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How to measure the color of honey?
After doing some research, I found that honey's color could be a factor of climate. I am very interested in this subject and I'm a high school student so we don't have as much apparatus as you would find in a chemistry school. The color of the honey can be determined using wavelengths which means a spectrometer is required and it is not available in our school. It would be very helpful if any of you recommended me with a way from which I can carry out this experiment and measure the color of honey.
Thanks for your valuable help!

A good statrting place for a literature search would be http://www.tecnofrom.com/moduloNotas/noti_67/archivosAdjuntos/AN53025-Analysis-of-Honey-Color-and-HMF-Content-FINAL-FHR_1_279.pdf

I am a high school student myself, so I understand your struggles! I have tried making a crude 'spectrophotometer' in the past when I had nothing but a camera and a computer. This is by no means a precise or accepted way of doing things, but it should definitely give you a start. Place your samples in a clear container of know volume and take a picture of them (a the same location) in front of a white background. You can then upload your photos to an online RGB color picker and use a chart to convert the RGB values to a wavelength. This has worked fairly well for me.

It is important to note that a diffraction grating is used to separate the constituent colors of your light. If you take photos of your samples, they will most definitely contain a mixture of light (thus you should use a diffraction grating); however, because you are only comparing samples relative to each other, this is probably not that relevant for your intents and purposes.

Hello @EliJones, sounds like a good method to use but since I'm doing this experiment from a chemistry perspective point, to what extent do measuring wavelengths in colors of honey involve chemistry?

I believe you said your goal was to perform a correlational study between the color of honey and the climate. In this case, it is irrelevant wheater or not your measurements are based on the wavelength of light absorbed or the color of honey. If you wanted to try to perform a correlational study between wavelength and climate, that would be much more difficult to interpret. This is because the light absorbed by honey is a mixture of colors. As you might imagine, using a spectrophotometer to obtain a complex mixture of colors won't provide the best data to interpret!

For the purpose of your experiment, I would probably just determine the color of the honey. The wavelength of light absorbed is generally used to determine structural properties of a certain compound -- such as the presence of multiple double bonds or a hydrated copper (II) complex. Because you are not actually analyzing molecular properties per say, spectroscopy is not really necessary.

Could you use a reference color chart? Here is one originally made for skin colors:

Wikipedia has other examples.
Paint stores might give you a nicely printed color chart for free.
There are also online versions of color selectors, for example the one from Pantone, but then you have to be very careful about color calibrating your screen, or at least not changing it between comparisons.
A third option, more realistic if you are only measuring color along one scale, would be to make your own comparison material. Start with a light honey colored liquid and mix in a darker color till it matches your honey. The amount of dark color used gives you the darkness of the honey. You probably have all supplies available in the art department.
Edit:
MaxW found a much better solution, a special color scale for honey. It is called the Pfund scale:

The Pfund scale mentioned in the link I provided is such a color reference for honey. From a image of the scale, and using a color picker on the computer, you could judiciously select the appropriate paint chips to make a reasonable facsimile.  You'd have to figure out the "standard" thickness of honey which to view the color.

If you haven't already, try asking a biology teacher if they have a spectrometer. High school biology departments often have a UV-Vis spectrometer.
You can also try making your own spectrometer using a CD as a diffraction grating. You can then use this site to upload a photograph of the spectrum for quantitative analysis.
Honey is a very good candidate for use in a simple homemade visible light spectrometer.

You wouldn't need a spectrometer. Good LED's are monochromatic enough for such a crude characterization.

hello @MaxW, sounds like a good method and I finally found a spectrometer to use but since I'm doing this experiment from a chemistry perspective point, to what extent do measuring wavelengths in colors of honey involve chemistry?

@chemistryenthusiast - UV-Vis Spectrometry follows Beer's Law which requires a monochromatic source.


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Measuring mole fraction in lab using Raoult's law
I was reading in my textbook of chemistry on the vapour pressure in liquid compounds, and came across a doubt regarding vapour pressure. If I have two liquids, one is A and another is B, and if I know their vapour pressure in a pure state and their initial mole fraction, would their mole fractions be the same in the vapour phase (after mixing them)? If the answer is yes, how can we prove it? Secondly, if it is practically not possible what other factors should I take into account while solving their value using Raoult's law?

The mole fractions in the gas phase also depend on the relative volatility of the two liquids, so they will not in general be equal to the mole fractions in the liquid mixture. For instance, if you have components A and B, then their vapor pressures will be
$$ \begin{align} p_A &= \chi_A p^\circ _A \\ p_B &= \chi_B p^\circ _B \end{align}$$
where $\chi_i$ is the mole fraction of component i in the liquid and $p^\circ_i$ is the vapor pressure of the component when pure. According to Dalton's law the mole fraction of a given component in the ideal gas is proportional to its contribution to the total pressure so that
$$\begin{align}  p&=p_A+p_B \tag{1}\\ &= \chi_A p^\circ _A + \chi_B p^\circ _B \\ &= \chi_A (p^\circ _A -p^\circ _B) + p^\circ _B \end{align}$$
From this last expression you can conclude that the total pressure above the mixture will vary linearly with the mole fraction of either component in the liquid.
You can also show that the mole fraction of either component in the liquid is not equal to that in the gas. For instance for component A you can write by combining Raoult's and Dalton's laws that
$$\chi_{A,g}=\frac{p_A}{p}=\chi_{A,l}\frac{p_A^\circ}{p} \tag{2}$$
where the mole fractions in the liquid and gas are labelled explicitly with subscripts $l$ and $g$. It should be clear that $\chi_{A,g} \neq \chi_{A,l}$ unless $p=p_A^\circ$, but the latter condition is only satisfied for the pure substance A.

is there any way we can calculate this?@Buck Thorn

@Yuvraj Please be more specific. Calculate what?

mole fraction of liquids in the vapor phase?@Buck Thorn

@Yuvraj Well, you can use the last equation I wrote, $\chi_{A,g}=\chi_{A,l}\frac{p_A^\circ}{p}$. It depends what information you have to start with. If you know the total pressure, the mole fraction in the liquid phase, and the vapor pressure of the pure substance, then you can use this equation.

The catch with the expression is that it assumes that evaporation to form the vapor phase does not significantly alter the composition of the liquid phase. It's a little more complicated otherwise, and I doubt you need to consider this.

ok as I have a question where i was given two liquids as I mentioned in the question I know the vapor pressure in their purest form, and their mole fraction liquid phase, so if I want to calculate mole fraction in the vapor phase,what will be total pressure value, using daltons law?@Buck Thorn

@Yuvraj you can compute the total pressure of the mixture using equation 1 and use the result to compute the mole fraction in the gas phase using equation 2 (I added labels to the relevant equations)

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Order of reactivity of substrates towards Wurtz reaction
In a book I refer, it is said that:

Both Ionic [carbanion] mechanism as well as free radical mechanism are widely accepted for Wurtz reaction.

When I read this statement, I was thinking, which of these mechanisms would be able to explain the order of reactivity of different substrates towards Wurtz reaction. My conclusion was that the rate of the reaction should follow the below mentioned order:

$1^\circ > 2^\circ > 3^\circ$ if it takes place through carbanion mechanism; and
$3^\circ > 2^\circ > 1^\circ$ if it takes place through free radical mechanism

where $1^\circ$ refers to primary alkyl halide, $2^\circ$ refers to secondary alkyl halide, and $3^\circ$ refers to tertiary alkyl halide.
So, my doubt is what is the correct consolidated order of reactivity of alkyl halides towards Wurtz reaction?
I know my question could be a possible duplicate of this.
However, I did not get a satisfactory explanation over there and hence have asked it as a new question.

I'd say free radical mechanism

@user93057  Eventhough 3° free radical is more stable,it does not give the expected product in the wurtz reaction.It is also reported that 3° alkyl halides undergo disproportionation reaction under conditions maintained for wurtz reaction.
So,will the order of reactivity be 2°>1°>3° (where this has the same meaning as what I have explained in my question)?

The Wurtz reaction is an organic reaction, in which two molecules of an alkyl halide (e.g., $\ce{R-Cl}$) are coupled in the presence of $\ce{Na, Mg}$, or other reactive metals  to form an alkane (original classical Wurtz reaction used $\ce{Na}$ metal). The mechanism begins with a single electron transfer (SET) from sodium metal to the alkyl halide, which dissociates to form an alkyl radical ($\ce{R^.}$) in a similar fashion to the formation of a Grignard reagent and sodium halide salt (e.g., $\ce{NaCl}$):
$$\ce{R-X + Na^\circ -> R^. + NaX} \; \text{ where } \ce{X = Cl, Br,} \text{ or } \ce{I}$$
A second molecule of sodium undergoes another SET to the alkyl radical to form a nucleophilic carbanion (e.g., $\ce{R-}$):
$$\ce{R^. + Na^\circ -> R^- + Na+} $$
This alkylmetal intermediate can be isolated (e.g., alkyllithium) and has been isolated in a several cases. The carbanion then attacks another molecule of alkyl halide in a nucleophilic substitution reaction ($\mathrm{S_N2}$) to form the final coupled product (e.g., $\ce{R-R}$) and another molecule of sodium halide salt:
$$\ce{R^- + R-X -> R-R + X-} $$
Thus, the mechanism is believed to be undergoing in both free radical and ionic pathways in particular steps. The depicted diagram of the mechanism is illustrated in below:

In a review of the Wurtz reaction (Ref.1) revealed that the reaction is severely limited in scope as demonstrated by the following observations:

In the synthesis of noncyclic systems (dimerization or cross-coupling) reasonable yields require the use of primary alkyl halides, with iodides giving the best results. Secondary halides give very poor results;
Cross-coupling reactions (using a mixture of two different alkyl halides) give approximately statistical ratios of cross-coupled : dimeric products, due to the high reactivity of the organosodium intermediates involved. Thus, in the synthesis of 3-methyluntriacontane (4 ($\ce{R-R'}$); see above diagram), coupling of the iodide (1; $\ce{R-I}$) with 1-iodooctadecane (2; $\ce{R'-I}$) under standard Wurtz conditions gave a mixture of 3 ($\ce{R-R}$) from dimerization of 1, 4 ($\ce{R-R'}$) from the desired cross-coupling, and 5 ($\ce{R'-R'}$) from dimerization of 2, in the ratio $21:50:29$.

However, This classical reaction is of value in the preparation of cyclic systems, notably [2.2]phanes and bicyclobutanes (e.g., bicyclo[1.1.0]butane), as well as in the preparation of symmetric dimers. The Wurtz reaction provides one of the most successful approaches to these class of compounds, by ring-closure reactions. The reaction may proceed either intra- or inter-molecularly, with intramolecular couplings giving better yields (Ref.1):

The modest yields of [2.2]phane products are due to competing oligomerization/cyclization reactions, which is demonstrated by the product distribution obtained on treatment of the dibromide (6) with sodium/tetraphenylethane (TPE) in THF at $\pu{-80 ^\circ C}$ (See above diagram). A homologous series of products has been obtained under these conditions, the approximate yields of the metacyclophanes being obtained were given in the scheme.
In conclusion, the Wurtz reaction is seldom used because of side reactions and it is intolerant of a range of functional groups. It has limited use to the synthesis of symmetric alkanes and cycloalkanes. If two dissimilar alkyl halides are taken as reactants in cross coupling reactions, the resulting product is a mixture of alkanes that is often difficult to separate. In the case of (1,3), (1,4), (1,5), (1,6) dihalides (when used as a reactant), the reaction leads to formation of cyclic products instead of intended dimerization. The Wurtz reaction fails in case of tertiary halides. In addition, since the reaction involves free radical species in one part of the mechanism, a side reaction occurs to produce an alkene (specially with 1,2-dihalides). This side reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon atom.
There is an organometalic compound in the mechanism at one stage, which undergoes a $\mathrm{S_N2}$ replacement of halide ion of available second $\ce{R-X}$ molecule. Whether these types of organometalic compounds are ionic (cabanionic nature of $\ce{R-}$) has been discussed and reviewed using quantum chemical calculations recently (Ref.2).
Thus, it is safe to say that the reaction is better for primary ($1^\circ$) halides, regardless of whether the mechanism is free radical oriented or carboanion oriented. Even more than 150 years after Charles Adolphe Wurtz introduced the reaction in 1855, the mechanism of the reaction is still up in air, regardless of involvement of both in the mechanism.
References:

David C. Billington, "Chapter 2.1: Coupling Reactions Between $\mathrm{sp^3}$ Carbon Centers (https://doi.org/10.1016/B978-0-08-052349-1.00068-8)," In Comprehensive Organic Synthesis, Volume 3: Carbon–Carbon $\sigma$-Bond Formation; Barry M. Trost, Ian Fleming, Editors-in-Chief; Elsevier Science Ltd.: New York, NY, 1991, pp. 413-434 (ISBN: 978-0-08-052349-1).
Christoph Lambert, Paul von Ragué Schleyer, “Are Polar Organometallic Compounds “Carbanions”? The Gegenion Effect on Structure and Energies of Alkali‐Metal Compounds,” Angew. Chem., Intl. Ed. 1994, 33(11), 1129-1140  (https://doi.org/10.1002/anie.199411291).

Should it be carbanion or free radical reaction

There are evidence for both.

I know that carbanion had been trapped by using ${CO_2}$  was there a proof for free radical as well?

It was mention that formation of alkene can be discussed by only free radical mechanism. I disagreed because any carbanion acting as a base could do that. However, SET can be done only by metal acing as a free radical.

All of this reminds me of the comment of a colleague about the utility of the Wurtz reaction. "If Wurtz comes to Wurtz, use the Wurtz reaction". :)

@user55119: That's a pretty smart joke, but it's a good metaphor.:-)

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Chemical Equilibrium - Le Chatelier's principle
If the concentration of reactants is increased such that one is a limiting agent, wouldn't the concentration of products also increase (because excess of a reactant will get carried to the other(product) side  of the equation considering that the limiting agent is in aqueous medium?
e.g.
  (Consider 1 mole of each reactant)
   $$\ce{H2O + NH3 <=> NH4+ + OH-}$$
Now, let us take another scenario in which  there are two moles of NH3 and 1 mole of H20. Then concentration of one reactant has increased (from the previous scenario). Wouldn't the concentration of the products also increase because one mole of NH3 does not react and gets carried to the products' side of the equation?
I am confused with how  Le Chatelier's principle is applied in the 2nd scenario. Why does the equilibrium shift to the right if the concentration of the reactants and products increase (if my assumption is right)
If my assumption is wrong (concentration of product increases to balance excess reactant), my concern is such an equation is not balanced. I am unable to reconcile how the Le Chatelier's principle is applied in the 2nd scenario.
(Based on the answers below, I would like to restate my question)
I understand that
$$\ce{2H2O <=> H3O+ + OH-}$$
Now if I add HCl, I have
$$\ce{H2O + HCl <=> H3O+ + Cl-}$$
The way I interpret this is : the concentration of reactants  is 1 + 1 = 2 moles and the concentration of products is also  1 + 1 = 2 moles. Since the concentration on both sides of the equation is same, why do we say equilibrium shifts left. What am I missing?

The reaction may be described an equilibrium constant (actually: two constants, one for the forward reaction and one for the reverse) which is/are independent of the concentrations of the reactants. If you change the concentration of one of the molecules involved in the equilibrium, the concentrations of the others will follow.

The equation for chemical reactions is typically written for the balanced equation. Typically the excess reactants are not carried over to the right side as products. Doing that would make creating the mathematical expression for the equilibrium from the chemical formula confusing. You could write for example $$\ce{NH3 + H2O <=>[excess NH3] NH4+ + OH-}$$

It does not make any sense to consider $1$ mole $\ce{H_2O}$ ($18$ g) and $1$ mole $\ce{NH_3}$ ($17$ g). This mixture will not be homogeneous. $18$ g water cannot dissolve more than $6$ g $\ce{NH_3}$ at room temperature and pressure. For the same reason, your second system, with $2$  moles $\ce{NH_3}$ and $1$ mole $\ce{H_2O}$ is even more impossible to reach.

@MaxW: Thanks for your response. I have a follow-up question to improve my understanding. What happens when HCl is added to water? Literature says equilibrium shifts to the left because the concentration of products increases. However I don't understand why concentration of the reactants doesn't increase due to HCl (on the side of reactants)

dah... my example above with "excess NH3" was brain dead. I should have said excess $\ce{H2O}$ in that example.// Another, correct example would be $$\ce{CaCl2 + Na2CO3 <=>[excess Na2CO3] CaCO3(s) + 2NaCl}$$ where $\ce{Na2CO3}$  is added in excess to quantitatively precipitate $\ce{Ca^{2+}}$.

Rose - Consider the reaction below: $$\ce{NH3 + HCl <=> NH4+ + Cl-}$$. A way to get the overall reaction when adding HCl is to add the three reactions below and cross out common species. $$\ce{NH3 + H2O <=> NH4+ + OH-}$$ $$\ce{HCl +H2O <=> H3O+ + Cl-}$$ $$\ce{H3O+ + OH- <=> 2H2O}$$

I understood excess reactants are not included/considered in the equation. Thanks a lot for this clarification.

A bit late to this discussion, but it appears to me that your big confusion is that Le Châtelier's Principle applies when the system is at equilibrium. In your first example, when you have 1 mol of each reactant, how much of each product is present? Now if you adjust the composition of reactants, what happens to the total balance of species in terms of both reactants and products?

For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.
When you define a reaction equation like: $\ce{aA + bB <=> cC + dD}$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $[Products](t) = [C](t) + [D](t)$ and not $[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$.
--
A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."
If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $\Delta G$.
For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $dG = \sum^{n}_{i=1}\mu_idn_i$. For an equilibrium case, which means that $\Delta G = 0$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$. This particular equation is defined for standard conditions (0 degC, 1 atm).
--
We're almost to an answer, let's look a little closer at $K_{eq}$. Let's use a gas phase system $\ce{aA + bB <=> cC + dD}$ as an example. If you look up the derivation of the $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$ in any standard thermodynamics textbook, you will see that $K_{eq}$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$. If one assumes that the pressure is fixed and that $a = b = c = d = 1$, then this expression would reduce down to: $K_{eq} = \frac{y_C y_D}{y_A y_B}$.
You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $\xi$ for a reaction system.
Where, $n_i = n_{io} + v_in_{io}\xi$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $n_{io}$). If you plug this into the expression for $K_{eq}$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $N_{total}$ will cancel out of the expression leaving:
$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$
For a given temperature and pressure, the Gibbs Free Energy ($\Delta G$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $\xi$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $\xi$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $\xi$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $\xi$.

While you probably mean good, please don't finish your posts with a tagline. I also recommend our guides on formatting, please have a look here and here.

Thank you for your feedback, I will definitely look into these.


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Products of o-aminobenzaldehyde thermal treatment with acetone and further with sulfuric acid

o-Aminobenzaldehyde is heated along with acetone in dilute $\ce{NaOH}$ solution to produce the major organic product (A), which on treatment with a catalytic amount of $\ce{H2SO4},$ produces another major organic product (B) formed via intramolecular reaction.
What will be the products A and B formed in this reaction?

I have tried solving it and I think A will be o-aminobenzoic acid. But I want to ask that what happens when o-aminobenzoic acid is reacted with catalytic amount of $\ce{H2SO4}?$

Why do you think A is a benzoic acid? What oxidant is present?

Do you ever heard of aldol condensation?

This is an example of the Friedländer synthesis of quinolines Wikipedia article  here.
Under catalysis by hydroxide acetone undergoes an aldol condensation with the aldehyde group to give 3, which is your compound A ($\ce{R2=Me, R3=H}$). Under the catalysis of sulfuric acid A loses water to give the quinoline 7 (for your B $\ce{R2 = Me, R3=H}$).
Several reviews of the Friedländer exist, this is a good one Chem. rev. here.

Image is from ref 1.

Which software do you generally use for making the structures?

this diagram was lifted from the Wikipedia entry. For drawing I resort to Chem Doodle https://web.chemdoodle.com/demos/2d-sketcher/  It's free but not as good as the packages I'm used to.


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Photochemical rearrangement of 4,4‐diphenylcyclohexa‐2,5‐dien‐1‐one

I can see how the option A and B are correct as it is just a standard dienone–phenol type reaction which involves a phenyl shift. However, the given answer is A, B, C and D.
I do not see how C and D are formed. Does the presence of light affect the reaction? If so, how?

For formation of D, electrophilic nature of double bond is considered.

@Zenix I just tried doing that now, it leads to the formation of that 3 mem-ring but for me the 3-mem ring's position was different. It came directly opposite to the c double bond o. Could u help me out pls?

C is a bogus result . Only five carbons. D is doable. If your 3-membered ring is the obvious one bearing two geminal phenyls, have the radical fragment it such that the diphenyl methyl radical is formed. Now couple this radical with the one on the oxygen resonance form.

@user55119 I cant believe that i didn't notice that c had only 5 c atoms. thanks. But the next part i dont get. Do you mind posting an answer? Thanks a lot!

The formation of structure 6 (your D) occurs as follows. Photolysis of dienone 1 affords excited state 2 that forms cyclopropane 3. In turn, 3 is capable of breaking a different cyclopropane bond to form 5-membered ring 4. Structure 5 is a resonance structure of 4. Structure 5 closes to the [3.1.0] product 6 (your D). I don't know how you formed A (12 here) but further photolysis of 6 can form 12 via a bridged phenyl migration as shown in structure 9. A similar migration in structure 2 leads to your B. Of course, structure C has only five carbons and was added to the choices as a trap. See reference 1 for an example of a dienone phenol rearrangement.

1) A. G. Schultz and S. A. Hardinger, J. Org. Chem., 1991, 56, 1105.

The above answer is totally Correct but for your information, if light was not there then there would have been a RETRO PINACOLE-PINNACOLON REARRANGEMENT.


text:
What should be the name of this compound containg both double and triple bond?
Which is a more correct name, according to IUPAC nomenclature: pent-2-ene-4-yne, or pent-3-ene-1-yne?

You should always draw alkynes with 180° bond angles. Unless you really have no choice (e.g. it's in a very small ring).

Sort of the same as this: https://chemistry.stackexchange.com/questions/67894/ but this is more concrete.

According to R-3.1.1 of IUPAC, Commission on Nomenclature of Organic Chemistry:

Numbers as low as possible are given to double and triple bonds as a set, even though this may at times give "-yne" a lower number than "-ene". If a choice remains, preference for low locants is given to the double bonds. Only the lower locant for a multiple bond is cited except when the numerical difference between the two locants is greater than one, in which case the higher-numbered locant is cited in parentheses (see R-0.1.4)

Numbering from the right side of the molecule as seen in the question figure will assign the locants (2,4) to the double and triple bonds respectively.
On the other hand, following a similar numbering process from the left side of the molecule as seen in the diagram will yield (3,1) as the locants for the double and triple bonds respectively.
Now, as per IUPAC Recommendations 1:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Therefore, since (1,3) is a lower set of locants as compared to (2,4) as per the above definition, the IUPAC name of the compound should be pent-3-en-1-yne
Note that this molecule can also exhibit stereoisomerism by the virtue of it's double bond, so depending on whether the hydrogens are cis or trans to each other, the complete name of the molecule could be (3 Z)-pent-3-en-1-yne or (3E)-pent-3-en-1-yne, respectively.

Note that this is a slightly outdated version of the IUPAC Recommendations (from 1993). The newest version is from 2013: https://pubs.rsc.org/en/Content/eBook/978-0-85404-182-4 Anyway, the corresponding rule is not changed very much, so the preferred IUPAC name is essentially the same. But the stereodescriptors should be written as (3Z)- or (3E)-.

@orthocresol Ok thanks. Edited accordingly


text:
What happens to chemical bonds if the nucleus changes?
Lets say I have some $\ce{PuF6}$ (with the Plutonium being $\ce{^{240}Pu}$). I take this $\ce{PuF6}$ and bombard it with neutrons. The $\ce{^{240}Pu}$ nuclei capture the neutrons, forming $\ce{^{241}Pu}$, which decays after some time to $\ce{^{241}Am}$. $\ce{AmF6}$ is not a stable molecule from what we know - we can't synthesize it in labs. So what exactly happens to the chemical bonds? $\ce{AmF4}$ seems stable - does a difluorine molecule just get "kicked" off?

https://chemistry.stackexchange.com/questions/111356/effect-of-spontaneous-transmutation-of-atom-in-crystal

https://chemistry.stackexchange.com/questions/65809/what-happens-to-a-radioactive-carbon-dioxide-molecule-when-its-carbon-14-atom-de


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How is carbon-halogen bond broken during Finkelstein reaction?
Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide:
$$\ce{R−CH2−Cl + KI -> R−CH2−I + KCl}$$
From an answer of a previous question:

As per Le chatelliar's principle the forward reaction is favored in
  presence of dry acetone which will dissolve $\ce{KI}$ but not
  $\ce{KCl}$ or $\ce{KBr}$. On account of insolubility of
  $\ce{KCl/KBr}$; they are not available for backward reaction[...]

This reaction proceeds with $\mathrm{S_N2}$ mechanism; i.e. it is a single step reaction. $\ce{C-Cl}$ bond breaks and $\ce{C-I}$ bond forms. Now we know that energy for bond breaking comes from bond making. However we know that
$\ce{C-I}$ bonding cannot supply enough energy to break $\ce{C-Cl}$ bond.
To put in other words: chlorine is a weaker leaving group than iodine.
From where does the energy to break $\ce{C-Cl}$ bond comes then because for $\ce{KCl}$ (precipitate) to form; in first place the $\ce{C-Cl}$ bond must break?

The solvent comes into play during the transition state (SN2 mechanism) ... Equilibrium ..where the possibility of formation of KCl is favoured due to the precipitation ( thanks to the solvent) and thus forward reaction occurs. Try understanding from the mechanism point of view..maybe it would help.

Think of the overall process and you will see where the energy comes from
Looking at just the reaction of iodide with the alkyl chloride doesn't give you the complete picture of the reaction. Sure, in that part of the reaction the bond-breaking and bond formation don't balance out. But in reality this doesn't mean there is no reaction: it just means that the equilibrium of that step strongly favours the alkyl chloride (there is still some dynamic formation of the iodide, perhaps in small quantities).
But the driving force here isn't that initial reaction. The real driving force is the precipitation of the KCl (it is very insoluble in the conditions of the reaction). From a mechanistic point of view this precipitation virtually eliminates the presence of free chloride in solution meaning that, if any is formed, it will rapidly disappear and not be available to drive the equilibrium back to the more stable alkyl chloride (even if the equilibrium of the halide exchange strongly favours the chloride).
From an energy or thermodynamic point of view, the overall reaction gets enough energy from the precipitation of the KCl (which releases energy) to compensate for the energy lost by substituting iodine for chlorine. So, if you look at the overall mechanism, the second step (precipitation) makes up for the apparent unfavorability of the first step.


text:
Why are pressure or volume not able to change the equilibrium constant?
It is stated in my textbook that temperature is the only factor that can change the equilibrium constant. According to the Ideal Gas Law, temperature is dependent on the pressure and volume. Thus, shouldn't pressure and volume change the equilibrium constant too?

Inasmuch as they change temperature, yes. Short of that, no.

Yes, it is dependant on pressure and volume, but a change in pressure or a change in volume doesn't necessarily mean a change in temperature. Thus, it doesn't affect the equilibrium constant.

I would add to the answer in the linked question, that you should be careful to distinguish between the standard equilibrium constant and equilibrium constants in general. The first do not depend on pressure because (as the linked answer explains, perhaps not entirely clearly) they are defined at a fixed standard pressure (each reagent and product is in pure form at the given T and at the standard pressure, typically 1 bar).

Just like any sweeping statements made by the General Chemistry textbooks, this statement is also not completely true. When you work at pressure extremes, as in modern day chromatography, such as 1000 or higher bar, large molecules can change their shape, in that case, one can see a change in retention factors which sort of indicates a change in equilibrium constant with pressure under constant volume.
Just for fun, some solvents can become solids at high pressures-but this happens at another high-pressure level.
Coming back to routine reactions, pressure will not affect equilibrium constants because whenever you try to adjust the pressure in a gaseous reaction, the equilibrium concentrations will change in such a way that their ratio remains constant-hence the equilibrium constant does not change.
One way to think about it as that the equilibrium constant is a ratio of forward and backward rate constant. Changes in pressure or volume will not change the kinetic energy of the molecules but only temperature can change the kinetic energy and hence affect the rate constants.

According to the Ideal Gas Law, temperature is dependent on the
  pressure and volume.

Mathematically, this is incorrect. You would rather say that pressure and volume of a gas are functions of temperature i.e., the converse of what you wrote.

Splitting hairs here but if you perform an adiabatic expansion on a gas its temperature drops.

Right, I was thinking about it but didn't want to confuse the student.

"the equilibrium concentrations will change in such a way that their ratio remains constant"How is it any different from temperature changing the equilibrium concentrations? Why do pressure change in the way described?

@Zratos, I edited the answer.

You should be careful not to confuse the student further, because it is likely that the question arises from a misunderstanding of the difference between equilibrium constants in general and standard equilibrium constants. Also, you invoke Le Chateliers principle as the reason that the equilibrium constant is invariant, but this is not the true reason why the constant does not change.


text:
Calculating ratio of conjugate base and acid when pH and pKa are given

Buffer solutions are used by biological mammalian systems to maintain the $\mathrm{pH}$ of blood plasma within a narrow range. In these systems, the compound from which this solution is obtained is $\ce{CO2}$, produced in cell respiration, which is converted into $\ce{HCO3-}$ and $\ce{H2CO3}$ inside the red blood cells. Using the Henderson-Hasselbalch equation, answer: $\mathrm{p}K_\mathrm{a} = 6.35$. What is the ratio between $\ce{HCO3-}$ and $\ce{H2CO3}$ in a blood sample with $\mathrm{pH} = 7.4$?

The problem is I try to make the exercise in this way:
$$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}} \right)$$
BUT in the solution they make $$\mathrm{pH} =\mathrm{p}K_\mathrm{a} - \log \dfrac{[\ce{HCO3-}]}{[\ce{H2CO3}]}$$
They make the inverse of $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \dfrac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$$
And this don't make sense for me. I don't understand.
The solution given is in the white box on the left side.

Recall the fundamental properties of the logarithm
$$+\log \left(\frac{a}{b}\right)$$
is the same as
$$-\log \left(\frac{b}{a}\right)$$
so Henderson Hasselbach can have plus or minus signs. The correct equation is
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{conjugate base}}{\text{acid}}\right)$$
or
$$\mathrm{pH}  = \mathrm{p}K_\mathrm{a} - \log \left(\frac{\text{acid}}{\text{conjugate base}}\right)$$
Now sort out whether the textbook solution is right or not.

I wish an image of the text would posted too in such cases. I'm guessing that by far the most probable explanation was that the OP just missed the flip in the log, but it is possible that the book was wrong.

Students can be further confused by the fact that $p=-log$. That is another headache.

Good night. The acid is H2CO3, right? And the conjugats base is HCO3-. In this reaction H2CO3 -> H+ + HCO3-

I put the image now

@Liz, The solution in the white box is correct. See the equations I wrote, with negative sign with log, acid will be in numerator.

"What is the ratio between HCO3− and H2CO3" don't mean HCO3-/H2CO3? Can you say which one is the acid?

The acid is H2CO3 and its conjugate base is HCO3-

Sorry for many questions, but I really don't understand. If the question is HCO3-/H2CO3 (CONJUGATE BASE)/(ACID), why is used ACID/CONJUGATE BASE in the resolution?

Did you read my answer once again?

Important is the knowledge how $K_\mathrm{a}$ is defined:
$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$
If we perform logarithmization, we get:
$$\log{K_\mathrm{a}}=\log{\ce{[H+]}} + \log{\frac{\ce{[A-]}}{\ce{[HA]}}}$$
If we apply the operator $\mathrm{p}X = - \log{X}$:
$$\mathrm{p}K_\mathrm{a}=\mathrm{pH} -\log{\frac{\ce{[A-]}} {\ce{[HA]}}}$$
respectively
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[A-]}} {\ce{[HA]}}}$$
In our case, $\ce{HA = H2CO3}$ and $\ce{A- = HCO3-}$.


text:
What is the product of reaction between 3,3-dimethyl butanal and SeO2 followed by conc. OH- ions?

So basically first step will be replacement of $\ce{-CH2 -}$ by a ($\ce{=O}$ bond) making a aldehyde and ketone side by side. Now what should I do after this? I tried attacking the aldehyde first as it is more electrophilic. I think both carbonyl groups will be attacked because of high concentration of hydroxyl ($\ce{OH-}$) ions. I even tried dehydration of  the formed product after attacking with $\ce{OH-}$ ions but I always end up with a ketone and a carboxylic acid by its side  after this.

First is Riley oxidation, then is cannizzaro reaction.

oh! damn that was simple thanks T-T

@Zenix Please avoid answering in comments, it is not aligned with our site policy. This is a good example of a homework question, so I don't see why you shouldn't try answering it. ;)

Is  answer  of this question B?

@WilliamR.Ebenezer Thanks for the encouragement, I am (and was) quite busy, will try to answer within a day.

@WilliamR.Ebenezer Answered within 23.30 hours ;)

In the first step, Riley Oxidation (selenium dioxide-mediated oxidation of methylene groups adjacent to carbonyls) is taking place. You will get A as:

A on reaction with conc. $\ce{OH-}$ (i.e. in strongly alkaline condition) undergoes redox reaction (famous Cannizzaro reaction). And terminal aldehyde will be oxidized to form the metal salt of acid, ketone will be reduced to form the alcohol, note that intramolecular reaction is taking place, to finally form the product B.


text:
How is octane rating defined for negative values and those over 100?
Based on my reading, the octane rating for a given fuel is calculated by finding the compression ratio at which it knocks and finding a mixture of $X\%$ isooctane and $(100 - X)\%$ n-heptane which knocks at the same compression ratio. The $X\%$ isooctane is then that fuel's octane rating.
But how does this work with negative numbers or those over 100?

According to the standards group ASTM International, octane numbers above 100 are defined using the formula:
$$ON=100+\frac{28.28T}{1.0 + 0.736T + (1.0 + 1.472T - 0.035216T^2)^{0.5}}$$
where $T$ is "mL tetraethyllead (TEL) per U.S. gal in isooctane" [1]. This is the standard used by the FTC to regulate automotive fuel ratings in the US [2]. It should be noted that this formula is not an extrapolation of the octane curve from 0 to 100 as many seem to think. In fact, if we were to continue the octane scale, "[the] increase in the size of the octane unit [would be] so rapid as to make the unit become infinitely large if the scale were extended to about 116." [3] We can see this graphically:

So how was this formula determined? I could not find this directly specified anywhere, but ASTM gives us a clue in another publication relating to another number usually reserved for aviation fuel, the Performance Number, via the following formula [4]:
$$ PN = 100 + 3(ON - 100) $$
Performance number on the other hand is very well defined for numbers above 100: "This scale related percentage-wise the knock-limited imep (indicated mean effective pressure) obtainable from the test fuel to that obtainable from isooctane, assigning a value of 100 to the imep obtainable from isooctane." So for example, a performance number of 130 would mean that fuel had 130% the in-cylinder pressure of a fuel containing 100% isooctane. We can see a graph relating ON, PN, and TEL here [5]:

So what does all this mean? It means that an octane number above 100 is simply 100 plus the knock-limited indicated mean effective pressure over that of 100% isooctane divided by 3. Why divide by 3 and not use PN / KLIMEP instead? @Poutnik makes the observation that the factor of 3 ensures the PN-ON curve remains a linear extrapolation past 100.
I could not find anything regarding negative octane numbers. I can only assume people are extrapolating from the existing octane curve, but as seen above, the nonlinear, asympototic nature of octane numbers certainly would give me pause about the information conveyed by such a number. Again, if anyone has more info, please post!

[1] ASTM D2699-04, Standard Test Method for Research Octane Number of Spark-Ignition Engine Fuel, ASTM International, West Conshohocken, PA, 2004, www.astm.org
[2] 16 CFR § 306.5
[3] BROOKS, DONALD B. “Development of REFERENCE FUEL SCALES for KNOCK RATING.” SAE Transactions, vol. 54, 1946, pp. 394–403. JSTOR, www.jstor.org/stable/44548282.
[4] “Standard Method of Test for Knock Characteristics of Motor and Aviation Fuels by the Motor Method.” ASTM Manual for Rating Motor, Diesel, and Aviation Fuels: 1973-74, American Society for Testing and Materials, 1973.
[5] MacAuley, J. B. “Advances in the Utilization of Tetraethyllead.” Third World Petroleum Congress, The Hague, 1951: Proceedings, E.J. Brill, 1951, p. 157.

So at least the aviation PN is an extrapolation of the ON  curve.

Not sure I'd call it extrapolation, @Poutnik. In the second picture, the right side uses a completely different x-axis (mL TEL) and we can see the curve, instead of continuing upwards to infinity, curves "down" instead. But it certainly continues off from where 100% isooctane ends ($100_{ON}=100_{PN}$)

But the picture on the right is just the extrapolation aviation formula PN = 100 + 3*(ON-100), for ON=100..120 with the coordinate X substitution ON=f(T) by the formula on the top, getting PN=100 + 3 * ( f(T) - 100). The factor 3 is then the empirical slope of the tangent line dPN/dON.

So they used the aviation linear extrapolation formula PN=g(ON), determined dependence of PN=h(T) and from that they derived ON=f(T).

Hmm... I see what you're saying. I couldn't find an explicit statement of motivation, but the desire to keep the same tangent line would explain the factor 3 and makes sense.

I recommend the @Stanley Yu own answer, as I admit I was not particular familiar with the shape of the ON number curve.

It is done by an extrapolation of the critical compression ratios of mixtures 0-100% of isooctane and 100-0% of n-heptane for ON values <0 and >100.
If, just illustratively, 100% isooctane had the critical compression ratio 1:20, 90% isooctane 1:19 ,... 90% n-heptane 1:11, 100% n-heptane 1:10, then the fuel with the critical compression ratio 1:21 would have the octane number 110.
The octane number/rating differs for different procedures. See Research octane number RON, Motor octane number MON,  Anti-knock index AKI=(RON + MON )/2 and Observed Road octane number RdON.

As seen in the OP own answer, the extrapolation of the octane number (ON) scale  above 100 is done by linear  extrapolation by the empirical aviation formula for performance number $$PN=100 + 3 * ( ON - 100 )
where on the graph is coordinate ON substituted by tetraethyllead addition.

Thanks. While that certainly makes sense in "completing the line", is it not actually defined? At that point these numbers bear no relation to the amount of n-heptane / isooctane. Seems dangerous to be comparing these numbers outside their domain.

What is the danger ?

My understanding is that motor vehicle fuel is not made with pure ratios of those chemicals anyway, so "octane rating" is just a proxy indicator or critical compression ratio. Through that lens, it seems benign having fuels that are not possible with reference reagents.

Maybe "dangerous" was poor word choice. I mean it is difficult to interpret its meaning with regards to its original definition. If we detach the octane number from its definition, and apply it to its intent, then I don't see any issue. It makes sense.

@Stanley Yu Why difficult ? Consider ON = f(x(IO)) as the calibration curve, not as the direct comparison of reference and test fuel with the same behaviour. It is common practice to extrapolate calibration curve if it behaves well and statistical errors are low.  BTW, even in IO+PbEt4 case, the performance number for the mixture is primarily defined by the above extrapolated function for n-heptane + isooctane. That mixture is just a practical, secondary standard to avoid testing the full calibration scale on an eventual non standard engine in ON testing context.


text:
absorption spectrum of hydrogen atom
In absorption spectrum of hydrogen atom, only one electron is present in its one atom which is in ground state, so it means that all electrons can only absorb energy of photon of wavelength which lies in UV region to get to a higher energy state (by calculation it can take max wavelength =122.55nm and minimum wavelength =91.9nm).Then why do we see dark lines in its absorption spectrum in visible region also, how can it absorb energy which is in visible region to get to a higher energy state i.e. how can it absorb wavelength like 656nm? How can the photon of this wavelength make the electron to get to a higher energy state? Please explain it intuitively (using simple mathematics)?

There is a misunderstanding in the question. Happy Sharma says that "we see dark lines in its absorption spectrum". This has no meaning. The absorption spectrum of H  is made of thin bright lines, and has no dark lines. Dark lines may only be seen in continuous spectra like in the spectrum of the sun.

thin bright lines are seen in emission spectrum ,in absorption spectrum there are some dark lines lines which are getting absorbed

NO ! There are no dark lines in the absorption spectrum of Hydrogen atom.

By principle, absorption spectra do have dark lines, as incoming light is absorbed, if it's energy equals the transition energy between 2 energy levels. Absorption spectrum is determined by the ratio of passing and coming light intensity ( possibly modulated ). If it had bright lines, it would mean hydrogen would absorb everything but these respective energies.

@Maurice, I am afraid you have a misconception regarding absorption spectrum when you state that "the absorption spectrum of H is made of thin bright lines".

@M. Farooq. You are right. I should have said "the emission spectra is made of thin bright lines". The absorption spectrum cannot be made of "bright" lines

Not a bad question but badly worded. Basically, you are wondering how can hydrogen atoms absorb 656 nm? The point is why not? Calculate the energy corresponding to 656 nm, this energy corresponds to a energy level difference in the electronic states of hydrogen atom.
Now note that you are interested in absorption spectrum. Now this is punch line. You have to do a thought experiment. Hydrogen is a diatomic gas, first you will have to provide enough energy to hydrogen that it be atomized. Now 656 nm line absorption corresponds to a transition from n=2 to n=3 as shown below.
Now you will only see this absorption when you have an electron in the hydrogen atom, which is already sitting at n=2 (this means you already have a very high temperature that that you are dealing with an excited hydrogen atom). Once this excited hydrogen atom absorbs 656 nm light, the electron will be excited to n=3.
Now no atom "absorbs" dark lines or emits dark lines. This is a poor choice of wordings. The dark lines indicate the absence or(better diminished intensity) of a certain wavelength from a continuous spectrum.


text:
what is a sugar unit?
How a sugar unit is defined?
I am trying to understand the structure of calicheamicins, and it is noted that They consist of an enediyne headpiece and a modified carbohydrate tail consisting of four sugar units and an unusually substituted aromatic ring. I can understand in the picture where sugar unit is referred, but I would like a general definition.
Thank you

The sugar unit features a linear chain of carbon carbon bonds. Distinct units are usually connected by glycosidic bonds (which may be hydrolyzed without breaking carbon carbon bonds).

A sugar unit is a monosaccharide, the Wikipedia entry is a good introduction  https://en.wikipedia.org/wiki/Monosaccharide

Calicheamicins are natural products, which were isolated from the fermentation broth of Micromonospora echinospora subsp. calichensis. These antibiotics exhibited significant activity against Gram-positive and Gram-negative bacteria in vitro. Calicheamicin $\gamma_1$ (Ref.1) demonstrated antitumor activity against P388 leukemia and B16 melanoma in vitro (Ref.2).
The structure of Calicheamicin $\gamma_1$ contains biologically active part of bicycloring system, which is described in the following diagram:

The rest of the structure contains four modified sugar units and one aromatic nucleus. The sugar 1 unit is a 6-deoxy-4-aminohexose unit (that's why it is called a modified sugar). Its anomeric oxygen (hemiacetyl $\ce{OH}$) bound to the $\ce{C}$9 position of the endiyne unit in $\beta$-linkage. The sugar 2 unit is a 2,6-dideoxy-4-thiohexose unit, hence it is also a modified sugar. This linkage is unusual among other sugar linkages in natural origin. The inkage is a hydroxyl amine bond ($\ce{-NH-O-}$) without eliminating $\beta$-$\ce{OH}$ of 2nd sugar unit. According to proposed biosynthetic pathway, these two sugars have been added to endiyne unit in first and second steps of the pathway:

The third step of the biosynthesis is addition of the aromatic nucleus (5-iodo-4-hydroxy-2,3-methoxy-6-methylbenzoate unit) to 4-thial group of sugar 2 as a thioester bond. The forth and fifth additions of other sugar units can be alternative. Nonetheless, one sugar (a 6-deoxy-3-O-methylhexose unit) attached to the free 4-phenolic $\ce{OH}$ group of aromatic nucleus in $\alpha$-linkage. The last sugar unit is attached with its $\alpha$-$\ce{OH}$ of the anomeric center to the free 2-$\ce{OH}$ group of the first sugar in $2 \rightarrow 1$ linkage. The unusual about this sugar is it is a pentose (a 2-deoxy-3-O-methyl-4-N-ethylaninopentose unit) instead of hexose among other three.
You can find a good description of mechanism of action of Calicheamicin $\gamma_1$ in Ref.3.
I hope this is what you want to know about this molecule.
References:

May D. Lee, Theresa S. Dunne, Marshall M. Siegel, Conway C. Chang, George O. Morton, Donald B. Borders, “Calichemicins, a Novel Family of Antitumor Antibiotics. 1. Chemistry and Partial Structure of Calichemicin  $\gamma_1$,” J. Am. Chem. Soc. 1987, 109(11), 3464-3466 (https://doi.org/10.1021/ja00245a050).
W. M. Maiese, M. P. Lechevalier, H. A. Lechevalier, J. Korshalla, N. Kuck, A.Fantini, M.J. Wildey, J. Thomas, M. Greenstein,  “Calicheamicins, a novel family of antitumor antibiotics: taxonomy, fermentation and biological properties,” The Journal of Antibiotics 1989, 42(4), 558-563 (DOI: 10.7164/antibiotics.42.558).
Miklós Simonyi, ”The Concept of Chiral Conformers and its Significance in Molecular Pharmacology,” In Advances in Drug Research: Volume 30; Bernard Testa, Urs A. Meyer, Eds.; Academic Press: San Diego, CA, 1997, pp. 73-111 (ISBN: 0-12-013330-X).

thank you!may I aks if you have such a clear explanation about the agent N-acetyl gamma calicheamicin dimethyl hydrazide (NAc-gamma calicheamicin DMH) as well?

Sounds like it it is  calicheamicin $\gamma_1$ with a linker. I didn't know biochemical manipulation of the natural product. Yet, it won't hurt if you ask the question.

yes it has been used for the production of a drug, gemtuzumab ozogamicin. thank you! https://www.ema.europa.eu/en/documents/assessment-report/mylotarg-epar-refusal-public-assessment-report_en.pdf

hello! could someone help me to decipher the structure of this molecule? I am trying to understand the structure of gemtuzumab ozogamicin: The anti-CD33 antibody is acylated with the bifunctional AcBut linker, which allows for hydrazone
formation with the calicheamicin derivative. This derivative, N-acetyl-gcalicheamicin
dimethylhydrazide (DMH), has a disulfide stabilized by both the two methyl groups and the calicheamicin bicyclic core, and an acetylated amino sugar. References: Joel C. Barrish, Percy H. Carter, Peter T. W. Cheng, Robert Zahler Accou...

@ marilu: You should post this as a new question for better results.


text:
Gibbs–Helmholtz equation
The Gibbs–Helmholtz equation was developed in the 1870's, while Nernst's heat theorem was developed in the early $20^{th}$ century (according to google/wikipedia). Nernst's heat theorem tells us that the change in entropy tends to be zero in absolute temperature. But did Nernst's heat theorem provide any new insight? Because if we put $T=0$ in the Gibbs–Helmholtz equation, then  $\Delta G=\Delta H$. Anyone could have thereby concluded that as temperature approaches 0, entropy change becomes zero already in the $19^{th}$ century. I am quite confused regarding it, those were my assumptions (probably wrong), what was new about Nernst's heat theorem ?

If you read the Wikipedia page for Nernst's Heat Theorem carefully, it explains that the novelty lies not in showing (as you point out) that as $T \rightarrow 0$, $\Delta G \rightarrow \Delta H$, but rather that as   $T \rightarrow 0$, the slope of $\Delta G$ as a function of T goes to 0, that is
$$\lim_{T\rightarrow0} \left( \frac{\partial \Delta G}{\partial T} \right)_p =0$$
Then, since that slope is given by $-\Delta S$, it predicts the experimentally confirmed conclusion that as $T \rightarrow 0$, $\Delta S \rightarrow 0$,


text:
Chemical bonding and crystal structure of silicon
I can understand that in Silicon Crystal a central Silicon atom shares its 4 valance electrons with 4 surrounding silicon atoms. And the chain continues. But then What happens to the outermost such Silicon atoms (in the drawing). They each will have incomplete so called octate.

What drawing? Have you seen What is the edge of a diamond like?

See: https://chemistry.stackexchange.com/questions/34971/why-do-we-not-see-silicon-in-a-structure-like-graphite

In a Silicon crystal, outermost Silicon atoms are probably oxidized yielding a monoatomic layer of Oxygen atoms, chemically bound by covalences like in $\ce{SiO_2}$, by analogy with Aluminium. Silicon and Aluminium are neighbors in the periodic table : they often have similar properties. And it is well known that an extremely thin layer of Aluminium Oxide $\ce{Al_2O_3}$ does cover Aluminium foils and pieces. It must be the same for Silicon.

Is there any reference for the claim?

Something like that is at least mentioned here on page 4 (15 in pdf) on the bottom.
https://pure.tue.nl/ws/files/3477173/148292.pdf

I did upvoted because beside references would be indeed beneficial the answer has the merit to point OP in the right direction. Every non molecular covalent lattice has boundaries at higher potential than the interior, so in many cases the atoms at the surface saturate their valence shell reacting with the surrounding. Just perfect vacuum would prevent reaction or adsorption, the latter doesn't fix the question but lower the potential as well. For Silicon, a layer of SiOx forms in instants in normal atmosphere. Part of the so called dampling bonds are saturated by OH group from moisture, too.


text:
Reaction involving Hydroperoxide group
THE GIVEN QUESTION

MY ATTEMPT

The answer to this question was not given in my book.I want to know if this is the correct reaction mechanism and these are the products which are formed.

Within minutes of asking the question, it has a down vote. Can anyone care to explain what is wrong with it so I can improve it?

definitely. your answer is right. this link may help you: https://en.wikipedia.org/wiki/Cumene_process

regarding to your query you raised in the form of a comment, I think it may be a standard reaction for graduates or post graduates. for undergraduate level, not so.

@user93057 Thank you. I had identified the question was related to the Cumene Process mechanism but was confused about the shifting of groups and therefore asked the question. Thanks for verifying that it is correct

You're welcome. it is based on just two main factors: the oxygen linked with hydrogen is more electrophilic as it has lesser steric hindrance than that beside it and that the phenyl group with OCH3 is more active towards EAS, which you've identified well.

Welcome to chem.SE. Have a look at the tour page  to have an idea what kind of questions are accepted on this site. Yes, downvotes can be discouraging and I assure you most of the users here(me including) has faced downvotes. It can't be helped. Downvoters remain anonymous unless specified. All you can do is improve your post, ask good, well researched question, format your question to look good, use software to draw chemical structure rather than clicking pictures. Hope, it helps


text:
Can a catalyst appear in the rate equation?
According to Wikipedia:

Chemical reactions occur faster in the presence of a catalyst because the catalyst provides an alternative reaction pathway with a lower activation energy than the non-catalyzed mechanism. In catalyzed mechanisms, the catalyst usually reacts to form a temporary intermediate, which then regenerates the original catalyst in a cyclic process.

Since catalysts react in intermediate reaction and hence in the rate-determining step, is it possible that they appear in the kinetic rate equation?
If so are there some examples?
Edit: For more clarity I encountered this doubt in a true or false question which went like this:" Catalyst can appear in kinetic rate equation".
Edit 2: As told in the comments, another similar question says specifically that catalyst are not included in rate equations, but the question I encountered says it might sometimes. So there is some irregularity in my question which seems highly unlucky to me as I have read some examples like hydrolysis of esters where rate is directly proportional to H+ concentration.
Am I wrong?

But that significantly says that catalyst is not included in rate equation. But the question I encountered says it might sometimes? Also I have seen that in hydrolysis of esters, rate of reaction is proportional H+ concentration?

In some cases the catalyst concentration will disappear in simplified rate equations especially when using the steady-state assumption to generate tractable rate equations. But some reactions are not well modelled by that assumption so the catalyst might need to appear in the equation.


text:
Where will halogenation take place?

In the following halogenation reaction:

At which position will halogenation take place?

It is obvious that the halogenation will take place at a 3° Carbon atom (the rates of halogenation for 3°,2°,1° atoms follow the ratio 1600:82:1 or thereabouts). The question is which 3° atom? I labeled the atoms as shown:

I believe that halogenation will take place at A because the cyclopropyl ring can stabilize the free radical formed at A better than the 6-membered ring. A similar question that deals with carbocations is here, which shows the stability of the cyclopropylmethyl carbocation. Can a similar logic also be applied to free radicals? Please confirm.

What do you know about the stability of cyclopropyl radicals?

@Waylander they are more stable than 3° radicals

And they ring open

@Waylander I did not know that.

I'll try to answer this on my own using the first page of this as reference: https://pubs.rsc.org/en/content/articlelanding/1993/cs/cs9932200347/unauth#!divAbstract

Cyclopropyl methyl radical is different from cyclopropyl radical. Why one C-H bond and not the other? Do bond angles play some role?


text:
Why is hofmann product prefered here?
In the following reaction, Hofmann elimination is preferred.

This reaction takes place by E1 mechanism. I thought the Zaitsev product would be major due to 1 extra hyperconjugating structure. In the book, it is given that it is due to sterically compressed alkene, so Hoffmann product is major. I am unable to understand why the sterical hinderance would lead to the instability in alkene as the three methyl groups is one bond away from alkene.

The -ch3 bonds are all connected via single bonds. This means that they are free to rotate. Even though they are one carbon atom away, they can swing and rotate causing steric hinderence thus leading to the hoffmann product.

@boomavijay but why is steric hinderance causing instablitity in the alkene?

No , steric hinderence causes the base oh- to not be able to approach and abstract that hydrogen. So that alkene is not formed, but is infact the more stable alkene.

I agree, this is all about the tBu group hindering the approach of the hydroxide ion

But this is E1,the base has nothing to do with this, the carbon gets differentiated in positive and negative ion

Regardless of the mechanism, there hast to be proton abstraction. That's what  Waylander and booma vijay are saying.

@MathewMahindaratne Oh okay,thanks.But is the former alkene more stable than latter due to latter being "sterically compressed alkene" or is the book completely wrong.The book is Organic chemistry part 1 by ks verma.If alkene is more stable,can you please tell why as I am unable to find any articles related to this concept of sterical compression.

Also if the base is very small,would the above be applicable(as it is not mentioned about the base at all)?

tert-Bu group is enormous. Even $\ce{OH}$ grop cannot penetrate through because your carbocation is also tertiary. Thus, Hofmann product is preferred.

why do you say that this reaction proceeds through E1 in the first place? I see we have a strong base and also a good leaving group. So shouldn't it be an E2? (also we have anti-periplanar hydrogen arrangement)

The Hoffmann product is formed as a major product generally in four cases. They are:

When the ${\gamma}$ carbon is 4°
When the leaving group is poor leaving group
When the leaving group is bulky leaving group
When the elimination is internal elimination (${E_i}$ mechanism)

Case 1:
It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product
Case 2:
It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)
Case 3:
It happens  in  the Hoffmann's exhaustive methylation reaction where the bulky leaving  group  is a 3° amine group
Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters
Lastly I want to add that these are general results the yields may vary based on reactant.

Look at the carbon which has more α-hydrogens and thus will be stabilized more by resonance when it is attacked by the $\ce{OH-}$ ions. So the carbon atom on the right hand side that has two methyl groups has more α-hydrogens with it and is stabilized by resonance and  $\ce{OH-}$ is attacked there. Hence the major product has carbon with the double bond.


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Covering conformers in a molecular modelling; consider them as a static entity or work with modelling shifts as well?
I am currently working with a molecular docking project, examining potential affinities of a solvent on a protein in an assay to evaluate potential error margin. The solvent is cyclohexanol (referred to as 'ligand'), and I am trying hard to wrap my head around the solution. I have a rotatable bond at the OH group, and this is quite easy to resolve by just rotating it around and trying to fit (most software does it for you, but it can be even calculated by hand, corresponds to left and middle pic). However, cyclohexanol can (theoretically) find itself in two conformations ('chair', 'boat', with latter being much less stable) within its structure (pic at the bottom for those unfamiliar). While one of them is higher energy and therefore not found as commonly in a regular situation, for all what we know, it may have much higher affinity and stability when confronted with aminoacids in the binding site, and therefore it would maybe be unreasonable to consider only the 'easiest' state should I wish to characterise potential energies of whole ligand-protein system. I did not manage to find as of now a way to model such transitions in the context of molecular docking, and I am wondering: should I just fit twice to 'protein 1', considering cyclohexanol-chair as 'ligand 1' and cyclohexanol-boat' as 'ligand 2', or is there a method used in computational chemistry I haven't thought of/managed to find? Or should I just discard the option of a much less energetically favourable 'boat' (third in the picture) ?

It sounds like you'd like to see "the smoking gun", that is, see a conformational change under the influence of the interaction with the protein. But this is quite unnecessary. Just run docking with the two different conformers and compare the energies you obtain to those of the "ligands" in the absence of protein.

It depends on how much higher in energy the 'unfavorable' conformer is (e.g., 30 kcal/mol can safely be discarded) .. or you can run some dynamics with multiple conformers. As @BuckThorn mentions, it's generally considered okay to dock with multiple conformers - often 'geometrically diverse' to examine a range of possibilities.


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Why do the first and second laws of thermodynamics not contradict each other?
We're learning about entropy right now in general chemistry, and I'm trying to understand something.
From the fact that entropy can be directly compared with enthalpy, free energy, electromotive force, etc., and its being measured in units of $\pu{J//mol*K}$ – including joules, the units of energy – it would seem that entropy is some form of energy. By the first law of thermodynamics, then, entropy can't be created nor destroyed, only transformed and transferred. From Gibbs' equation we see that entropy is being transformed out of the molecular potential energy, and from Clausius' equations we see that entropy can be transferred between a system and its surroundings, with $\Delta S_\mathrm{system}=-\Delta S_\mathrm{surroundings}$. Since $\Delta S_\mathrm{universe}=\Delta S_\mathrm{system}+\Delta S_\mathrm{surroundings}$, it follows that $\Delta S_\mathrm{universe}=0$ – as one would expect from the first law.
Yet the second law of thermodynamics states that $\Delta S_\mathrm{universe}>0$.

If entropy is a form of energy, then how can universal entropy tend to increase?
If entropy is not a form of energy, then how can it be compared with actual forms of energy and measured in energy units?

Whoever downvoted and voted to close as unclear: which part is unclear that you'd like me to clarify?

Entropy is not energy. Entropy times temperature has energy units. Entropy can be regarded as a statistical property of thermodynamic systems. Although such a definition is not necessary to apply the concept, it is difficult otherwise to grasp what entropy represents. It is a measure of the number of ways you can configure a system of fixed volume, energy and composition.

@Buck Okay, so that’s where my confusion lies. Maybe you can flesh that out into a fuller answer?

Note also that it's never the case for any real process that $\Delta S_\mathrm{system}=-\Delta S_\mathrm{surroundings}$

Just because the units match up doesn't mean it's the same quantity, it just means it is measured in the same units.

To those who voted to close as homework question: This is very clearly not a homework question, and I've shown my effort to understand the underlying principles. To those who voted to close as unclear: once again, what is unclear that you'd like me to clarify?

The 1st law says that total energy is conserved. The 2nd law says that USEFUL energy is never conserved.

It is something of a historical accident that entropy has units of J/K.  It came out of the fact that the connection between heat, temperature, and energy was not obvious to early scientists, and so they effectively picked different units for measuring temperature and for measuring energy.
In the more modern statistical interpretation of entropy, the entropy of a system is simply a number.  Specifically, if the number of microstates associated with a given macrostate is $\Omega$, then $S = k \ln \Omega$.  The number of microstates ($\Omega$) is just a number, without any units, and therefore so is $\ln \Omega$.  You can see that we actually have to insert Boltzmann's constant, with its units of J/K, to make the units "come out right".
Arguably, a more natural way to define entropy would be to just make it it a dimensionless quantity:  $S = \ln \Omega$, without the factor of $k$.1  This would be equivalent to measuring temperature in units where $k$ is equal to 1 exactly, rather than defining our unit of temperature such that $k= 1.380649 \times 10^{-23}$ J/K exactly.
If we did this, we'd be effectively measuring temperature in units of energy as well;  for example, in an ideal monatomic gas with a "temperature of 1 J", the average KE of each molecule would be $\frac{3}{2}$ J.  Quantities such as Helmholtz free energy would still have units of energy, since we'd still define $F = U - TS$, with $T$ having units of energy and $S$ being dimensionless.
Of course, in this parallel universe where entropy is defined as a dimensionless number, another gen. chem. student would be asking why temperature is not the same thing as energy, even though they're measured in the same units.  But that's another question and another answer.

1 In fact, entropy is defined exactly this way in information theory, since there's not really a notion of energy (or temperature) to speak of in such contexts.

"Here's Boltzmann's equation, but you don't need to know it" – my chemistry professor, when she just briefly discussed it in class. No, ma'am, skipping over that is much of what led to my confusion. Seems like it would be more intuitive if $k$ were separated from $S$, defining it the same way in chemistry as it is in information theory, and with Gibbs' equation being rewritten as $G=H-kTS$, thus leaving $S$ dimensionless as it seemingly should be.

"RMS speed of each molecule would be $\frac32$ J" — this doesn't make sense. Joule is mass times velocity squared, while RMS speed has the units of velocity. They can't be equal even if we somehow take mass as dimensionless.

@Ruslan:  You're quite right;  I should have said "average energy".  Corrected.

"It is something of a historical accident that entropy has units of J/K. It came out of the fact that the connection between heat, temperature, and energy was not obvious to early scientists."  I don't believe that's giving those who developed the foundations of thermodynamics enough credit. I would say entropy historically had those units precisely because they did understand the relationship between heat flow, temperature, and entropy change:  $dS = \frac{\text{đ}q_{rev}}{T}$ => $\text{absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$

I'll add that the framework that represents entropy as dimensionless isn't a better framework, it's simply a different one.  Your choice of which to use is based on which you find more convenient and meaningful for the specific type of problem you're trying to solve.  E.T. Janynes may have argued otherwise, but there are those on both sides of this.

Incorrect assumptions

[OP] we see that entropy can be transferred between a system and its surroundings, with $\Delta S_\mathrm{system}=-\Delta S_\mathrm{surroundings}$

This equation is usually not correct, except when you have a reversible process (an ideal situation where something happens even though everything is at equilibrium). For an equilibrium, you are right: The entropy does not increase.

[OP] If entropy is a form of energy, then how can universal entropy tend to increase?

Entropy is not a form of energy. It does not even have the same dimensions. Also, there are forms of energy that increase without breaking the first law. You could have a space heater turning electrical energy into thermal energy. The first law can't be applied separately to electrical energy ("the electrical energy in the universe is constant" is not true).

[OP] If entropy is not a form of energy, then how can it be compared with actual forms of energy and measured in energy units?

It is not measured in energy units. The term $T \Delta S$ is measured in energy units. Consider speed vs time and speed vs distance. They share units, but that does not mean speed is the same thing as time, or as distance. And laws about distances or time do not automatically apply to speed.
Simple counter example
If two bodies of different temperature are brought into thermal contact, they will reach thermal equilibrium (same temperature). The thermal energy lost by the hotter body is equal to the thermal energy gained by the colder body (first law). The entropy lost by the hotter body is less than the entropy gained by the colder body (entropy increases, second law).
Why there is no contradiction
Lots of quantities have the same dimensions as energy (work, heat), and there are a lot of forms of energy. The first law does not apply to any of those, just to all energy combined. So applying the first law directly to entropy does not make any sense, and entropy is not a form of energy.

Entropy is not energy. Entropy times temperature has energy units. Entropy can be regarded as a statistical property of thermodynamic systems. Although such a definition is not necessary to apply the concept, it is difficult otherwise to grasp what entropy represents. It is a measure of the number of ways you can configure a system of fixed volume, energy and composition.
It is worth repeating: entropy is a statistical property. So why can you transform it into a property with energy units through simple multiplication with temperature? Temperature is a measure of the ease with which new confígurations become available through  an increase in energy, or rather the inverse temperature is:
$$\left(\frac{\partial S}{\partial U} \right)_V=\frac{1}{T}$$
At low T it takes just a bit of energy to expand the number of possible configurations of the system (logarithmically scaled). On the other hand, when a system is "hot" you have to add a lot of additional energy in order to significantly expand the number of available configurations (on a logarithmic scale). Borrowing the definition $S \propto \log W$ provided by another answer  we can rewrite this last equation as
$$\frac{1}{W} \left(\frac{\partial W}{\partial U} \right)_V\propto\frac{1}{T}$$
to emphasize that $T^{-1}$ measures the relative scale of the change, that is, the rate of change of the logarithm of the number of configurations.


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Spraying difluoroethane into an enclosed space
I'm working on developing physics lab activities that my students can do at home during the covid quarantines. For an activity involving waves and thermodynamics, I came up with an experiment at home in which I blow over the top of a bottle (beer bottle or 2-liter soda bottle), measure the frequency, and then spray a can of "dust-off" (difluoroethane) into the bottle and measure the frequency again. The frequency gets lower because difluoroethane is much more dense than air.
Environmentally, it looks like difluoroethane is fairly benign. It's used as an alternative to substances that are worse for our planet's atmosphere. People can huff it, and if you do, it's dangerous. It's flammable, so you want to stay away from flames and sparks. All of those things seem like they're not huge problems for my purposes.
But the label also says, "Never spray into an enclosed space, such as a trash can or paper shredder." What's up with this? That's exactly what I need to do for this experiment. Why is this dangerous?
Would there be any safety or environmental advantage to using 2,3,3,3-tetrafluoropropene ("Opteon" or R-1234yf)? It doesn't seem as readily available in small packages.
[EDIT]
Following up a comment by @MaxW: Interesting, thanks. I found a video of the liquid exploding when it's poured into water. Also a demonstration that it's flammable in gas form if you collect it in an open-topped container. Burning produces toxic $\ce{HF}$ and $\ce{COF2}$. So I'm hypothesizing (tell me if I'm wrong) that it's safe for my application if you (1) stay away from any source of ignition, (2) do it in a well ventilated area, and (3) in case of ignition, get away and let the fumes dissipate thoroughly. I'd be interested in answers as to whether this seems sufficient.

Making an educated guess, I'd say that the danger is explosive limits in a container with an open top would be the worrisome problem. Explosive limits, vol% in air: 3.7-18

That 1,1,-difluoroethane (physically?) explodes in contact with water is rather surprising. Unfortunately the explanation given in the video is rather unsatisfying. Have you considering using other gasses and volatile liquids (saturating the air with vapour), like the butane in the video, or diethyl ether, or acetone? As you say, flammability isn't a huge issue for you, with the proper precautions. I do understand sourcing the material is a major issue, but I can't think of clear winner from a hardware store (e.g. dichloromethane would probably work really well, but it's more toxic, etc..).

I don't think the potential explosion of liquid difluoroethane is a risk if you are just working with the gas. And not using vessels filled with water.

@NicolauSakerNeto: I'd be happy to find an alternative material, but it has to be something cheap that we can buy and mail to students. The technique involves blowing over the top of the bottle, so I don't want to use acetone, etc. -- it has to be an inert gas.


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Why is the heat of combustion of sulphur equal to the heat of formation of SO2 and not SO3
By definition the heat of combustion for compound is equal to the energy that is released on complete oxidation of one mole that substance to its highest oxidation state. In case of sulphur, online I found that ∆H for $\ce{S8 + O2->SO2}$ is heat of combustion. Clearly + 4 is not the highest oxidation state of sulphur, then why is ∆H for $\ce{S8 + O2->SO3}$ not considered as heat of combustion

Exceptions. Chemistry has lots of exceptions when making sweeping statements.

Because you get SO2 not SO3 in combustion, as simple as that.


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Is there an adhesive that is not dissolved by acetone?
I am trying to glue glass sheets together to make a sealed 13" x 13" x 13" box. Inside this box there would be acetone gas used to melt 3D printed parts to make them more durable. Acetone will remove just about any type of adhesive I know and I was wondering if there is anything out there that would not react to acetone?
If you are going to send me the questions that was posted on this website titled "Is there a paper glue that is NOT dissolved by acetone?" don't send it, because I've seen it and the thing he suggested was to use tape, which wouldn't work because not only would it be very difficult to cover all the inside corners with tape to seal off the acetone gas from escaping, the acetone gas would either way slowly remove the tape away from the glass because it would seep under the edges of the tape.

It would be very difficult to put tape on the inside corners of the glass to completely prevent the acetone from seeping through and even if I do manage to do so, the acetone would remove the tape slowly over time.

Ohk, then I am retracting my flag. Still, if it helps, Is there a paper glue that is NOT dissolved by acetone?

How big a part are you working with? I was wondering if a gallon jar would work.

How about a one-piece box made of fused glass sheets? See https://www.sigmaaldrich.com/labware/glassware-catalog/custom-glassware.html, for example. You might also test epoxy cements.

fusing the glass together would be a bit difficult to do. Acetone removes epoxy so I would imagine it would still do the same to epoxy cement, but I will still run some tests. For the sizing of the 3d printed parts I have to be able to fit things that are at most 12x12x12". And could people please stop posting the other post titled "is there a paper glue that is not dissolved by acetone" because that does not help for two reasons. first of all his suggestion is for short term use, over time acetone will remove tape and second, it would be extremely difficult to cover all that cracks with tape.

@daniel Some epoxies are not rapidly damaged by acetone.

look for a small one-piece fish tank or a chromatography jar and cover it with a glass plate. Or use a can made of metal.

It is a notably different question to the flagged duplicate not least because there is little overlap between glues used for paper and glues used for glass. I used to make filtering devices for air-sensitive chemistry in a lab by gluing glass to long metal tubes. The glue survived exposure to most solvents, though perhaps not permanently.

You can use JB Weld, as once it is fully cured, acetone shouldn't dissolve it. I haven't tested it personally, but according to the JB Weld FAQ, it states that not much can dissolve it.


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Calculating Gibbs free energy for the reaction of burning methanol

Calculate $Δ_\mathrm{r}G^\circ$ for the following reaction:
$$\ce{2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l)}$$
using $Δ_\mathrm{r}H^\circ$ and $Δ_\mathrm{r}S^\circ$ at $\pu{298.13 K}$ $(\pu{25 °C}).$ The values for $Δ_\mathrm{f}H^\circ$ and $Δ_\mathrm{f}S^\circ$ can be found in Appendix G of the textbook.
Also state if this reaction is spontaneous at this temperature.

In Appendix G of the textbook (Openstax Chemistry), I found the $Δ_\mathrm{f}H^\circ$ to be $\pu{-239.2 kJ mol-1}$ and the $Δ_\mathrm{f}S^\circ$ to be $\pu{126.8 kJ mol-1}.$
The values of the Gibbs free energy of formation for both the reactants and products were also listed in the Appendix G, this is where I got them. So I plugged those values into the equation, leaving out the $\ce{O2}$ because it is just an element, and multiplied the appropriate values by the numbers in front of the compounds, and that is how I got the Gibbs free energy of the reaction:
$$
\begin{align}
Δ_\mathrm{r}G^\circ &= \sum Δ_\mathrm{f}G^\circ(\text{products}) - \sum Δ_\mathrm{f}G^\circ(\text{reactants})\\
  &= [2 × (\pu{-394.4 kJ mol-1}) + 4 × (\pu{-228.6 kJ mol-1})] - [2 × (\pu{-163.3 kJ mol-1})]\\
  &= \pu{-1379 kJ mol-1}
\end{align}
$$
However, I don't know how to get the Gibbs free energy by using the $Δ_\mathrm{f}H^\circ$ and $Δ_\mathrm{f}S^\circ,$ and I don't know if it was spontaneous or not. Can anybody point me in the right direction, please?

I'm guessing that, in the same way that you've calculated the Gibbs Free Energy based off of the energy of formation for each component, you can also calculate the Entropy & Enthalpy of the reaction. Since the Gibbs Free Energy is defined as: dG = dH - TdS you would need to plug in the result of the Entropy / Enthalpy of reaction, and the temperature of interest (which happens to be 298.15K).

I took a liberty to correct formatting, notations and fix missing links and units (please check and correct, if necessary). Do note that omitting units is never permitted, and that "k" is not a unit, it's a decimal prefix — never shorten $\pu{kJ mol-1}$ to $\pu{k},$ this is plain wrong. As for the formatting, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I also cleaned up comments section and added your essential comment as an edit to the question.

Like @samp stated, you can calculate the enthalpy and entropy of the reaction using the individual enthalpies and entropies of formation for each compound in the reaction. Then you can use the derived definition of Gibbs Free Energy as above. An equation is spontaneous if Gibbs is negative.

You need to look up $\Delta S_f$ for the elements as well because they are non-zero at that temperature.

That makes sense, thank you! And for that equation would it still be the summation of ΔH°rxn and TΔS°rxn for the reactants and products?

How did you compute $\Delta H$? The value you obtained looks wrong.


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Where should I publish my formula related to Aufbau Principle?
I have made a formula for determining the sequence of orbitals which is currently obtained using the Aufbau diagram. So, I wanted to know where should I publish it (if it is new). I searched the same on internet but did not get any such formula.

Hey there! It's nice to play around with ideas and discover new things. An important part of publishing is scoping the relevant background in the literature. This process alone can take months and often involves reading many dozens of articles, some often multiple times over. It seems you're still at too early a stage to consider publication. Do the reading, and you'll have a much better understanding of how to fit your work within our body of knowledge, including where you might be able to publish.

@NicolauSakerNeto Thank You for your sincere help, and concern. Actually I am a student of 11th grade and I have to acquire lots of knowledge. But who knows this formula may be new, and may help in the subject. Although I have also heard about the limitations of Aufbau principle. But still we have to study it and this formula will help atleast me and my friends. Thank you once again for your concer.

No problem. I can even suggest a place to start. Eric Scerri has written multiple times about the aufbau principle, from a critical standpoint. Looking at his work, as well as the sources he cites, will give you a much broader view.

Better idea would be just write your ideas in question and see if they a) make sense b) aren't obvious c) are useful.

@AdityaAsthana As Mithoron set, if you write the idea here first, we can check if it makes sense/is useful. Also, if it does wind up being useful, there will be a visible record of when you came up with the idea in case someone else publishes similar work.

Yes  you both are right. I will post it tomorrow morning here in the comments. Thank You for your response.

Have you searched the Journal of Chemical Education? There are so many articles on this topic you are interested in. If you don't have access to formal databases, start from Google Scholar and read abstracts.
Journal of Chemical Education is the right platform for such "teaching" tricks. However, you have to justify with some theory as to why your idea works. What is the logic behind it? As far as I can recall (>10 years ago), there was an article by a high-school student in that journal. So, students can also publish but you have to be quite good at that subject - similar to a professional.
However, sometimes I receive such articles for review but the tricks or analogies are so horrible that there is no option but to reject them. Similarly, make sure you know where your approach fails. Can it predict the electron configuration of all the elements? If you are a student share the idea with a chemistry teacher. See their reaction (no pun intended). Do they find it useful or have they seen it before? If yes, certainly consider writing a short note.

Thank You sir for your kindly response. I have searched on Google Scholar for the same formula but did not get anything, even close to it. Would you mind telling me how to publish on Journal of Chemical Education? The least which will happen is that it will be rejected for not being something new. Thank You.

All you have to find now is an expert, who can give you more insight. What about your own college teacher to begin with?

Another possibly useful step is to consider writing a draft and submitting to a preprint server such as ChemRxiv. The publishing sphere in Chemistry is slowly but surely moving towards accepting preprints, and indeed this topic was specifically raised and defended by the now editor-in-chief of JCE, Thomas Holme.


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Does pH affect physical property of agarose and poly(acrylamide) gel?
We want to change pH in hydrogels of agarose and poly(acrylamide) by soaking into an acidic solution after gelation on pH 7.0.
Then, the physical property is changed? Now, we consider the condition of pH range from 7.0 to 4.0.
Please answer, and if you could, please inform the important reference.
Best, regards.

To understand the process, read this.


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Can MCM-41 (mesoporous materials) be used for ion exchange?
Wikipedia, and this journal article talk of a microporous and mesoporous material MCM-41. Although MCM-41 has dynamic sized pores which can be changed, but it is hydrothermally unstable.
Can this microporous material be used for ion-exchange similar to zeolites? What is the property that typically allows a substance to act as an ion-exchange resin?

Ion-exchange resins need to have charged groups that can bind your ions of interest. If a certain ion-exchanger will work for your problem depends on what kind of ions you want you to exchange or separate. The pores must be larger than your molecules of interest. E.g. polystyrene resins which are fine for water desalination (to remove e.g. calcium and magnesium ions) won't work well for separating proteins, as their pores are too small for biomolecules to enter the beads.


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Methyl group and acidic nature

After reading various answers here on StackExchange (like Why is the methyl group electron-donating via the inductive effect? and Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups), I am repeatedly seeing that -CH3 is electron donating through hyperconjugation. Otherwise is is electron withdrawing. So when comparing the acidic nature of the above two compounds, should we use electron withdrawing or electron donating nature of the methyl group?
Also will there be a difference if there is a normal cyclohexane instead of the conjugated benzene.

yes there will be a difference,when it's benzene due to orthoeffect 2-methylbenzoic acid will be more acidic than benzoic acid,but in case of cyclohexane as methyl is electron donating droup it's acidic nature will be less than unsubstituted one.

but I read here: https://chemistry.stackexchange.com/questions/65106/strongest-negative-inductive-effect-group-between-trimethylammonium-ammonium-an/65129#65129 that it is necessary for the neighbouring atom to have an empty p-orbital

as far as i know methyl group is an electron donating group and it is can stabilize by hyper conjugation.But in this case due to bulky nature of methyl group the COOH group adjacent to it will be out of plane.This inhibits the resonance of carboxyl group with phenyl ring which increases the acidity of carboxyl group which was otherwise reduced due to destabilizing cross conjugation.(this is orthoeffect).

In this question https://chemistry.stackexchange.com/questions/65106/strongest-negative-inductive-effect-group-between-trimethylammonium-ammonium-an    which you have posted is not telling anywhere that methyl is an electron withdrawing group.

Ortho effect which states that 'When a group is present ortho to carboxylic acid group in substituted benzoic acid, the steric hindrance forces the carboxyl group to twist out of the plane of the benzene ring. This inhibits the resonance of carboxyl group with phenyl ring which increases the acidity of carboxyl group which was otherwise reduced due to destabilizing cross conjugation'