What single, pure compound would best replace petrol? Petrol for use as automotive fuel is produced by blending of different product streams of oil refining. It consists of hundreds of different compounds, and the exact composition can vary widely. Certainly, not all components are absolutely necessary. Is there a single pure compound that could be used instead and that is similar or even superior to typical petrol?

In order to limit the scope of this question, I only consider pure hydrocarbons ($\ce{C_xH_y}$), i.e. no alcohols, ethers, amines, nitro compounds, or other compounds containing hetero atoms. I am only asking about fuel for spark ignition (Otto) engines, i.e. not about diesel fuel.

The relevant criteria for the comparison shall be only the knock resistance and the energy content of the compounds. Other properties may be ignored.

For similar compounds, there may be a tendency for fuel with a higher knock resistance to have a lower energy content. The challenge is to find a single pure compound with a suitable balance of knock resistance and energy content, with better values than found for typical petrol.

The knock resistance shall be expressed as research octane number (RON). Currently, the most important petrol blends in most countries have an RON of 95. However, blends with an RON of up to 102 are widely available. Therefore, the candidate compound should have at least an RON of 102.

The energy content shall be expressed as specific enthalpy of combustion $\Delta_\mathrm ch^\circ$ at a temperature of $T=25\ \mathrm{^\circ C}$. If necessary, this value can be directly calculated from the molar enthalpy of formation $\Delta_\mathrm fH_\mathrm m^\circ$ or the molar enthalpy of combustion $\Delta_\mathrm cH_\mathrm m^\circ$. For this purpose, the generated water shall be assumed to be in liquid form (although this assumption is not representative for vehicle engines and may favour small saturated compounds). The energy content of petrol can vary. For example, typical average values for $\Delta_\mathrm ch$ given by Aral in Germany are

Regular (RON = 91): $-44.2\ \mathrm{MJ/kg}$
Super (RON = 95): $-43.5\ \mathrm{MJ/kg}$
Super Plus (RON = 98): $-42.7\ \mathrm{MJ/kg}$

Therefore, the candidate compound should have at least an energy content corresponding to about $\Delta_\mathrm ch^\circ=-44\ \mathrm{MJ/kg}$.

In order to limit the choice, the candidate compound shall be a liquid at room temperature. The boiling point should be below $210\ \mathrm{^\circ C}$, preferably between $70\ \mathrm{^\circ C}$ and $150\ \mathrm{^\circ C}$. The melting point should be well below $0\ \mathrm{^\circ C}$.

By way of comparison, the following table shows the relevant parameter values for the reference compounds that are used for the octane number (heptane and 2,2,4-trimethylpentane) as well as for one candidate compound (2,2,3-trimethylbutane) that I have already found.

$$
\small\begin{array}{llllllll}
\hline
\text{Compound}&\text{Formula}&\Delta_\mathrm fH_\mathrm m^\circ&\Delta_\mathrm cH_\mathrm m^\circ&\Delta_\mathrm ch^\circ&\mathrm{RON}&T_\mathrm{mp}&T_\mathrm{bp} \\
&&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{kJ/mol}&\text{in}\ \mathrm{MJ/kg}&&\text{in}\ \mathrm{^\circ C}&\text{in}\ \mathrm{^\circ C} \\
\hline
\text{Heptane}&\ce{C7H16}&-224.2&-4817&-48.1&\hphantom{00}0&\hphantom{1}{-90.55}&98.4
 \\
\text{2,2,4-Trimethylpentane}&\ce{C8H18}&-259.2&-5461&-47.8&100&-107.3&99.22\\
\text{2,2,3-Trimethylbutane}&\ce{C7H16}&-236.5&-4805&-48.0&112&\hphantom{1}{-24.6}&80.86 \\
\hline
\end{array}$$

I would be interested in answers that search through literature data and find better candidate compounds as well as in answers that use a theoretical approach to estimate where to find the best compounds.

Higher $\mathrm{RON}$ seems possible. Yet the boiling point rises accordingly while heat of combustion remains roughly the same. Here are two compounds that fit all posed criteria.

\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{Molecule} & \mathrm{mp\ \mathrm{(^\circ C)}}  & \mathrm{bp\ \mathrm{(^\circ C)}} & \Delta_\mathrm cH_\mathrm m^\circ\ (\mathrm{kJ/mol}) \ \text{@}25\ ^\circ \mathrm{C} & \mathrm{RON} & \mathrm{MON}\\
\hline
\ce{1,3,5-TMB^{[a]}} & -44.7^{[1]} & 164.65(15)^{[1]} & -5193.1(1.3)^{[2]} & 137^{[3]} & 124^{[3]} \\
\ce{2,2,3,3-TMH^{[b]}} & -54^{[4]} & 160^{[5]} & -5405^{[6]} & 112.8^{[3]} & 92.4^{[3]} \\
\hline
\end{array}

[a] 1,3,5-trimethylbenzene, [b] 2,2,3,3-tetramethylhexane

Theoretical predictions of $\mathrm{RON}$

Tareq A. Albahri$^{[7]}$ summarises in his paper the usage of statistics $\mathrm{RON}$ and $\mathrm{MON}$.

  The research octane number $\mathrm{RON}$, which is representative of the fuel performance during low-speed city driving, is more often reported in the literature than the motor octane number $\mathrm{MON}$, which is representative of the fuel performance during high-speed highway driving.$^{[7]}$

Theoretical intercalculation between the two characteristics is possible.$^{[7]}$ This work was done by Jenkins$^{[8]}$. First predictions of $\mathrm{RON}$ itself were and are still largely empirical$^{[7]}$. These models use regression techniques to assign effective octane numbers to various organic groups. Albahri examplifies the procedure via Anderson, Sharkey, and Walsh$^{[7]\ [9]}$.

  Anderson et al.,$^{[9]}$ for example, developed an empirical model for calculating the RON based on chromatographic analysis of gasoline.$^{[7]}$

  In their model, the gasoline is divided into 31 hydrocarbon groups or pseudocomponents, all of which are assigned an “effective” octane number that is estimated by regression of experimental data. The octane number of gasoline is calculated by adding the contribution of octane number[s] from each group.$^{[7]}$

Many similar methods include those proposed by Van Leeuwen et al.$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$, Ramadhan and Al-Hyali$^{[13]\ [14]}$, Nelson$^{[15]}$, Baird$^{[16]}$,  Twu and Coon$^{[17]\ [18]}$, Rusin et al.$^{[19]}$, Habib$^{[20]}$, Cotterman and Plunkee$^{[21]}.$ Some of these fail to differentiate between isomers, or are valid in specific mixtures.$^{[7]}$ Additionally,

  Although  these  techniques [by Van Leeuwen et al$^{[10]}$, Sasano$^{[11]}$, and  Lugo  et  al$^{[12]}$]  give reasonably accurate results, they are usually too time consuming for planning studies and often the compositional data are not available.$^{[7]}$

Tareq A. Albahri's method of SGC (structural group contribution)

After considerable testing, Albahri found that the best equation to take into account contributions of different groups is$^{[7]}$

\begin{split}
  \mathrm{ON} &=
  f\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^{-1}+ a + b\left(\sum_{i}\mathrm{\left(ON\right)_i}\right) \\
  &\quad + c\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^2 + d\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^3 + e\left(\sum_{i}\mathrm{\left(ON\right)_i}\right)^4
\end{split}

where $\mathrm{ON}$ is either $\mathrm{RON}$ or $\mathrm{MON}$, and $\left(ON\right)_i$ is the contribution from the $i$th group. Note how the powers increase from left to right.

The coefficients still have experimental basis.$^{[7]}$ This equation is a modified version of what Albahri put forward a year earlier.$^{[22]}$ For many compounds the difference between the estimate and experimatal data is just $5$.$^{[7]}$ If the $\left(ON\right)_i$ are known, one only needs the knowledge of the structure of the molecule to apply Albahri's method.

I suggest reading both papers by Albahri since they include tables of data and an example calculation.

Here is a selection of $\left(ON\right)_i$ values.$^{[22]}$

QSPR Model for octane number prediction

In a research article by Al-Fahemi, Albis, Gad, a quantitative structure-property relationship (QSPR) is performed. This includes establishing a correlation relation between $\mathrm{ON}$ and various physical parameters.$^{[23]}$ Again the model is based upon regression rather than derivation from first principles.

\begin{align}
\mathrm{ON} =
    & - (193.53 \pm 319.19) + (1.47 \pm 1.01)M\\
    & - (53.06 \pm 31.47)E_H - (8.67 \pm 2.73)B_P\\
    & - (24.94 \pm 19.44)M_R - (50.52 \pm 26.92)\log P\\
    & + (4.33 \pm 3.09)C_P + (3.72 \pm 2.04)C_V + (5.17 \pm 2.08)C_T\\
\end{align}

where

$M$ is molecular mass,
$E_H$ is hydration energy,
$B_P$ is boing point,
$\log P$ $-$ octanol/water distribution coefficient,
$M_R$ is molar refractivity,
$C_P$, $C_V$, $C_T$ is the critical point (pressure, volume and temperature, respectively).$^{[23]}$

TL; DR

There is no model to date that estimates or derives a result for $\mathrm{ON}$ from first principles. Different models depend more or less on multiparameter regression analysis and experimental data. Nevertheless, the methods of Albahri and others have predictive capability. Often the difference beteen estimated and experimental octane numbers is less than $5$.

Or as Albahri mentions,

  Octane number is one of the most difficult properties to estimate or correlate because of its complex dependency on the molecular structure of the compound.$^{[7]}$

  An estimation technique of the octane rating of pure hydrocarbons, though  essential, is nonexistent.$^{[7]}$

So for your second question, I suggest applying Albahri's method because

widest applicability among hydrocarbons,
relatively easy to use,
few $\mathrm{RON}$s are known from experiments,
Al-Fahemi et al.'s procedure requires many parameters which will be a pain to track down (if known at all).

Qualitative considerations$^{[3]\ [7]}$

shorter alkane chain $\ce{->}$ higher $\mathrm{ON}$ (but higher volatility)
branching of alkane $\ce{->}$ higher $\mathrm{ON}$
aromaticity in hydrocarbon $\ce{->}$ higher $\mathrm{ON}$ (but tends to burn sooty with carcinogenic byproducts)
effects on $\mathrm{RON}$ and $\mathrm{MON}$ can be very different

Arithmetic average of $\mathrm{RON}$ and $\mathrm{MON}$ $-$ anti-knock index or $\mathrm{AKI}$ $-$ is also used.

$$\mathrm{AKI} = \frac{\mathrm{RON} + \mathrm{MON}}{2}$$

$[1]$
T.F. Ardyukova, I.K. Korobeinicheva, A.I. Rezvukhin. Atlas of
Spectra of Aromatic and Heterocyclic Substances. $(1973)$, 7.

$[2]$
W. H. Johnson, E.J. Prosen, F.D. Rossini. 'Heats of combustion and
isomerization of the eight $\ce{C9H12}$ alkylbenzenes', Journal of
Research of NIST, $(1945)$, 35, pp 141$-$146.
$[3]$ A.
Demirbas, M. A. Balubaid, A. M. Basahel, W. Ahmad, M. H. Sheikh.
'Octane Rating of Gasoline and Octane Booster Additives'. Petroleum
Science and Technology, $(2015)$, 33(11), pp 1190$-$1197. DOI:
10.1080/10916466.2015.1050506.

$[4]$
Jean-Claude Bradley Open Melting Point Dataset. (January 29, 2017)

$[5]$
R. Stenutz. 2,2,3,3-tetramethylhexane. A collection of tables for
chemistry. (January 29, 2017)

$[6]$
Pure Components Data. (Excel file.) Petroleum Engineering, Texas A&M
University. (January 29, 2017)
$[7]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbon Liquids'. Industrial & Engineering Chemistry Research $2003$, 43(24). DOI: 10.1021/ie020306+.
 $[8]$ G. I. Jenkins. 'Calculation of the motor octane number
from the research octane number'. Journal of the Institute of
Petroleum, $1968$, 54, 14.

$[9]$
Anderson, P. C.; Sharkey, J. M.; Walsh, R. P. 'Calculation of  the
Research  Octane  Number  of  Motor  Gasoline  from  Gas
Chromatographic Data and a New Approach to Motor Gasoline Control'.
Journal of the Institute of Petroleum, $1972$, 58, 83.

$[10]$
J. A. Van Leeuwen, R. J. Jonker, R. Gill. 'Octane Number Prediction
Based on Gas Chromatographic Analysis and Nonlinear Regression
Techniques'. Chemometrics and Intelligent Laboratory Systems,
$1994$, 25(2), pp 325$-$340. DOI: 10.1016/0169-7439(94)85051-8.

 $[11]$  Y. Sasano. 'Measuring Research Octane Number of
Gasoline by Gas Chromatograph'. JP Patent 09138613, $1997$.
 $[12]$ H. J. Lugo, G. Ragone, J. Zambrano. 'Correlations
between Octane Numbers and Catalytic Cracking Naphtha Composition'.
Industrial & Engineering Chemistry Research, $1999$, 38, 2171.
 $[13]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'New Experimental and
Theoretical Relation to Determine the Research Octane Number (RON) of
Authentic Aromatic Hydrocarbons that Could be Present in the Gasoline
Fraction'. Petroleum Science and Technology, $1999$, 17, 623.

 $[14]$ O. M. Ramadhan, E. A. S. Al-Hyali. 'Aromatic
Hydrocarbons in Some Iraqi Gasoline and Their Influence in the Value
of the Research Octane Number'. Petroleum Science and Technology,
$1999$, 17, 607.
 $[15]$ W. L. Nelson. 'Octane Numbers of Naphthas'. Oil and Gas
Journal, $1969$, 67, 122.
 $[16]$ C. T. Ch Baird. Address: De  la  Haute-Belotte  6,  1222
Vezenaz, Geneva, Switzerland. Unpublished results.
 $[17]$ C. H. Twu, J. E. Coon. 'Predict Octane Numbers Using a
Generalized Interaction Method'. Hydrocarbon Processing, $1996$, 71,

$[18]$
C. H. Twu, J. E. Coon. 'Estimate Octane Numbers Using an Enhanced
Method'. Hydrocarbon Processing, $1997$, 76, 65.
 $[19]$ M. H. Rusin, H. S. Chung, J. F. Marshall. 'A
Transformation  Method  for  Calculating  the  Research  and  Motor
Octane Numbers of Gasoline Blends'. Industrial and Engineering
Chemistry Research. $1981$, 20, 195.
 $[20]$ E. T. Habib. 'Effect of Catalyst, Feedstock, and
Operating Conditions  on  the  Composition  and  Octane  Number  of
FCC Gasoline'.  ACS  Symposium  Division  of  Petroleum  Chemistry,
Miami, FL, Sept $1989$.
 $[21]$ R. L. Cotterman, K. W. Plunkee. 'Effects  of  Gasoline
Composition  on  Octane  Number'.  ACS  Symposium  Division  of
Petroleum Chemistry, Miami, FL, Sept $1989$.
$[22]$ Tareq A. Albahri. 'Structural Group Contribution Method for Predicting the Octane Number of Pure Hydrocarbons and Their Mixtures'. Fuel Chemistry Division Preprints, $2002$, 47(2), 532
$[23]$ Jabir H.
Al-Fahemi, Nahla A. Albis, Elshafie A. M. Gad. 'QSPR Models for
Octane Number Prediction'. Journal of Theoretical Chemistry,
$2014$, vol 2014, article ID: 520652, 6 pages. DOI: 10.1155/2014/520652.


Supercooling, freezing point depression, and finding the right freezing point Pure liquid can be supercooled below its freezing point. Conversely, impurities in the liquid will cause a freezing point depression. How then is the freezing point accurate?

Melting (and boiling) points are only defined for pure substances. Rather than defining them as ‘the temperature when pure substance $\ce{X}$ melts/boils’ they are better defined as:

The melting point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and solid phases of a pure substance are at equilibrium.
The boiling point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and gaseous phases of a pure substance are at equilibrium.

(I didn’t copy that from anywhere, I made it up. Apologies to everyone who I was unintentionally quoting without attribuition.)
Thus, you can supercool pure liquids but if you allow the supercooled liquid to reach equilibrium at its supercool temperature (and standard pressure), it will solidify. Similarly, you can overheat solids but if you let the overheated solid reach equilibrium at the overheated temperature, it will liquidify. Thus, neither the supercool liquid’s nor the overheated solid’s temperature can be considered the melting point, because neither phase is stable at equilibrium.


Is there a difference between a chelate ligand and a polydentate ligand? I'm confused because my lecture notes state that chelate ligands are a subset of polydentate ligands.

A chelating ligand should bind more than once to the same metal atom. Therefore, it must be polydentate. However, a regular polydentate ligand does not have to ligate to the same metal, and therefore, it is not necessarily chelating.


Detecting the concentration of chlorpromazine Is there a way to detect the concentration of chlorpromazine in a small sample ; ex: human saliva, using electro-chemical analysis? Does the sensor necessarily have to be a bulky piece of lab equipment or are there smaller sensors available for that kind of test?
Thanks in advance!

A cursory online search furnished

Graphene paste electrode for detection of chlorpromazine by Mohammad Hadi Parvin in Electrochemistry Communications, 2011, 13, 366-369 (DOI).
A multipurpose Voltammetric Sensor for the Determination of
Chlorpromazine in Presence of Acetaminophen, Uric Acid,
Dopamine and Ascorbic Acid by Binesh Unnikrishnan, Pei-Chi Hsu, and Shen-Ming Chen in Int. J. Electrochem. Sci., 2012, 7, 11414-11425 (PDF).


How can 2-indanone be prepared? Would adding phosgene to ortho-xylene lead to the formation of 2-indanone?

If you need to start with ortho-xylene, then you should use fazekazs's approach. I searched through SciFinder and found that the most commonly reported synthesis of 2-indanone involves oxidation of indene using a variety of conditions.

The following Organic Synthesis prep highlights this approach, although it certainly uses some outdated methodology (distillation by aspirator vs. rotary evaporation, for example). Newer reports use a variety of oxidations, including oxone or the the Wacker process.

In this example, indene 1 is added to a mixture of formic acid and hydrogen peroxide to afford the monoformate ester of indene 1,2-diol 2. The monoformate ester was hydrolyze with sulfuric acid to give 2-indanone 3.

However, when I first saw your question, I wanted to propose a synthesis based on a Dieckmann condensation on diethyl phenylene-1,2-diacetate 4 followed by decarboxylation of the $\beta$-ketoester 5. However, I did not find this approach in the literature, and both the diacetate ester and its parent diacid are of comparaple price to 2-indanone.

Another interesting idea that I did not find in the literature is a pericyclic approach based on photolysis or thermolysis of beznocyclobutene 5 in the presence of carbon monoxide. Following electrocyclic ring opening, intermediate 6 can react with CO in 4+1 cycloaddition. An isocyanide could be used instead of CO, and hydrolysis of the imine would give 2-indanone. Alternatively, there may be a transition metal species that could mediate this transformation.


What solvents can get rid of oil stains? What kind of solvent would you use to remove an oil based stain?

I need help for a chemistry lab report and I tried to look it up but nothing seems to be helping me

Depending on where you live different dry-cleaning solvents are indeed the best choice. Perchloroethylene $\ce{Cl2C=CCl2}$ is probably still the easiest one to get.


Photoconductivity of Selenium Selenium is a good conductor of electricity in presence of light. This phenomenon is called photoconductivity. How does this phenomenon occur? Are there any other elements or compounds in which this phenomenon is noted?

A very handy diagram and explanation of photoconductivity is shown below, from the American Physical Society page Viewpoint: Holey Intrinsic Photoconductivity:

The caption explains:

  Excitation process leading to photoconductivity in a condensed matter system (upper panel) and its cold-atom counterpart (lower panel). In the condensed matter system, incoming laser light ($hv$) excites an electron to the valence band, leaving a positively charged hole behind. In the cold-atom analogy, the optical lattice is tuned to allow an atom to move into the excited band, also leaving a hole behind.

From the Wikipedia page Photoconductivity, states that some 'classic' examples of photoconductive materials (asides from selenium) include:

The polymer polyvinylcarbazole
Lead sulfide

Some more examples exist in the paper Organic photoconductive materials: recent trends and developments include the chemical classes:

Phthalocyanines
Squaraines
Azo pigments
Perylene pigments


Calculation of molarity using dilution equation Using the formula $C_1V_1 = C_2V_2$ I am supposed to calculate the molarity of dilutions.

The original solution is $\pu{0.100 M}$ at $\pu{10 mL}$. I diluted it by mixing a $\pu{7.50 mL}$ stock solution with $\pu{2.50 mL}$ water, and made another one by mixing a $\pu{5 mL}$ stock solution with $\pu{5 mL}$ water.

Intuitively, I should get a higher molarity for my first dilution since it is higher in concentration, but when I calculate using
$$0.1(10)=C_2(7.5)$$
for the first dilution and
$$0.1(10)=C_2(5)$$
for the second, the second dilution's molarity is higher than the first. I know I am doing something wrong but I just can't figure out what.

You're mixing up the terms and formulae. Instead of rote learning the formulae, remember that the moles of the solute will remain the same in a solution - before and after dilution.

With that in mind, let's inspect your $\pu{7.5ml}$ stock solution. Before dilution, it had $\pu{7.5ml}\times\pu{0.1M}$ moles of the solute. After dilution, it has an unknown molarity. Let that molarity be $x ~\pu{M}$. Then, the new moles are $\pu{10ml}\times x~\pu{M}$. Equating and solving, you get $x=0.075$.

If you do that again for the second solution (I give it you as a homework), you'll get $x=0.05$. This easily verifies your own intuition that you should get "a higher molarity for the first dilution since it is higher in concentration".


Why is the temperature graph sigmoidal in simple distillation of components with very different bps? I'm writing an organic chemistry lab report on distillation and I thought I understood this, but the more I look at my lab textbook, I realize I don't.  My textbook 1 explains that the temperature of a mixture will rise continuously while it boils and distills if it is composed of two components with similar boiling points, but take on a sigmoidal shape when they are

When a liquid mixture is distilled, often the temperature does not
remain constant but increases throughout the distillation. The reason
for this is that the composition of the vapor that is distilling
varies continuously during the distillation (see Figure 14.3B).

I don't understand this sentence: "The reason for this is that the composition of the vapor that is distilling varies continuously during the distillation." I understand why it is true, but not why it is an explanation for the shape of these graphs. Let's say I'm distilling volatile component B from non-volatile component A. If the boiling point of the mixture is the point at which the vapor pressure of the mixture equals the atmospheric pressure, what does the composition of the vapor have to do with anything?  The relevant pressure is the atmospheric pressure, right?  I would think that as component B is boiled off, the vapor pressure of the mixture would decrease, because the mixture has less B, by Raoult's law, Ptotal = PA + PB = PA°NA + PB°NB, and as its vapor pressure goes down, its boiling point goes up.  I don't see why this isn't happening continuously.
Is the relevant pressure not the atmospheric pressure, but some additional pressure exerted by the vapor in the flask?  (It's been a long time since I took general chemistry.)

Pavia, D. L., Lampman, G. M., Kriz, G. S., & Engell, R. G. (2011). A small-scale approach to: Organic laboratory techniques. Belmont, CA: Thomson Brooks/Cole

You must understand... in an experiment, the sigmoidal shape will never have infinite slope... This would be a discontinuity... the sigmoidal shape (rightmost) and the "curve" (middle image) are essentially representing identical phenomena... there is a transition from one pure component to another. The only difference is how abrupt the change is... and this is determined by the endpoints/boiling points for each component. The more abrupt the change from one to another pure component, the steeper the slope of the change. I hope this helps. I can speak to this further, if necessary.


Bromination Pathways with alkane, alkene, and alkyne substrates I'm confused about halogenation pathways. From what I understand, there are multiple halogenation pathways. I'll try summarizing them below; let me know if I do anything incorrectly.

Halogenation with the diatomic halogen - i.e. $\ce{Br2}$. This reaction proceeds through a halonium intermediate.
Halogenation with the diatomic halogen in the presence of light or heat. This reaction proceeds via radical intermediates. From what I understand, the light or heat cleaves the weak halogen-halogen bond and we get two radicals, which can initiate the radical reactions. Also the cool thing about this reaction is that it doesn't require an nucleophilic pi bond, so if we were starting from an alkane, we don't even need to first form an alkene or alkyne; we can free-radical halogenate that alkane and then eliminate a beta proton as to make the alkene. All the other reactions mentioned here require some sort of nucleophile to start with.
Halogenation in the presence of water - i.e. $\ce{Br2}$ with $\ce{H2O}$. Again, we go through a bromonium intermediate. The $\ce{Br-Br}$ bond is instantaneously polarized and at that point we can form the bridged halonium ion and that gives water an entrance - the ring is highly strained and has partial charges all over the place - perfect for water to do something.
Are these all the halogenation pathways I need to know for an elementary understanding of organic chemistry? Did I miss any? I think I finally clarified all of them to myself through typing up this post ...
Oops yes I missed a few. NBS for example is used for allylic (or benzylic) bromination, and this directly implies the need for a starting alkene ... so this isn't as synthetically useful as free-radical bromination in the presence of light.
Addition of $\ce{HX}$ - the pi bond acts as a simple nucleophile and attacks the proton. Then the anionic halogen adds to the resultant carbocation.
Addition of $\ce{HX}$ in the presence of peroxides - the pi bond again acts as a simple nucleophile but this reaction proceeds via a radical intermediate and as a result the reaction is overall anti-Markovnikov.

I wanted to offer some commentary on the points you've enumerated:

You mentioned halogenation of alkenes in the presence of $\ce{H2O}$. This reaction is known as halohydrin formation. It's worth mentioning that this reaction can also be performed in the presence of other nucleophiles, perhaps most commonly alcohols and salts of other halogens to give mixed dihalogenated products. (Other types of nucleophiles may also successfully react, but I can't recall seeing any specific examples.) The reaction isn't always efficacious. Frequently, addition of both halogen atoms (i.e., ordinary halogen addition) will occur; the proximity of the halide leaving group to the halonium bridge will often result in a rapid reaction before any other nucleophile can intercept.
Radical halogenation can be very difficult to control, even when dealing with substrates that should exhibit high regioselectivity. NBS (and its chlorinated counterpart, NCS) is actually an extremely useful and versatile reagent which mitigates some of the practical pitfalls of radical halogenation reactions. Under photocatalytic conditions, or with sufficient heating and a radical initiator, it can react by free radical pathways. It's often preferable to using the diatomic halogen directly (unless exhaustive halogenation is really the goal) because it generates lower concentrations of $\ce{Br2}$ in situ, so the reactions are somewhat more controllable and side-products from direct addition to $\pi$-bonds are minimized. It can also react analogously to $\ce{Br2}$ with alkenes to give a halonium ion, which makes it preferable for halohydrin formation (since there is no $\ce{Br-}$ leaving group present to immediately attack the bromonium bridge).

Overall, my opinion is that you have a good core knowledge base re: halogenation reactions within the confines of introductory organic chemistry. Of course, there are a multitude of other halogenation reactions operating on myriad functional groups, but I presume those aren't germane to this particular question.


What defines incomplete combustion? I know that complete combustion is a process where a fuel reacts with $\ce{O2}$ to become $\ce{CO2}$ and $\ce{H2O}$. But what is incomplete combustion?

According to the Internet, $\ce{CH4 + O2}$ and $\ce{C6H12O6 + O2}$ are examples of incomplete combustion. What makes them so?

Complete combustion is, as the name suggests, complete. That on its own doesn’t mean much, but I can tell you that the completeness implies all atoms that were put into combustion leave it fully oxidised (or fully reduced in the case of oxygen, fluorine, chlorine; or unchanged in the case of nitrogen).

An incomplete combustion is one where molecules remain that may still be oxidised, i.e. can react with further oxygen. Examples would be carbon monoxide or elemental carbon (colloquially ‘soot’). Incomplete combustion often results in a yellow flame stemming from carbon particles, while complete combustion typically yields hotter, colourless flames.

In your average Bunsen burner, you can choose between complete and incomplete combustion by regulating the amount of air (meaning: oxygen) that reaches the flame. Complete combustion would follow the equation $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$. There is no unique equation for incomplete combustion, but one possibility would be $\ce{CH4 + O2 -> C + 2 H2O}$.

By the way, the way living beings burn carbohydrates in cellular respiration is, by definition, a complete combustion since every carbon atom ends up in $\ce{CO2}$ and a set of two hydrogen atoms always ends up as $\ce{H2O}$. The overall equation is $\ce{C6H12O2 + 6O2 -> 6CO2 + 6H2O}$. If glucose is combusted with less than 6 equivalents of $\ce{O2}$, then incomplete combustion results.


Auto-ionisation and Dilution When you dilute an acidic solution, pH increases towards 7 and pOH decreases. However, the decrease of pOH indicates that the concentration of hydroxide ions increases. This seems to contradict my understanding of how equilibrium systems respond to external disturbances, as it suggests the auto-ionisation equilibrium of water shifts to favour the ions (according to Le Chatelier's principle) to such an extent that the hydroxide ion concentration increases from prior to the dilution. However, I was under the impression that equilibrium systems can never fully undo the change due to the bi-directional nature of equilibria, meaning the reaction is using up ions as they are replaced. Please help to clear up my confusion! Thank you.

Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$
Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$.

Suppose a solution has $10^{-4}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-4}]$ which is equal to $4$.

So as you can see, lower the $\mathrm p \ce{OH}$, the higher the concentration of $\ce{OH-}$ ions. Le Chatelier's principle is not violated.
(How I like to remember this is - the strongest acids have the lowest $\mathrm pK_\mathrm a$. So that extends to $\mathrm p\ce{OH}$ too. This is all due to the fact that $\mathrm p\ce{OH}$ is the negative logarithm of the $\ce{[OH-]}$ concentration in solution.)


Do you break the metallic bond when you break a metal into two pieces? As we all know, metallic ions are surrounded by a sea of electrons. If we continually bend or stretch a metal, say, iron, it will break. Does that mean we break the metallic bond in subatomic level?

In addition to ManishEarth’s answer, I'll add that this is only strictly true if you take a pure single crystal of metal and break it. If you take a real-life piece of iron, it is not a pure crystal but contains a large number of crystalline defects. The most common (and easiest to think about) in a polycrystalline material is the presence of grain boundaries between the many crystallites that make up the metal particle:

  Micrograph of a polycrystalline metal; grain boundaries evidenced by acid etching (from Wikipedia)

So, your metal piece is not a single crystal, and crystalline defects are key in the understanding of mechanical response to applied stress, including the case of fracture in which you are interested. So, by breaking a piece of metal in two, you may not actually break so many metallic bonds :)


Does ethanol react with potassium to form potassium ethanolate and hydrogen gas? I'm finding information in Google with the keyword (C2H5OH + K) to find the reaction between $\ce{C2H5OH}$ and $\ce{K}$ and found very little information about it. Just 2-3 related results with contradictory pieces of information.

The first one gave

$$
\ce{K + C2H5OH -> KOH + C2H5},
$$

the second one gave

$$
\ce{2C2H5OH + 2K -> 2C2H5OK + H2},
$$

and the third one gave

$$
\ce{C2H5OH + KOH -> C2H5OK + H2O}
$$

There are no trusted public sources so I'm very baffled. I think that $\ce{C2H5OH}$ will react with $\ce{K}$ to form $\ce{C2H5OK}$ because it looks like the reaction with $\ce{Na}$. But I'm not sure and just want to find a trusted source. Can anyone confirm the true result for this reaction?

You can view water as the simplest alcohol, and indeed, water reacts with all of the Group 1 metals to form hydrogen gas and the corresponding metal hydroxide according to the following equation

$\ce{2M + 2HOH -> 2MOH + H2}$

where $\ce{M}$ is the metal.

The same reaction occurs with simple alcohols to produce the corresponding metal alkoxide and hydrogen gas

$\ce{2M + 2ROH -> 2MOR + H2}$

As you increase the hydrocarbon portion of the alcohol, or reduce accessibility to the $\ce{OH}$ portion of the alcohol (e.g. make the alcohol less like water), you slow down the rate of the reaction.

Here is a nice link that compares and discusses the relative reactivities of the Group 1 metals with water.


What is the product of a Dieckmann condensation of dimethyl adipate? My attempt: $\ce{MeO-}$ preferably gives SN2 or E2, being a strong base and good nucleophile. But which reaction proceeds in this case?

I've given a strong hint by changing the title of your question!

Methoxide as a base can abstract a proton from an acidic $\alpha$-position of an ester, that is from $\ce{-C\color\red{H}_2COOMe}$.
The resulting carbanion/enolate can add to the other ester group of your starting material.
The cyclic intermediate eventually loses methoxide to form ...

This Dieckmann condensation is the intramolecular variant of a Claisen condensation.

I suggest to do it on paper!


Is "propanol" a correct name in IUPAC nomenclature? My teacher said that propan-1-ol can also be named propanol in the latest IUPAC recommendations. I doubt this because Wikipedia only mentions propan-1-ol as the IUPAC name (propanol being relegated to the "other names" section).

Is "propanol" an acceptable name?

The last IUPAC nomenclature recommendations were published in 1993. The current draft documents (release date uncertain) are far more lenient in their acceptance of semi-systematic naming of molecules. Please make sure you understand the differences between the Preferred IUPAC Name (PIN), the retained name, and trivial name.

Propanol is not be the preferred IUPAC name, but significantly, neither is it listed in the tables of Trivial and semisystematic names retained for naming organic compounds. It is, however, a widely accepted name for this chemical, and you are far more likely to encounter propanol in industry than propan-1-ol. Nobody uses propan-1-ol in the real world, even if it is the preferred IUPAC systematic name.


Conversion of Hessian from internal coordinates to cartesian coordinates I have performed a CASPT2 frequency calculation using the MOLCAS8 program package. I need the full Cartesian Hessian matrix as input for a subsequent program package.

MOLCAS8 performs CASPT2 frequency calculations numerically. The program MCKINLEY is called for this purpose, however, it is now my opinion that all numerical second derivative calculations in MOLCAS are handled not by MCKINLEY but by the barely documented module supermac. Anyway, the Hessian is fully printed in internal coordinates once the calculation is finished. For a system of 10 atoms the internal Hessian is of size 24 x 24, which suggests non-redundant internal coordinates (MOLCAS calls them "primitive internal coordinates").

Does anyone know of a software capable of transforming a Hessian matrix from internal coordinates to Cartesian coordinates? I am aware that this will probably be complicated as internal coordinates are not as unambiguous and straight-forward as Cartesians. However, my initial search efforts did not yield a single software package capable of transforming a calculated Hessian in internal coordinates to Cartesians. Any help would be appreciated!

Alternatively, does someone know a quantum-chemical program package using which I might be able to obtain a full Cartesian hessian at CASPT2 level?

I don't know if this helps, but MOLCAS also seems to print out the force constants of each atom in Cartesian coordinates. Furthermore a .molden file is produced where the normal modes are included. The full Hessian does not seem to be written to this Molden file though.

BAGEL (http://nubakery.org/ and https://github.com/nubakery/bagel) has CASPT2 analytical gradients and can run frequency computations.

It's currently being prepared for 1.0 release, so it might still have bugs. On the other hand, it is freely available under the GPL.

As for converting the Hessian, it might be useful to look at the INTDER program, which is (among others?) part of the PSI3 suite of programs available here: https://sourceforge.net/projects/psicode/files/psi/3.4.0/psi-3.4.0.tar.gz/download - PSI3 is under the GPL as well.

The manual page for INTDER is here: https://manned.org/intder and it says:

"Force field transformations between Cartesian and general internal
coordinate spaces up to fourth order, including nonstationary
reference structures.  Both forward and reverse transformations
are possible; hence, by using intermediate Cartesian coordinates,
force fields can be transferred among different representations in
the internal space."

Whether or not it will be easy to coerce the MOLCAS Hessian into a format INTDER likes and then transform it as you want I cannot say as I don't have any direct experience with INTDER - it is typically run as a step in a chain of commands automatically generated by the PSI3 driver routines.


How can two electrons lie together in an orbital? Two electron of opposite spin can lie in a single orbital.. But what about the electron-electron repulsion.
Okay! I got that the nuclear charge rather the large Z-effective overcome this repulsion by pulling them together towards the nucleus.

One thing more turned out in mind that it may be the attraction of the two unlike pole of the magnet developed due to opposite spin of the electrons of an orbital, which may be reason for the two opposite spin electron lying in an orbital..

Am I right ?

Preludium

In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain discrete energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel.

Main answer

But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be

\begin{equation}
  \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .
\end{equation}

If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| > \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look here: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given here (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does "cost" energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.

Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases.


How does Gypsum Concrete work? When mixed with water how does gypsum act as  a binder in creating concrete?

The setting of cement involves the reaction of clinker minerals with water, to form a system of interlocking crystals, which lock the materials together.
The setting of cement is primarily due to the reaction of tricalcium aluminate(C3A), which is the most of the four clinker minerals present in concrete. The set regulating action of gypsum is due primarily to its reaction with C3A.
In the absence of gypsum, the reaction of C3A with water is very rapid. The C3A is converted to calcium aluminate hydrate ($\ce{4CaO.Al2O3.19H2O}$) which forms interlynking bridges of solid material between the cement particles. These interlinking bridges reduce the mobility of the cement particles and produce a firm structure, which results in setting of cement mixture. The rapid setting is called a flash set and its irreversible.
The presence of gypsum in cement slows the reaction of the C3A. When mixed with water gypsum dissolves and reacts with C3A to form ettringite ($\ce{3CaO.Al2O3.3CaSO4.32H2O}$). This ettringite is initially formed as very fine-grained crystals, which form a coating on the surface of the C3A particles.

These crytals are too small to bridge the gaps between the particles of cement. The cement mix therefore remains plastic.
After the initial hydration reactions of cement, a dormant period occurs during which the rate of hydration of cement is greatly reduced. During this dormant period, ettrinite crystals continues to grow and eventually are large enough to impede the mobility of cement particles. At this stage setting occurs.


What is the underlying chemistry behind the triboelectric effect? I remember briefly studying the triboelectric effect in Physics but as I now approach the domain of Chemistry, I'd like to know what processes are involved (at the molecular level) when the the triboelectric effect comes into play. What variables (if any) would have have to be considered when conducting an analysis of the triboelectric effect?

You would've probably heard of the commonly given example, rubbing a glass rod with silk. The glass rod becomes positively charged, and the silk becomes negatively charged.
But why? That's your question I believe, and for that we need to zoom in a little further, until we can see the molecular structure.
Glass, is a polymer mainly composed of silicates. It is a network polymer, with a bunch of $\ce{Si-O}$ groups, as evident in it's structure:

(source: mechanicalengineeringblog.com)
You can observe that some of the oxygens are negatively charged, and are countered by sodium ions.
Now, lets observe silk,
Silk, like most natural fibers, are composed of proteins. These proteins exist as $\alpha$-helices. Sericin is the outermost protein in the silk structure, so I'll show a bit of it's structure,

Now, let's get to the rubbing part!
What happens when you rub them?
You're agitating the two molecules and bringing them in contact. Loose bonds can be broken off. Consider the $\ce{N-H}$ bond in the Sericin structure. It can be easily removed as a $\ce{H+}$ ion, since the resulting negative charge can be shared with the nearby oxygen in the adjacent carbonyl group. This leaves silk with a net negative charge.
This $\ce{H+}$ can now bind to the negatively charged oxygen atoms on glass, giving the glass rod, a postive charge.

I shall end my answer with a note:
Whenever in doubt, look deeper into the structure.


How can I interpret the CH4+ structure? I am trying to teach myself about Electron Ionization, and am very confused about the formation of the CH4+ Ion.

Do the dotted lines between the C and the H, and between the two Hydrogens indicate a resonance structure of sort? I have taken organic chemistry, but cannot wrap my head around this.

The dotted lines in this structure could be thought of as partial bonds. In other words, none of those three bonds has a bond order of 1 with two electrons attributable to that bond.

Another way to look at this structure is that the dashed lines represent multicenter bonding. A common example of multicenter bonding is diborane $\ce{B2H6}$. The structure of diborane features 3-center-2-electron bonds (often abbreviated 3c2e bonds - in this notation, a typical single sigma bond is a 2c2e bond) with two hydrogen atoms shared by both boron atoms:

These bonds in $\ce{CH4+}$ could be considered a 3-centered-3-electron bond as originally each C-H bond would have had two electrons. The radical cation is one electron short, so there are only three electrons shared between those three atoms.


How to increase equilibrium of Cl₂ $\ce{Heat\ +\ PCl5(g)\ \rightleftharpoons\ PCl3(g)\ +\ Cl2(g)}$

According to the process above, which of the following can be done to increase the number of moles of $\ce{Cl2}$ in a sealed container at equilibrium?

Reducing temp
Increasing the pressure at a constant temp
Adding more moles of $\ce{PCl5}$

I'm pretty sure it's not 1. but I'm stuck between 2. and 3. Are they both true or only one of them?

Hint:

For 2. Try to apply Le Chatelier's principle. Increasing pressure will make the system try to decrease it and there is one side of the equilibrium with an unequal number of moles.


Does changing the concentration of reactants alter the equilibrium constant $$\ce{S(s) + O2(g) -> SO2(g)}$$
The reaction of sulfur with oxygen is written in equation form above. This equation can be interpreted in all of the following ways except
(A) either $\ce{S(s)}$ or $\ce{O2(g)}$ will be completely used up
(B) $Q$ must be close to $1.0$ since there is  one mole of gas on each side of the equation
(C) this reaction goes to completion
(D) adding $\ce{O2}$ will change the equilibrium constant

I do not know how to extrapolate $Q$ from just an equation with no other data. However, I also learned that the equilibrium constant is, well, constant, and adding O$_2$ should change the equilibrium concentrations, not the equilibrium constant.
The correct answer is (B). How would (D) be a valid interpretation?

D is an invalid statement because the ''equilibrium constant'' for a certain reaction is constant as long as the temperature of the system is constant.
When the concentration of a species Is changed according to Le Chatelier's principle the equilibrium system will change it's rates of the forward and backward reactions so as to balance the concentrations thus keeping the equillibrium constant the same.


Why should the product of the normality of the acid or base and its volume be constant? $$\ce{N1V1 = N2V2}$$
 where, N is the Normality of the Acid / Base &
 V is the volume in litres.
We ask you to give a scientific justification or derivation to the above equation
i.e. Why must the product of Normality & Volume be equal for Acid & Base ?

This is a question I found online and would greatly appreciate an answer.

$$\ce{N1V1 = N2V2}$$ The equation is used basically for dilution of solution and determining one of the parameters of acid or base.

In the case of dilution of solution, the equation holds true because the no of gm equivalents of the solute remains constant in the solution even the solution is diluted.

Initial no of gram equivalents of solute = Final no of gram equivalents of solute ....................eq (1)

And we know, $$\text{Normality} = \frac{\text{No of gram equivalents}}{\text{Volume in Litre}}$$
$$\text{No of gram equivalents}=\text{Normality}\times\text{Volume in Litre}$$
Now lets consider,

Initial Normality =$\ce{N1}$

Initial Volume=$\ce{V1}$

Final Normality=$\ce{N2}$

Final Volume = $\ce{V2}$

Then , eq (1) becomes

$$\text{Initial Normality}\times\text{Initial Volume}=\text{Final Normality}\times\text{Final Volume}$$
$$\ce{N1V1 = N2V2}$$

In case of acid and base reaction,
No of gram equivalents of acid will always react or consume equal no of gram equivalents of base and vice versa

No of gram equivalents of acid = No of gram equivalent of base .......eq (2)

Normality of acid = $\ce{N1}$

Volume of acid =$\ce{V1}$

Normality of base =$\ce{N2}$

Volume of base =$\ce{V2}$

Then eq (2) becomes,
$$\text{Normality of acid}\times\text{Volume of acid}=\text{Normality of base}\times\text{Volume of base}$$

$$\ce{N1V1 = N2V2}$$

The number of grams of a chemical element or compound equal to the weight of its chemical equivalent. The gram equivalent of a chemical element is numerically equal to the quotient resulting from dividing the element’s atomic weight by its valence. For example, the gram equivalent of iron (atomic weight 55.847) with a valence of 2 is 27.92 g and with a valence of 3, 18.62 g. The gram equivalent of acid is that amount of the acid in grams containing one gram equivalent of hydrogen capable of being displaced by metal with the formation of salt. The gram equivalent of a base is that amount of the base in grams that is necessary for its complete reaction with one gram equivalent of acid. The gram equivalent of salt is that amount of the salt in grams containing one gram equivalent of metal. The concept of gram equivalent is widely used in volumetric analysis.

http://encyclopedia2.thefreedictionary.com/Gram+Equivalent


Monosaccharides configurations (alpha beta, D L) identification I am really finding it difficult to identify the correct configuration(s) of a given monosaccharide, in fact my friends and I have been trying to solve this, much to our annoyance.

In one of our classes, our teacher said, in cyclic form:

  If $\ce{CH2OH}$ is below the ring its configuration is  L, and if its above the ring its D.
   If $\ce{C1}$'s $\ce{OH}$ is below the ring its alpha configuration and if its above its beta configuration .

My friends seem to agree with the above explanation, however my understanding is:

  If all groups of $\ce{OH}$ and $\ce{-CH2OH}$ is reversed from D form, its L.
  If $\ce{C1}$'s $\ce{OH}$ is on the same side of the ring as $\ce{-CH2OH}$ its beta else its
  alpha.

So I want to know if these assumptions are right:

D and L in linear form:

If the last chiral carbon's OH (one before the last carbon) is on the right its D and if its on the left its L. so if you want to make for example L glucose from D, you should reverse ALL groups.

D and L in cyclic form:

It's hard mainly but can be deduced by making linear form into cyclic and cyclic into linear and comparing them or reversing all carbon branches.

Alpha and Beta in cyclic form:

If the carboxyl carbon's $\ce{OH}$ is on the same side of last ring carbons branch ($\ce{CH2OH}$ in glucose, $\ce{COO-}$ in glucoronate) it's neta, else it's alpha. So in D form, if $\ce{OH}$ is above it's beta else it is alpha. In L form, if $\ce{OH}$ is above its alpha else its beta.

Also when i see 3D linear form of Glucose it doesn't match with 2D at all: 2D view of this site isn't like what I find in books I can't find the match between 2D and 3D view neither.

I'm still confused because of following examples:

1-Alpha-L-Glucose

2-Alpha-D-Glucose

3-Beta-D-Glucuronate

4-Beta-L-Iduronic Acid

these just don't match.

also Harper biochemistry image is this:

but lehningers is this:

It’s easy to understand your frustrations, however you need to understand the basic stereochemistry and conformations of sugars.

A monosaccharide is designated D if the hydroxyl group on the highest numbered asymmetric carbon is drawn to the right in a Fischer projection, as in D-glyceraldehyde:

Note that the designation D or L merely relates the configuration of a given molecule to that of glyceraldehyde and does not specify the sign of rotation of plane-polarized light.

As I covered this explanation already on one of my posts:Are glucose and galactose cis-trans isomers of each other?

  D sugars have the same absolute configuration at the asymmetric
  center farthest removed from their carbonyl group as does
  D-glyceraldehyde. The L sugars, in accordance with this convention,
  are mirror images of their D counterparts

Cyclic forms

The designation α indicates that the hydroxyl group at the anomeric center (hemiacetal or hemiketal carbon)  is, in a Fischer projection, on the same side as the hydroxyl attached at the farthest chiral center, whereas  β indicates that these hydroxyl groups are on opposite sides:

In the α anomer, the $\ce{OH}$ substituent to the anomeric carbon is on the opposite side of the sugar ring from the $\ce{CH2OH}$ group at the chiral center that designates the D or L configuration ($\ce{C5}$ in hexoses). The other anomer is known as the β form.

The Haworth projections represent the stereochemistry of sugars more realistically than do the Fischer projections, which is why they are preferred:

For a D sugar, any group that is written to the right of the carbon in a Fischer projection has a downward direction in a Haworth projection; any group that is written to the left in a Fischer projection has an upward direction in a Haworth projection.

The terminal $\ce{-CH2OH}$ group, which contains the carbon atom with the highest number in the numbering scheme, is shown in an upward direction.

  Similarly  in the  α-anomer, the hydroxyl on the anomeric carbon is on the opposite side of the ring from the terminal $\ce{-CH2OH}$ group (i.e., pointing down). In the β-anomer, it is on the same side of the ring (pointing up).

References

Lehninger Principles of Biochemistry
Voet and Voet biochemistry
Biochemistry Grisham
Biochemistry Campbell and Farrel


A Book on Preparation and Reactions of organic compounds I have to learn many methods of preparations of various hydrocarbons and their reactions. For this, I went through many articles and some books but didn't found the details of reactions anywhere (There are details for only some of these) For example, why do we chose Nickel as the catalyst other than anything else to prepare ethane by catalytic hydrogenation of ethene? And why this reaction has to take place at 300°C and not at a lower or greater temperature?

I have 'a lot of hydrocarbons and a lot of methods to prepare them and a lot of their reactions' to learn so I want a book which has been designed especially to cover this topic so it could provide the details I want. Basically my course/syllabus includes Methods of preparation and reactions of methane, ethane, ethene, ethyne, benzene + alkyl-halides, alcohols, phenols, ethers, Aldehydes, ketones, carboxylic acids and Esters.

For example, why do we chose Nickel as the catalyst other than anything else to prepare ethane by catalytic hydrogenation of ethene?

Palladium is expensive ($690~\$\mathrm{US}/\mathrm {oz.tr}; 1~\mathrm{oz.tr = 31.1 g}$).
Nickel is cheap ($11~\$\mathrm{US}/\mathrm {kg}$).

Catalytic hydrogenation of ethene is the model reaction performed to understand the mechanism of this type of processes, but I'd be astonished if it had any technical relevance.

There is enough ethane available from the processing of natural gas by distillation (demethanizer, deethanizer).
Ethene is the more interesting starting material for polymerization, oxidation, etc.

  And why this reaction has to take place at 300 °C and not at a lower or greater temperature?

Does it really? In earlier mechanistic studies, such as  Review of Physical Chemistry of Japan, 1937, 11, 153-165, or Discuss. Faraday Soc., 1950, 8, 152-159, the hydrogenation of ethene on a nickel catalyst was performed at much longer temperatures (-78 to 0 °C).

  preparation and reactions of alkyl halides, alcohols, phenols, ethers, aldehydes, ketones, carboxylic acids and esters

For lab-scale reactions, any decent introductory book on Organic Chemistry  will do.

If you're looking for books that focus on experimental procedures (cook books), there are two which I can recommend (Note that have only used the original German versions):

Organikum – Organic Chemistry Basic Practical Course
Reactions and Syntheses: In the Organic Chemistry Laboratory


Problems with thermodynamic derivatives How is the following true?

$$\left ( \frac{∂S}{∂P} \right )_{T}=\frac{-C_{P}}{T}\left ( \frac{∂T}{∂P} \right )_{S}$$

I know that

$$C_{P}=T\left ( \frac{∂S}{∂T} \right )_{P}$$

which means the right-hand side of the first equation becomes

$$-\left ( \frac{∂S}{∂T} \right )_{P}\left ( \frac{∂T}{∂P} \right )_{S}$$

but how is that equal to $\left ( \frac{∂S}{∂P} \right )_{T}$ ?

It's a consequence of the triple product rule for partial derivatives of thermodynamic state variables.

In summary, if you have three interdependent (not independent) variables, then this is true:

$$
\left ( \frac{\partial{X}}{\partial{Y}} \right )_Z \left ( \frac{\partial{Z}}{\partial{X}} \right )_Y = - \left ( \frac{\partial{Z}}{\partial{Y}} \right )_X
$$

Substituting in your variables, if we let $X = T$, $Y = P$, and $Z = S$, we get:

$$
- \left ( \frac{\partial{S}}{\partial{T}} \right )_P \left ( \frac{\partial{T}}{\partial{P}} \right )_S =  \left ( \frac{\partial{S}}{\partial{P}} \right )_T
$$

Note that the order of multiplication of the two partial derivatives doesn't matter.

Now, why this rule holds for thermodynamic variables is a little bit more complicated - but I think the wikipedia article that I linked to has a pretty good derivation if you are interested.


Electrodeposition of Ni How can one electrodeposit Nickel on a substrate using a liquid crystal template?

substrate is Indium Tin Oxide coated plate and the template is cetyl trimethylammonium bromide(CTAB)+WATER+ETHANOL

This seems to be a very specific case. I would say that you should start with a standard Ni electroplating solution and see the result. You could then adjust the parameters of your plating operation and the concentrations of the additives in the solution.

However, I have a bigger concern. How stable is your liquid crystal template under applied potential in an electrolyte? Before trying to deposit Ni, you should try to do simple electrochemistry find the potential window where your template stays intact.


Any way to see "cubic-active" vibrational modes in molecules? For molecular complexes with a moderate amount of symmetry there are usually vibrational modes that are neither infrared nor Raman active. Sometimes these modes transform as some cubic polynomial in x,y,z. For example the $B_{1u}$ irrep in the $D_{4h}$ group which transforms as $xyz$.

My question is: it possible to measure these kinds of vibrations using some sort of non-linear optics? For example, measuring the absorption of three photons instead of just one like in infrared spectroscopy/reflectivity. In the case of $B_{1u}$ above, this would translate to 3 photons polarized with x,y,z respectively and whose total wavenumber matches a $B_{1u}$ vibration. Or perhaps a more complicated multi-photon version of Raman spectroscopy could do this?

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon is used so as not the excite the sample. Also, as the transition is multiphoton the focus area is smaller than a 1 photon transition and this improves spatial resolution).

As the product $xyz$ can be thought of as $xy$ multiplied by $yz$ then this is like a double Raman transition. Raman involves a change in polarisability. Polarisability is proportional to a volume but the projection for the radiation is 2D hence $xy$ type products in the point group tables. A  Raman type transition may occur between 4 levels. Levels 1 and 4 are  'real' and 2 and 3 'virtual'. The steps are 1 up to 3 , 3 down to 2, and 2 up to 4. This has been demonstrated in doppler free geometry between Na $3^2S_{1/2}\to 3^2P_{1/2}$.


Colour change of fluorescein in acidic and basic medium I prepared fluorescein for my chemistry project in school and I realised that this molecule and its preparation steps are very closely related to that of phenolphthalein. I have studied briefly about the colour change in phenolphthalein in different mediums.

What I have thought is that the colour change is due to increase or decrease in density of quantised states due to increase or decrease in pi conjugation (Not sure about that though). I tried to observe colour change of fluorescein in strongly acidic and basic medium.

What I noticed was that in acidic medium, addition of small amount of fluorescein gave red solution whereas when the same amount was added in basic medium, it gave fluorescent green solution.

One form of fluorescein is shown below (1); in basic media it cyclises to form a lactone (2):

But I couldn't figure out how this relates to the colour. I thought that the acidic form 1 would have greater π-conjugation, the energy gap between the electronic states would be smaller, and light of a lower frequency (longer wavelength) would be absorbed.

However, the opposite is observed. What is actually going on?

The lactone form is not stable in basic medium, so the cycle opens and gives conjugation to flow and unite all four rings. The problem is you really swithched the acidic and basic forms of fluorescein, so the left one exists at basic pHs (except for the H atom at carboxylate - it goes away), and acid medium favors the existence of the right form.
You may check the forms of phenolphthaleine - they are pretty much the same.


How to determine the order of nucleophilicity for given chemical species? How to determine the order of nucleophilicity for given chemical species?

Like I came across this question, to rearrange $\ce{RCOO-, OR-, OH-, H2O}$ (alkyl acetate, alkoxide, hydroxide, and water) in decreasing order of nucleophilicity and it ranked $\ce{OR-}$ first.

Shouldn't it be $\ce{OH-}$ since $\ce{OR-}$ would be more sterically crowded?

Within a group of nucleophiles with the same atom, the nucleophilicity decrease with decreasing basicity of nucleophile.

Decreasing basicity means the decreasing affinity of electron pair for a proton. The decreasing order of nucleophilicity is shown in the figure.

But the relationship between nucleophilicity and basicity can be reversed by steric effect. Less basic but steric unhindered nucleophile therefore have higher nucleophilicity than strong basic but strically hindered nucleophile.


Does resorcinol give positive iodoform test? It's given in my book that other than the usual compounds that answer iodoform test, resorcinol (benzene-1,3-diol) also gives positive iodoform test.

Is this because of tautomerization which gives a diketone with an active methylene group? But why should a ring lose its aromaticity by doing so?

There is an equilibrium between the keto enol forms, however small the keto form may be as a contributor. The rest is as you say. The precipitation of iodoform drives the equilibrium further to the products, as per Le Chatelier's principle.Of course, it is not in all cases where the equilibrium is controlled by such factors. The net gibbs energy change for the reaction must be negative.


Why is there only one bond length for sulfur hexafluoride? In sulfur hexafluoride, there are two 3-centre 4-electron bonds, according to the book I am reading (Essential Trends in Inorganic Chemistry by D. P. M. Mingos). The other two fluorines are bonded to sulfur, by interacting with its sp-hybrid orbitals. Therefore, we should expect to observe two sets of bond lengths, one set for those involved in the 3-c 4-e bonds and one set for those involved in the bonds involving the hybrid orbitals. However, only one set of bond length (156.4 pm) is reported on Wikipedia. Why is this the case?

In comparison, we observe two different sets of bond lengths in sulfur tetrafluoride, one set of values for the axial bonds and another set of values for the equatorial bonds.

This is an old question and is largely answered in the comments, but here's a full answer for the sake of completeness.

There are really two questions here:
1) is the given bonding orbital description correct for SF6?
2) why is SF6 octahedral?

The first question is easier: no, your description is incomplete as you've described it, but perhaps correct as intended by the author (I don't have the textbook to see the exact wording). It is correct if you explain further that there are three possible representations with two $sp$ bonding orbitals and two 3c/4e bonds. The three different representations look identical except that the $sp$ orbitals are aligned on different Cartesian axes (ie one for $x$, one for $y$ and one for $z$). The true structure is a composite of these three, meaning that each S-F bond is 1/3 $sp$ and 2/3 3c/4e (which is essentially 1/2 of a $p$ orbital) which adds up to a total of 2/3 of a bond with a contribution of 1/6 from $s$ and 1/2 from $p$.

I suspect that the textbook uses this representation because it connects readily to earlier discussions of integer hybridization, i.e. "$sp$" orbitals are familiar and easy to visualize. And presumably the concept of a 3c/4e bond has also been previously introduced. Making a composite of familiar concepts is perhaps easier than introducing the idea of delocalized molecular orbitals, but in this case they are simple enough that it's worth going through them to show that the result is the same.

First, the $s$ orbital is spherically symmetric, so it necessarily interacts the same with all six identical ligands. The lowest energy molecular orbital is thus represented by the $s$ orbital on sulfur and the six fluoride ligand orbitals all in the same phase (images below). That's where the 1/6 $s$ comes from. The $p$ orbitals each are able to interact with two of the ligands, one on each end of orbital. The other four ligands do not contribute at all, because they are exactly on the nodal plane of the $p$ orbital. Since each $p$ interacts with two ligand orbitals, each S-F bond gets 1/2 $p$ to go with the 1/6 $s$, exactly the result we got from the composite of hybrids. Below are drawings of the four bonding molecular orbitals:

Now on to the second question: why is SF6 octahedral if it only makes a total of 4 bonds?

Given that there is a lot of negative charge on the ligands and positive charge on S, a simplistic representation is to make the atoms solid charged spheres. A simple electrostatic potential energy minimization would then lead to an octahedral arrangement of the ligand spheres around the sulfur sphere. That arrangement minimizes the repulsive interactions between the ligands and maximizes the attractive interaction between ligand and sulfur.

Without a lot of empirical data about atomic interactions, though, we cannot rule out the possibility that there could be favorable bonding interactions with a strong distance dependence that would make it more favorable to, for example, push four of the ligands even closer to the sulfur and push the other two farther away. However, it turns out that the distance dependence of the bonding interaction is similar enough to the distance dependence of the electrostatic interaction that the geometric shape of the minimum energy conformation is the same - all six ligands at equal distance from the sulfur and spaced evenly around it.


If I have a chiral substrate, should I always assume that both enantiomers are present? Considering the Sn1 reaction of (S)-2,3-dimethyl-3-pentanol, knowing that this molecule has a chiral carbon, should I always assume that both enantiomers are present, and therefore the products of a reaction will always have $2^n$ products for $n$ stereocenters on the substrate, assuming that each stereocenter has equal chance of reacting? Is there two products of this Sn1 reaction due to the fact that the molecule is chiral, and therefore its enantiomer is present and has equal chance of reacting?

In this SN1 reaction there would be two products even if only one enantiomer reactions.

Enantiomerically pure (S)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

ALSO, enantiomerically pure (R)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

When this kind of phenomenon happens, there must be an achiral intermediate that can react to produce either enantiomer with equal propensity. In this case, what is the structure (geometry and everything) of the carbocation formed when the leaving group leaves?


How do valence shell configurations affect attraction and repulsion forces between atoms? It seems clear that the electrons of one atom are attracted to the protons of another and that this is the source of attractive forces between atoms. However, it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms. Are the electrons not attracted to protons in the same way? Is it because the repulsion between the electrons wins out over the electron-proton attractive forces. Furthermore, why do valence shell arrangements with few electrons bring about attractive forces?

Firstly, why should covalently bonded atoms be incapable of bonding? Bond changes are the very basis of quite a huge amount of chemical reactions. Oxygen occurs in the $\ce{O2}$ form, which is covalently bonded, but readily breaks its O–O bond in order to form other bonds.

Now, to your question. Why are elements in group 18 (generally: think of the xenon fluorides) inert? Why is sodium so willing to give up its electron, while neon doesn’t form any compounds at all, when they only differ by one proton in the nucleus?

The answer lies in the orbital energies. You can find a table listing the outer shell orbital energies page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.). Let us compare the outer orbital energies of Ne and Na. Neon has two 2s electrons with an energy of $-48.47\ \mathrm{eV}$ and four 2p electrons with an energy of $-21.59\ \mathrm{eV}$, while the 3s electron in the sodium atom lies at $-5.14\ \mathrm{eV}$. The difference is overwhelming. When a ionic compound is formed, an electron is removed from the less electronegative atom and added to the more electronegative atom, forming a cation and an anion. Now, the Na electron is much easier to remove than the Ne electron – you just need $-5.14\ \mathrm{eV}$ to do the task, while eliminating the Ne electron is much more difficult. The electron is then placed into an atom such as fluorine, whose outer 2p subshell lies at $-18.65\ \mathrm{eV}$: you needed energy to remove the Na electron, but energy is released when you place it in the F atom (since it is put into a low energy orbital). Placing an electron from F to Na wouldn’t release energy because it is much more stable in the fluorine atom, due to their orbital energies (if you want to simplify things my anthropomorphizing the electrons, you could say they’re “happier” in the fluorine atom). Removing an electron from the neon valence shell is very difficult because it has very low energy; adding one doesn’t bring much because since the second shell is full, you would have to add it to the third shell – its energy is similar to the Na 3s shell, so it’s not convenient.

I have to note here that while the first ionization energy corresponds more or less to the orbital energy (Na: $495.8\ \mathrm{kJ/mol}$ which corresponds to $5.139\ \mathrm{eV/atom}$), the electron affinity doesn’t (F: $-3.399\ \mathrm{eV/atom}$), because of electron-electron repulsion in the fluorine p subshell. That’s why an isolated $\ce{Na+ - F-}$ compound cannot exist: the additional energy released when forming the lattice is needed to form a stable entity.

Having said this, we can move on. You may ask: Yes, but why are the F electrons more stable than the Na ones, when Na has more protons? The reason is that the outer Na electrons are in the third shell, and thus they have higher energies. In a hydrogen-like atom (an atom with only one electron, which is clearly an oversimplification, since it doesn’t involve screening) the orbitals have the following energies:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

You can see that raising the $Z$ (atomic number) value from 9 (F) to 11 (Na) isn’t enough for lowering the energy, since $n$ (the principal quantum number, i.e. the shell) changes from 2 to 3.

So, this is why sodium forms positive ions, fluorine forms negative ions, and neon doesn’t form any ions at all (chemically). Nota bene: as I said, the formula above doesn’t take screening into account – when going down a group, it (usually: see d-block contraction) prevails over the growing nucleus. That’s why Ar is less inert than Ne, even though the formula above would suggest the contrary.

But what about covalent compounds? According to VB theory, bonding occurs through orbital overlap. Lower energy orbitals mean smaller overlaps, which is inconvenient if you want a stable bond: Carbon forms bonds while neon doesn’t because its orbitals are less stable (see table) and can overlap with other orbitals. Its orbitals are less stable because neon has more protons with the same principal quantum number – see the equation above to understand the concept.

In short: noble gases, especially those positioned higher in the periodic table, don’t bond because their outer orbitals have the lowest energies. Ultimately, this is the result of having the largest number of protons possible for a given outer shell in the ground state.


Is it correct to say that active carbon is an enhancer of either acid or base? As a non chemist I tried to read to understand if active carbon acts in solutions as an acid or a base but I understood that it can act as both, depends on the material which used to make the solution.

For example, mixing a sodium bicarbonate with active carbon in a solution will make that solution with higher PH (hence a stronger base, per my understanding).

Is it correct to say that active carbon is an enhancer of either acid or base?

Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions.  But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the rate of the forward reaction : $$\ce{HCO_3^- <=> CO_2 + OH^{-}}$$ and this reaction is favored if the air in the room contains a low concentration of $\ce{CO_2}$, according to Le Chatelier's principle. As a consequence, the pH of the solution has slightly increased, due to this supplement of $\ce{OH^-}$ ions.


Open Source SDF (Chemical Table File) Parser in any language I'm looking to parse a large .sdf file and put it into a .csv file.

The only software I could find that parses .sdf files is OpenBabel's Pybel, but it won't properly install for my El Capitan, for either python2.7 or 3.+. (It's an open issue, and might hopefully be fixed with the next release). Are there any other languages/OSs that can easily parse this file?  If there aren't any, is there a recommended operating system I could spin up with VirtualBox that will easily run OpenBabel/Pybel?

The free, open-source RDKit chemoinformatics software can read .sdf files.  This software is most often used via it's Python API.

You can learn more about RDKit at the project's home page.

I recommend installing it via conda.  (If you use python regularly but don't use conda for controlling environments and package distribution, you probably should.)  The installation command is conda install -c https://conda.anaconda.org/rdkit rdkit
.

The python code to parse an sdf file and write to a csv file is very simple.

import pandas as pd
from rdkit import Chem
from rdkit.Chem import PandasTools

my_sdf_file = '/Users/curt/Desktop/sdf-isothiocyanates.sdf'

frame = PandasTools.LoadSDF(my_sdf_file,
                            smilesName='SMILES',
                            molColName='Molecule',
                            includeFingerprints=False)

frame.to_csv('/Users/curt/Desktop/excel_isothiocyanates.csv')


How do we calculate N factor and why is it useful in calulations? I don't understand the concept of N factor at all. Why is it important?

I know that it is something dependent upon the oxidation number, for example N factor of Na would be 1 since it has oxidation number of +1.

For acid and bases we look at the replaceable $\ce{H+}$ or $\ce{OH-}$ present in them.

Apart from this, I didn't really understand how I can use N factor for calculation or how to find the N factor for whole chemical reactions. Please correct me where I'm wrong.

There are some different applications.
For one, it tells you how many times an acid can lose a proton. For instance, you can have a diprotic ("with $\ce{2 H+}$") molecule, with $n=2$. For instance, $\ce{H2SO4}$, deprotonates to $\ce{HSO4-}$, which can then go to $\ce{SO4^{-2}}$.
Why is that important? Take an acidic reaction with $\ce{H2SO4}$. If the extent of reaction is strongly forward, you would treat $\ce{H2SO4}$ as if it had twice its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.
It's also known as equivalent concentration, which might make it sound less foreign to you.
In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.
Sources: Wikipedia


Balancing REDOX reactions in a basic solution, when no H+ ions are present I am trying to balance the following redox reaction, in basic solution. From what I have learned, to balance a redox reaction in basic solution you simply balance it as you normally would in acidic solution, then make sure the H+ ions are neutralized by adding OH- ions. In the example below, is my answer balanced in basic solution already? Or do I need to somehow cancel out the H present in H2O?

The bottom equation looks balanced to me. There are the same number of atoms of each element on each side of the equation and charge is conserved. The hydrogen in the $\ce{H2O}$ is balanced by the hydrogen in the $\ce{HAsO2}$ on the other side so it looks fine.


Can Cl(g) cause quick rusting? I'm actually digging in something that comes from Syria...
How does Chlorine cause metal to get rusted?
It's full of pictures of rockets (allegedly filled with chlorine gas) that are all rusted. In my mind i find it pretty obvious that a strong acid like HCl can cause rust, but what's actually the precise mechanism?
Does HCL even matters oj simple Cl(g) can cause rusting?

PS: consider all the reactions taking place in open air

REPHRASED:
Is it possible for Cl(g) to cause some metal to get rusted in a matter of minutes?
What are the chemical reactions that make it possible/impossible?

Neither hydrogen chloride ($\ce{HCl}(g)$) nor hydrochloric acid ($\ce{HCl}(aq)$) are capable of oxidising iron to the ferric state ($\ce{Fe^3+}$) but in the presence of oxygen (air) the ferrous ions initially produced can then be further oxidised to ferric ions:

$$\ce{4 Fe^2+ + O2 + 2 H2O -> 4 Fe^3+ +4 OH-}$$

For reasons explained here, that process is slow in acidic conditions, though.

Dry chlorine does oxidise iron to $\ce{Fe(III)}$ but at room temperature this is likely to be a very slow process.

However, due to the equilibrium:

$$\ce{Cl2 + H2O <=> HCl + HOCl}$$

moist chlorine is likely to be far more aggressive towards most metals including iron, because $\ce{HOCl}$ is a very strong oxidiser, easily capable of oxidising iron (or $\ce{Fe(II)}$) to $\ce{Fe(III)}$. Due to $\ce{Fe(OH)3}$'s extreme insolubility, the hydrated ferric oxide, i.e. rust, would then precipitate.


Nomenclature of aldehydes with double bonds and stereoisomerism I'm struggling to name this compound shown below:

I'll post my chain of thought here so it can be corrected!

longest chain is marked as below:

this is because the functional group with the highest priority is the aldehyde (HCO) functional group, so the chain should contain this functional group
the chain continues down to to bottom-right to obtain the maximum length with the greatest number of substituents (although actually looking at it now, could the carbon chain go left and skip the double bond?)
there's a butyl side chain on carbon 3
there's a double bond at carbon 3
there's an ethyl side chain on carbon 4
there are two methyl groups on carbon 5
following CIP rules, the groups with highest priority are arranged opposite each other, so the molecule is E

Therefore, would the molecule be (E)-3-butyl-4-ethyl-5,5-dimethylhex-3-enal?
Or have I got myself completely mixed up...
Thanks for any help!

IUPAC has changed some of the rules of nomenclature as of 2013. In aldehyde 1 both the aldehyde group and double bond are included in the base name (noun) and the methyl group is the substituent (adjective) because both chains are $\ce{C4}$. In structure 2 the methyl group of 1 has been changed to ethyl. Now the longest chain is $\ce{C5}$ containing the aldehyde group and the double bond, as a substituent is the modifier methylene. In structure 3 the branch at $\ce{C3}$ has two $\ce{C3}$ units. The one containing the double bond takes priority as the "noun". In structure 4 the chain containing the double bond is shorter, which becomes the "adjective". In your compound 5, the noun is heptanal (in black) and the adjective, which contains the double bond, is in red. The two entities are numbered independently.


Have P-acyl phosphines ever been synthesized? I've done some amount of searching and can't seem to find any evidence in the literature that there exists a phosphine analogue of carboxyl amides. I'm looking for compounds of the form $ \ce{ R(C=O)PR_{2}} $; basically a moeity similar to a primary or secondy amide but with a phosphorous atom instead of a nitrogen.

I would also wonder what the proper name for these functionalities would be. Taking a shot in the dark I might guess "phosphomide", "acyl phosphine", or "acyl phosphinide".

I would like to specifically disqualify Acyl-phosphine oxides or any compounds where the acyl-connected phosphorous is also connected to an oxygen.

When I saw this question, I realized I couldn't remember ever seeing an acylphosphine, but I was curious and went looking. I found that they're actually quite easy to make. One common thing I saw was the use of trimethylsilylphosphines with acyl chlorides to form acylphosphines.

The easier method, however, was simply to react a substituted phosphine in the presence of triethylamine with an acyl chloride. The analogous reaction can be done with amines to form amides. In this paper, they describe the synthesis of P-acetyldiphenylphosphine using diphenylphosphine, triethylamine and acetyl chloride. They have some other examples in there, but this suggests that these compounds are typically fairly trivial to synthesize.

One more note: In my searches, I saw "acylphosphine," "acyl phosphide," and "phosphomide" as named. The paper I linked uses the term "phosphomide."


How to experimentally determine the elemental composition in fingernails? I am doing a project in my highschool about analysing C/N and C/S ratio of finger nails of various ages and sexes.
Please suggest me a way to do this. I don't know where to start, I know the various methods of estimating the amount of N or S present theoretically. Please provide info that how should I dissolve the nails etc.

I will be doing my experiment in a nice lab which has got all necessary equipment that I might need, so please answer without considering the limitations of my lab.

I got the idea by this website: https://pubmed.ncbi.nlm.nih.gov/18424895/

These analyses are difficult to carry out at a high school level. Nails are not soluble in any solvent. They must be burned and the gases coming out are analyzed and weighed. To burn them, they have to be mixed with a great excess of copper oxide, put in a long horizontal glass tube, heated from below. A current of oxygen gas should be sent through the glass tube, and the gas should be absorbed first by solid $\ce{CaCl2}$ which absorbs $\ce{H2O}$, then by $\ce{NaOH}$ which absorbs $\ce{CO2}$. By weighing the $\ce{CaCl2}$ and $\ce{NaOH}$ flasks before and after the operation, you can calculate the amount of $\ce{C}$ and of $\ce{H}$ contained in your original sample. To measure the amount of nitrogen of your nails, you have first to destroy them by sulphuric acid, then add them to $\ce{NaOH}$ and some Dewarda alloy (mixture $\ce{Al}$, $\ce{Zn}$) have it boiled to liberate $\ce{NH3}$ gas, and then absorbed this gas containing $\ce{NH3}$ into a solution of an acid like $\ce{HCl}$ or $\ce{H2SO4}$, and then titrate the remaining acid to deduce the percentage of $\ce{N}$ in your original sample. This can not be easily done in a high school lab, slowly.


Do ions in a salt bridge come from the half cells in electrochemistry? I understand the basics of electrochemsitry and galvanic cell, where there are half cells where either an oxidation or reduction reaction occur, and electrons moving between the wires provide the current.

What confuses me is the salt bridge. In this diagram here, nitrate ions, which are from the solution of a half cell, go into the salt bridge.

However, other diagrams (and answers like those to Do the salt bridge ions have to match the ions in the half-cells solutions?) do not state that the ions have to be the same (you don't need nitrate ions in the salt bridge if you have a silver nitrate solution for example).

So do ions like nitrate from the solution in a half cell enter the salt bridge?

You must think electrically, without thinking first to the chemistry. In the anodic solution first, the reaction produces positive ions $\ce{Mg^2+}$ or $\ce{Zn^2+}$. It does not really matter here. The problem is that this cannot stay like this. Either the positive ions have to move where negative ions are available. Or new negative ions must arrive in the solution. Or both.

The simplest way for solving this difficulty is to mix the two solutions. Because a similar difficulty appears on the cathodic region : positive ions ($\ce{Ag+}$ or $\ce{Cu^2+}$) are here discharged, leaving an excess of positive ions around the cathode. By mixing the solutions the excess negative ions produced around the cathode move to the anode where they are needed. But mixing the solutions is the least thing to do, because now $\ce{Ag+}$ or $\ce{Cu^2+}$ ions can touch the $\ce{Zn}$ or $\ce{Mg}$ anode. And the chemical reaction will go on in the contact between positive ion and metal. The same electrons of the half-reaction are produced but they will not go through the long external circuit form the anode to the cathode. They prefer the shortest circuit direct from metal to ion in solution, and metallic $\ce{Ag}$ will  soon be formed on the $\ce{Mg}$ anode (or $\ce{Cu}$ on the $\ce{Zn}$ plate). No current will be emitted in the outer circuit. It is the end of the cell. So it is absolutely necessary to prevent the $\ce{Cu^2+}$ or $\ce{Ag+}$ ions to touch the anode, so the two compartments must be separated.

That is why we introduce the salt bridge. It must contain ions that do not react with any ion contained in the two compartments. Their nature is not important. Usually it is $\ce{K+}$ and $\ce{NO3-}$.

So let's go back to what's happening electrically in solution when the cell is working. There is an excess of positive ions around the anode, and an excess of positive ions around the cathode. To compensate this situation, the negative ions of the bridge are attracted by the positive ions in excess at the anode, and the positive ions of the bridge are attracted by the cathode. Both quit the bridge and get mixed in the solutions. This allow the cell to work.

After some time, the bridge is loosing all its stuff, so some positive ions of the anode ($\ce{Zn^2+}$or $\ce{Mg^2+}$) may enter the bridge and replace the initial positive ions $\ce{K+}$that have moved to the cathode. And symmetrically some negative ions ($\ce{NO3-}$ or $\ce{SO4^2-}$) from the anode will enter the bridge to replace the initial ions $\ce{NO3-}$.


Why is molybdenum that important in inorganic synthesis as well as biosynthesis of ammonia? We know that Mo (Molybdenum ) is used as  a catalyst for Haber's process, during synthesis of Ammonia from Nitrogen. Again, in biological systems, Nitrogen fixation (ALSO forming Ammonia from Nitrogen) uses Mo as activator of Nitrogenase enzyme. My question is "why Mo is so special for Nitrogen?"

Why Molybdenum is Used in the Manufacture of Ammonia?
Molybdenum tubing has particularly good resistance to corrosion by
mineral acids, provided oxidizing agents are not present. It is also
resistant to many liquid metals and to most molten glasses. Molybdenum
is relatively inert in hydrogen, ammonia, and nitrogen up to about
1100 C, but a superficial nitride case may be formed in ammonia or
nitrogen.(Ref. 1)

Also, molybdenum has a special capability to form stable complexes with nitrogen atoms and hence molybdenum goes very well in biological reactions as well(See the link @user79161 has posted in the comment section). Recently, a group of scientists has developed a method for synthesis of ammonia using a dinitrogen-bridged dimolybdenum complex as catalyst and under very mild conditions of room temperature and at atmospheric pressure. However, the reaction mechanism of this new method is still unclear.

Reference

https://samaterials.podbean.com/e/why-molybdenum-is-used-in-the-manufacture-of-ammonia/
https://www.u-tokyo.ac.jp/focus/en/articles/a_00259.html (Picture)


Why do closo-boranes have a 2− charge? In a question I was solving, it was mentioned that:

  [...] closo-boranes are compounds having the general formula $\ce{B_nH_n^2-}$ [...] an example is $\ce{B6H6^2-}$.

Why do these boranes possess a (2−) charge? I am aware that in electron-deficient compounds such as these, there are different types of bonds between boron atoms (such as three-centre two-electron bonds), but I don't know why this should lead to a negative charge.

The application of MO theory to (electron-deficient) cluster compounds such as boranes is generally summarised in a set of rules, known as Wade's rules. We can look at the closo structure $\ce{[B6H6]^2-}$ in more detail to see where the two negative charges come from.
In closo boranes, each BH fragment points outwards from a vertex of a polyhedron. For $\ce{[B6H6]^2-}$, it turns out that this polyhedron is the familiar octahedron, which has six vertices.
What we must do now is to count how many electrons and how many orbitals each BH fragment contributes to the entire cluster. This can be done fairly easily:

Here we have neglected the boron 2s orbitals. One 2p orbital on boron has the correct symmetry to overlap with the hydrogen 1s orbital, forming bonding (blue) and antibonding (red) orbitals; the other two 2p orbitals remain non-bonding (gold).
So, each BH fragment contributes two valence electrons to the cluster. Since there are six BH fragments, the total valence electron count is (so far) 12. Saying this implies that the boron 2s electrons aren't valence electrons, which is, in a sense, true; they do not participate much in the delocalised bonding in the entire cluster.
If we use the three bonding and nonbonding orbitals to form properly symmetrised MOs across the whole cluster, it turns out we get seven bonding MOs across the entire cluster. I'm not keen to ChemDraw these MOs, so I've taken a picture from Housecroft's Inorganic Chemistry (4th ed.):

The lowest energy MO ($\mathrm{a_{1g}}$) comes from an in-phase combination of all six bonding fragment MOs. The three next lowest ($\mathrm{t_{1u}}$) come from overlap of two bonding fragment MOs with four nonbonding 2p orbitals on the other boron atoms. Three more ($\mathrm{t_{2g}}$) come from "sideways", or "tangential", overlap between four boron 2p orbitals.
In general, it turns out that for a polyhedron with $n$ vertices (at least, where $n \geq 6$ — see comments), there will be $n+1$ bonding MOs which can be made using the fragment MOs at each vertex.
Now, in order to get a stable cluster, we need to be able to fill all these $n+1$ bonding MOs with electrons. In the case of $n = 6$, we have 7 MOs to fill, and so we need 14 electrons. However, as established previously, we only have 12 valence electrons!
The solution is therefore to tack on two extra electrons, and that's why the closo borane requires a (2−) negative charge. If it didn't have this negative charge, it wouldn't be able to fill all of the bonding MOs in an octahedron, and it's quite likely that it would choose to adopt a different geometry that is no longer closo.


How to make something solid that dissolves faster than bar soap in water What is a soap-like material that holds a solid shape, but dissolves completely in water over 10-15 minutes. Think disappearing Rubber Ducky.

Potassium soap is soft.  Vary the sodium-potassium ratio to get what you want.  Leave some glycerin in.  Make the soap from polyunsaturated fats.


Why is phenol more acidic than aniline? I find that there are two contradicting factors at play here:

1) On removing proton, negative charge on oxygen is more stable than it is on nitrogen (because of greater electronegativity of oxygen). Therefore, phenol is more acidic than aniline.

2) +M effect of nitrogen is more than oxygen, so negative charge (after deprotonation) in the case of aniline will be more delocalized making the anion more stable. Therefore, aniline should be more acidic than phenol.

So is that a factual question or am I missing something?

Taken from my answer here:

Comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.

Here methane and acetylene form anions where the negative charge on an $\mathrm{sp^3}$ and $\mathrm{sp}$ carbon respectively. Here, there is an electro negativity difference between the two carbons which leads to a very high difference in $\mathrm p K_\mathrm{a}$.

The difference between their electronegativities too large to explain the $\mathrm p K_\mathrm a$ difference of $15$. Nitrogen is $3$ and oxygen is $3.5$ on Pauling scale of electronegativity.

This small amount makes a big difference in $\mathrm p K_\mathrm a$ values. I say this because, if methane and acetylene have a difference of 20 units where the only difference is their hybridization and therefore electronegativity, oxygen and nitrogen will have a big difference in $\mathrm p K_\mathrm{a}$ values.
This can be seen in the actual values as referenced in Evan's $\mathrm pK_\mathrm a$ table, where aniline has a $\mathrm pK_\mathrm a$ value of $30.6$ and phenol has a $\mathrm p K_\mathrm a$ value of $18.0$ both referenced in DMSO.