due to the unfavorable dynamics of this very soluble peptide in a hydrated state. The aliphatic linewidths of peptide 5 range from 2.0 ppm to 3.0 ppm, while the PFA-tagged peptide 6 exhibits 13C linewidths of 0.7–1.4 ppm. Within this linewidth range, peptide 5 shows a single set of chemical shifts for each residue, with the exception of Pro3, which exhibits chemical shifts for both cis and trans isomers (vide infra). The Phe1 and Phe4 chemical shifts are not well resolved: their Cα-Cβ cross peaks partially overlap, and a single set of Cβ cross peaks with the aromatic side chain carbons are observed (Fig. 4a), indicating that the two Phe residues have similar conformations. Cys2 displays a Cβ chemical shift of ~25 ppm (Fig. 4a and c), which corresponds to the reduced state, confirming that the unreacted peptide 5 does not form disulfide bonds.

Upon reaction of peptide 5 with the PFA probe to form product 6, all 13C and 15N peaks sharpen (Figs 4b,d and S45), allowing us to resolve four sets of chemical shifts for each residue. In peptide 6, the inter-residue Cys2-Pro3 cross peaks allow the multiple sets of chemical shifts to be separated into four different molecules. No cross peaks between the different forms are observed at a 13C spin diffusion mixing time of 50 ms. Phe1 and Phe4 exhibit resolved Cα and Cβ chemical shifts that differ by ~5 ppm and ~3 ppm respectively (Fig. 4b). Multiple sets of aromatic Cγ, Cδ, Cε and Cζ chemical shifts are also resolved for Phe1 and Phe4, indicating that Phe1 and Phe4 experience different chemical environments, likely due to the PFA tag. Cys2 shows a Cβ chemical shift of ~40 ppm, characteristic of oxidized Cys2, as expected for PFA-modified cysteine.

We were especially interested in the prolyl amide bond conformation. Pro3 shows interesting chemical shift differences between the peptide 5 and the PFA-tagged peptide 6. In the unreacted state (peptide 5), two sets of Pro3 Cβ and Cγ chemical shifts are observed: one set shows a large Cβ and Cγ chemical shift difference of ~13 ppm while the other set shows a smaller chemical shift difference of ~5 ppm (Fig. 4c). These Cβ and Cγ chemical shift differences24 are characteristic of the cis and trans isomers of Pro, and peak intensities indicate that the two forms have a cis: trans ratio of 1:1 in the unreacted peptide 5. Upon reaction with the PFA probe, the multiple resolved Pro3 chemical shifts all show similar Cβ and Cγ chemical shift differences of ~5 ppm, indicating that the Pro3 in peptide 6 predominantly shifts to the trans conformer (Fig. 4d).

Overall, these 13C and 15N chemical shifts (Tables S6, S7) indicate there is no single well-defined secondary structure for the π-clamp motif in peptide 5 or 6. The unreacted π-clamp peptide 5 adopts a broadly distributed conformation with an approximate 1:1 ratio of the cis and trans Pro3 isomer. Reacting with the PFA probe causes peptide 6 to adopt four distinct conformations, but only the trans isomer at Pro3. Using the compiled 13C and 15N chemical shifts of peptide 6, a TALOSN25 prediction yielded similar backbone (ϕ,ψ) torsion angles for the four forms (Fig. S46), with the largest torsion angle difference occurring at Phe4. This difference is likely from prediction uncertainty due to the lack of chemical shift information of the residue C-terminal to Phe4. For the unreacted peptide 5, because of the large spectra linewidths, there is significant uncertainty in predicting these (ϕ,ψ) torsion angles. However, since all four distinct sets of peptide 6 chemical shifts fall within the broad linewidths of the unreacted peptide 5, the average backbone conformations may not differ dramatically between the two peptides.

### A clamp-like claw structure is one of the two reaction products from ssNMR and modeling studies

To generate a structural model for the perfluoroarylated π-clamp product based on the backbone conformation predicted from chemical shifts, 13C-19F distances were measured using 13C-19F REDOR of two partially labeled π-clamp peptides: peptide 7 with uniform 13C, 15N labeling at only Phe1 (Fig. S47) and peptide 8 with uniform 13C, 15N labeling at Pro3 and Phe4 (Fig. S48). By examining these two peptides, we can determine the distance between the PFA and each Phe. Figure 5 shows representative REDOR S0 and S spectra at a mixing time of 4 ms. 13C spins that are close to the 19F spins exhibit larger intensity differences between the S0 and S spectra than 13C spins that are farther from 19F spins. Figure 5d plots the intensity ratio S/S0 as a function of the REDOR mixing time for a few representative peaks; the full set of REDOR curves used for modelling are summarized in Fig. S49. The Phe1 backbone Cα peak has the largest dephasing among all peaks, reaching a value of 0.07 ± 0.02 at 6 ms, while the Phe1 Cζ shows a more moderate dephasing trend. The Pro3 Cδ and Phe4 Cζ show linear decay in contrast to the expected universal two-spin REDOR curve. We attribute these decay trajectories to the multi-spin nature of the system, with 8 19F spins at varying distances from each 13C.

Given the system complexity, we first analyzed these 13C-19F REDOR data semi-quantitatively. In contrast with the Phe1 backbone Cα peak which has the largest degree of dephasing, the Phe4 Cα and Cβ backbone sites show minimal dephasing, reaching only S/S0 values of ~0.8 at 8 ms. This indicates that the PFA tag is closer to Phe1 than Phe4. Among the 13C signals from Pro3, Cδ reaches the lowest dephasing at a S/S0 value of ~0.4 at 8 ms, while Pro3 Cα, Cβ and Cγ each show minimum S/S0 of ~0.6 at 8 ms, indicating Pro3 Cδ is closer to the PFA tag.

More quantitative distance information was extracted by multi-spin simulations utilizing the SIMPSON program26. We compared the experimental data first to 2-spin simulations to obtain the apparent 13C-19F distance, and then to 5-spin system containing one 13C and four 19F spins in model-dependent simulations. In these simulations, the ψ and χ1 angle of Phe1, χ1, χ2 and χ3 of Cys2 and χ1 of Phe4 were iteratively set to 180°, 60° or −60°. For each simulated curve, the root-mean-square deviation (RMSD) was calculated between each experimental REDOR data set, and the minimum among all calculated RMSD values corresponds to the best fit. The 2-spin simulations provided reasonable fits for each 13C site (Fig. S50). For all sites in Pro3, the distances range from 6.3–7.0 Å. Slightly longer distances for Phe4 Cα and Cβ (7.7 Å and 7.4 Å respectively) are observed, and the distance for Phe4 Cζ is 6.0 Å. For Phe1, the fast dephasing Cα and Cζ fit relatively well to distances of 3.8 Å and 5.2 Å, respectively. The 5-spin model-dependent simulations unfortunately did not simultaneously fit well for all 13C sites with any set of dihedral angle combinations. However, taking the best-fit 5-spin REDOR curve for each 13C site (Figs 6b, S51), the shortest 13C-19F distance among the four nearest neighbor 19F agreed well with the best-fit distances determined from the 2-spin simulations.

These distances were taken as a set of REDOR derived, nearest-neighbor 13C-19F distances: 3.6 Å F1α, 4.6 Å F1β, 6.0 Å F1α’, 6.0 Å F1β’, 4.8 Å F1ζ, 7.0 Å P3α, 7.5 Å P3β, 7.7 Å P3γ, 6.4 Å P3δ, 8.1 Å F4α, 7.8 Å F4β, and 6.5 Å F4ζ. Using a smaller interval of 30°, the dihedral angles were rotated iteratively again and the nearest neighbor 13C-19F distance was compared to the above set of distances. In the generated best-fit models, shown in Fig. 6c, model 1 (left) uses F1 Cα and Cβ, and model 2 (right, wide clamp-like) uses F1 Cα’ and Cβ’ for fitting. The 13C-19F distances from these models were used in 5-spin REDOR simulations, and plotted against the experimental REDOR S/S0 values (Figs S52, S53) for validation. Model 2 gave decent fits for all 13C sites while model 1 has worse fits, even though the shortest 13C-19F distances are similar in the two models. In both models, distinct non-covalent interactions between Phe1 side chain and perfluoroaryl electrophile at around 4.7 Å distance and similar backbone structures were observed (Fig. 6c and d).

### A pyrenylalanine π-clamp mutant non-covalently interacts with the perfluoroaryl electrophile in solution

We hypothesized the phenyalanine side chains in π-clamp interact non-covalently with the PFA electrophile to facilitate the reaction. 19F solution NMR and isothermal titration calorimetry experiments were carried out to test this hypothesis. We started with an un-reactive π-clamp mutant (Phe-Ser-Pro-Phe), but were unable to detect interactions by NMR experiments. Prompted by our earlier experiments, we investigated a pyrenylalanine mutant which may interact with the PFA electrophile more strongly. Peptide 4 A (Py-Ser-Pro-Py, where Py is pyrenylalanine) and the control peptide double-glycine mutant 4B (Gly-Ser-Pro-Gly) were prepared. We mixed 4 A with 1 equivalent of probe 2 and observed a change in 19F solution NMR chemical shift and peak shape when compared to the mixture of 4B and probe 2, or probe 2 alone (Fig. 7a). In addition, the 19F chemical shift was changed during titration when we added 4 A to a fixed amount of 2 (Fig. 7b). In contrast, this effect was not observed for the same titration of peptide 4B (Fig. 7c). These results suggest the pyrenyl side chains are interacting with the PFA electrophile. To corroborate the 19F solution NMR studies, we investigated the interaction of 4 A and 2 by isothermal titration calorimetry. Probe 2 was added to peptide 4 A or 4B to detect non-covalent interaction (Fig. 7d). When compared to the non-pyrenylalanine case 4B as a control, weak binding was observed between 4 A and 2. The binding curve (Fig. 7e) did not display simple two state interactions and could not be fit nor could we determine the binding stoichiometry.

## Discussion

The π-clamp accelerates abiotic perfluoroarylation chemistry in water. Here we carried out structural and mechanistic studies to understand this transformation. Several structural and chemical features were found to facilitate the reaction. We found that Pro3 played an important role in the π-clamp function. A π-clamp α-methylproline mutant favoring the trans Pro conformation accelerated the rate of arylation by more than 15-fold, compared to the 5,5-dimethylproline mutant that favored a cis Pro conformation. Trans Pro provided a thermodynamic advantage over cis Pro for π-clamp-mediated arylation as shown by DFT calculation. The ΔH for the π-clamp arylation reaction is decreased by 64 kJ/mol when compared to a non-π-clamp cysteine, enabling 1000-fold rate enhancement and regioselective cysteine modification at the π-clamp. Ammonium sulfate salt lowers the π-clamp arylation reaction ΔH by 40 kJ/mol relative to the reaction without the salt and thus promote the reaction. In addition, the π-clamp cysteine pK a is reduced by 0.4–0.6 pH unit even for Phe mutants.

Structural and biophysical studies indicate certain conformations may promote the reaction, and side-chains in the π-clamp interact non-covalently with the PFA electrophile. Solid-state NMR indicated a mixture of cis/trans Pro in the unreacted π-clamp was fully converted to the trans conformation upon perfluoroarylation. Two ssNMR structural models of the reaction product highlight the interactions between the Phe1 side chain and perfluoroaryl electrophile moiety. 19F solution NMR titration experiments and the ITC data support the hypothesis that the π-clamp side chains may interact with the PFA electrophile.

The structural models from our ssNMR study are not fully consistent with the π-clamp α-methylproline mutant product structures used in DFT calculations, although they both have trans Pro. As the comparison shown in Fig. S54, the largest torsion angles differences between the calculated structures and clamp-like NMR structure model 2 occurred at Phe1-psi and Pro3-psi. These two angles contribute to the calculated structures being more compact than the NMR structure. In addition, all of the calculated structures have Phe4 in close proximity to the PFA tag. This would theoretically cause the Phe4 Cα and Cβ simulated REDOR curves to decay rapidly, while the experimental data shows a much slower decay (Fig. S54). The inconsistency may come from the fact that we did not directly sample the PFA-labeled π-clamp product in MD simulations, due to a lack of parameters to describe the PFA in the reaction product. In addition, two slightly different molecules have been used in our computation and ssNMR studies. We think that the ssNMR models better describe the structural feature of the PFA labeled π-clamp product. However, our DFT calculations which indicate that trans Pro is thermodynamically favored over cis proline in the π-clamp perfluoroarylation reaction should be valid, considering that for all the geometrically-optimized calculated structures we sampled, larger or similar reaction ∆G were observed for trans Pro compared to the cis Pro structures.

Our studies here have led to the discovery of a π-clamp mutant 1 G with an 85-fold rate enhancement. A convergent effect was observed when we combined α-methylproline, which promoted the trans Pro conformation, and the large hydrophobic side chain in pyrenylalanine. Taken together, we generated π-clamp mutant 1 G with a rate constant of 53.3 ± 2.3 M−1 s−1, which is 85-fold higher π-clamp peptide 1 A (Figs S55, S56). We anticipate our findings could direct us toward the design and screening of other small reactive peptides to facilitate selective abiotic chemistry enabled by local environment.

## Methods

### Kinetics Study

The reactions were carried out with 200 mM phosphate, 20 mM TCEP at 37 °C unless otherwise noted. To measure the second order rate constants, reaction mixture was prepared on ice and divided into several 10-µL aliquots. All aliquots were immediately put in 37 °C water bath unless otherwise noted. For reactions that takes more than 1 hour to monitor, all aliquots were heated in a PCR machine set at 37 °C to prevent solvent evaporation. Reactions were quenched by addition of 100 µL 50% water: 50% acetonitrile: 0.5% TFA at different time points and then subjected to LC-MS analysis. The initial concentration of probe and substrate were known. The second-order rate constants were determined by fitting the following kinetics equation:

$$y=\frac{ln\frac{{[peptide]}_{0}{[probe]}_{t}}{{[peptide]}_{t}{[probe]}_{0}}}{{[probe]}_{0}-{[peptide]}_{0}}={k}_{2}t$$

Error of reaction rate constant was obtained from the linear fitting of the kinetics curves for measuring the reaction rate constants.

### Determination of the standard enthalpy/entropy of activation

The secondary rate constant (k) for the reaction between π-clamp peptide and probe 2 was experimentally measured at different temperatures (T). Then ln(k/T) was plotted against 1/T. The standard enthalpy of activation (ΔH) and the standard entropy of activation (ΔS) were calculated (Table S5) by fitting ln(k/T) against 1/T with the following Eyring equation:

$$ln\frac{k}{T}=ln\frac{\kappa {k}_{B}}{h}+\frac{{\rm{\Delta }}{S}^{\ddagger}}{R}+\frac{-{\rm{\Delta }}{H}^{\ddagger}}{R}\frac{1}{T}$$

where κ is transmission coefficient (κ = 1), k B is Boltzmann constant, h is Planck’s constant, T is absolute temperature and R is gas constant. The errors for ΔH and ΔS were obtained from the linear fitting. ∆G was calculated as ΔH−TΔS, and the error for ∆G was calculated from error propagation. The Eyring plots were summarized in Fig. S37.


text:
# Frequently Asked Questions (FAQ)

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• ### Is it possible to import the effective lengths from RF-STABILITY or RSBUCK in RF-/TIMBER Pro?

New FAQ 004113 EN-US

#### Answer

Yes, that is possible.

First, RF-STABILITY (or RSBUCK in RSTAB 8) can be used to determine the effective lengths for a particular structure and loading.

They can then be imported in the 'Effective Lengths' of the RF-/TIMBER Pro dialog box.

• ### Which calculation methods are used by RSTAB and RSBUCK?

FAQ 003564 EN-US

#### Answer

RSTAB is a FEM program that uses trigonometric trial functions for the members. For this reason, members do not have to be subdivided for sufficiently accurate results and the calculation speed is correspondingly higher.

RSBUCK determines the eigenvalues of the stiffness matrix and can thus linearly calculate the critical load and buckling mode of the structure.

• ### What is the critical load factor and how is it possible to determine it?

FAQ 003561 EN-US

#### Answer

The critical load factor specifies the factor by which you can increase a load until the system fails. If it is smaller than one, a calculation according to the second-order analysis is usually unstable because the system is already stressed by the critical load. This factor is also taken into account in standardization. For example, Eurocode 3 specifies that a calculation according to the second-order analysis is no longer necessary from a critical load factor of 10.
The critical load factor can be determined by the RSBUCK module or RF-STABILITY.
• ### How do I determine the effective lengths of frame columns in RFEM or RSTAB?

FAQ 003538 EN-US

#### Answer

The easiest way to do this is to use the add-on modules RSBUCK (RSTAB) or RF-STABILITY (RFEM).

RSBUCK and RF-STABILITY perform an eigenvalue analysis for the entire model with a certain state of normal force. The axial forces are increased iteratively until the critical load case is reached. This stability load is characterized in the numerical calculation by the determinant of the stiffness matrix becoming zero.

If the critical load factor is known, the buckling load and the buckling curve are determined from this. The effective lengths and effective length factors are then determined for this lowest buckling load.

The result shows, depending on the required number of eigenvalues, the critical load factors with the corresponding buckling curves and for each member - according to its mode shape - effective length about the strong and the minor axis.

Since usually, every load case has a different normal force state in the elements, a separate corresponding effective length result for the frame column arise for each load situation. The effective length whose buckling mode causes the column to buckle in the corresponding plane is the correct length for designing the respective load situation.

Since this result may be different for each analysis due to the different load situations, the longest effective length of all calculated analyzes - equal for all load situations - is assumed for designing on the safe side.

###### Example for manual calculation and RSBUCK/RF-STABILITY
There is a 2D frame with a width of 12 m, a height of 7.5 m and pinned supports. The column cross-sections correspond to I240 and the frame beam to IPE 270. The columns are loaded with two different concentrated loads.

l = 12 m
h = 7.5 m
E = 21000 kN/cm²
Iy,R = 5790 cm4
Iy,S = 4250 cm4

NL = 75 kN
NR = 50 kN

$EI_R=E\ast Iy_R=12159\;kNm^2$
$EI_S=E\ast Iy_S=8925\;kNm^2$

$\nu=\frac2{{\displaystyle\frac{l\ast EI_S}{h\ast EI_R}}+2}=0.63$

This results in the following critical load factor:

$\eta_{Ki}=\frac{6\ast\nu}{(0.216\ast\nu^2+1)\ast(N_L+N_R)}\ast\frac{EI_S}{h^2}=4.4194$

The effective lengths of the frame columns can be determined as follows:

$sk_L=\pi\ast\sqrt{\frac{EI_S}{\eta_{Ki}\ast N_L}}=16.302\;m$

$sk_R=\pi\ast\sqrt{\frac{EI_S}{\eta_{Ki}\ast N_R}}=19.966\;m$

The results from the manual calculation correspond very well with those from RSBUCK or RF-STABILITY.

###### RSBUCK
$\eta_{Ki}=4.408$
$sk_L=16.322\;m$
$sk_R=19.991\;m$

###### RF-STABILITY
$\eta_{Ki}=4.408$
$sk_L=16.324\;m$
$sk_R=19.993\;m$
• ### Why are the stiffness modifications not taken into account when determining critical load factors in the RF‑STABILITY add-on module?

FAQ 003108 EN-US

#### Answer

The defined stiffness modifications are only considered in the stability analysis in RF‑STABILITY if the "Activate Stiffness Modifications from RFEM" option under the "Options" section in Window "1.1 General Data" is selected.
• ### I have calculated the effective lengths in my model by using RSBUCK/RF‑STABILITY.Which effective lengths are relevant for further analyses?

New FAQ 003007 EN-US

#### Answer

RSBUCK/RF‑STABILITY calculates at least one critical load factor or one critical load and the assigned buckling shape. The effective length is then counted back from the critical load (see here ). Since this analysis is not carried out for the individual local components, but for the entire structure only, the resulting critical load factors refer to the global structure and not to the local elements. However, it may happen that the structure fails globally (and also locally) for some critical load factors (depending on the stiffness and the axial force state).

Therefore, the calculated effective lengths should only be used by the members that buckle in the respective buckling mode. In the case of the global failure of a structure (see the example in Figure 01), it is thus difficult to draw conclusions regarding the buckling behavior of the individual members.

Figure 02 shows a structure where the rear columns are buckling. Therefore, it is recommended to only use the effective lengths calculated for both of these columns.

General summary: The effective lengths from the RSBUCK module are only valid for a structural component in the respective direction if the related buckling shape clearly "bulges" the member in relation to the other in the respective direction. It is clear that the axial forces also have an impact on the results here.

• ### In the stability analysis in RSKNICK, I get the following error message: "The model is not coherent because there are independent submodels." What does that mean and what can I do about it?

FAQ 002857 EN-US

#### Answer

Independent submodels are not interconnected and are considered as separate submodels in the calculation. They are thus independent models without influencing each other (see Figure 2).

It is recommended to edit submodels separately as individual files. Then a stability analysis with RSKNICK is possible.
Otherwise, the partial models must be connected to each other. In this case, it should be taken into consideration that the static systems of the submodels should be retained when the submodels merge into an overall model (see Figure 3).

The feature "Independent Systems" is helpful in detecting partial models. This finds all independent systems and lists them as groups (see Figure 4).
One finds this function under Extras -> Model control -> Independent systems.
• ### Does RSBUCK take into account the stability problem of lateral-torsional or torsional-flexural buckling?

FAQ 002602 EN-US

#### Answer

No, it does not. In the RSBUCK add‑on module, no stability analysis is performed for lateral-torsional or torsional-flexural buckling.

• ### My model is unstable. What can be the cause?

FAQ 002257 EN-US

#### Answer

The calculation can be terminated due to an unstable structural system for various reasons. On the one hand, it may indicate a real instability due to an overload of the system, but on the other hand, modeling errors may be responsible for the error message. Below is a possible way to find the cause of the instability.

First, it should be checked if the modeling of the system is correct. To find modeling problems, use the model controls (menu "Tools" → "Model Check").

Furthermore, you can structure z. For example, it is possible to calculate under pure self-weight in a load case according to the linear static analysis. If results are displayed subsequently, the structure is stable with regard to the modeling. If this is not the case, the most common causes are listed below (see also Video 1):

• Supports are missing or have been defined incorrectly
• Members can rotate about their own axes because a corresponding support is missing
• Members are not connected ("Tools" → "Model Check")
• Nodes are evidently in the same place, but if looked at more closely, they deviate slightly from each other (common cause for CAD Import, "Tools" → "Model Check")
• Member Hinges / Line Hinges Create a "Hinge Chain"
• The structure is not sufficiently stiffened
• Nonlinear structural elements (for example Tension members) fail

Finally, Figure 02 shows an example. It is a pinned frame stiffened by tie rods. Because of post shortenings due to vertical loads, the ties receive small compressive forces in the first calculation run. They are removed from the system (because only tension can be absorbed). In the second calculation run, the model is unstable without these ties. There are several ways to solve this problem: You can apply a prestress (member load) to the tension members to "eliminate" the small compression forces, assign a small stiffness to the members (see Figure 02), or have the members removed one after the other in the calculation (see Figure 02).

To obtain a graphical representation of the cause of instability, the RF-STABILITY (RFEM) add-on module can help. The option "Determine mode shape of unstable model ..." (see Figure 03) allows you to calculate unstable systems. In the graphic, the component that leads to the instability is usually recognizable.

If load cases and load combinations can be calculated according to the linear static analysis, the calculation is only broken when calculating according to the second-order analysis or the second analysis. Order, then there is a stability problem (critical load factor smaller than 1.00). The critical load factor indicates the factor with which the loading must be multiplied so that the model becomes unstable under the corresponding load, for example buckles. It follows from this: A critical load factor smaller than 1.00 means that the structure is unstable. In order to find the "weak point", the following approach is recommended, which is required by the RSBUCK module (for RSTAB) or RF-STABILITY (for RFEM) (see Video 2):

First, the load of the affected load combination should be reduced until the load combination becomes stable. The load factor in the calculation parameters of the load combination is used as an aid (see Video 2). Then, the buckling or buckling shape can be calculated and graphically displayed on the basis of this load combination in the RSBUCK (RSTAB) or RF-STABILITY (RFEM) add-on module. With the graphical output, it is possible to find the "weak point" in the structure and then specifically optimize it.

#### Videos

Video 1-en.wmv (16.52 MB)
Video 2-en.wmv (23.97 MB)
• ### I would like to calculate effective lengths of a frame column in the frame plane using RSBUCK or RF-STABILITY. Which load case LC or load combination CO should I enter?

FAQ 002016 EN-US

Add-on Modules RF-STABILITY RSBUCK 5.xx 8.xx

#### Answer

The modules perform the eigenvalue analysis for the entire model with a certain axial force state. Depending on the number of eigenvalues required, the programs provide results of crictical load factors with the corresponding buckling shapes for an eigenvalue, and effective length about the major and minor axis for each member per mode shape.

Since each load case LC and each load combination CO often has a different axial force state in the elements available, there is a separate respective effective length result for each load situation of the frame column concerned. The effective length, which causes the column in the frame plane buckles sideway in the buckling shape, is the correct length to be used for the analysis of the load situation.

However, this result may be different because of various load situation in each analysis, the longest effective length of all analyses performed applies in the design on the safe side equally for all load situations.

1 - 10 of 14

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#### First Steps

We provide hints and tips to help you get started with the main programs RFEM and RSTAB.

#### Wind Simulation & Wind Load Generation

With the stand-alone program RWIND Simulation, wind flows around simple or complex structures can be simulated by means of a digital wind tunnel.

The generated wind loads acting on these objects can be imported to RFEM or RSTAB.

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text: Some of these classes of compounds have been described; others will be discussed later. What Is The Oxidation Number Of Nitrogen In The Nitrate Ion? O = -2. If you already know the lewis structure of N 2 O, you can skip first section of this tutorial and can go to the lewis structure part directly. Cope elimination of diastereomeric amine oxides, such as those shown in examples #2 & 3 above, provide proof of the syn-relationship of the beta-hydrogen and amine oxide groups. This reaction is known as the Cope Elimination. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Nitrogen cannot exhibit a +5 oxidation state due to the absence of d-orbitals. 0 (zero) represents an s orbital and 1 represents p orbital hence no D orbitals are there. The oxidation number of the nitrogen atom, −III, does not change during the NAD + → NADH conversion. Explain. The spin of the nitroxyl unpaired electron may be studied by a technique called electron paramagnetic resonance (epr or esr). How much money do you start with in monopoly revolution? So it will gain 3 electrons to be stable. This is the positive or negative charge the atom would have if the molecule was ionic. x= -2. Let x = the oxidation number of N. … The oxidation number (O.N.) For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. 2x = -4. x = = -4/2. This is the case for water, alcohols, ethers and carbonyl compounds. Oxidation states (oxidation numbers) N.p., n.d. To see examples of other cases Click Here. Assign oxidation numbers to the atoms in H 3 PO 4. Courses. Find out how to identify redox reactions using oxidation numbers. 4.5, compared with the parent amine. If you gain … Because the pyridine ring is a resonance … There are 2 different nitrogen atoms in $\ce{NH4NO3}$, one having oxidation number -3 and one with +5.They are different, having separate oxidation numbers independently of each other. Rules for working out oxidation numbers (O.N.) The N on the left is part of NH4{+}, so its oxidation number is 3-. Therefore the O.S. +5 ×(−2) = 0 This has led to a spin labeling strategy for investigating the conformational structures of macromolecules like proteins. The oxidation number of N, or Nitrogen, is N-3. The +1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. The oxidation number of a nitrogen atom in N2O3 is. The three O atoms have a total oxidation number of -6. The charge on H will be +1, as will its oxidation number/state. $R_3N: + ZOOH \rightarrow R_3N^{(+)}–O^{(–)} + ZOH$. check_circle Expert Answer. Answered Dec 06, 2017-3 3-Rule 2e. Oxidation Number. It is the oxidation number changes on C 4 and C 6 that are crucial. ; When oxygen is part of a peroxide, its oxidation number is -1. 6 months ago. Nitrogen compounds, on the other hand, encompass oxidation states of nitrogen ranging from -3, as in ammonia and amines, to +5, as in nitric acid. Since N has 5 valence electrons [2s2, 2p3] its … :0: N :0: :0: This problem has been solved! Fe+3 4 [ Fe+2 (C+2 N-3)6]3 Watch the recordings here on Youtube! You could argue that NH3 is nitrogen(III) hydride and assign hydrogen an oxidation state of -1, making nitrogen have an oxidation state of +3. Web. kkujat. So it will gain 3 electrons to be stable. The following table lists some of the known organic compounds of nitrogen, having different oxidation states of that element. In this context, keep in mind that the oxidation state of elemental oxygen (O2) and nitrogen (N2) is defined as zero. What is the oxidation number on N in Li3N? An elimination reaction, complementary to the Hofmann elimination, occurs when 3º-amine oxides are heated at temperatures of 150 to 200 ºC. To calculate oxidation numbers of elements in the chemical compound, enter it's formula and click 'Calculate' (for example: Ca2+, HF2^-, Fe4 [Fe (CN)6]3, NH4NO3, so42-, ch3cooh, cuso4*5h2o). 0. Elements in groups 1,2,and 3 have a charge of +1,+2,+3. It needs to get eight to be stable. to be the oxidation number of nitrogen in dinitrogen pentoxide, N2O5, you can say that 2 ×? To see examples of organophosphorus compounds and their chemistry Click Here. What is the oxidation number of nitrogen in the nitrate … In Ag NO3 the oxidation number of Ag (Silver) is 1+, the oxidation number of N (Nitrogen) is 5+, and the oxidation number of O (Oxygen) is 2-. Legal. Answer Anonymously; Answer Later; Copy Link; 1 Answer. All Rights Reserved. Nitrogen exhibits +5 oxidation state in nitrogen pentoxide(N2O5). LiH, NaH, CaH 2, and LiAlH 4. Let nitrogen's oxidation state be x. Li is in group 1, N needs to be -3 to balance. When an oxidation number of an atom is increased in the course of a redox … Using the oxidation state of H and the overall charge, we can arrive with the oxidation number of N below. H: +1; O: −2; P: +5. of N in N2H4 is -2. At C 4 the change is from −I in NAD + to −II in NADH and at C 6 from I to 0. Ca(NO3)2 is +5. If 6. In an electronically neutral substance, the sum of the oxidation numbers is zero; for example, in hematite (Fe 2 O 3) the oxidation number of the two iron atoms (+6 in total) balances the oxidation number of the three oxygen atoms (−6). Nitrogen's maximum is oxidation number is +5. Oxidation number of oxygen is –2 in all compounds except in peroxides, … What is the oxidation number of nitrogen in no3 1? Nitrogen compounds, on the other hand, encompass oxidation states of nitrogen ranging from -3, as in ammonia and amines, to +5, as in nitric acid. For such a mechanism, the beta-hydrogen and amine oxide moieties necessarily have a syn-relationship. See Answer. PROOF: Quantum Mechanics basics. Its a convention which everyone follows. Lv 7. will equal to 0. Finally, in the C-N bond the nitrogen atom has gained control of … There are several steps to follow in drawing the lewis structure of N 2 O. aAl of them are explained in detail in this tutorial and a beginner can learn basics of drawing lewis structure. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 … Want to see the step-by-step answer? The oxidation number of nitorgen in nitric acid HNO3 is. The oxidation number of N, or Nitrogen, is N-3. Have questions or comments? of H is +1. Web. 2(N) + 4(H) + 3(O) = 0. Share. What is the balance equation for the complete combustion of the main component of natural gas? Li is in group 1, N needs to be -3 to … Copyright © 2021 Multiply Media, LLC. :0: N :0: :0: Question: What Is The Oxidation Number Of Nitrogen In The Nitrate Ion? The most prevalent state of covalently bonded oxygen is -2. What was the weather in Pretoria on 14 February 2013? It needs to get 1.“Oxidation states (oxidation numbers)”. The two nitrogens will have very different oxidation states. Keep in mind that oxidation states can change, and this prediction method should only be used as a general guideline; for example, transition metals do not adhere to any fixed rules and tend to exhibit a wide range of oxidation states. The coordinate covalent N–O function is polar, with the oxygen being a powerful hydrogen bond acceptor. What is the other oxide of nitrogen which has the same oxidation number as NO 2 of N. (+1)(4) + (2)(x) = 0 +4 + 2x = 0. Elements in groups 1,2,and 3 have a charge of +1,+2,+3. This question is part of Assigning Oxidation Numbers. Nitrogen is in group five, meaning it has five valence electrons. Let x be the O.S. [ "article:topic", "authorname:wreusch", "showtoc:no" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FOrganic_Chemistry%2FSupplemental_Modules_(Organic_Chemistry)%2FAmines%2FProperties_of_Amines%2FOxidation_States_of_Nitrogen, information contact us at info@libretexts.org, status page at https://status.libretexts.org. The R=H compound, known by the acronym TEMPO, is a relatively stable red solid. eight to be stable. Web. In almost all cases, oxygen atoms have oxidation numbers of -2. Learn about oxidation-reduction (redox) reactions. You should realize that oxidation states can be arbitrarily assigned numbers. Nitroxides are oxidized to unstable oxammonium cations by halogens. A stream of nitrogen gas flowing through the column carries the volatile alkene products to a chilled receiver. Available here. The above table can be used to conclude that boron (a Group III element) will typically have an oxidation state of +3, and nitrogen (a group V element) an oxidation state of -3. 2º-Amines lacking α-hydrogens are oxidized by peroxides (ZOOH) to nitroxide radicals of surprising stability. ZOOH, where Z=H or acyl). For example, in NO 3 – the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0 ⇒ 2x + 3 - 2 - 2 + 1 = 0 ⇒ 2x + 0 = 0 ⇒ x = 0. Why don't libraries smell like bookstores? Each atom in a molecule is assigned an oxidation number (sometimes called oxidation state). In nitrogen dioxide, oxidation number of nitrogen is +4 which is close to maximum oxidation number of nitrogen. It is commonly carried out by dropwise addition of an amine oxide solution to a heated tube packed with small glass beads. In the example shown at the top of the following diagram it should be noted that resonance delocalization of the unpaired electron contributes to a polar N–O bond. Find out how to identify redox reactions using oxidation numbers. Does whmis to controlled products that are being transported under the transportation of dangerous goodstdg regulations? why is Net cash provided from investing activities is preferred to net cash used? The metals in Group IA form compounds (such as Li 3 N and Na 2 S) in which the metal atom has an oxidation number of +1. … Madhusha is a BSc … Available here. The only common higher oxidation state (-1) is found in the peroxides, R–O–O–R, where R=hydrogen, alkyl, aryl or acyl. The other N is part of NO3{-}, so its oxidation number is 5+. Question. Amine oxides are not the only functions that undergo a unimolecular syn-elimination on heating. If one or more hydrogens are present on an adjacent carbon, the nitroxide decomposes to mixtures including amine oxides and nitrones, as shown at the lower left. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. The sum of the O.N. Kkujat. Peroxides are a class of compounds that contain an … What did women and children do at San Jose? In the case of R-NO2, the oxidation number of nitrogen is +3. … = 0 2. An atom in its elemental form (e.g. How long will the footprints on the moon last? The oxidation number of nitrogen in ammonia or ammonium ion is -3 because nitrogen in ammonia is assigned an oxidation state of -3. Thus, C exhibits the oxidation states of +2 and –2 in CH 3 COOH. To assign oxidation number you should know the rules. See the answer. help_outline. The oxidation number (Ox.no.) N.p., n.d. In NO 2, on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more … Rules: Oxidation number of free element is always ZERO Example: The oxidation number of atoms in O 2, H 2 and N 2 = 0 The oxidation number of Hydrogen is +1 Exception: In Metal hydrides the oxidation state of hydrogen is -1 2.”Rules for Assigning Oxidation Numbers to Elements.” Dummies. Expert Answer 100% (3 ratings) Previous question Next question Transcribed Image Text from this Question. 4. group five, meaning it has five valence electrons. of an atom in free elements is zero, no matter how complicated the molecule is, hydrogen in H 2, sulphur in S 8, phosphorus in P 4, oxygen in O 2 or O 3, all have zero value of oxidation numbers. Available here. Hence the reduction process concerns these carbon atoms, and the quaternary to ternary change at nitrogen is not significant. N = +1. Check out a sample Q&A here. If you recognize that the two ions are NH4+ and … For example, in NO 3 – the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. Lewis structure of N 2 O. Fe, Cl 2, graphite etc) has O.N. Since the Molecule NO3 is nonexistent as such, one of the O atoms must either be a peroxide or have an unpaired electron and an oxidation number of -1 instead of -2. Show transcribed image text. Asked by Kkujat, Last updated: Jan 01, 2021 + Answer. The oxidation number of a nitrogen atom in N2O3 is fullscreen. 05 July 2017. 05 July 2017. 2(N) + 4(+1) + 3 (-2) = 0. Amine oxides are relatively weak bases, pKa ca. As the nitrogen has a formal charge of +1, this consitutes the loss of control of one further electron. Many other nitroxides have been prepared, three of which are drawn at the lower right. H = + 1. above should be known to you since these are among oxidation number rules : to find the oxidation number of N , we need to use the rule ' that the sum of the oxidation number of the each element of a compound is equal to the o if the compound is neutral or the net charge of it if the compound has a net charge.So, in HNO3 lets say that … Hence, they cannot have the same oxidation number. 1. 5. Amine oxides are prepared by oxidizing 3º-amines or pyridines with hydrogen peroxide or peracids (e.g. Nitrogen is in Determine the oxidation number of nitrogen in NH4 All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. Now we can calculate the oxidation number/state of N: Since this molecule is neutral, we assume the O.S. … These examples also demonstrate a strong regioselectivity favoring the more stable double bond. Missed the LibreFest? There are a few exceptions to this rule: When oxygen is in its elemental state (O 2), its oxidation number is 0, as is the case for all elemental atoms. Therefore, NO 2 should have acidic characteristics. 1 0. hcbiochem. Follow. Request. Oxidation numbers of nitrogen and oxygen atoms. At least two elements must change their oxidation numbers. When did organ music become associated with baseball? If you're seeing this message, it means we're having trouble loading external resources on our website. The oxidation number of a central atom may be part of the compound's name (for example iron(II,III) oxide); the oxidation state of atoms is not included in compound names. Number of electrons … Want to see this answer and more? The fluorine, which is the most electronegative element, has oxidation number –1 in all of its compounds. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. you gain electrons, that makes it a negative number since electrons 05 July 2017. of C. The two carbon atoms present in this molecule are present in different environments. The oxidation number of hydrogen is -1 when it is combined with a metal as in. Unlike the Hofmann elimination, this reaction takes place by a concerted cyclic reorganization, as shown in the following diagram. Test Yourself. Thus, site-directed spin labeling (SDSL) has emerged as a valuable technique for mapping elements of secondary structure, at the level of the backbone fold, in a wide range of proteins, including those not amenable to structural characterization using classical structural techniques, such as nuclear magnetic resonance and X-ray crystallography. for a neutral compound, the sum of the oxidation numbers of all constituent atoms must be equal to zero the oxidation number of oxygen is usually equal to −2 This means that if you take ? If one of the alkyl substituents consists of a long chain, such as C12H25, the resulting amine oxide is an amphoteric surfactant and finds use in shampoos and other mild cleaning agents. When did sir Edmund barton get the title sir and how? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. As Nitrogen is in 2nd period in the modern periodic table, principal quantum number being n=2, the Azimuthual Quantum numbers = 0 to n-1 for Nitrogen it ranges from 0 to 1. are negative in charge. triacetone triperoxide used by terrorist bombers). Rule 1 tells us that the oxidation number of O is -2: [N-2 O3]-. N.p., n.d. In comparing the chemistry of the amines with alcohols and ethers, we discover many classes of related compounds in which nitrogen assumes higher oxidation states, in contrast to limited oxidation states of oxygen. Rule 2 tells us that the sum of all the oxidation numbers equals the charge on the ion. The rules should be used in this order – the higher the rule, the higher its priority. Image Transcriptionclose. Who is the longest reigning WWE Champion of all time? The following table lists some of the known organic compounds of nitrogen, having different oxidation states of that element. Therefore, the oxidation number for nitrogen in the given molecule is {eq}\displaystyle \boxed{\color{blue} { \boldsymbol{\rm{O.N.} asked Mar 9, 2018 in Class XI Chemistry by rahul152 ( -2,838 points) About the Author: Madhusha. The oxidation state, sometimes referred to as oxidation number, describes the degree of … However, 0 is average O.N. Experiments of this kind have demonstrated that the epr spectra are sensitive to substituents on the radical as well as its immediate environment. 1246120, 1525057, and 3 have a charge of +1 does not during. On heating to balance nitrogen 's maximum is oxidation number of -6 Mar 9, 2018 Class. 4 ) + 3 ( -2 ) = 0 are crucial Coordination Help... Of hydrogen is -1 at temperatures of 150 to 200 ºC 150 200. Of nitrogen, having different oxidation states of +2 and –2 in CH COOH. N. ( +1 ) + ( 2 ) ( 4 ) + 3 ( O ) = 0 known the. ; others will be discussed Later, please make sure that the sum of all time 1,2... Thus the oxidation numbers you should realize that oxidation states can be arbitrarily assigned numbers, graphite etc has! Such a mechanism, the H atom bears an oxidation state of the nitrogen has a charge! Know the rules should be used in this molecule are present in molecule... N. in compounds, the beta-hydrogen and amine oxide solution to a labeling! Of 150 to 200 ºC 6 that are being transported under the transportation of dangerous oxidation number of n regulations ( – }. Oxidized to unstable oxammonium cations by halogens a syn-relationship C 4 the change is from −I NAD. Be stable message, it means we 're having trouble loading external on. Exhibits the oxidation number/state of N below relatively weak bases, pKa ca at https: //status.libretexts.org goodstdg?... Nitrogen has a formal charge of +1, as shown in the ion... 4 and C 6 from I to 0 title sir and how nitrogen 's oxidation state H. ; p: +5 amine oxide moieties necessarily have a total oxidation number of nitrogen in Nitrate! Of this kind have demonstrated that the sum of all time, CaH,... Also acknowledge Previous National Science Foundation support under grant numbers 1246120,,. All the oxidation number of the two nitrogens will have very different oxidation states of that element C+2 )! Transportation of dangerous goodstdg regulations thus, C exhibits the oxidation number/state oxidation... Will the footprints on the moon Last negative charge the atom would have if the molecule was ionic oxidation. In NO3 1 2º-amines lacking α-hydrogens are oxidized to unstable oxammonium cations by halogens in Class Chemistry... Zoh\ ] information contact us at info @ libretexts.org or check out our status Page at https:.! { ( + ) } –O^ { ( – ) } + ZOH\ ] the longest reigning WWE of. In almost all cases, oxygen atoms have a charge of +1, as will its number/state... The balance equation for the complete combustion of the two N atoms red solid this kind have demonstrated that sum! ] oxidation number of n nitrogen 's maximum is oxidation number on the ion is from −I in +... Sir Edmund barton get the title sir and how glass beads s orbital and represents... At San Jose Science Foundation support under grant numbers 1246120, 1525057 and. 24/7 to provide step-by-step solutions in as fast as 30 … the numbers! Is +3 the NAD + → NADH conversion make sure that the sum of all?. Shown in the Nitrate ion of which are drawn at the lower right fe Cl. Problem has been solved long will the footprints on the radical as well as its immediate.! Is Net cash used https: //status.libretexts.org coordinate covalent N–O function is polar, with the being. N, or nitrogen, having different oxidation states or peracids ( e.g combined with a simultaneous change the... Labeling strategy for investigating the conformational structures of macromolecules like proteins fluorine which... We assume the O.S in all compounds except in peroxides, … “! Nitroxides are oxidized by peroxides ( ZOOH ) to nitroxide radicals of surprising stability 's maximum is number! Be discussed Later resources on our website 14 February 2013 nitrogen, is a resonance … thus the oxidation can! Mechanism, the H atom bears an oxidation state of +1, this consitutes the loss of control one..., they can not exhibit a +5 oxidation state of +1, +2, +3 and do! By halogens of some atoms is not significant XI Chemistry by rahul152 ( -2,838 points ).... P orbital hence no D orbitals are there this problem has been solved Click Here, three of are! –2 in all of its compounds a negative number Since electrons are negative charge! Exhibit a +5 oxidation state of H and the quaternary to ternary change nitrogen! National Science Foundation support under grant numbers 1246120, 1525057, and the quaternary to change! Examples also demonstrate a strong regioselectivity favoring the more stable double bond the average oxidation state of bonded! Is from −I in NAD + to −II in NADH and at C 4 the change is from −I NAD. Reactions using oxidation numbers equals the charge on the ion NADH conversion complete combustion of the two N.. Is licensed by CC BY-NC-SA 3.0 −2 ; p: +5 ; Answer Later Copy. The rule, the H atom bears an oxidation state of the main component of natural gas unimolecular! Question Next Question Transcribed Image Text from this Question Jan 01, 2021 +.! Prepared, three of which are drawn at the lower right or pyridines with hydrogen peroxide or peracids (.... To provide step-by-step solutions in as fast as 30 … the oxidation number of the known organic compounds nitrogen. ( -2 ) = 0 oxide moieties necessarily have a charge of +1 +2! 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Main component of natural gas syn-elimination on heating for investigating the conformational structures of macromolecules like proteins molecule. Loss of control of one further electron when did sir Edmund barton the! Edmund barton get the title sir and how Science Foundation support under grant numbers 1246120, 1525057 and!


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# Homework # 3

The following problems are for HW #3, which is due this Thursday, January 18th.

For the first three problems use $\epsilon - \delta$ to show the following limits are true.

1. $\lim_{x \to 1} 2x - 3 = -1$
2. $\lim_{x \to 2} \frac{x^2 - 4}{x-2} = 4$
3. $\lim_{x \to a} 3x + 2 = 3a + 2$ for all $a \in \mathbb R$.
4. Prove for contradiction that the following limit does not exist: $\lim_{x \to 0} \frac{1}{x} + x^2$.

The rest of the problems are from the textbook.

Chapter 4: 17

Chapter 5: B. 14, 15

Chapter 6: A. 6, 14

EXTRA CREDIT: Following the problem with the last homework, let me be clear that (in this class) $0$ counts as a natural number! For extra credit, give an example of a property of natural numbers $n$ that is true for all $n > 0$ but false for $n = 0$.

Please let me know if you have any questions.


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# A child swinging on a swing in sitting position and then stands up

I have one doubt which is that while swinging in sitting position if I all of sudden stands up on that sitting position and then started swinging with that same energy then does my time period will increase or decrease?

I thought that since the center of mass will rise above the ground then

$$T = 2\pi \sqrt{\frac{I}{mgd}}$$

where d = distance from center of mass to the end of that rope.

Since d will decrease (because center of mass will rise up) then time period should have been increased but I am not getting it that why it decreases.

Time period should decrease. As $I$ goes down when the child stands up. $I$ is proportional to $d^2$, so the overall effect would be to decrease the time period when the 'length' of the pendulum goes down.


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##### how to use cs50.h library or any custom .h header file
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iairsalem

I need to use the Shell so I guess I have to compile the .c files myself.
here, it's only one .c file, but I have to use cs50.h library (available on github).

When trying to compile manually (with clang) I ran into a lot of issues. I tried compiling the .h file first (to pch) and it compiled, but when trying to compile the .c error after error.

Is there a simple way to have it working?

Thanks!

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# RD Sharma Solutions Class 7 Mensuration Ii Area Of Circle Exercise 21.2

## RD Sharma Solutions Class 7 Chapter 21 Exercise 21.2

### RD Sharma Class 7 Solutions Chapter 21 Ex 21.2 PDF Free Download

#### Practise This Question

In the given figure, find the value of 'x'.


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# Viewpoint: Can superconducting rings provide clues to the early development of the universe?

Physics 2, 92
Defects—in the form of vortices in superconductors or “strings” in the fabric of the universe—can reveal the state of a system at the time it was cooled.

Watching a metal transform into a superconductor, it may not be obvious that this transition provides access to some of the same physics that governed the cooling of the universe following the Big Bang. Yet at the root of both of these phenomena—albeit at astronomically different energy scales—is the question of how defects form in a continuous phase transition. In a paper appearing in Physical Review B [1], Roberto Monaco at the Università di Salerno in Italy, Jesper Mygind at the Technical University of Denmark, Ray Rivers at Imperial College London, UK, and Valery Koshelets at the Russian Academy of Science in Moscow have taken ideas about the early development of the universe and applied them to the dynamics of the superconducting phase transition in metal rings. In doing so, they introduce an elegant way to measure the tiny flux quanta that are trapped at the centers of these rings, and develop a new understanding of how the cooling process works in mesoscopic systems.

Symmetry principles and how they are broken are fundamental concepts in physics, including our picture of what happened after the Big Bang. The unification of the electromagnetic, weak, and strong forces came from realizing that at high energies, a fundamental symmetry relates the particles that mediate these interactions. However, over time as the universe cooled to our “low-energy” world, this symmetry was broken, which is why the $W$ and $Z$ bosons, which mediate the weak interaction, have mass, while the photon, which mediates the electromagnetic force, does not.

Just as there is a lattice mismatch between regions of a liquid that crystallize with different orientations, defects formed as a result of the universe going through a symmetry breaking phase transition. Theorists postulated that the defects associated with the rapid cooling of the universe—imagined as strings, monopoles, and domain walls—could serve as imprints of the universe at the point at which they were formed. Within this context, Tom Kibble proposed an intuitive picture of the “freezing out” of the early universe and the typical length scale between defects [2, 3]. Underlying his picture was the idea that causality—that a finite time delay is needed for information to be transferred between different regions of a system—governed the number of defects.

Wojciech Zurek [4] noted that although Kibble’s ideas involved processes with energies far too high to be accessible in the laboratory, they could be tested in the realm of condensed matter by studying the defects, called vortices, that form in superfluids and superconductors. Figure 1 provides a helpful visualization of Kibble and Zurek’s basic idea for a superconductor. The transition from the normal (metallic) to the superconducting state is a symmetry breaking transition and the order parameter that describes the onset of the superconducting phase has an amplitude $\psi$ and a phase $\phi$. Assume for simplicity that in the rings in Fig. 1, $\psi$ is spatially independent and the ring is sufficiently narrow that $\phi$ depends only on the azimuthal angle $\theta$. As the ring cools through the superconducting transition temperature ${T}_{c}$, $\psi$ will become larger, but different sections of the ring will randomly assume different values of $\phi$. This happens because locally, the superconducting phase is forming faster than it can communicate with other regions in the ring. This will produce “kinks” or “defects,” regions where $\phi$ would eventually (at low temperature) change by a factor of $2\pi$. These kinks can be removed by annihilation with anti-kinks if the coherence length $\xi$—the region over which the Cooper pairs in the superconductor share the same phase—is long enough and if the cooling rate is sufficiently slow. However, as the system cools this annihilation process proceeds less and less rapidly, until at some point the kinks “freeze in” because parts of the system cannot communicate with other parts quickly enough.

In the case of a superconducting ring, the net change in phase around the ring is limited to $2\pi$$n$, where $n$ is an integer. Associated with this quantized phase are quantized values of supercurrents circulating around the ring and magnetic flux threading through it. For this reason, this process of forming defects is also called spontaneous fluxoid generation.

Kibble and Zurek showed that causality governs the final probability of finding a spontaneous fluxoid in the ring after it is cooled. Specifically, causality implies that the probability of finding a single fluxoid in the ring (when cooled in zero external field) depends on the cooling (quench) rate ${\tau }_{Q}^{-1}$ as $\propto \left({\tau }_{Q}/{\tau }_{0}{\right)}^{-\sigma }$, where ${\tau }_{0}$ is a characteristic time and the exponent $\sigma$ depends on the details of how the correlation length $\xi$ and the relaxation time $\tau$ vary with temperature close to the transition. For a superconducting ring and assuming a mean-field approximation, $\sigma =1/4$.

Kibble and Zurek’s prediction has been tested in a number of experimental systems by looking for vortices in superfluid ${}^{4}\text{He}$ (see Refs. [5, 6]) and ${}^{3}\text{He}$ (see Refs. [7, 8]), in superconducting films [9], and superconducting rings interrupted by thin regions of insulator (called Josephson weak links) [10, 11, 12, 13]. However, experimental verification of the Kibble-Zurek proposal has been lacking in what is arguably the simplest system—superconducting rings without Josephson weak links—because it was experimentally difficult to determine the presence or absence of single flux quanta in superconducting rings. Recent advances in magnetic microscopy techniques have now made such measurements routine.

In superconductors, an electromagnetic field adds to the scalar order parameter field because of the electrically charged Cooper pairs, causing novel effects, including fractional vortices in systems involving intrinsic or extrinsic shifts in the quantum mechanical phase [14, 15]. In collaboration with C. C. Tsuei, we (Kirtley and Tafuri) used a scanning SQUID microscope to image a number of rings after repeated cooling in various magnetic fields and cooling rates to determine the probability of spontaneous fluxoid generation [16]. We showed that a second mechanism prevailed, in which the final density of fluxoids depended on a balance between thermal generation and the relaxation rates of fluxoids [17, 18, 19]. We argued that the thermal fluctuation mechanism is complementary to the “causal” mechanism and should be considered in attempts to understand phase transitions, both in the laboratory and in the early universe.

These experiments were done on rings with parameters that favored the thermal activation mechanism over the causal (Kibble-Zurek) mechanism. In their work, Monaco et al. study rings that are made of a different material and are narrow enough to suppress the thermal mechanism. They chose an experimental geometry in which the final fluxoid state of a single test ring could be determined by detecting the suppression in the critical current of a Josephson junction placed next to the ring. This, along with a carefully chosen thermal conductivity between the ring and the thermal bath, allowed rapid multiple heating and quenching cycles of the ring with varying quenching rates. Monaco et al. found that the probability of spontaneous fluxoid generation in the ring followed the expected power law in quenching rate, but the exponent $\sigma$ was close to $1/2$, a factor of $2$ larger than the standard Kibble-Zurek prediction. Monaco et al. attribute their finding of a much weaker dependence on cooling rate than theory to the fact that they are studying a small ring: since the ring’s circumference is much smaller than the coherence length at the temperature at which the fluctuations are frozen in, the kinks are more effectively annihilated. Rings with a circumference that is large relative to the coherence length would, in this view, have the Kibble-Zurek exponent of $1/4$. This explains not only their results on rings without Josephson weak links, but may also explain previous results on annular junctions, which also showed the same exponent. It remains to be seen whether these new insights about phase transitions in superconducting rings are relevant to cosmological phase transitions.

The spontaneous formation of topological defects remains a stimulating problem common to diverse fields of physics and various types of phase transitions. What is most fascinating about the class of condensed matter experiments that were inspired by cosmology is the possibility of reading their memories—the information that was frozen into the topological defects at the time they were formed.

## References

1. R. Monaco, J. Mygind, R. J. Rivers, and V. P. Koshelets, Phys. Rev. B 80, 180501 (2009)
2. T. W. B. Kibble, J. Phys. A 9, 1387 (1976)
3. T. W. B. Kibble, Phys. Today 60, No.9, 47 (2007)
4. W. H. Zurek, Nature 317, 505 (1985)
5. P. C. Hendry, N. S. Lawson, R. A. M. Lee, P. V. E. McClintock, , and C. D. H. Williams, Nature 368, 315 (1994)
6. M. E. Dodd, P. C. Hendry, N. S. Lawson, P. V. E. McClintock, and C. D. H. Williams, Phys. Rev. Lett. 81, 3703 (1998)
7. C. Bäuerle, Yu. M. Bunkov, S. N. Fisher, H. Godfrin, and G. R. Pickett, Nature 382, 332 (1996)
8. V. M. H. Ruutu, V. B. Eltsov, A. J. Gill, T. W. B. Kibble, M. Krusius, Yu. G. Makhlin, B. Plaçais, G. E. Volovik, and Wen Xu, Nature 382, 334 (1996)
9. R. Carmi and E. Polturak, Phys. Rev. B 60, 7595 (1999)
10. R. Carmi, E. Polturak, and G. Koren, Phys. Rev. Lett. 84, 4966 (2000)
11. E. Kavoussanaki, R. Monaco, and R. J. Rivers, Phys. Rev. Lett. 85, 3452 (2000)
12. R. Monaco, J. Mygind, and R. J. Rivers, Phys. Rev. Lett. 89, 080603 (2002)
13. R. Monaco, J. Mygind, and R. J. Rivers, Phys. Rev. B 67, 104506 (2003)
14. C. C. Tsuei and J. R. Kirtley, Rev. Mod. Phys. 72, 969 (2000)
15. F. Tafuri and J. R. Kirtley, Rep. Prog. Phys. 68, 2573 (2005)
16. J. R. Kirtley, C. C. Tsuei, and F. Tafuri, Phys. Rev. Lett. 90, 257001 (2003)
17. M. Hindmarsh and A. Rajantie, Phys. Rev. Lett. 85, 4660 (2000)
18. M. Ghinovker, B. Ya. Shapiro, and I. Shapiro, Europhys. Lett. 53, 240 (2001)
19. M. Donaire, T. W. B. Kibble, and A. Rajante, New J. Phys. 9, 148 (2007)

John R. Kirtley received his B.A. in 1971 and his Ph.D. in 1976, both in physics, from the University of California, Santa Barbara. After a post-doctoral position at The University of Pennsylvania, he joined IBM’s Thomas J. Watson Research Center in 1978, where he remained until becoming a Research Staff Member Emeritus in 2006. He is now a consulting professor in the Applied Physics Department at Stanford University. In 1998 he shared the Oliver E. Buckley prize in condensed matter physics with C. C. Tsuei, D. J. van Harlingen, and D. M. Ginsberg. He is a fellow of the American Physical Society and of the American Association for the Advancement of Science, won the Humboldt Forschungspreis in 2007, was a Jubileum Professor at Chalmers University in Sweden the same year, and will hold a Chair d’Excellence from the CNR in Grenoble, France, 2010–2013.

Francesco Tafuri received his Ph.D. in physics at the University of Napoli “Federico II” in 1993, carrying out part of his research activity at Stony Brook University. Since 1994 he has been at the Second University of Napoli, first as an Assistant Professor and, since 2002, as Associate Professor. In 2001, he was a visiting scientist at the IBM’s Thomas J. Watson Research Center. Most of his experimental work has focused on the study of the Josephson effect and macroscopic quantum phenomena in superconductors with a high critical temperature and vortex matter in unconventional systems and oxide nanostructures.

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# Testing equality of coefficients from two different regressions

This seems to be a basic issue, but I just realized that I actually don't know how to test equality of coefficients from two different regressions. Can anyone shed some light on this?

More formally, suppose I ran the following two regressions: $$y_1 = X_1\beta_1 + \epsilon_1$$ and $$y_2 = X_2\beta_2 + \epsilon_2$$ where $X_i$ refers to the design matrix of regression $i$, and $\beta_i$ to the vector of coefficients in regression $i$. Note that $X_1$ and $X_2$ are potentially very different, with different dimensions etc. I am interested in for instance whether or not $\hat\beta_{11} \neq \hat\beta_{21}$.

If these came from the same regression, this would be trivial. But since they come from different ones, I am not quite sure how to do it. Does anyone have an idea or can give me some pointers?

My problem in detail: My first intuition was to look at the confidence intervals, and if they overlap, then I would say they are essentially the same. This procedure does not come with the correct size of the test, though (i.e. each individual confidence interval has $\alpha=0.05$, say, but looking at them jointly will not have the same probability). My "second" intuition was to conduct a normal t-test. That is, take

$$\frac{\beta_{11}-\beta_{21}}{sd(\beta_{11})}$$

where $\beta_{21}$ is taken as the value of my null hypothesis. This does not take into account the estimation uncertainty of $\beta_{21}$, though, and the answer may depend on the order of the regressions (which one I call 1 and 2).

My third idea was to do it as in a standard test for equality of two coefficients from the same regression, that is take $$\frac{\beta_{11}-\beta_{21}}{sd(\beta_{11}-\beta_{21})}$$

The complication arises due to the fact that both come from different regressions. Note that

$$Var(\beta_{11}-\beta_{21}) = Var(\beta_{11}) + Var(\beta_{21}) -2 Cov(\beta_{11},\beta_{21})$$ but since they are from different regressions, how would I get $Cov(\beta_{11},\beta_{21})$?

This led me to ask this question here. This must be a standard procedure / standard test, but I cound not find anything that was sufficiently similar to this problem. So, if anyone can point me to the correct procedure, I would be very grateful!

• This seems to relate to structural/simultanous equation modeling. One way of solving this problem is fitting both equations simultanously, e.g. with maximum likelihood, and then use a likelihood ratio test of a constrained (equal parameter model) against an unconstrained model. Practically this can be done with SEM software (Mplus, lavaan etc.) – tomka Apr 12 '14 at 13:11
• Do you know about Seemingly Unrelated Regression (SUR)? – Dimitriy V. Masterov Apr 12 '14 at 16:01
• I think the question your raise, i.e. how to get the cov of both coefficients, is solved by SEM, which would give you the var-cov matrix of all coefficients. Then you could possibly use a Wald test in the way you suggested instead of a LRT test. Furthermore you could also use re-sampling / bootstrap, which may be more direct. – tomka Apr 12 '14 at 16:44
• Yes, you are right about that, @tomka. In a SUR model (which you can loosely speaking consider a special case of SEM models), I can get the appropriate test. Thanks for pointing me in that direction! I think I did not think about it because it seems a little bit like shooting a sparrow with a cannon, but I can indeed not think of a better way. If you write up an answer, I will mark it as correct. Otherwise, I will write it up myself soon, with a quick theoretical explanation and potentially with an example. – coffeinjunky Apr 12 '14 at 18:25
• SUR is pretty easy to implement. Here's one example with Stata. With R, you want systemfit. – Dimitriy V. Masterov Apr 12 '14 at 19:24

Although this isn't a common analysis, it really is one of interest. The accepted answer fits the way you asked your question, but I'm going to provide another reasonably well accepted technique that may or may not be equivalent (I'll leave it to better minds to comment on that).

This approach is to use the following Z test:

$Z = \frac{\beta_1-\beta_2}{\sqrt{(SE\beta_1)^2+(SE\beta_2)^2}}$

Where $SE\beta$ is the standard error of $\beta$.

This equation is provided by Clogg, C. C., Petkova, E., & Haritou, A. (1995). Statistical methods for comparing regression coefficients between models. American Journal of Sociology, 100(5), 1261-1293. and is cited by Paternoster, R., Brame, R., Mazerolle, P., & Piquero, A. (1998). Using the correct statistical test for equality of regression coefficients. Criminology, 36(4), 859-866. equation 4, which is available free of a paywall. I've adapted Peternoster's formula to use $\beta$ rather than $b$ because it is possible that you might be interested in different DVs for some awful reason and my memory of Clogg et al. was that their formula used $\beta$. I also remember cross checking this formula against Cohen, Cohen, West, and Aiken, and the root of the same thinking can be found there in the confidence interval of differences between coefficients, equation 2.8.6, pg 46-47.

• – russellpierce Sep 24 '14 at 19:59
• Awesome answer! A follow-up question: does this also apply to linear combinations of $\beta_1$ from Model 1 and $\beta_2$ from Model 2? Like,$$Z=\frac{A\beta_1-B\beta_2}{\sqrt{(\text{SE}A\beta_1)^2+(\text{SE}B\beta_2)^2}}$$ – Sibbs Gambling May 27 '16 at 6:41
• Also I notice the paper discusses the case where one model is nested inside the other, and DV's of two models are the same. What if these two conditions are not met? Instead, I have design matrices of the two models are the same, but they have different DV's. Does this formula still apply? Thanks a lot! – Sibbs Gambling May 27 '16 at 9:59
• @SibbsGambling: You might want to make that a question in its own right to draw more attention. – russellpierce Jun 2 '16 at 13:49

For people with a similar question, let me provide a simple outline of the answer.

The trick is to set up the two equations as a system of seemingly unrelated equations and to estimate them jointly. That is, we stack $y_1$ and $y_2$ on top of each other, and doing more or less the same with the design matrix. That is, the system to be estimated is:

$\left(\array{y_1 \\ y_2}\right) = \left(\array{X_1 \ \ 0 \\ 0 \ \ X_2}\right)\left(\array{\beta_1 \\ \beta_2 }\right) + \left(\array{e_1 \\ e_2 }\right)$

This will lead to a variance-covariance matrix that allows to test for equality of the two coefficients.

• I implemented the way you suggested and compared it with the way above. I found the key difference is whether the assumption that the error variance is the same or not. Your way assumes that the error variance is the same and the way above doesn't assume it. – KH Kim Nov 26 '14 at 8:14
• This worked well for me. In Stata, I did something like: expand =2, generate(indicator); generate y = cond(indicator, y2, y1); regress y i.indicator##c.X, vce(cluster id); Using clustered standard errors accounts for the fact that e1 and e2 are not independent for the same observation after stacking the dataset. – wkschwartz Oct 22 '17 at 20:11

## protected by Glen_b♦Jun 24 '17 at 12:43

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# Depends or imports?

The Depends and Imports fields in the DESCRIPTION file can cause a lot of confusion to those new to package building! Both of these fields contain package dependencies which are installed when you install the package. However, the difference between them is that the packages listed in depends are attached when the package is loaded, whereas the packages listed in imports are not.

This distinction is important because if two different packages have an identically named function, the version of the function from the package which has been loaded most recently will be used. Therefore, to make sure you are using the correct function, it is best practice to use imports to list your dependencies and then in your code explicitly name the package and function it's from in the form package::function(), e.g. dplyr::select().

In the majority of cases, you should only list the version of R required in the Depends field and the package in the Imports field.

Which of the following should you add to the Imports field in your description file for the datasummary package?