# Category Archives: Science

Science subjects. Is quite general.

## Choosing an Acetophenetidin Solvent

When synthesizing Acetophenetidin (Phenacetin) in an Organic Laboratory, you might end up with a cake of crude product filled with impurities from the synthesis process.  A giveaway sign of this is a calculated percent yield in excess of one hundred percent.  This is not high-quality product.  It must be recrystallized in order to refine the product for larger crystals and purer yield.

In order to recrystallize a product for better purity, a solvent must be found that will solvate everything at high temperatures, but will precipitate the pure product at cooler temperatures while keeping the impurities solvated and out of the crystal structure of the reforming product.

Perhaps you are given three options: deionized water, ethanol, and hexane.  To test which would be the better solvent, add a bit of the crude product to small amounts of the potential solvents in test tubes.  Observe solubility.  Then heat the solvents in a boiling hot-water bath and observe solubility at that state.  Finish the mock recrystallization by removing the test tubes from the bath and letting them cool to room temperature before sticking them in an ice bath.  Observe solubility once again.

The better solvent will preferably not solvate the acetophenetidin at room temperature.  This means that the recrystallization will begin earlier in the cooling phase.  Everything must be dissolved at boiling.  A solvent with a low boiling point would not help here, as it will evaporate away leaving everything as an impure solid stuck to the sides of the test tube.  When cooled in ice, crystals must form for the solvent to be worth anything.

Here is some sample data:

Water

• Room Temperature – Crude product appeared to be insoluble
• Boiling – Completely solvated the crude product
• Freezing – Crystals reappeared

Ethanol

• Room Temperature – Completely solvated
• Boiling – Still solvated
• Freezing – Remained solvated

Hexane

• Room Temperature – Insoluble
• Boiling – Solvent evaporated
• Freezing – The solvent evaporated away in the previous step

The hexane is out as a solvent; its boiling point is too low.  Ethanol does not work because the acetophenetidin never precipitates out of it.  This leaves only water, and the observations associated with it prove it to be a useful solvent.

It is important to know what to look for when observing the Freezing solubility.  Even though the impure cake may enter the test tube as a lump, it will not precipitate that way.  Once the test tubes are sitting in the ice bath, let them sit undisturbed for five minutes.  Then remove and observe.  Wipe the condensation off the tubes and look very closely at them.  From farther away, the water and ethanol tubes look much the same.  Upon closer inspection and perhaps a swirl, one can see very tiny white particles floating in the water that are absent in the ethanol.  That is the pure acetophenetidin and a good sign.  These particles may be even smaller than any dust in the tube; close observation is important here.

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## Getting Tripped By Markovnikov

Markovnikov’s rule is a decent tool to predict some simple organic reactions. It states that in simple electrophilic additions of alkenes, such as Hydrogen Chloride to n-Pentene, the electrophile will attach itself to the carbon of the double bond that has more Hydrogen atoms already attached; the less substituted carbon. It can trip a user up, however, if they do not understand the actual process taking place. Example:

1-phenyl-2-methylpropene has two “sp2” hybridized carbons in its Propene backbone. In a reaction with Hydrogen Chloride, we will only consider these two points and will leave out attachments to the Benzene ring itself for simplicity.

When the PI bond reattaches itself to a passing Hydrogen (HCl), it creates an intermediate carbocation and a Chloride ion. Which of the two carbons the intermediate forms on is determined by stability considerations. Markovnikov’s rule would predict the hydrogen attaching to the Phenyl side, as there is only one substituent there. The reverse actually happens, creating a “non-Markovnikov” scenario where the Chloride ion attaches to the molecule close to the Phenyl substituent.

Instead of relying on Markovnikov’s rule for simple additions which create carbocations, use the idea of hyperconjugation. This boils down to the idea that carbocations (positively charged) can be stabilized by the sigma bonds of nearby substituents (usually carbon to hydrogen bonds) because the electrons involved in sigma bonding are negatively charged. The more substituents, the more stabilized a present carbocation becomes. The stability of carbons with positive charges becomes:

$1^{\circ} < 2^{\circ} < 3^{\circ}$

Primary carbocations are less stable than tertiary because tertiary carbocations have two more sets of hyperconjugation sigma bonds to stabilize the positive charge.

However, sigma hyperconjugation is nowhere near as stabilizing as the effects of lone pairs of nearby electrons such as those found in “sp2” hybridized carbons. The negative charge is less diluted in the lone pair. This makes Allyl and Benzene groups even more effective than tertiary carbons. The order stands at:

$1^{\circ} < 2^{\circ} < 3^{\circ} < Allyl < Benzene$

The point of this is that when a carbocation intermediate forms, it will form in the most probable place most of the time. The most probable is the most stable. Carbocations will form on carbons attached to Benzene before they attach to carbons with only two other carbons (methyl groups, for instance) attached to themselves.

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## Spring Pushing a Block Off Another Block Involving Simple Harmonic Motion

Two blocks, the small one of mass 3.5 kg and the large one of mass 7.3 kg are stacked, small block on top. The large block is connected to a horizontal spring of spring constant 210 N/m. The large block sits on a frictionless floor. The coefficient of static friction between the two blocks is 0.492. What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?

The maximum acceleration is what is important here. The simple harmonic motion is merely a spin on an old problem. So lay out some Newtonian Force equations:

$F=ma$ – Standard Force Equation
$F_{s} = -kx$ – Hook’s Law for Spring Motion
$F_{N} = mg$ – Normal Force
$F_{f} = \mu_{s} F_{N}$ – Frictional Force, tailored for the small block.

Start with the Spring equation, treating it as an external force acting on the two block system:

$F_{s} = -kx$

Because it is the singular external force acting on the system, we can set the general force equation equal to this, reminding ourselves that we must add the two block masses together as they are one system currently:

$M$ – Mass of the small block and the mass of the large block.

$-kx = Ma$

$\frac{-kx}{M} = a$ – Acceleration of the two block system.

With that, we concentrate on the two block system internally now. We find the maximum force that friction exerts between the two blocks:

$m$ – Mass of the small block.

$F_{f} = \mu_{s} F_{N}$ – Frictional Force
$F_{f} = \mu_{s} mg$ – Substitute Normal Force
$F_{f} = \mu_{s} mg = ma$ – Friction is the only force acting on the block.

$F_{f} = \mu_{s} g = a$ – Small mass cancels

Substitute one acceleration for the other:

$\mu_{s} g = \frac{-kx}{M}$

Solve for x:

$x = \frac{M\mu_{s}g}{-k}$

We can solve directly for x here because we know that the maximum acceleration that the system can undergo without sliding is always the negative of the maximum x value. This is a property of simple harmonic motion.

$|x| = x_{max}$

And we solve:

$x = \frac{(3.5[kg] + 7.3[kg]) * 0.492 * 9.8[m/s^2]}{-210[N/m]}$

$x_{max} = 0.248[m]$

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## Silver Phosphate and Acetic Acid

Silver phosphate has the formula:

$Ag_{3}PO_{4}$

Acetic Acid has the formula

$CH_{3}COOH$

Mix them together and you get the equation:

$Ag_{3}PO_{4(s)} + 3\; CH_{3}COOH_{(aq)} \rightleftharpoons 3\; AgCH_{3}COO_{(s)} + H_{3}PO_{4(aq)}$

The equilibrium equation from this formula is:

$\kappa = \frac{[H_{3}PO_{4}]}{[CH_{3}COOH]^3}$

To compute the equilibrium constant from this, multiply by values that are equal to one, but that allow you to use known constants:

$\kappa = \frac{[H_{3}PO_{4}]}{[CH_{3}COOH]^3} * \frac{[H_{3}O^{+}]^3}{[H_{3}O^{+}]^3} * \frac{[CH_{3}COO^{-}]^3}{[CH_{3}COO^{-}]^3} * \frac{[Ag^{+}]^3}{[Ag^{+}]^3} * \frac{[HPO_{4}^{2-}]}{[HPO_{4}^{2-}]} * \frac{[PO_{4}^{3-}]}{[PO_{4}^{3-}]}$

$\kappa = \frac{(\kappa _{a(CH_{3}COOH)})^3 * \kappa_{sp(Ag_{3}PO_{4})}}{(\kappa _{sp(AgCH_{3}COO)})^3 * \kappa _{a(H_{2}PO_{4} ^{-})} * \kappa _{a(HPO_{4}^{2-})}}$

$\kappa = \frac{(1.8\cdot 10^{-5})^{3}(1.3\cdot 10^{-20})}{(2.0\cdot 10^{-3})^{3}(6.3\cdot 10^{-8})(4.2\cdot 10^{-13})}$

$\kappa = \frac{[H_{2}PO_{4}^{-}]}{[CH_{3}COOH]^2} =5.0\cdot 10^{-5}$

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## Two Trains (Constant Velocity) and a Flying Creature

Two trains are headed toward one another on the same track. Train A has a constant speed of 40[km/hour] and Train B has a constant speed of 20[km/hour]. A Tengu that can fly at 60[km/hour] is perched on top of Train A. When the trains are 90[km] apart, it hops off and begins to fly toward Train B. When it reaches Train B, it instantaneously reverses direction and begins flying toward Train A. How far does the Tengu fly before the trains collide?

Do not make this harder than it is. A series of increasingly small distances would fit this problem perfectly, but that is needlessly complex.

The trick here is to understand that the “Flying Creature” is always moving at a constant velocity and merely bounces off of an imaginary plane in front of the oncoming train before heading back to the other train. Because of this assumption, the Tengu will cover the same amount of ground in the time before the trains collide flying between them as if it had been flying in a straight line for the same amount of time.

First, calculate the time between the start and when the trains collide. The idea is that at some time in the future, the distance traveled by Train A and Train B added together will equal the original distance between them at the start.

$t$ = Time Before the Trains Collide [hour].
$V_{A}$ = Velocity of Train A [km/hour].
$V_{B}$ = Velocity of Train B [km/hour].
$D$ = Distance Between the Trains [km].

$(V_{A}*t)+(V_{B}*t)=D$
$(40[km/hour]*t)+(20[km/hour]*t)=90[km]$
$t=1.5[hour]$

Now with time in hand, it is simple to calculate how far the Tengu flew in the alloted time.

$V_{T}$ = Velocity of the Tengu [km/hour].
$H$ = Distance Traveled by the Tengu [km].
$V_{T}*t=H$
$60[km/hour]*1.5[hour]=90[km]$