text:
How to get a precise answer to a textbook precipitation problem?
This question concerns a textbook problem* on precipitation. Briefly, there are $2\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.003$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.1 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3\cdot 10^{-14}$, K$_{sp}$ for CuS is $8\cdot 10^{-37}.$  K$_1$ and K$_2$ for H$_2$S are given as $1.0\cdot 10^{-7}$ and $1.2\cdot 10^{-13}$ respectively.
In part one of the question it is found that Mn remains in solution and  Cu$^{2+}$ will precipitate because the product [Cu][S] = $2.6\cdot 10^{-20}$ far exceeds K$_{sp}$ of CuS. In part two the problem is to determine how much Cu remains in solution.
The solution begins by stating that "most of the Cu$^{2+}$ will precipitate" and explains that a corresponding increment of [H$^+$] will be added to the solution from the H$_2$S which has lost S to precipitate, requiring a correction which leads to $8\cdot 10^{-21}$ moles Cu remaining in solution vs. $6\cdot 10^{-21}$ moles without the correction.
To make the correction the text assumes all the Cu precipitates, so here is my question.
The question being asked is: How much Cu remains in solution? To answer the question, it is assumed that none remains in solution...(!) This leads to a slight correction (from $3\cdot 10^{-21}$ % to $4\cdot 10^{-21}$ %) Cu remaining. The text does ignore the small change in [S] that accompanies the change in [H$^+$].

If there is a situation in which this very small difference matters, what is the right way to find it?

My guess is that there would be a simple ODE governing this problem--but I'm not sure.
The details above are included for completeness but the answer I think does not depend on the numbers here.
*The source of this problem is Shaum's College Chem. (10 ed.), 18.13-14.

Please correct your text - does the problem refer to Mn or Mg? You have both in the text.

I'm afraid your K2 of H2S is many order of magnitude too high, common thing in books using outdated data.

@Mithoron - The OP must use whatever constants are given in the problem, regardless if they are outdated values. However having a value that is "off" should get the OP to double check that the values were copied correctly from the problem.

@daniel - Could you add a picture of the problem and solution as given, or rewrite the problem and answer exactly as given?

@MaxW: I will post more of this when I get a moment, yes.

To get pH bump due to formation of CuS, it is assumed that an insignificant amount of copper remains in solution, not literately none. Since $2\times10^{-4}\gg \text{  }\approx 10^{-21}$ this is true. Using this very reasonable simplification greatly reduces the complexity of the problem.

Ok let's rewrite the problem for four significant figures.

There are $2.000\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.003000$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.1000 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3.000\cdot 10^{-14}$, K$_{sp}$ for CuS is $8.000\cdot 10^{-37}.$  K$_1$ and K$_2$ for H$_2$S are given as $1.000\cdot 10^{-7}$ and $1.200\cdot 10^{-13}$ respectively.

Now let's solve this using significant figures in a reasonable fashion.

Carry six significant figure in intermediate calculations to try to avoid rounding errors and then round the final result to four significant figures.

First let's assume that the $\ce{[H^+] = [ClO4-]}$, in other words the perchloric acid completely dissociates, and all the acid comes from the perchloric acid.

For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.20000\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.20000\cdot10^{-20})(0.10)}{(3.000\cdot10^{-3})^2} = 1.33333\cdot10^{-16}$$

Now using the two $K_{sp}$ values we can calculate the final concentrations of the  $\ce{Cu^{2+}}$ and $\ce{Mn^{2+}}$ cations
$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.33333\cdot10^{-16}} = 6.00002\cdot10^{-21}$$
$$\ce{[Mn^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{3.000\cdot10^{-14}}{1.333\cdot10^{-16}} = 225.001$$
So virtually all of the $\ce{Cu^{2+}}$ will ppt, but none of the $\ce{Mn^{2+}}$ will ppt.
Now knowing that the $\ce{Cu^{2+}}$ combining with $\ce{S^{2-}}$ frees $\ce{H+}$ according to the overall reaction:
$$\ce{Cu^{2+} + H2S -> CuS + 2H+}$$
a pH correction is needed. We can ignore $\ce{H+}$ contributions of $\ce{[HS-]_f}$ and $\ce{[S^{2-}]_f}$ as being insignificant. Likewise the $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$) that the acid neutralizes is insignificant. So
$$\ce{[H]_f = [ClO4-] + 2[Cu^{2+}]} = 3.40000\cdot10^{-3}$$
and
$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.200\cdot10^{-20})(0.1000)}{(3.40000\cdot10^{-3})^2} = 1.03806\cdot10^{-16}$$
$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.03806\cdot10^{-16}} =7.70668\cdot10^{-21} \ce{->[rounding]} = 7.707\cdot10^{-21}$$
$\text{* * * * * * * * * * * * }$
$\text{* *}\quad\text{CHECKS}\quad\text{* *}$
$\text{* * * * * * * * * * * * }$

$\ce{[HS-]_f \ll [H+]_f}$
$$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.000\cdot10^{-7})(0.1000)}{3.40000\cdot10^{-3}} = 2.94118\cdot10^{-6}$$
$$0.00294\cdot10^{-3} \ll 3.40000\cdot10^{-3}$$THIS FAILS
$\ce{[S^{2-}]_f \ll [H+]_f}$
$$\ce{[S^{2-}]_f} = 1.333\cdot10^{-16} \ll 3.40000\cdot10^{-3}$$
$\ce{10^{-7} - [OH-]_f \ll [H+]_f}$
$$\ce{10^{-7} - [OH-]_f} = 1.0\cdot10^{-7} - \dfrac{K_w}{\ce{[H+]_f}} =1.0\cdot10^{-7} - \dfrac{1\cdot10^{-14}}{3.40000\cdot10^{-3}} = 9.99971\cdot10^{-8}$$
$$9.99971\cdot10^{-8} \ll 3.40000\cdot10^{-3}$$

This fails in the last of the six digits being used for intermediate calculations. Since we are going to round to four significant figures this should be OK, but it is right on the ragged edge of being acceptable.

$\text{* * * * * * * * * * * * * * * * * *}$
$\text{* *}\quad\text{RECALCULATION}\quad\text{* *}$
$\text{* * * * * * * * * * * * * * * * * *}$
The pH correction made was not quite correct. We can ignore $\ce{H+}$ contributions of $\ce{[S^{2-}]_f}$ and likewise the $\ce{H+}$ consummed in the neutralization $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$).  However the contribution of $\ce{[HS-]_f}$ is not insignificant. So
$$\ce{[H]_f = [ClO4-]_f + 2[Cu^{2+}]_i + [HS-]_f}$$
Now the correction for $\ce{[HS-]_f}$ is small. So we can probably just calculate the correction needed form the $K_{a1}$ equation using a guess that the $\ce{H+}=3.40294\cdot10^{-3}$. So let's check that for consistency.
$$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.000\cdot10^{-7})(0.1000)}{3.40294\cdot10^{-3}} = 2.93864\cdot10^{-6} = 0.00294\cdot10^{-3}$$
We don't need to match all six figures for the $\ce{[HS-]}$, but just the first three. So the small correction works.
Recalculating $\ce{[S^{2-}]}$ :
$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.20000\cdot10^{-20})(0.1000)}{(3.40294\cdot10^{-3})^2} = 1.03627\cdot10^{-16}$$
Recalculating $\ce{[CuS^{2+}]}$ :
$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.03627\cdot10^{-16}} = 7.72000\cdot10^{-21} \ce{->[Rounding]} 7.720\cdot10^{-21}$$
Note that we still haven't even needed to use a quadratic equation for the solution.

Safe to say this would never show up on a gen. chem. test but very satisfying to see the calculation. We didn't use a quadratic... but the mental process of adjusting the pH correction is not trivial.

The point of course is to use significant figures to your advantage by simplifying the problem. The calc for $\ce{HS-}$ could easily be setup as a quadratic to get a "better" solution, but why go  to the effort when a quick and dirty trick will work?

Ok, let's do the "complete" and "exact" problem.
First, I'm just going to ignore the $\ce{Mn^{2+}}$. Using a first pass with the simplifying assumptions would be enough to show that MnS will not ppt.
Second the problem statement notes that:

This solution is then saturated with H2S. The solubility of $\ce{H2S}$ is given as 0.1 mol/L and assumed independent of the presence of other substances in solution.

Chemically $\ce{H2S}$ would be bubbled through the solution to saturate it. While the $\ce{H2S}$ was bubbling $\ce{CuS}$ would be ppt'ing. So when the final concentration of $\ce{H2S}$ is 0.1 molar the ppt of $\ce{CuS}$ is complete.
Let:

The final $\ce{[H2S]_f = 0.1\bar0}$ molar
This means that the final concentration of all the sulfur species ($\ce{[H2S]_f, [HS-]_f,}$ and $\ce{[S^{2-}]_f}$) will be a bit more than $0.1\bar0$ molar.
The initial and final $\ce{[HClO4]} = 3.\bar 0\times10^{-3}$ molar
From $K_{sp}$ for CuS we know that:
$$K_{sp} = [Cu^{2+}]_f[S^{2-}]_f$$ or $$\ce{[S^{2-}]_f} = \dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}}$$
For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\bar0\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f}}{\ce{[S^{2-}]_f}}} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}$$
and
$$\ce{[HS-]_f} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H+]_f}} = K_{a1}\ce{[H2S]_f}\sqrt{\dfrac{\ce{[S^{2-}]_f}}{K_t\ce{[H2S]_f}}}=K_{a1}\sqrt{\dfrac{\ce{[H2S]_f[S^{2-}]_f}}{K_t}}= K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}}$$
$\ce{[Cu^{2+}]_f}$ be the final concentration of copper.
The initial $\ce{[Cu^{2+}]_i}=2\times10^{-4}$ molar which we will assume to be $2.\bar0\times10^{-4}$ molar. Thus the amount of copper that is removed from solution is $2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}$
For the final $\ce{[H+]}$ we know that
$$\ce{[H+]_f = [HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}) - (10^{-7} -[OH-])}$$ or $$ 0 = \ce{[HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
substituting for $\ce{[HClO4]}$
$$0 = 3.0000\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
and collecting terms
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f - 2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[S^{2-}]_f}$
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[HS^-]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
and finally substituting for $\ce{[H+]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +K_w\sqrt{\dfrac{K_{sp}}{K_t\ce{[H2S]_f[Cu^{2+}]_f}}} - \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}}$$

Since $\ce{[H2S]_f = 0.10}$ per problem statement, the only variable left is $\ce{[Cu^{2+}]_f}$. I'll leave it to the reader to solve this mess for $\ce{[Cu^{2+}]_f}$, but to 2 significant figures the answer should be $7.7\cdot10^{-21}$.
The whole point in using significant figures to simply the problem is not to get the "wrong" answer, but to avoid a computational mess that doesn't yield any better answer.

As I begin to look at this, I notice that of the six terms on the r.h.s. of the final equation, just taking [H2S] as 1 and using [Cu] from the previous answer (which is surely close), all the terms are < 9x10[-6] but the sixth term is about -.0107...

If the answer changes it will only be because I made a mistake. // $\ce{[H2S]_f = 0.10}$ as noted in specification of the problem. So the only variable left is $\ce{[Cu^{2+}]_f}$ // Notice in equations above the following equation was derived $$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]f[Cu^{2+}]}}{K{sp}}}$$

Bother--my mistake. I think this is correct and will be looking at it more carefully.

Totally agree with your last point. One lingering confusion--solving your last eq. for [Cu] gives about 7.71954x10[-21]. This corresponds to [H] =0.0034029 and I think we can't exceed 0.0034. I can't find any problem with your reasoning, so I may have made an error calculating--but I am prepared to call it a day and accept your upvoted answer--with much appreciation for the help .

Well to 2 significant figures $7.71954\cdot10^{-21} = 7.7\cdot10^{-21}$ and $0.0034029 = 0.0034$. That is the whole point. So do you want to do the "easy" calc and get $7.707\cdot10^{-21}$ which rounds to $7.7\cdot10^{-21}$, or do you want to do the "hard" calc which yields $7.71954\cdot10^{-21}$ and rounds to $7.7\cdot10^{-21}$?

PS - Checked calcs and my last formula does yield $7.71954\cdot10^{-21}$

Ok let's rewrite the problem for two significant figures.

There are $2.0\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.0030$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.10 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3.0\cdot 10^{-14}$, K$_{sp}$ for CuS is $8.0\cdot 10^{-37}.$  K$_1$ and K$_2$ for H$_2$S are given as $1.0\cdot 10^{-7}$ and $1.2\cdot 10^{-13}$ respectively.

Now let's solve this in a reasonable fashion.
First let's assume that the $\ce{[H^+] = [HClO4]}$

For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.2\cdot10^{-20})(0.10)}{(3\cdot10^{-3})^2} = 1.333\cdot10^{-16}$$
carrying two extra digits in intermediate calcs

Now using the two $K_{sp}$ values we can calculate the final concentrations of the  $\ce{Cu^{2+}}$ and $\ce{Mn^{2+}}$ cations
$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.0\cdot10^{-37}}{1.333\cdot10^{-16}} = 6.002\cdot10^{-21}$$
$$\ce{[Mn^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{3.0\cdot10^{-14}}{1.333\cdot10^{-16}} = 225$$
So virtually all of the $\ce{Cu^{2+}}$ will ppt, but none of the $\ce{Mn^{2+}}$ will ppt.
Now knowing that the $\ce{Cu^{2+}}$ combining with $\ce{S^{2-}}$ frees $\ce{H+}$ according to the overall reaction:
$$\ce{Cu^{2+} + H2S -> CuS + 2H+}$$
a pH correction is needed. We can ignore $\ce{H+}$ contributions of $\ce{[HS-]_f}$ and $\ce{[S^{2-}]_f}$ as being insignificant. Likewise the $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$) that the acid neutralizes is insignificant. So
$$\ce{[H]_f = [HClO4] + 2[Cu^{2+}]} = 3.4\cdot10^{-3}$$
and
$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.2\cdot10^{-20})(0.10)}{(3.4\cdot10^{-3})^2} = 1.038\cdot10^{-16}$$
$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.0\cdot10^{-37}}{1.038\cdot10^{-16}} =7.707\cdot10^{-21} \ce{->[rounding]} = 7.7\cdot10^{-21}$$
$\text{* * * * * * * * * * * * }$
$\text{* *}\quad\text{CHECKS}\quad\text{* *}$
$\text{* * * * * * * * * * * * }$

$\ce{[HS-]_f \ll [H+]_f}$
$$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.0\cdot10^{-7})(0.10)}{3.4\cdot10^{-3}} = 2.94\cdot10^{-6} \ll 3.4\cdot10^{-3}$$
$\ce{[S^{2-}]_f \ll [H+]_f}$
$$\ce{[S^{2-}]_f} = 1.333\cdot10^{-16} \ll 3.4\cdot10^{-3}$$
$\ce{10^{-7} - [OH-]_f \ll [H+]_f}$
$$\ce{10^{-7} - [OH-]_f} = 1.0\cdot10^{-7} - \dfrac{K_w}{\ce{[H+]_f}} =1.0\cdot10^{-7} - \dfrac{1\cdot10^{-14}}{3.4\cdot10^{-3}} = 9.99971\cdot10^{-8}$$
$$9.99971\cdot10^{-8} \ll 3.4\cdot10^{-3}$$

The point here being that $$2.0\cdot10^{-4} - 7.7\cdot10^{-21} = 1.999999999999999923\cdot10^{-4} \ce{->[rounding]} = 2.0\cdot10^{-4}$$

It is just an algebra problem, but I'm lost as to exactly how the significant figures should be handled. The gist is that significant figures are used to simply the problem by assuming that certain aspects can be ignored. Without the simplifications you'll end up with a cubic or higher equation that can't be easily solved by hand.
For example, the perchloric acid will completely dissociate so the $\ce{[H+] = 3\times10^{−3}}$. The $\ce{Cu^{2+}=2\times10^{−4}}$, so if "all" the $\ce{Cu^{2+}}$ ppts, then the boost to the $\ce{[H+] = 4\times10^{−4}}$. So is the final $\ce{[H+]}$ supposed to be $3\times10^{−3}$ or $3.4\times10^{−3}$?
RANT - If the problem wanted the perchloric acid to be 0.00300 molar, then the !@#$%^& problem should state that.
The point here is that if we start adding extra significant figures, then how many should be added, and to which species? 1? 2? 3? 4? ...
So if we get slightly different answers assuming that:

The perchloric acid is 0.003 molar, the copper is $2\times10^{-4}$ molar and the hydrogen sulfide is 0.1 molar
The perchloric acid is 0.00300 molar, the copper is $2\times10^{-4}$ molar and the hydrogen sulfide is 0.10 molar
The perchloric acid is 0.00300 molar, the copper is $2.00\times10^{-4}$ molar and the hydrogen sulfide is 0.100 molar

I did double-check $K_1,K_2$ and they are as given. And yes, the final [H] is assumed 3.4 x 10$^-3$ in the text. But we know this is approximate because all the Cu does not precipitate. I am interested in how to find the answer without assuming an artificially high boost of [H]. I will post more of the original question when I get a moment but you touch on what interests me in your first point--how do we formulate the "not easily solved by hand" version--

How crazy do you want to get? You gave the initial concentration of copper as $2\times10^{-4}$ but the final is about $10^{-21}$. So do you want to work out problem "exactly" to say 18 significant figures?

Let me put it this way--I put in a lot of time trying to understand a solution that was essentially an exercise in tail-chasing (almost all of the Cu precipitates... first we assume that all the the Cu precipitates...). So if  you get crazy I will put in a lot more work to make sure I understand it.  It's obviously a more subtle point than the authors wanted to address.

The reason I thought ODE was that this is not instantaneous and it seems like a "coroner's equation" equilibrium situation with d[Cu]/dt=k f ([Cu]) and after integrating w'd get a function of t that will reach some eq. point. The [H] will fall somewhere between 0.003 and 0.00034. But I am not sure.

A ODE would be useful to examine the kinetics. You could use kinetics and extrapolate to infinity time, but that is the very hard way to solve the equation. The problem also doesn't give any data for forward and backward reactions so there really isn't way to solve this problem using kinetics.


text:
Reaction of nitrobenzene with sodium metal in ethanol
The reactant is nitrobenzene(only) and reagent is Na metal in solvent liquid ammonia along with ethanol (which i am assuming is playing the role of proton donor please correct me if i am wrong).
The first thing that comes in mind is "alkyne" but there is not any alkyne in nitrobenzene( not even in its resonance structures).How this reaction will proceed?
In answer, the DBE is just reduced by one.I was also trying what if it was toluene?Can i have a explanation based on mechanism?

Birch Reduction will occur with Nitrobenzene.

The ethanol makes little sense, unless it is added in the end only, to destroy remaining Na metal.

Benzene Rings or Substituted Phenyl rings undergoes reduction in presence of $\ce{Na}$ in liq. $\ce{NH3}$ in presence of $\ce{EtOH}$ solvent, and generally forms substituted Cyclohexa-$1,4$-dienes.
This reduction generally follows Single Electron Transfer (SET) Mechanism.The single unpaired electron of Sodium is transferred to the Benzene ring forming a radical which subsequently forms a Carbanion. This carbanion eventually takes the Proton from $\ce{EtOH}$ and becomes Cyclohexa-$1,4$-diene.
But in the first step, the transfer of electron by $\ce{Na}$ occurs in that place, where the intermidiate formed after this step gets maximum stabilisation. In case of Electron-withdrawing group, such as $\ce{-NO2, - COOH, -SO3H, -CHO, -COOR, -COR }$ etc., that transfer occurs para to that group, as the negative charge generated is highly stabilised by that electron-withdrawing group. The mechanism thus in case of Nitrobenzene is as follows,

But if the group is electron donating such as $\ce{-CH3, -OH, -OMe, -NMe2 }$ etc. The transfer will occur in such a way to stabilise the generated negative charge by keeping it in meta to the group (where the effect of electron donation by the group is least). For those compounds, the mechanism and corresponding product will be as follows,

Thus, in your case, the product will be $\text{3-nitrocyclohexa-1,4-diene}$ as described in the first mechanism.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
It is possible to create any metal?
This question just popped up in head. There are many things which we are good at. We are good at:

constructions.
build stuff.
destroying things.
repairing things.
cleaning.

But I want also to know, if it is possible to create for example: Beryllium, Gold, Iron, Silver and Platinum.
Question: Are we humans able to create natural metals? Or,it is impossible?
If it is possible for us to create any kind of metals, then, can you grant me informations of how or what substances I need to use for creating each metal? I have found in here that shows you can create gold, but what about other metals? Are we capable of doing so? If we create metals, then do they have the same value and resistance just like the metals which are found in mountains and underground?
Just in curiosity.
NOTE: I’ve used on list “destroying” because, that’s how I see. Because, I have seen on a video about pollution, that we humans almost brought our home world to an end due to our ignorance. And that’s from where I got this idea. We’re incapable of taking care of our world.

Not sure if much Be was formed in the Big Bang (Physics SE has good questions/answers on nucleosynthesis in the Big Bang), but all the rest were formed later through nuclear reactions. Yes, they can all be made artificially at ridiculous cost.

Yes, can be done. After all this is how they make the super-heavy elements

Related: Turn lead into gold via radioactive decay, Would it be possible to destroy gold?, Can we goldify metals?, Why can't we use fusion to make ourselves rich?, Will adding up protons and electrons (without neutrons) create a new element?

I'm not migrating this. It's on topic for Chemistry, possibly not for Physics. Just because Physics has questions on the Big Bang doesn't mean that this should be migrated, that's a bit of an overreaction - your question isn't even remotely about the Big Bang. Jon was just describing how the elements were made naturally, the point being that humans are capable of mimicking the nuclear reactions which formed the heavier metals.

The idea of creating metals is from alchemy, which was the ancient forerunner to chemistry. This belief started to become debunked after Robert Boyle published "The Sceptical Chymists" in the 1600's.
In 1785,  Antoine Lavoisier, discovered the law of conservation of mass which states mass is not created or destroyed, it merely changes form. This law is one of the foundations of modern chemistry.  Even in the video, it states the gold is not being created it is merely changing form.  In the discipline of chemistry, it is considered not possible to create any metal.
Outside the discipline of chemistry in the discipline of physics it was discovered mass and energy are interchangeable.  This has given rise to the idea of nuclear transmutation.  Nuclear transmutation is certainly possible; however, it is an incredibly expensive feat which is often considered too impractical for consideration in the practice of practical chemistry.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Bromination on aromatic compound

So guys, very quick question: Why does bromination occur at exactly this position on the aromatic ring? Has it something to do with the substituents and their respective electronic effects?

Bromination (in fact any Electrophilic aromatic substitution) occurs at ortho/para to electron donating groups. Generally ortho-product is lesser (except special stabilisation casese like Hydrogen Bonding etc.) due to steric factor. Para is generally the major product which is the case here also due to electron donating $\ce{N}$ directly attached to Benzene ring.

To follow up @SoumikDas, you've got essentially an N,N-dimethylamino group which is making it difficult to access the unsubstituted ortho position.


text:
Shape of urea molecule around N atom
I was doing a question where I was supposed to predict the shape of the urea molecule $[\ce{H2N(CO)NH2}]$ using Lewis and VSEPR. I identified that the N atoms will have a lone pair on them and so I assigned a trigonal pyramidal shape around N, having the ammonia structure in mind. However when I looked up some 3D structures of the urea molecule it had a trigonal planar shape around N making the whole molecule lie in the same plane. I did some further research and read on wikipedia that
"The urea molecule is planar in the crystal structure, but the geometry around the nitrogens is pyramidal in the gas-phase minimum-energy structure."
What causes this change in geometry when the molecule crystallizes? Does the same thing go for other molecules? Would there have been a way of predicting this as I was solving the question?

Its an interesting question and I would enjoy seeing an answer, but I doubt it possible to predict that change in geometry beforehand. The structure around N in the crystal would depend a lot on the packing together of molecules of urea, and I don't think you could get a good sense of that at the level of VSEPR. Also, I see you are new to the site. If you haven't yet, take the [tour] and visit the [help]. Welcome to chem SE! @Louloulelle

I was supposed to predict the shape of the urea molecule $[\ce{H2N(CO)NH2}]$ using Lewis and VSEPR.

Don't forget that there are assumptions in VSPER to simplify the model. From Wikipedia:

Valence shell electron pair repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules...

Essentially VSPER is for molecules in the gas phase where the only electrostatic interactions are within the molecule. Once the molecule is in a condensed phase, the assumption that it does not interact with external forces breaks down.

What causes this change in geometry when the molecule crystallizes?

For one let's not forget that urea is a resonant structure, and does not have fixed bonds which also goes against assumptions of VSPER.

Further you have neighboring molecules with the hydrogens of those molecule taking some electron density from the resonance structure which is shown in the unit cell representation shown below:

In summary you have resonance which will flatten out the molecule and other molecules taking some electron density that combined will contribute further causing the molecule to flatten out.

Does the same thing go for other molecules?

Yes, other molecules can condense and form structures that deviate from VSPER. Water in ice for example has a $109.5 ^\circ$ bond angle (tetrahedral angle) where as VSPER would predict $104.5^\circ$ (bent angle).

Would there have been a way of predicting this as I was solving the question?

No, you would have to know how urea crystallizes and many other factors, then used a computer simulation to calculate (estimate) such.


text:
How to drain engine oil by understanding oil viscocity issues
I asked this on the stackexchange automotive site but didn't get much response perhaps because it entails some chemistry knowledge. I could have asked without mentioning the automotive details however I think it puts the question into context making it easier to give a good answer.
When you drain an engine of oil in some engines about 12.5% of old oil remains because it coats engine parts, sits in galleys, and some is at the bottom of the sump container(below the drain plug) and won't come out unless you tilt the car which is quite impractical.
The general approach is to remove the older oil by dilution\further oil changes. As an example if you have a 4 liter capacity and an oil change allows you to fill 3.5l new oil, this means 0.5l\12.5% of old oil remains and so a further oil change will reduce the initial 12.5% to 1.5% and a further oil change to 0.19%.
However from my understanding this won't work when you mix oils of different viscosity because oils of different viscosity do not mix when they settle at cold\low temperature(which is where it will be at point of drain). But they do mix when they are at hot\engine temperature.  Is this correct?
Do I need to be using the same viscosity oil to ensure dilution happens at point of drain e.g. 5w30 needs to be mixed with 5w30 but adding 5w40 to 5w30 won't get the dilution effect? What would happen if you added 10w40 to 5w30, would you get dilution at point of drain?
Also how do the oils settle in practice? would the thinner oil sit on top of the thicker oil?  Therefore if you are trying to dilute with a thinner oil the thicker oil may settle below the drain plug and sit below thinner oils in galleys so it may not dilute\get drained with successive oil changes?
Suppose I had a container with 0.5 liter of 5w30 and 3.5 liters of 5w40 would the latter sit below the former? What if you had 5w30 mixed with 10w40?
Also can factors besides viscosity e.g. additives in oil or any other factors affect dilution?
My issue is that I have 0.5 liters of black and thick looking oil in my engine(which I think is affecting performance) and I don't really know the viscosity of the oil they put in.  I don't want to do countless and costly oil changes trying to get it out so approaching this in an educated way may help.
Thanks.

Modern multiweight oils do not work the way you think they do.

I agree with Jon. I'd expect the oils to be completely miscible. I would expect some sludge in the oil pan, but that would be oil adhering to particulates.

@airhuff I can remove all references to engines and automotive.

@JamesWilson, no, don't do that. After reading your question a few more times I decided to change my vote to close. With a small stretch of the imagination, it could be considered a chemistry question so I should just let it be. Best of luck, and welcome to chemistry.se.

If you've been changing oil according to the car manufacturer's recommendation, there's no need to "rinse". It would be a waste of time and oil. Oil has only a tiny effect on performance -- if concerned, used the lowest-temperature multigrade oil recommended (e.g. 5W-30 as opposed to thicker 10W-30).

If you drain the oil when the engine is hot or at least warm, it is much less viscous and will drain pretty efficiently. Oils of different viscosity or even different density will surely mix well while the engine is running, so if you still want to do a flush, be sure to run the engine for a while after you add the new oil and before you do your second oil change.


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Why does a catalyst increase the forward and reverse reactions equally
With reference to the picture below, how is it that when there is a lower activation energy (due to the alternate reaction pathway provided by the catalyst) that the catalyst will increase the forward and reverse reactions exactly the same amount.
It only makes intuitive sense that when the $Ea$ is lowered, the forward reaction increases more than the reverse reaction because the decrease was an absolute amount (that is, the same decrease for both forward and reverse) and thus proportionately, the forward $Ea$ has been reduced far more than the reverse $Ea$. So why should the reverse reaction instantly be raised such that it is still equal to the forward reaction (if from equilibrium)?

Because the amount of particles in state "reactant" or "product" only depends on the energy difference, period. In QM, you have the Boltzmann distribution for that, in chemistry it's the same thing, just a bit more complicated because you usually have more than two species involved.

Perhaps what you overlooked: There is a third thing in your system, the thermal energy reservoir $kT$. The relative energy difference from it to the product and educt state changes matters. The transition state only make a quick loan from the reservoir.

Related: Do catalysts shift the equilibrium constant towards 1?

Because of microscopic reversibility principle.

https://chemistry.stackexchange.com/questions/58846/wont-the-net-effect-of-a-catalyst-be-zero-if-it-creates-a-new-path-with-lower-a

The equilibrium is unaffected because the barrier to both forwards and back reaction is reduced, and even though the amount is constant it works out that this effect cancels out. This is a consequence of the fact that the population of energy levels is exponentially distributed from the Boltzmann distribution. This also leads to Arrhenius type rate expressions.
If the energy in your figure from reactants to transition state is $E_a$ and the difference in products to reactants $G$ and we lower the energy $E_a$ by $E_0$ then using Arrhenius type rate expressions the equilibrium constant is $K_e=k_{forward}/k_{back}$
$\displaystyle K_e=\frac{\exp(-E_a/RT)}{\exp(-(E_a+G)/RT)}$,  and for the catalyst  $\displaystyle K_e^C= \frac{\exp(-(E_a-E_0)/RT)}{\exp(-(E_a-E_0+G)/RT)}$  which is the same as $K_e$if you expand the terms in the exponentials.


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How to assess sunscreen effective via blueprinting?
How do I assess the effectiveness of sunscreen creams by using the blueprinting process? I know blueprinting is done when something is sensitive to light and this may cause reactions to produce a product that causes Prussian Blue to appear. However I am unsure how to connect sunscreen and blueprinting - does it have to do anything with the colour in relation to its wavelengths and how it is connected to UV light?

From Wikipedia page on the blueprint process:

The blueprint process is based on a photosensitive ferric compound. The best known is a process using ammonium ferric citrate and potassium ferricyanide. The paper is impregnated with a solution of ammonium ferric citrate and dried. When the paper is illuminated, a photoreaction turns the trivalent ferric iron into divalent ferrous iron. The image is then developed using a solution of potassium ferricyanide forming insoluble ferroferricyanide (Prussian blue or Turnbell's blue) with the divalent iron. Excess ammonium ferric citrate and potassium ferricyanide are then washed away. The process is also known as cyanotype.

Exposing blue print paper to UV light will cause it to develop blue. Sunscreen is meant to block UV light so applying sunscreen over blueprint paper then exposing to UV light and developing will give you a semi-quantitative way to evaluate the sunscreen based on the intensity of the blue.
See: Colorimetry


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Does silicon have catenation property?
We know that there are countless organic compounds of carbon chain because of carbon's catenation. Silicon also is in the same group of carbon. Does it also have the catenation property? If not, why?

Somewhat related question with useful answer: https://chemistry.stackexchange.com/q/47432/544

https://chemistry.stackexchange.com/questions/18843/what-makes-carbon-special-and-versatile


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Understanding the partial derivative of temperature with respect to thermodynamic beta (coldness)
I am trying to prove that the specific heat is related to the fluctuations in the energy:
$$c_V = \frac{\langle E^2 \rangle - \langle E \rangle^2}{k_\mathrm BT^2}$$
Where:
$$\beta = \frac{1}{k_\mathrm BT}$$
$$\langle E \rangle = -\frac{ \partial \log(Z)}{\partial \beta}$$
I did:
$$\frac{\partial^2}{\partial \beta^2}\ln Z = \frac{\partial}{\partial \beta}\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial T} \frac{\partial T}{\partial \beta}$$
My issue is that I do not understand why:
$$-\frac{\partial T}{\partial \beta} = k_\mathrm BT^2$$

You know that:

$$\beta = \frac{1}{k_\mathrm BT}$$

Rearranging yields:
$$\color{red}{\beta} = \frac{1}{k_\mathrm B\color{blue}{T}}\implies \color{blue}{T} = \frac{1}{k_\mathrm B \color{red}{\beta}}$$
Thus:
$$\frac{\partial \color{blue}{T}}{\partial\beta} = \frac{\partial}{\partial\beta}\left(\frac{1}{k_\mathrm B \beta}\right) = \color{red}{-}\frac{1}{k_\mathrm B \beta^\color{red}{2}}$$
Substituting $\beta = \color{red}{\frac{1}{k_\mathrm BT}}$ yields:
$$\frac{\partial T}{\partial\beta} =  -\frac{1}{k_\mathrm B \color{red}{\left(\frac{1}{k_\mathrm BT}\right)}^2} = -\frac{k_\mathrm B^2T^2}{k_\mathrm B} = -k_\mathrm BT^2$$
$$-\frac{\partial T}{\partial\beta} = k_\mathrm BT^2$$
Q.E.D.


text:
Synthesis Golf X: TAK-457
I shall try to revive this!
A full FAQ post has been written on chemistry.meta.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

TAK-457 is a drug (more accurately, an injectable prodrug of TAK-456) that displays antifungal activity:[1,2,3]

Some rules, as usual:

At most one chiral centre may be bought in your starting material(s). [Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.]
Your synthesis must feature a method for construction of either the triazole or tetrazole ring (i.e. do not buy both).
The counterion doesn't matter.
Bonus points for synthesis that can be adapted to produce the other three stereoisomers of the drug (preferably via late stage diversification as opposed to, say, using a different enantiomer of starting material). These stereoisomers were of interest.[4]

References

Ichikawa, T.; Kitazaki, T.; Matsushita, Y.; Yamada, M.; Hayashi, R.; Yamaguchi, M.; Kiyota, Y.; Okonogi, K.; Itoh, K. Optically Active Antifungal Azoles. XII. Synthesis and Antifungal Activity of the Water-Soluble Prodrugs of 1-[(1R,2R)-2-(2,4-Difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone. Chem. Pharm. Bull. 2001, 49 (9), 1102–1109. DOI: 10.1248/cpb.49.1102.
Hayashi, R.; Kitamoto, N.; Iizawa, Y.; Ichikawa, T.; Itoh, K.; Kitazaki, T.; Okonogi, K. Efficacy of TAK-457, a novel intravenous triazole, against invasive pulmonary aspergillosis in neutropenic mice. Antimicrob. Agents Chemother. 2002, 46 (2), 283–287. DOI: 10.1128/AAC.46.2.283-287.2002.
Tsuchimori, N.; Hayashi, R.; Kitamoto, N.; Asai, K.; Kitazaki, T.; Iizawa, Y.; Itoh, K.; Okonogi, K. In Vitro and In Vivo Antifungal Activities of TAK-456, a Novel Oral Triazole with a Broad Antifungal Spectrum. Antimicrob. Agents Chemother. 2002, 46 (5), 1388-1393. DOI: 10.1128/AAC.46.5.1388-1393.2002.
Ichikawa, T.; Yamada, M.; Yamaguchi, M.; Kitazaki, T.; Matsushita, Y.; Higashikawa, K.; Itoh, K. Optically Active Antifungal Azoles. XIII. Synthesis of Stereoisomers and Metabolites of 1-[(1R,2R)-2-(2,4-difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone (TAK-456). Chem. Pharm. Bull. 2001, 49 (9), 1110–1119. DOI: 10.1248/cpb.49.1110.

Those heterocycles are harder than they look... :/

Adopting this synthesis for late-stage diversification was definitely a challenge, but the Jacobsen epoxidation fit the problem perfectly! I am actually surprised the Jacobsen epoxidation is not taught more in organic chemistry. It is much more versatile than the Sharp Asymmetric Epoxidation.

@orthocresol♦: I will have to admit, this synthesis was probably the most fun I have had in the last month! Will you be posting more synthesis-golf questions in the future? I enjoy seeing the various approaches to total synthesis people come up with!

@EliJones thanks, I am glad you liked it. In principle I am not opposed to it. In practice I don't have all that much time (and truthfully there hasn't been that much interest in recent years). You're more than welcome to copy the format.

Fragment #1 Preparation

1) An exact preparation of an aryl imidazolidinone (54% Yield)1
2) A modified preparation of a tetrazole from an aniline derivative (90% yield with pure aniline)2
3) A modified acetal formation (protocol modifications based on urea) (no reported yield)3
Fragment #2 Preparation

1) An exact preparation of an alpha-bromophenone (Hell-Volhard-Zelinski Reaction) (97% yield --likely prior to purification)4
2) Modified coupling of 1,2,4-triazole and alpha-bromophenone (100% yield with chloroacetone -- likely prior to purification)5
3) Wittig reaction modification (73% yield with phenyl methyl ketone)6
4) A modified triazolium formation (82% yield with methyl iodide)7
5) Stereoselective Jacobsen epoxidation (92% yield with 1-phenylcyclohexene)8
Final Fragment Coupling

1) Deprotonation of the amine with LDA
2) Epoxide ring opening followed by the protonation of the alkoxide and hydrolysis of the acetal to yield the final product
Stereochemical and General Considerations
The use of the Jacobsen epoxide formation allows for the use of the (L, L) Jacobsen catalyst in the final steps of the synthesis to yield the enantiomer of TAK-457. Synthesis of the other stereoisomer would require modification of earlier steps in the reaction.
Longest Linear Sequence: 5 steps
Total Number of Steps: 9 steps (I consider the final two steps as one since they are performed without subsequent purification)
Comment on 1,2,4-Triazole Ring: Yes, I did not prepare the triazole ring directly; however, this seems to be the more efficient route.
Comment on Wittig Reaction: This is the step I am probably the most worried about. The steric bulk from the phenyl group may be enough to slightly favor the desired enantiomer; however, this cannot be determined until further experimentation is performed.
References

Stabile, P.; Lamonica, A.; Ribecai, A.; Castoldi, D.; Guercio, G.; Curcuruto, O. Mild, Convenient and Versatile Cu-mediated synthesis of N-Aryl-2-imidazolidinones. Tetrahedron Lett. (2010) 51 (24). 3232–3235. DOI:10.1016/j.tetlet.2010.04.064
Kundu, D.; Majee, A.; Hajra, A. Indium Triflate-Catalyzed One-Pot Synthesis of 1-Substituted-1H-1,2,3,4-Tetrazoles Under Solvent-Free Conditions. Tetrahedron Lett. (2009) 50 (22). 2668–2670. DOI:10.1016/j.tetlet.2009.03.131
Shuai, F.; Wang, X.; Zhang, J. Preparation of Dipyridamole Active Pharmaceutical Ingredient. CNIPA Patent 108069972, Nov 16, 2016
Alonso-Alija, Cristina; et al. Preparation of Amino(monocyclic aroyl)pyridinones that Inhibit p38 Map Kinase for use as Antiinflammatory Agents. WIPO Patent 2003076405, Sep. 18, 2003
Sun, Y.; Huang, N.-Y.; Ding, M.-W. Efficient Synthesis of 6-(1H-1,2,4-Triazol-1-yl)-thieno[2,3-d]pyrimidin-4(3H)-ones via an Iminophosphorane. Synth. Commun. (2010) 40 (13). 1985–1991. DOI: 10.1080/00397910903219351
Pratsch, G.; Overman, L. E. Synthesis of 2,5-Diaryl-1,5-dienes from Allylic Bromides Using Visible-Light Photoredox Catalysis. J. Org. Chem. (2015) 80 (22). 11388–11397. DOI:10.1021/acs.joc.5b01962
Myles, L.; Gathergood, N.; Connon, S. J. The Catalytic Versatility of Low Toxicity Dialkyltriazolium Salts: in Situ Modification Facilitates Diametrically Opposed Catalysis Modes in One Pot. Chem. Commun. (2013) 49 (46). 5316. DOI:10.1039/c3cc41588k
Brandes, B. D.; Jacobsen, E. N. Highly Enantioselective, Catalytic Epoxidation of Trisubstituted Olefins. J. Org. Chem. (1994) 59 (16). 4378–4380. DOI:10.1021/jo00095a009

Wouldn't protecting carbonyl also break tetrazole?

The protecting step is not necessary for protecting the carbonyl from reacting as much as it is for activating the amide nitrogen. If I tried deprotonating the amide, it would just result in the formation of an alpha-amino alkoxide when I treat it with LDA. It would be difficult to avoid this acetal formation and the tetrazole ring from interfering. I have not necessarily seen any literature claiming catalytic p-TSA would break the tetrazole, but I suppose this is something that would need to be determined experimentally!

Hmm, prophylactically it probably could be done before forming tetrazole. Let's call it a hunch but this unstable ring might be best made in the end, just to be sure ;)

Thanks for the advice! I thought about forming it before making the tetrazole; however, I think that would render the NH too active. This might then result in a mixture of products when I try to form the tetrazole from aniline. It might be best to form the tetrazole at the end, but my synthesis would be less suitable for late-stage diversification (for multiple stereoisomers).

If your Wittig isn't selective enough, that's an easy way to access two diastereomers at once. *taps forehead*

Interesting ... I did not think of it that way! I did not expect the Wittig reagent to be particularly selective and expected to have to separate a bunch of regioisomers with a loss of yield; however, this might not be a total loss -- as you mentioned -- since I could use the 'unwanted' isomer to prepare TAK-457's other diastereomer.

Body must be at least 30 characters.

"Body must be at least 30 characters." Yes, and shoes should be tied. Please clean up your answer, as it stands it's not suitable for prime-time.

@ToddMinehardt I am sorry the answer is not typed up. I have not had time to type it up and since I left university this year, I do not have access to ChemDraw. Please let me know if any of the reactions are not clear and I will try to elaborate.


text:
What citrate is a common anticoagulant?
When I underwent plateletpheresis, something that the staff called "citrate" was added to my blood as an anticoagulant. Everything I can find about this product online refers to it as just "citrate". But from what little I know about chemical names, citrate is an ion, normally found not by itself but rather as magnesium citrate, potassium citrate, calcium citrate, vel sim.
What's the other half of this anticoagulant: which citrate is it?

Per the NIH:

ANTICOAGULANT CITRATE DEXTROSE A...
...
Citric Acid, anhydrous, USP      0.073 g
Sodium Citrate, dihydrate, USP   0.220 g
Dextrose, monohydrate, USP       0.245 g
Water for Injection, USP         q.s.
pH: 4.5 – 5.5

They used the citrate ion in the form of citric acid and sodium citrate. The citric acid is for chleating calcium while the sodium citrate does this and acts as a buffer to prevent the citric acid from raising your blood pH too much when injected as this would be painful and hazardous. Sodium is naturally in your blood and will not interfere with action potentials and is a abundant and cheap ion to use in manufacturing so it makes a logical counter ion for the citrate buffer.

Actually citric acid and sodium citrate form a buffer. Calcium chelates with citrate anion not citric acid per sey. However decreases citrate anion causes citric acid to ionize restoring the equilibrium between pH, citric acid and sodium citrate.


text:
Shape of the container and the number of possible energy states
A question came up when studying the formula for calculating the total number of possible energy states of a particle within a container.
Why is the number of possible energy states independent of the shape of the container and merely dependent on the number of states in only 3 perpendicular container dimensions ($x$, $y$ and $z$), while a momentum vector can be directed at any direction within the container? Shouldn't the length of the container in that same direction as the momentum vector also determine the number of energy states in that direction?

The number of possible energy states is infinite; why would it be dependent of anything?

Hi Johnny, can you provide more details regarding the model you’re working with, including a citation to a textbook or other reference? It is a lot easier to answer questions of this type when the model is fully laid out. Also, this question is quite similar to your last one, and I would probably recommend editing your last question than posting a new one.

@a-cyclohexane-molecule I have made an illustration but in 2D momentum space along with the question, found here: https://www.dropbox.com/s/wsomw9odrwbrsec/Question.jpg?dl=0  Regarding my last question, the question is down the list by now and I thought it would not be bumped up and noticed by others if it is edited.

Note that all edits will automatically bump questions.

@a-cyclohexane-molecule So shall I put my illustration in my previous question and ignore this one for now?

For a sufficiently small container, the size and shape of the system is significant, e.g. some energy states are prohibited. For example, CdTe quantum dots behave differently from the bulk material.
However, the kinetic theory is based on the assumption that, "The number of molecules is so large that statistical treatment can be applied." The momentum is treated as a vector having x, y and z components and does not depend on the shape of the container (though a cube is convenient for illustration of three orthogonal directions). The equations can be derived using spherical coordinates, and they still must integrate collisions in all directions, so are equivalent.

Within a momentum sphere, the total number of energy states is based on the number of states in each of the 3 container length dimensions multiplied by each other and fractionized according to the sphere volume, right? This implies that the density of states within the momentum sphere is homogeneous. But why isn't it so that each momentum vector within that sphere would have a number of states according to the container's length dimension in that SAME direction of that momentum vector, since the number of states is dependent on dimensions in the first place?


text:
Basicity of singlet and triplet carbene
Which form of n-heterocyclic carbene is more basic and stable? ( Singlet or triplet)

Singlet carbene is sp2 hybridised whereas triplet carbene is sp Hybridised. More s character, more nuclear force, more Electro negativity so more acidic hydrogen and hence less basicity.
So singlet carbene is more basic.


text:
Question regarding mole calulation
A 10.00g, mixture of potassium carbonate and potassium hydrogen carbonate is heated to constant mass. The final mass was found to be 8.90g. calculate the % composition of the original mixture.
My values were 35.5% KHCO3 and 64.5% K2CO3
Thanks in advance

Well, show your calculation and we show you how to simplify it. Or we can start with your ansatz. Two givens, two unknowns, what are the equations?

As @OscarLanzi noted $\ce{K2CO3}$ does not decompose when heated. Also when heated $\ce{KHCO3}$ decomposes to $\ce{K2CO3}$.
$$\ce{K2CO3 ->[\Delta] K2CO3}$$
$$\ce{2KHCO3 ->[\Delta] K2CO3 + CO2 ^ + H2O ^}$$
MW $\ce{K2CO3}$ = 138.205
MW $\ce{KHCO3}$ = 100.115
$x$ = mass $\ce{KHCO3}$
$\dfrac{200.230-138.205}{200.230} =\dfrac{10.00 - 8.90}{x} =\dfrac{1.10}{x}$
$ x = \dfrac{200.230\times1.10}{62.025} = 3.55$
% $\ce{KHCO3}$ = $\dfrac{3.55}{10.00}\times 100\% = 35.5\%$
You must write really really large...

Purely documentation:  The reference to me is in a now deleted answer.  That answer originally assumed that potassium carbonate decomposes to the oxide upon heating; but except for lithium, alkali metal oxysalts do not really do that.  Potassium carbonate should be considered stable.


text:
Is iron(III)nitrate changed and if so how?
We have the solution:
$\ce{Fe^3+(aq) + SCN-(aq) <=>Fe(SCN)^2+}$
When we add more $\ce {SCN-}$ to the solution above, the solution produces more product.
My question is, how does it affect $\ce {Fe^3+}$? Do the concentration of iron(III)nitrate change or something?

If a stoichiometric amount of $\ce{SCN-}$ is added then the reaction is mostly
$$\ce{Fe^3+(aq) + SCN-(aq) <=>Fe(SCN)^2+(aq)}$$
If an excess of of $\ce{SCN-}$ is added then the reaction is mostly
$$\ce{Fe^3+(aq) + 2SCN-(aq) <=>[excess SCN-] Fe(SCN)2^+(aq)}$$

It's Le Chatilers principle.If you increase the reactant,the product will also increase to maintain equilibrium.This has nothing to do with Fe ions as they are already in dynamic equilibrium,provided all the reactants are present in excess.

Can anyone explain the downvote?


text:
How is the reaction rate constant affected by concentration?

Consider the concentration-time dependence graph for two first-order reactions.

Which reaction has a larger rate constant?

Unable to determine
The reaction represented by the upper curve
The reaction represented by the lower curve correct

The above question is confusing me.
It gives two concentrations vs. time curves, and asks which reaction has the larger rate constant. The answer is (3) the reaction represented by the lower curve.
I know that the reaction is first-order, hence
$$\mathrm{rate} = k[A].$$
Thus, algebra reveals that
$$[A] = \frac{\mathrm{rate}}{k}.$$
Thus, it makes sense that increasing the value of $k$ means dividing rate by a larger number, yielding a smaller concentration at any given time. This makes sense why the lower curve is the correct answer.
However, I also understand that the value of $k$, the rate constant, does not depend/is not correlated with the concentration of the reactant(s). Thus, is my logic not applicable? How else can you reason your way to the correct answer?

The rate constant, k, does not depend on concentration. The reaction rate itself, $k\ce{[A]}$, does depend on concentration. For graph $k_{lower} > k_{upper}$ since the starting concentration of A is the same.

@MaxW That is a very clear explanation. Thank you so much! Because we're comparing two different reactions, and we know that the initial concentrations are the same BUT the rates are different, k must be a different value in the two reactions. Thank you!

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.
Anyway, you are correct, $k$ is a constant.
$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$
thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as:
$$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$
(the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.
Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$
$$\mathrm{rate}_{upper}  = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\
\text{and }\\
\mathrm{rate}_{lower}  = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$
Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:
$$\mathrm{rate}_{1}  = \pu{2 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}\\
\text{and }\\
\mathrm{rate}_{2}  = \pu{4 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}$$
Since a reaction of $\mathrm{rate}_2$ has a more negative slope, it must correspond to a curve  that decreases faster than a reaction of $\mathrm{rate}_1$ and the concentration would be lower sooner, thus $\mathrm{rate}_2$ is the lower curve and $\mathrm{rate}_1$ is the upper curve.


text:
How to interpret a singlet term symbol for an s¹p¹ configuration?
I am interested in the $^1P_0$ singlet state of $\ce{Hg}$ ($\ce{5d^10 6s^1 6p^1}$).
I interpret this to be the two following electron configurations below, because for an $S = 0$ state the two electrons have to have different spins.

Are these two both a valid state? Do they both have the same energy? How would I tell if it's only one of them?

The two states you have drawn correspond to the (antisymmetrised) wavefunctions
$$\begin{align}
\Psi_i &= \mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
\Psi_j &= \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2)
\end{align}$$
Neither of the two states above are valid, as they do not obey the requirement for indistinguishability. For fermions such as electrons (spin-1/2), this is achieved by making the wavefunction antisymmetric upon permutation of any two electron labels. However, neither $\Psi_i$ nor $\Psi_j$ obey this:
$$\begin{align}
\mathcal{P}(\Psi_i) &= \mathrm{s}(2)\alpha(2) \cdot \mathrm{p}(1)\beta(1)  \neq -\Psi_i \\
\mathcal{P}(\Psi_j) &= \mathrm{s}(2)\beta(2) \cdot \mathrm{p}(1)\alpha(1)  \neq -\Psi_j
\end{align}$$
The appropriate wavefunction is a linear combination of these states (and a couple more):
$$\Psi_\text{singlet} = \frac{1}{\sqrt 2}[\mathrm{s}(1)\mathrm{p}(2) + \mathrm{p}(1)\mathrm{s}(2)] \cdot \frac{1}{\sqrt 2}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]$$
If you expand these you can see the constituent wavefunctions
$$\begin{align}
\Psi_\text{singlet} &= \frac{1}{2} [\mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
&\qquad + \mathrm{p}(1)\alpha(1) \cdot \mathrm{s}(2)\beta(2) \\
&\qquad - \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2) \\
&\qquad - \mathrm{p}(1)\beta(1) \cdot \mathrm{s}(2)\alpha(2)] \\
\end{align}$$
You need four "simple" wavefunctions in total because you have to account for the spatial indistinguishability of the two electrons (hence need a linear combination of two spatial wavefunctions) as well as the spin indistinguishability (linear combination of two spin wavefunctions).


text:
Regarding acid-base reactions, how does one know when a hydrophobic base substance is fully protonated?
When there's a weak base substance (hydrophobic powder), and it needs to be completely protonated by adding an acid (eg: HCl), how does one know when it's completely protonated? Since the powdered base substance isn't soluble until protonated, doesn't that mean that it won't be reflected when measuring the pH (via the commonly used pH test strips)?
I'm under the impression that when the base is suspended in water, as HCl is added and given time to react, the pH would float round 7 or 8. Then when it's fully protonated, the pH will start to dip below 7, which would mean there's now an excess of HCl since there's nothing for it to bind with, which would be too much.
I don't exactly have a specific substance base substance in mind. This is more a general question about acid-base reactions involving a solid base.
Thank you

Yes, that is correct.
As you add HCl, the pH will remain more or less constant at around 5, since in solution there is a salt of a strong acid and a weak base.
Then when all the base has been protonated, the pH will fall rapidly.
If you need to know how much insoluble base you have, it must be done through back titration. You add an excess of HCl to the sample and then titrate the remaining free HCl.

Ah!.. "back titration", exactly what i was looking for! Thank you very much.


text:
Would a red or a white light be more effective at charging a calculator battery?
White light contains the entire visible spectrum including red light, so would it be more effective at charging a battery?

There is a theoretical and a practical issue here (and smaller ones, too).
The theoretical issue is the energy of a photon depends on its frequency (ie color), therefore a blue photon carries more energy than a red one. If you try to charge a battery, then you need more than a photon, so both the color and the intensity of the light/time of irradiation will be a factor.
The practical issue here is that solar cells generally can grab the energy of photons is given energy packages. If photons cannot be harvested at all or only with low efficiency,  you cannot charge a battery even upon extended irradiation. As the comment pointed it out, it is not an issue for red light for regular silicon solar cells.

Red is certainly above the band gap of silicon, the most common solar cell material.

@JonCuster  ok, my bad, you are correct


text:
What's wrong with this gas law problem?
I recently took a chemistry test in which the following problem was offered:

a 137 mL gas sample is collected over $\ce{H2O}$ at $753\ \mathrm{mmHg}$ and $22\ \mathrm{^\circ C}$. What is the volume of the dry gas at STP if the vapor pressure of $\ce{H2O}$ at $22\ \mathrm{^\circ C}$ is $22\ \mathrm{mmHg}$?
a) $106\ \mathrm{ml}$
b) $117\ \mathrm{ml}$
c) $123\ \mathrm{ml}$
d) $126\ \mathrm{ml}$

Here's my work:
To start with, I used Dalton's Law of partial pressures:
$$p_\text{total}=p_\text{gas}+p_{\ce{H2O}}$$
$$p_\text{gas}=731\ \mathrm{mmHg}\ \left(\text{at}\ 22\ \mathrm{^\circ C}\right)$$
Then I used the following proportion to find the volume of the gas:
$$\frac{p_\text{gas}}{p_\text{total}}=\frac{V_\text{gas}}{V_\text{total}}$$
$$V_\text{gas}=133\ \mathrm{ml}\ \left(\text{at}\ \mathrm{^\circ C}\right)$$
Which made sense to me because if the total number of moles of the gas mixture is not changing, and the temperature is not changing the ratio of the volumes should be the same as that of the pressures (shouldn't it?)
Finally I did a bit of dimensional analysis to change from $22\ \mathrm{^\circ C}$ to STP:
$$133\ \mathrm{ml}\ \cdot\ \frac{731\ \mathrm{mmHg}}{760\ \mathrm{mmHg}}\ \cdot\ \frac{273\ \mathrm K}{295\ \mathrm K}$$
$$V_\text{gas}=118\ \mathrm{ml}\ \left(\text{at STP}\right)$$
I feel like I'm missing/forgetting a key concept here because each individual step made sense in my head (and still makes sense now). The answer the prof gave was d. This is bugging me a lot and I would appreciate any help.

Let's derive the appropriate realtionship...
$$PV = nRT\tag{1}$$
so
$$\dfrac{PV}{T} = nR\tag{2}$$
but $nR$ is constant for the given sample of gas, thus
$$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$
For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K.
Now here is a bit of a wrinkle. What is STP? Until 1982 IUPAC defined STP as 760 mm of mercury at 0 °C = 273 °K. After 1982 IUPAC has defined STP as  750.06 mm of mercury at 0 °C = 273 °K.
For old IUPAC STP
$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{760} = 121.95 = 122\text{ ml}$$
For current IUPAC STP
$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{750} = 123.57 = 124 \text{ ml}$$
In looking at some possible variations, it seems the prof is using the wrong initial pressure, and the old IUPAC definition.
$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{753\times137}{295}\times\dfrac{273}{760} = 125.62 = 126\text{ ml}$$
NOTE
When collecting a gas over water you must subtract the partial pressure of water at the collection temperature from the atmospheric pressure, but you don't adjust the volume too.
Also this assumes that the atmospheric pressure given, 753 mm Hg, is for dry air.
If you were measuring the pressure with a laboratory barometer then you'd have to adjust the measured laboratory pressure for the relative humidity (partial pressure of water).

So in other words you are saying that the mixture of gases occupies the same volume (137 mL) as each individual gas? I thought gas volumes were additive. Are they not?

For formula 3 you need $P_1$ and $V_1$ of the dry gas to get the appropriate value for the product $P_1V_1$. Since there is water vapor in the gas you can correct either the pressure or the volume, but not both. Pressure is the easier thing to correct since you have to start there anyway.

In other words $P_1V_1 = 100147$. So it can either be $137\times731$ or $133\times753$.

See, that's where I get stuck. Why don't you need to correct both? As in, what is the physical reason? The way I picture it, you have two individual gasses, each of which makes an individual contribution to the total pressure, the total volume and the total moles. So to calculate the number of moles of the dry gas you would need to find the pressure of the dry gas (which was the first thing I did) and the volume of the dry gas (which was the second thing I did). Why is this wrong? Are the volumes not independent/ additive? Does each component have the same volume as the total volume?

$t$ = total, $w$ = water, $g$ = gas $$P_tV_t = P_wV_w + P_gV_g$$ $$P_gV_g = P_tV_t - P_wV_w = 753\times137 - 22\times137 = (753-22)\times137 = 100147$$

So, seeing how you're using 137 mL for $V_t$, $V_w$ and $V_g$ I assume you are saying that each of the component gases has the same volume as the total volume of the mixture, yes?

Yes, but as I said before I could also think of the three pressures as being 753 mm Hg. Then the $V_t = 137$ ml, $V_w = 4$ ml, $V_g = 133$ ml. So in this case I'm thinking that there is an "Star Trek Force field" separating the 137 volume into two parts. a 133 ml part with dry gas and a 4 ml part with pure water vapor at 753 mm Hg. Of course in the real world water vapor at 22 C and 753 mm Hg would condense so this abstraction is purely a Gedanken-experiment.

Thank you! After a bit more research I discovered this: http://www.departments.bucknell.edu/chem_eng/cheg200/CinChE_Manual/Ch06/examples/idgas.pdf
Which is a more nuanced discussion of the topic than we had when I learned about this. I think I have the right mental picture of it now.

The number of moles of gas collected is $$n=\frac{PV}{RT}=\frac{(0.137)(753/760)}{(0.08206)(295)}=0.00561\ moles$$
The mole fraction of dry air is $$x=\frac{731}{753}=0.9708$$So the number of moles of dry gas is (0.00561)(0.9708)=0.00544.  At STP, 1 mole of dry gas occupies 22400 ml.  So the volume of dry gas is (22400)(0.00544)=122 ml.

Since 1982 STP is  750.06 mm of mercury at 0 °C so the molar volume is 22.711 liters.

OK.  Then 123 ml.


text:
What is the electrode potential at the equivalence point if concentrations of titrant and titrand are going to be zero?
Let's assume that the titrand is $\ce{Fe^2+}$ and the titrant is $\ce{Cr2O7^2-}$. Then
$$E = E^\circ_\ce{Fe^{2+}/Fe^{3+}} - \frac{RT}{nF}\ln\frac{\ce{[Fe^{2+}]}}{\ce{[Fe^{3+}]}}$$
$$E = E^\circ_\ce{Cr^{6+}/Cr^{3+}} - \frac{RT}{nF}\ln\frac{\ce{[Cr^{3+}]}}{\ce{[Cr^{6+}]}}$$
Adding both the above equations, we get
$$2E = E^\circ_\ce{Fe^{2+}/Fe^{3+}} + E^\circ_\ce{Cr^6+/Cr^3+} - \frac{RT}{nF}ln\frac{\ce{[Fe^{2+}][Cr^{3+}]}}{\ce{[Fe^{3+}][Cr^{6+}]}}$$
Since $\ce{[Fe^2+]=[Cr^6+]}$ and $\ce{[Fe^3+]=[Cr^3+]}$
They would all be getting cancelled, ending in
$$E = \frac{E^\circ_\ce{Fe^{2+}/Fe^{3+}} + E^\circ_\ce{Cr^{6+}/Cr^{3+}}}{2}$$
But, at equivalence point, the titrant ions would have entirely reduced and the titrand ions would have entirely oxidised, leaving
$$\ce{[Fe^{2+}]=[Cr^{6+}]}=0$$
And as we know, in mathematics, $\frac00$ is infinite or not defined.
So, how do can we actually mathematically derive the electrode potential at the equivalence point?

"ions would have entirely reduced" is not true, there eq. constants are finite.

@Mithoron  So does that mean there will be a little (close to 0) amount of unreacted Fe 2+ ions even after the supposed complete oxidation?


text:
Does 3,6-di-X-cyclohexa-1,4-diene have stereo centers?
The following question has multiple correct answers:

The book out of which the question is taken  shows that a, b and d are correct answers.
According to me c should also be correct, as it shows cis-trans isomerism, both X could go to the front (wedge bond), or one could go to the front (wedge) and one to the back (dash bond).  After all geometrical isomerism is also a sterioismersim.
I came to this conclusion considering the definition given on Wikipedia:

A stereocenter or stereogenic center is any point in a molecule, though not necessarily an atom, bearing groups, such that an interchanging of any two groups leads to a stereoisomer.

Good point. The (c) compound has two tetrahedral stereogenic carbon atoms, with four “substituents”

$\ce{-H}$,
$\ce{-X}$,
$\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$,
and the same one, $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$ but from the other side, thus with the opposite configuration, or its mirror image.

(‘*’ denotes (the other) stereogenic carbon atom)
They are called pseudo-asymmetric carbon atoms.
A different relative configuration at these two pseudochiral centers leads to these two ‘cis’ and ‘trans’ isomers.
The definition of stereogenic center (for more complete one, see e.g. this answer) is fulfilled.
(Maybe this topic is too early, for this textbook stage, and the author did not notice this complication.)

Now, the key to understanding stereoisomerism is that we have to look for elements of symmetry of which:

Plane of Symmetry(PoS)
Centre of Symmetry(i)
Alternating Axis of Symmetry(AAoS)
are the prominent ones.
Again, please keep it in mind that chiral centres may or may not enough to decide the chirality of the entire molecule(examples are meso compounds with two or more chiral centres, or "pseudo chiral" molecules which are chiral in one configuration but not in another)
Now, since you are clear on the other three options, I am going to jump straight to the third option.

Now, as you can notice, the molecule does possess a centre of symmetry. What this means is that now, when you will try to make those cis-trans configurations, you will always find a plane to cut through to make identical halves. Also, since they are present on a ring,so there is no scope of "weird" configurations like keeping both the X and H on cis or trans. Hence, it is safe to say that as soon as you have located even one symmetry element, you can call the molecule achiral.
If the molecule turns out to be achiral, so any change at any group of atoms won't give you a steroisomer. So it also won't possess any stereocentres

Yes, we have looked upon chirality elements so as to find whether optical isomerism will be shown or not. Well, we jump on  the conclusion that the molecule is a chiral. OK. But why there is no scope of geometrical isomerism?

@Saifuddin Have you seen that there is a centre of symmetry in part (c)?

Yes the molecule is achiral, but what about geometrical isomerism? Why cis or trans can't be shown on the ring?

@ Saifuddin  Just try to draw the stereoisomer you are talking about and visualize it properly. You will find that you will always be able to send a plane of symmetry passing through the two opposite substituted carbons and perpendicular to the ring. Whenever you will try to make the trans isomer, the other X and H will pick up a symmetric positioning with it. And as I have already explained in my answer, both the groups cannot be kept on cis as the ring will not have such flexibility due to steric reasons. So ultimately, the molecule will exist is one form only.

@Martin-マーチン Can you explain what really is the difference between chirality and stereocentres?

@Saifuddin: The difference between chirality and stereocenters? See my answer at chemistry.stackexchange.com/questions/91539/… and the Mislow reference cited. In your example, only a, b and d have chirotopicity but all four have stereogenicity. C has a cis and trans isomer. Arguments as to whether or not a cis isomer can exist (YUSAF HASAN) are irrelevant.


text:
Is graphite a better lubricant on the Moon than on the Earth?
My book says that it is a better lubricant on the Moon because of absence of gravitational pull on the Moon. However, I also read that it's lubricant nature is because of a film of moisture or gas molecules absorbed on the surface of its layers and in vacuum, it gets dried up and would be a bad lubricant. Which is correct?

Googled enough to convince myself that "regular" graphite would be a bad lubricant in the vacuum of space, but couldn't find a good authoritative reference.

Gravitation pull has no impact on lubricity (contact forces perhaps).

By curiosity:  please state your source (here: your book) you cite. Perhaps SE Space Exploration has contributors who happens to know about the lubricants used for lunar rover and lunochod by NASA and pre-Roscosmos and their experiences.

Graphite, like most solid lubricants, has a layered structure which makes it a good lubricant.
The atoms of a layer are strongly bonded to each other, while the interlayer (van der Waals) forces are much weaker.
The layers align themselves between the sliding surfaces and the layers slide one over the other providing lubrication.
Since it are the shear forces that come into play, gravity has no influence on the system.
From that point of view, there is no difference between its use here or on the Moon.
But, there are also extrinsic factors which influence its performance.
Graphite works best in a humid environment.
This is probably due to the weakening effect of the water molecules on the interlayer forces.
So, graphite should not work well in vacuum.
Another solid lubricant, $\ce{MoS2}$, which has also a layered structure like graphite, works much better in vacuum and should be the lubricant of choice on the Moon.


text:
How to explain the excited states in the dinitrogen cation?
The $\ce{N2^+}$ molecule has the molecular orbitals
$$\ce{\sigma(1s)^2\; \sigma^*(1s)^2\; \sigma(2s)^2\; \sigma^*(2s)^2\; \pi(2p_x)^2 \; \pi(2p_y)^2 \; \sigma(2p_z)^1,}$$
which can be seen on the following picture:

So, the ground state has $\mathrm{A_g}$ symmetry in $D_\mathrm{2h}$ point group, i.e. $\Sigma_\mathrm{g}^+$ in $D_\mathrm{\infty h}$.
But what will the excited states look like?
I'd expect, that the first excited electrons will be the ones, which "need" the smallest amount of energy to get into a higher state.
So, the first electron will be excited from $\ce{\pi(2p_x)}$ or $\ce{\pi(2p_y)}$ orbitals to $\ce{\sigma(2p_z)}$ like this:

The first excited state will have $\mathrm{B_{3u}}$ or $\mathrm{B_{2u}}$ symmetry in $D_\mathrm{2h}$.
The second excited state would be

I.e. it would have the $\mathrm{B_{3g}}$ or $\mathrm{B_{2g}}$ symmetry.
Is my expectation correct or will the electrons be excited in a different order? And what will the 3rd and 4th excited states look like?

Not a full answer, but strictly speaking its only an approximation that you could obtain an excited state from exciting from an occupied to a virtual orbital. If we consider a model where we do promote an electron, that promotion should be followed by a step where all the orbitals relax to better conform to the new configuration of electrons. So you really can't excite into a different orbital because as soon as you move an electron, you change the shape/energy of all the orbitals of the molecule.

Just a note on convention. Capital letters are reserved for the spectroscopic state/wavefunction symmetry, while we label orbitals/electronic configuration with lowercase letters. (e.g. a_g rather than A_g )

Excited states make things a bit more complicated. The electron configuration only represent the HF ground state, and changing the occupied orbitals will have an effect on all the orbitals (they are correlated). So this picture becomes more and more inaccurate the higher lying the electronic state. What you do to resolve this is expand the electronic state in many configurations (which all most belong to the same IRREP).
For diatomic molecules there is a lot of spectroscopic data available. So let's check it: $\ce{N2+}$ in the NIST Chemistry WebBook
The states in that table are ordered by energy, with the ground state at the bottom. So scrolling down shows us the 3 lowest lying states: $\rm X ^2\Sigma_g^+$, $\rm A^2\Pi_u$, $\rm B^2\Sigma_u^+$.
Basically, this means you are right about the first state. To explain why we need a bit of symmetry. The first excited state has $\Pi_u$ symmetry, in the $D_{\infty h}$. Since we are working with $D_{2h}$ here, we will need to reduce it. Sparing the details, the result is $\Pi_u$ becomes $B_{2u}\oplus B_{3u}$, which means it will show up once in each IRREP, which is fine because $\Pi$ states are doubly degenerate anyway (just as $\pi$ orbitals are doubly degenerate).
The other state ($\Sigma_u^+$) becomes $B_{1u}$. This one you can for example get by excitation from the anti-bonding $2\sigma_u$ into $3\sigma_g$, with respect to the ground state configuration (see this reference). So there is an electron "jumping over" the $1\pi_u$ orbitals.
Note, that the ordering of the states can change if you start to move away from the equilibrium distance towards dissociation.

By "the other state ($\Sigma_u^+$)" you mean that one electron from $\sigma^*(2s)$ will get excited to $\sigma(2p_z)$, while both electrons in $\pi(2p_x)$ stay in place? I.e. the electron will "jump over" $\pi(2p_x)$ and $\pi(2p_y)$ orbitals?

Yes, this is indeed what happens here. Not really straight-forward to spot. Excited states are more complex ...


text:
Why is the active space for the dinitrogen cation defined this large in my example?
I have a Molpro script which computes $\ce{N2^+}$ potential energies.
The significant part is the input for the wave function:
 {multi;
   occ,3,2,2,0,3,2,2,0;
   closed,1,0,0,0,1,0,0,0;
   wf,13,5,1;
   state,3;
 }

Why is the active space specified by OCC flag so large? It specifies 14 molecular orbitals, which means that both nitrogen should habe 7 atomic orbitals. Which atomic orbitals are additionally included?
The ground states of $\ce{N2^+}$ molecule has this electron configuration:

Thich can be specified as occ,3,1,1,0,2,0,0,0;.
What advantage does occ,3,2,2,0,3,2,2,0; have in the script and how can I understand and visualize those additional orbitals?

Explanation of OCC card
The OCC card in Molpro specifies symmetries of the orbitals in active space. In this case, they're specified in $D_\mathrm{}$ point group, according the following table:
$$D_\mathrm{2h}\\
\begin{array}{ccc} \hline
\text{No.} & \text{Name} & \text{Function}\\
1& \mathrm{A_g}    & s  \\
2& \mathrm{B_{3u}} & x  \\
3& \mathrm{B_{2u}} & y  \\
4& \mathrm{B_{1g}} & xy \\
5& \mathrm{B_{1u}} & z \\
6& \mathrm{B_{2g}} & xz \\
7& \mathrm{B_{3g}} & yz \\
8& \mathrm{A_u}    & xyz \\
\hline
\end{array}
$$

The active space is a truncation of the full CI space. Hence, including more virtual orbitals will lower the energy and eventually approach the FCI limit. The larger the active space, the more accurate the results.
The smaller active space will give you a good qualitative description.
But if your results turn out to be quantitatively wrong, increasing the active space is one way to improve them.
To get an idea about the higher orbitals, take a close look to the Molpro output. You might want to check the Print options, to get all desired energies and orbital coefficients. Furthermore, you can use for example put to create molden files and create 3D isosurface plots of all MOs.
Possible active spaces
The smallest meaningful would be to include the singly occupied $3\sigma_g$ orbital, together with its anti-bonding pendant $3\sigma_u$. Furthermore all the $1\pi$ orbitals should be included. This makes a CAS(5,6):
occ,   3,1,1,0,3,1,1,0
closed,2,0,0,0,2,0,0,0

Your occ,3,1,1,0,2,0,0,0 card actually excludes the anti-bonding $3\sigma_u$ and $1\pi_g$ orbitals, which corresponds to a ROHF calculation.
The next larger space would be the full valence active space, covering the whole $2s2p$ AO shell:
occ,   3,1,1,0,3,1,1,0
closed,1,0,0,0,1,0,0,0

Addtionally one could also include the $1s$ core orbitals with closed,0
The next thing to include would be the next higher lying virtual orbitals coming from the $3s3p$ AO shell: occ,5,2,2,0,5,2,2,0. After that one could go on with the $3d$ and $4s4p$ shells.
The occ,3,2,2,0,3,2,2,0 card would include the bonding $2\pi_u (3p_{x/y})$ and anti-bonding $2\pi_g (3p_{x/y})$ orbitals, but not the $4\sigma_{g/u} (3s)$ and $5\sigma_{g/u} (3p_z)$ which are similar in energy.


text:
Is it possible to have a ferrocenol?
I'm specifically talking about having a hydroxy- group on the ferrocene ring.
Is it simply impossible because the ring is anionic?
I haven't seen any hydroxyferrocene molecules available online. (unless they are a methanol group)

Ring isn't really anionic.

Yes we can have hydroxyferrocene.  See here for a brief summary description of possible synthesis and a nucleophile substitution reaction forming alkoxyferrocenes.


text:
According to MO theory, do 1s orbitals in dioxygen interfere?
According to MO theory, are the 1s orbitals in $\ce{O2}$ too radially contracted to interfere with each other and thus cannot form bonding and antibonding orbitals, or do they still do that?

It is always possible to define a bonding and antibonding orbital regardless of how small the overlap is. Because we solve the homonuclear diatomics in the symmetry of the molecule ($D_\mathrm{\infty h}$), the MOs we choose must respect that symmetry. It wouldn't matter if I put the oxygen atoms 10 light years apart, if I am treating them as part of the same quantum system, there are bonding and antibonding molecular orbitals. (Of course, at large distances bonding and antibonding orbital pairs becomes very near degenerate.)
That said, because oxygen has a doubly occupied 1s orbital, the corresponding sigma 1s bonding and antibonding orbitals will both be filled. Within MO theory we are always free to 'rotate' any two occupied MOs and the observable properties will all be the same.
So, there's no difference between saying that each oxygen has a non-bonding 1s orbital, and saying that there is a filled bonding and antibonding orbital. When both orbitals are filled you can't even ask the question of what the electrons are 'doing' down there, because the concept of a distinct orbital has no meaning.


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What is the substance with the highest specific heat?
I always learned that water has the highest specific heat, but I recently saw that hydrogen has a specific heat as high as $\pu{14 cal g^-1 °C^-1}$ and helium has a specific heat of $\pu{5 cal g^-1 °C^-1},$ which would be much higher than water? Is this true? and how can it be if there are so many “proofs” in nature, showing that water has the highest specific heat?

related: https://chemistry.stackexchange.com/q/26651/23561

Yes, that's true, except that the figures are in Joules, not calories. Still, they are a good deal bigger than that for water. There is nothing strange or anomalous about it.

You would be better to compare heat capacity in joules/mole/K rather than mass, in that case hydrogen is approx 28 , water  approx 75, but for example benzene 117  and anthracene  211.

I bet the interior of the sun has a very high specific heat.

Since we are really interested in the heat effects of a liquid at somewhat normal temperatures. I am going to disregard the extremes displayed by diatomic gaseous elements and look at liquids at stp. For years water was considered the champion, but I reasoned that the more massive heavy water would show an even greater resistance to the transfer of heat motion. Not only is duterium based water higher but it appears that tritium based water is even higher.

+1 But giving the values would make the answer better.

What do you mean by “resistance to transfer of heat motion”? An explanation of why the heavier waters have higher specific heat would make this a much better answer. (Hint: think about vibrational frequencies)

This may not be the absolute highest, but on a mass basis hydrogen gas has more than three times the specific heat as water under normal laboratory conditions. Diatomic gases under ambient conditions generally have a molar specific heat of about $\pu{7 cal mol^-1 K^-1},$ and one mole of hydrogen has only $\pu{2 g}$ mass. Thus $\pu{3.5 cal g^-1 K^-1}$ for hydrogen versus $\pu{1 cal g^-1 K^-1}$ for water.
Helium, contrary to the question, is $\pu{5 cal mol^-1 K^-1},$ not $\pu{7 cal g^-1 K^-1}.$


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Oxidation No. of Fe and NO in Nitroprusside ion
For the complex $\ce{[Fe(CN)5NO]^{2-}}$ wikipedia cites that the oxidation no. of Fe is +2 and that of NO is +1. CN being -1 the net charge thus becomes -2. However at certain places I read that NO is neutral and Fe is in its +3 oxidation state. In both cases the net charge is -2. However which of the two is correct ?

Related: Is iron in the brown ring compound in a +1 oxidation state?

The Wikipedia articles states that by epr experiment the Fe is $ d^6 $ so $\ce{Fe^{2+} }$. If NO can be formally either 0 or +1 you cannot determine which without some experimental data.

The thing about nitrogen monoxide ($\ce{NO}$) is that it can be both a neutral compound and an ionic compound, depending on the context. This problem is about complex ions, which consist of a metal cation and supporting 'ligands.' Ligands are entities/molecules which bind to a metal cation to form a complex or coordinate ion.
For more info: https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.03%3A_Nomenclature
According to this article, in $\ce{[Fe(CN)_5NO]^{-2}}$, $\ce{Fe}$ is the metal cation, and $\ce{CN^{-1}}$ and $\ce{NO}$ are the supporting ligands. Given that the ion CN has a charge of -1, that will be its oxidation number. NO is neutral, so it has no oxidation number. Putting this all together:
$\ce{Fe} + 5\cdot(-1) + 0 = -2\ \ \ \ $ - - - - - - -> $\ \ \ \ \ce{Fe} = +3$
Therefore, iron has a charge and an oxidation number of +3. ($\ce{Fe^{+3}}$)

Check signs in your equation.


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How could I produce/rig this flame-colour lottery?
Not sure if this is the wrong SE for this, happy to move it if not...
In the Netflix TV series Sabrina, there is a scene where people take part in an appropriately-gothic-and-ominous lottery, which proceeds as follows:

Fourteen participants draw, one at a time, pieces of paper from a box.  The papers are all blank and have no differences visible to the human eye in the (rather moody) lighting.
The participants burn their paper in the flame of a candle which is in front of them.  There is a separate candle for each participant, but they all appear to burn with the same yellow flame.
Twelve of the papers burn up quickly, like flash paper.  The other papers burn more slowly, and release red (for the 'runner up') or white (for the 'winner') smoke, respectively.

 The runner up waits on the winner hand and foot for three days, and then the winner is killed and eaten cannibalistically, with a strong religious 'ascension' dogma.

Depending on your point of view, this is a lottery you may or may not want to win, and also one which there would be great advantage in being able to rig to ensure a particular winner.  So my questions are:
What chemistry could you use in the papers and candles to achieve the effects shown?

Don't think this will be answered here. Pictures may help

I removed the follow up questions as they are too broad to answer.

@A.K. some sort of "identify" tag.  Sci-fi-and-fantasy SE, for instance, has story-identification, which is "I remember this story where X happened, can anyone identify what it was?".  I guess I'd be looking for a tag which meant "suggest chemistry which could result in X non-scientific effect".

As another example question in the same style, I remember from my school days a demonstration with two clear beakers of liquid, a white rag and a bottle of dark pigment.  He splattered the pigment on the rag, then dunked the rag in the first beaker, causing the whole beaker to turn black.  Then he removed the rag, which had also become black, and dunked it in the other beaker; the liquid bleached the pigment from the rag, and when he then moved the rag back to the other beaker, the rag bleached the pigment from the liquid.  The whole demonstration was accompanied by a religious narrative.

I believe part of the reason we don't have an [tag:identify] tag is it is not helpful for drawing attention from needed individuals. not a bad suggestion though.

There is a tag [tag:identification], but we have already talked about removing it, because it is a meta tag. It simply doesn't help in determining which chemistry it would belong to. cc @A.K.

I think the question (as it is now) is fine and there are probably a few (but narrow in scope) legit answers. I would recommend keeping it open and see what happens.

Smokeless and quick burning isn't hard to achieve, just use flash paper. For white smoke, I would use flash paper again but with a white oxide powder such as titanium oxide with a flame retardant to slow the burn. For red, again flash paper but, you need a compound that releases bromine when burned. I think copper(i) bromide or silver(i) bromide would be ideal as both are easy to decompose and tolerant to moisture in air, though silver bromide is not tolerant to prolonged light exposure. These salts may discolor the paper and you may need to color the non-red smoke papers to make them all look the same.

I don't think bromine is a good way to get red in a flame and is certainly not safe (though given the context this might not be an issue). Surely a strontium salt would be better?

@matt_black red smoke, not flame. And yes bromine is not the best, but given the context I would not worry about the irritating smoke.


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Is it possible to generate electricity by using two inert electrodes in a galvanic cell?
The Daniell cell has two active electrodes, meaning that both the copper and the zink electrodes participate in the redox reaction. However, in a lemon battery for example, the copper electrode is inert - it doesn't participate in the redox reaction. When is it possible to only have one active electrode? Is it possible to generate electricity by using two inert electrodes?

a Hydrogen fuel cell has two inert electrodes. Inert in the sense that matter from neither electrode is oxidized or reduced.

A non-chemical way to generate electricity using two inert electrodes in salt water is through magnetohydrodynamics. If the salt water moves past the electrodes in a magnetic field, it generates an electric current without consuming the electrodes or electrolyte.
This could make an interesting "magic trick" or puzzle: use a moving magnetic field, e.g. a magnet attached to a motor, in a box below a beaker filled with salt water, and insert two carbon electrodes connected to an AC current meter. Of course, make sure the motor is very quiet.

There are many half-reactions where the electrode does not participate in the redox reaction (and no metal is deposited on the electrode, which arguably would change the electrode such that it now does participate). The prime example is the standard hydrogen electrode, where hydrogen gas is bubbled into a solution in a way that the bubbles come in contact with a platinum electrode.
Half reactions where all species are soluble
Any half reaction where a metal changes oxidation states without reaching oxidation state zero (i.e. elemental metal). Iron(II) and iron(III) would be an example, or various species involving manganese.
Half reactions involving an element that is a liquid
At least at normal pressure and room temperature, only bromine and mercury are liquids. In these cases, you would dip the platinum electrode into the liquid while also making contact with the solution containing the soluble species of the half reaction.
Half reactions where one of the species is a gas
This would include the hydrogen electrode, but also other gaseous elements like oxygen or chlorine. The hydrogen fuel cell mentioned by MaxW in the comments is an example where both electrodes are inert.

When referring to a chemical reaction that generates electricity we talk in general about 2 coupled half-reactions : one in which something gets an electron and one in which something gives an electron. Now the electrode can act as part of the redox reaction but that is not necessarily , it can only act as a way of moving the electron generated by one half-reaction to the site where the other half-reaction takes place (takes an electron).This generates a movement of electrons from one site of reaction to the other , in other words generates electricity.

Technically you could, just place two graphite inert anode in a Zinc Sulfate electrolyte, charge the battery with about 2 volts and you will end up with a Carbon/Zinc rechargeable battery


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Mechanism of oxidation of phenol to quinone by chromium(VI)
In the supplementary information for Org. Lett. 2005, 7 (19), 4297–4300, the following experimental method for the synthesis of 2-bromo-6-methylcyclohexa-2,5-diene-1,4-dione is given:

To a suspension containing 20.0 g (75.8 mmol) of 2,4-dibromo-6-methylphenol in 50 mL of 80% acetic acid in water and 25 mL of acetonitrile, was added 8.34 g of $\ce{CrO3}$ in 25 mL of water. The reaction mixture was heated to 60℃ for 1.5 h, cooled to room temperature, diluted with 400 mL of water and extracted with three 200 mL portions of chloroform.

I've summarised the reaction here:

How does this work, mechanistically? My first guess would be a nucleophilic substitution of $\ce{Br}$, followed by an oxidation by $\ce{CrO3}$; but I can't imagine there are many free hydroxide ions floating around in a solution of acetic acid.

A possibility: the phenol OH attacks CrO3, which creates a Cr leaving group on the phenol oxygen. Then some O-based nucleophile attacks the para position. You can push the double bonds around the ring onto the phenolic O and expel the Cr leaving group. Now the para position has a Br and a OR group, which can collapse to a carbonyl. Not sure about the finer details, but some phenol -> quinone oxidations (e.g. with hypervalent iodine) work like this mechanistically.


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Do all sigma bonds are stronger than pi bonds?
Many textbooks refer that sigma bonds are stronger than pi bonds . But each individual bond ( between different atoms) have different bond energy. So how we know for sure that every possible sigma bond is stronger than every possible pi bond ?

That's because, typically, these textbooks are comparing X–Y sigma bonds with X–Y pi bonds. Otherwise, as you say, it doesn't make much sense to compare them.

Every possible sigma bond is not stronger than every possible pi bond.
From this table at Wikipedia, we can see that the bond dissociation energy for the F-F bond in $\ce{F2}$ is 157 kj/mol. Meanwhile, the bond dissociation energy for the C-C bond in ethane $\ce{H3C-CH3}$ is on the order of 360 kJ/mol. The bond dissociation energy of the C=C bond in ethene $\ce{H2C=CH2}$ is approximately 710 kJ/mol. The note for this compounds suggests that the pi bond is around 270 kJ/mol (65 kcal/mol), leaving the sigma bond at 440 kJ/mol. Thus, the carbon-carbon pi bond is stronger than the fluorine-fluorine sigma bond.
A general statement of "sigma bonds are stronger than pi bonds" must thus be interpreted in the "all else equal" mode. The relative strength of the carbon-carbon pi bond is less than the relative strength of the carbon-carbon sigma bond. However, just as the strength of the carbon-carbon sigma bond has no bearing on the strength of the fluorine-fluorine sigma bond, the comparison of pi bonds to sigma bonds involving different elements is not helpful. The relative strength of the bonds depends as much or more on the atoms in the bond as it does the type of orbital overlap used to form the bond.


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Synthesize the trisubstituted benzene
I am trying to synthesize 1-(2-amino-4-chlorophenyl)ethan-1-one from benzene. The answer I came up with is that the benzene must undergo nitration first, then halogenation, then hydration for $\ce{NO2}$ to be $\ce{NH2}$, then acylation. But from the readings I've made acylation cannot happen with $\ce{NH2}$. Is there another way to synthesize this to benzene?

@Mithoron, https://www.sigmaaldrich.com/catalog/product/aldrich/a37804?lang=en&region=GB

Well, I overstated it, but even if you get aminoketone, it may react with itself, here two or more molecules could condense especially if catalyst is present.

You can synthesize your desired compound from Benzene in the following way.
First, you can synthesize m-dinitrobenzene, and perform selective reduction to have 3-nitroaniline. Now, you can acylate the ring by first protecting the amine group. After acylation you can convert the amide to amine by acidic hydrolysis, diazotize the amine group and perform Sandmeyer's reaction. Then just reduce the remaining nitro group to amine to have your product.
The synthetic scheme is given below,

is the acylation of the benzene ring possible with the presence of NO2?

You are correct that acylation will be trouble with the amine there as similar reagents acylate benzene and amines.

Could this not be used in your favor? Once converted to an amide, the $\ce{NH2}$ is no longer trouble, and it is still an ortho-para director. In this regard the amide is acting as something of a protecting group: it prevents the amine from misbehaving. Hydrolysis to the amine requires strong acid or base are higher temperatures and concentrations.

There are better amides out there than the acetyl amide for this purpose.  The tert-butyloxycarbonyl (BOC) group can be removed with dilute acid.

However, if this is more than an academic exercise, please note Mithoron's comment. Amines and ketones react to form imines under the conditions of some of these reactions, so oligomerization of your target compound is likely.

I would like to clarify based on what I have understood in your reply. So, benzene must go through nitration first, then halogenation, then hydration to form NH2. Then NH2 must be converted to amide (by what type of reaction?). Then it will be able to undergo acylation then lastly hydrolysis (?) to convert amide back to NH2. Am I correct? Thank you.

Note - your BOC group is removed by Lewis acids such as AlCl3

@Ben Norris, When you are doing the acylation in the ring, isn't it preferable for the $\ce{-CO-CH3}$ group to go to the para of $\ce{NHCOCH3}$ moiety (ortho to $\ce{Cl}$) as there is too much steric repulsion. Won't the para product be majorly formed ?

@SoumikDas - You are right! Good catch. I see you have submitted an answer that addresses this issue.