Grignard Is The Child Of A Sail Producer Essay

Grignard Is The Child Of A Sail Producer Essay
Introduction:
Grignard was the child of a sail producer. In the wake of concentrating on arithmetic at Lyon he exchanged to science and found the manufactured
response bearing his name (the Grignard response) in 1900. He turned into an educator at the University of Nancy in 1910 and was granted the Nobel
Prize in Chemistry in 1912. Amid World War I, he studied chemical warfare agents, especially the produce of phosgene and the identification of
mustard gas. His partner on the German side was another Nobel Prize–winning chemist, Fritz Haber. (2) The Grignard reagent is exceptionally
responsive and responds with most natural mixes. It likewise responds with water, carbon dioxide and oxygen. (2) Grignard reagents are set up by the
response of magnesium metal with fitting alkyl halide in ether dissolvable. The halogen might be Cl, Br, or I. A standout amongst the most imperative
employments of the Grignard Reagent is the response with aldehydes and ketones to frame liquor. A related blend utilizes ethylene oxide to plan
alcohols containing two more carbon molecules than that of the alkyl halide. (2)
Grignard is most noted for creating another procedure for delivering carbon–carbon bonds utilizing magnesium to couple ketones and alkyl halides.
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Conversion of Alcohol to Alkyl Halides Essay
Conversion of Alcohols to Alkyl Halides Ankita Patel August 6, 2013 Introduction This lab consisted of the conversion of alcohols into alkyl halides
through common substitution methods. These methods include SN1 and SN2 mechanism, both of which can occur for this type of reaction. For both
reactions, the first step of protonation will be to add hydrogen to the –OH group and then the rest of the reaction will proceed according to the type of
mechanism. SN1 reactions form a cation intermediate once the H2O group leaves, then allowing a halide (such as Br) to attack the positively charged
reagent1. On the other hand, SN2 reactions are one–step mechanism in which no intermediate is formed and the halide attaches as the leaving... Show
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Thus, one could safely assume the product from 1–propanol was 1–bromopropane. This is mainly due the C–H wag around 1260cm–1 indication it was
a terminal alkyl halide. This reaction went through SN2 mechanism not only because the alcohol was primary but also because there were no
rearrangements. If a rearrangement would have occurred, it would have indicated that it was a SN1 mechanism. Further analysis was then done to
determine the exact identity of the product and the chemical makeup. B B C C A A B B C C A A Figure 2: NMR Spectrum for 1
–propanol The
results from the NMR of 1–propanol showed 3 different prominent peaks with the peak at 2.2 cm–1 being the acetone. Because 1–bromopropane has
three non–equivalent hydrogens it was found to represent this set of NMR data. The other product, 2–bromopropane only had 2 different types of
hydrogens and would have only had 2 peaks. Further analysis of the structure of 1–bromopropane showed that the hydrogens closest the bromine group
were an indication of peak A in the graph. Because of the electronegativity of the bromine, this peak was located further downfield. There were 2
neighboring hydrogens so using the n+1 rule gave the 3 peaks. Going down peak B showed the next carbon which had 5 neighboring hydrogens thus
giving 6 peaks. Finally, the carbon furthest away from the bromine was found at peak C. It had 2 neighboring hydrogens and provided 3 peaks.

Advantages Of Supercritical Fluids
Supercritical fluids (SCFs) exist at temperatures above the liquid–vapor critical temperature and have densities between liquid and vapor states. The
advantages of supercritical fluids as environmentally benign solvents arise from their non–toxicity, non–combustibility, availability and possibilities for
modulating physical and chemical properties through minor changes of temperature or pressure. These fascinating features of SCFs lead to their
important potentials as green alternatives to toxic organic solvents and there have been many industrial and technological applications using SCFs
[1–7]. During the last two decades, there has been renewed interest in using supercritical water (SCW) in pyrolysis [8], hydrolysis [9], oxidation [10],
electrochemical reactions [11] and in material synthesis [12]. Physical properties... Show more content on Helpwriting.net ...
These resulting differences have a direct consequence on the solvation properties of SCW. Ionic association occurs to a larger extent in SCW because
of the lower solubility of ions in water under extreme conditions. Aqueous electrolyte solutions constitute an integral part of a large number of
biological and geological processes. Particularly, the thermodynamic properties of ions in supercritical water (SCW) have crucial roles in many
geological processes [13]. Therefore, it is important to study and model the solvation structure and dynamics of aqueous electrolyte solutions in
supercritical conditions at the microscopic level for the purpose of understanding the chemical processes and properties of such hydrothermal systems.
Ion hydration and ion–pair association in SCW have been the important subjects for hydrothermal technologies [14]. Aqueous fluids which occur
ubiquitously in the earth's crust and upper mantle at higher temperatures and pressures are involved in magmatic processes, metamorphism,

Compare The Reaction Between Tert-Butyl Chloride And Alkyl...
The data reveals that reactions occurred in all four test tubes. This means that all four test tubes turned out to be positive reactions. The data collected
regarding the reactions did not meet the expectation. The expectation was that two out of the four tubes would yield a reaction. To be specific, the
expectation was that only one of the first two tubes would have a reaction, and only one of the second two tubes (tubes 3&4) would have a reaction.
This was the expectation because the goal of the reactions was to verify the structure of the alkyl halides. The structure of the tert–butyl chloride and
chlorobutane were in question. If they reacted with NaI, it could be concluded that they were primary alkyl halides. If they reacted with AgNO3, it
could be concluded that they were tertiary alkyl halides.... Show more content on Helpwriting.net ...
AgNO3 easily undergoes reactions with tertiary alkyl halides (SN1). The four tubes should not all have undergone reactions because that would give
an unclear indication whether or not the alkyl halides were primary or tertiary. Two tubes were supposed to undergo reaction and two should not have.
It is likely that the error that made the extra two test tubes react was contamination. If the test tubes contained traces of both solvents, it would be
possible that the alkyl halides would react in all the test tubes. This is because the alkyl halide would react with its preferred solvent, even if the
solvent is present in small quantities. This contamination could have been due to using the same pipette to transfer the solvents. It also could have
been possible that the test tubes used were not properly cleaned and still contained traces of

Police Crime Analysis Essay
Police Crime Data Management & Reporting: Software/user licenses needed for crime analysis by the Morrisville Police Department (MPD). Data
driven/predictive policing is an integral component in crime prevention and reduction. Currently, MPD lacks the necessary software to effectively
identiy and address current crime trends within the town. The software will serve as a tool to facility crime mapping for targeted preventive and
enforcement activities and enable our residents to track crime within their communities.
Louis Stephens Drive Quarterly Maintenance: Use of town staff and rental equipment to maintain the portion of Louis Stephens Drive from Polar Pike
Lane to Gray Marble Drive that is the back entrance/access connector for the Walnut Woods Town Home Community and Breckenridge. This is an
NCDOT road, but its budget for gravel road maintenance has been limited. Funds should be used to perform maintenance quarterly to provide a
smoother, longer lasting drive surface than what can currently be provided by NCDOT or the Town within existing budget resources. TOTAL
Onetime or Capital Impacts (Non–Routine)
Fire Apparatus – Ladder 3: Purchase of the third Quint for Station 3. Quints are already in place at Stations 1 and 2. This purchase coincides with ...
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Currently, the stream is eroding the adjacent private property, causing a wall/patio to potentially fall into the stream. The project will include enhancing
the existing riparian bugger that is being overwhelmed with invasive vegetation, stabilizing the stream banks, restoring the buffer and removing
invasive plants, and replacing an existing failing storm drain outfall from the parking lot with a new outfall that the future MAFC facility enhancement
can tie into. This project if fully funds by the Stormwater Fund and using of existing nutrient funds collected in prior

Grignard Reaction Lab Report
Side reactions are reactions that compete with one another that produce unwanted products. One of the competing reactions with the Grignard reaction
was the Grignard reagent reacting with oxygen to form peroxide which is very reactive. The second most electronegative element is oxygen. The
second competing reaction was the Grignard reagent reacting with carbon dioxide to form a carboxylic acid. The carbon dioxide contains an
electrophilic carbon. The products from these two side reactions that are problematic but are not important because there is a very finite amount of
oxygen and carbon dioxide that is dissolved in the solvent. If performing the experiment without air completely, then oxygen and carbon dioxide would
be eliminated. The third competing reaction is when the Grignard reagent reacts with a halide to form a C–C bond that is not needed. ... Show more
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However, it can be minimized by having a larger concentration of magnesium than halide. With this, it will cause the halide to react with magnesium
instead of other atoms in the solution. When the halide encounters magnesium atoms, the formation of the Grignard reagent occurs. By adding a small
amount of halide, dropwise, at a time this lowers the ratio of halide to magnesium. The magnesium concentration decreases while the Grignard reagent
increases. The halide at the end of the reaction will mainly react with the Grignard reagent since most of the magnesium will be used in forming the
Grignard reagent. Therefore, the side reactions are preferred to occur at the end of the experiment, so the production of them are minimized as much as

Identify And Distinguish Between Ions In The Halide Family...
Andrea Moreno
9 February 2016
Block A
Calculations:
There are no calculations.
Analysis/ Critical Thinking Questions:
1. Explain what information you found from each of the first three (or four) parts of this experiment. Make separate clear, concise and grammatically
correct statements for each part. Someone else should be able to take your answers and do the unknowns successfully. Only fluoride precipitates with
calcium ion, not with silver. All three of the others precipitate with silver: chloride is white at first, then turns gray, bromide is sort of ivory colored
and iodide is yellow. Silver chloride is soluble in ammonia and in sodium thiosulfate solution. Silver bromide dissolves in ammonia, if you stir it, but
only a little bit in ... Show more content on Helpwriting.net ...
(The problem/objective, materials, procedures and safety notes above are taken from "World of Chemistry" by John Little, 2013)
Hypothesis: If the solutions react with a particular halide ion, then it will be able to identify each halide as an unknown and subsequently identify two
halide ions mixed together in solution.
Hypothesis Supported or Not Supported: This hypothesis is supported.
Data to Support/ Not Support the Hypothesis:
The first unknown substance could be fluoride because the reaction it had with Ag+ was almost the same reaction.
The second unknown substance is bromide because both of them had the same reaction of turning darker yellow.
Error Analysis:
There is no error analysis.
Improvements:
There are no improvements.
Major Scientific Principle Discussion: The scientific principle that was supported by the lab was to conduct tests on solutions of the four halide ions.
The tests consisted of comparing the responses of each of the halide ions to each test and to identify two unknowns based on the results obtained using
known solutions of the anions. This activity relates to informations from the textbook and class discussions because of the anion analysis we are
learning. The outcome of this activity connects to what occurs in the world around us because we can learn to make observations. This lab was

important, because of the small size, students

Relativities of Alkyl Halides in Nucleophilic Substitution...
Title: Relativities of Alkyl Halides in Nucleophilic Substitution Reactions
Introduction:
The purpose of this lab was to perform a comparison of relative reactivities of various alkyl halides with two different reagents, sodium iodine in
acetone and silver nitrate in ethanol. (Below are the reaction equations). We used different substrates, which were primary, secondary, and tertiary.
These substrates included 2–bromobutane, 2–bromo–2–methylpropane, 1–bromobutane в€
ћ–Bromotoluene, bromobenzene, and I–bromoadamantane.
This lab helped discover what kind of mechanisms (either SN1 or SN2) are involved in the performed reactions. 1. Sodium Iodine in Acetone:
RX + NaI –––––Acetone––––> RI + NaX (X=Br) 2. Silver Nitrate in ... Show more content on Helpwriting.net ...
| Bromocyclohexane| 2В°| RT: NO changeAfter heating: Evaporated slightly, made crystals, light yellow| SlowerSN1 reaction due to secondary carbon.|
1–bromoadamantane| 3В°| RT: No changeAfter Heating: Evaporated slightly, clear crystals formed| SN2 |
Sliver nitrate in ethanol reactions: Alkyl Halides| Classification of RX| Observations| Conclusion| 2–bromobutane| 2В°| RT: Precipitate and densely
cloudy.No heating required| Slower SN1 reaction due to secondary carbon.| 2–bromo–2–methylpropane| 3В°| RT: ClearAfter heating: No change| Fast
SN1 reaction due to it being a tertiary carbon.| 1–bromobutane| 1В°| RT: Precipitate and slightly cloudy.NO heating required| Heating required due to
primary carbon SN2 reaction| в€
ћ–Bromotoluene| 1В°Benzylic| RT: Precipitate and very cloudy.No heating required| Heating required due to primary
carbon SN2 reaction| Bromobenzene| 1В°Allylic Vinylic| RT: ClearAfter heating: No change| Heating required due to primary carbon SN2 reaction|
1–bromoadanabtane| 3В°| RT: Precipitate and cloudyNo heating required| Fast SN1 reaction due to it being a tertiary carbon.|
Conclusion:
In conclusion, out of the reactions for NaI in acetone, 1–bromobutane reacted the fastest, and did not require

Nucleophilic Preparation Of 1-Bromobutane Lab Report
Nucleophilic Addition of Bromine Discussion: The goal of this lab is to create 1–bromobutane through the nucleophilic addition of bromine to
1–butanol. Sulfuric acid is used on this lab to help protonate the alcohol, thereby transforming it into water thus making a better leaving group. The
bromide ion can then attack and allow for the formation of 1–bromobutane. The reaction starts through a reflux. A reflux allows for a mixture or
compound to be continuously heated and condensed, so that no product is lost to evaporation, while still providing enough activation energy for the
reaction to occur. The apparatus is simply a round bottom flask attached to a condenser with a continuous flow of water in and out. In this experiment
the reaction... Show more content on Helpwriting.net ...
The confirmation that 1–bromobutane was the final product obtained was confirmed by a infrared spectroscopy. The IR spectra produced by the
product being tested aligned with the expected functional groups of 1–bromobutane. A C–H stretch, and CH2 bend were observed at 2961.17 cm–1
and 1464.7 cm–1 respectively. The fingerprint region of the the IR spectra of the product being tested also aligned with the reference IR spectra for
1–bromobutane. However, the peak a 740.cm–1 had a larger transmittance or amount of light absorbed than expected which may be due to scattering
of the sample and the small presence impurities. However, all the expected wavelength of peaks in the fingerprint region were approximately the same
as the reference IR spectra for 1–bromobutane. Alkyl halide tests were performed to further confirmed that a halogenated product was produced. Both
Sodium iodide and silver nitrate tests formed precipitate thus confirming a halogenated product. Furthermore, the sodium iodide reaction with the
product occurred much more rapidly than the silver nitrate test, therefore, it can be assumed that the Sn2 route is preferred because the product has a
lack of steric hinderance. The combination of the IR spectra of the isolated product, which showed the expected functional of a 1–bromobutane sample,
as well as a fingerprint region that

Summary: Comparison Of The Behavior Of Halides
Comparison of the Behaviour of Halides Purpose: to compare the properties of halides and relate them to the position of halogens in the periodic
table. To use these properties to identify an unknown halide. Hypothesis: The halogens that are further down the group are more reactive than those
that are further up. Also, Unknown substance C will be sodium iodide. Materials: 4 test tubes 1 dropper 1 test tube rack Safety goggles Sodium
chloride Sodium bromide Potassium iodide Unknown substances A, B or C Hydrogen peroxide Ammonia silver nitrate Nitric acid Sodium thiosulfate
Potassium permanganate Method: Please refer to SCH3U Learning Guide 6– Periodic Table "Comparison of the Behaviour of Halides" worksheet....
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The silver nitrate disintegrates less effectively the fact that it is further down inside of the column. Chloride is the second from the most inside of the
column. At the point when the nitric acid and hydrogen peroxide was included there was not any precipitate. Iodide is at the base and there was a lot
of precipitate remaining. 2.The ease of oxidation from hydrogen peroxide builds the further down the element is in the column.Chloride is closer to the
highest point of the column and was still clear when the peroxide was

Conversion of Alcohol to Alkyl Halides
Introduction In this experiment, Conversion of Alcohol to Alkyl Halides and alcohol is converted to an alkyl halide through SN1 or an SN2
mechanism. This is done by using 1–propanol and 2–pentantol with HBr, Hydrobromic acid. Only half of the groups will use 1–propanol, and
2–pentantol. All results are analyzed using NMR and IR. AnSN1 reaction, requires two steps. The first step, using an alcohol as an reactant, is the
pronation of the –OH group from the R group. This produces a cation intermediate. The cation intermediate is attacked by the –Br group, from HBr.
This is the second step. An SN2 mechanism, is only one step, unimolecular , and spontaneous. In an alcohol, the –Br group will attack at the water,
H2O, is leaving the reaction. The R group attacks to the Bromine leaving an excess water. Figure 1 Mechanism for 2–Pentantol SN1
Figure 2 2–Pentantol SN2 Mechanism Nuclear Magnetic Resonance, NMR, and Infrared, IR, Spectroscopy are used to determine the structure of
unknown compounds. Nuclear Magnetic Resonance allows the student to see the nuclei in a molecule by the usage of light (1). The spectrum shows
how carbon can form a number of different bonds hydrogens present in the molecule, double and single. IR spectroscopy gives an idea of the
frequency of the molecules through the vibrations of the molecules. Infrared shows how molecules can perform like springs which connects with
Hooke's Law. Hooke's Law is used to describe the vibrations of springs.
Table

Grignard Reagent Lab Report
the purpose of this experiment is to prepare a Grignard reagent by reacting with alkyl or aryl halide and to ultimately react the Grignard reagent with
carbon dioxide in order to produce carboxylate. The formed carboxylate is then protonated with an acid to produce carboxylic acid that could be used
with liquid–liquid extraction to isolate the unknown acid from the other products from side reactions. The final unknown product is identified by
measuring the melting point and calculating the molecular weight obtained from titration.
Organometallic compounds are compounds that contain carbon–metal bonds (C–M bonds), in which carbon bears a partial negative charge because
metal is less electronegative than the carbon. This partial negative charge of the carbon atom allows it to be a good nucleophile that attacks the
electrophile to make a new carbon–carbon bond. There are several examples of organometallic compounds, such as organolithium, Gilman reagents,
and Grignard reagents (organomagnesium reagent). In this experiment, Grignard reagents are prepared and reacted with other electrophilic carbon to
form a new carbon–carbon bond. Victor Grignard discovered Grignard reagent around 1890s and received a Noble Prize in 1912 with his discoveries.
In this experiment, an alkyl halide or aryl halide is reacted with magnesium metal to prepare a Grignard reagent (R–MgX). In this Grignard reagent
preparation reaction, halide is typically used with bromine (sometimes with chlorine, not

Sn1 Reaction Lab Report
SN1 reactions only occur with certain reagents under very specific conditions. These reactions have an intermediate and occur in two distinct steps.
The first step is the formation of the carbocation once the leaving group departs from the molecule. The type of leaving group is vital in determining if
an SN1 reaction will occur. In general, the efficiency of the leaving group increases as the size of the halide ion increases. However, p–Toluensulfonate
is the most effective and reactive leaving group. As this step forms a carbocation, it is important to note that the type of carbocation that will form. SN1
reactions will not occur if the intermediate is not stable. Therefore, the degree of substitution of the leaving group, and consequently the carbocation, is
vital. ... Show more content on Helpwriting.net ...
Correspondingly, primary alkyl halides are the least likely to react. Additionally, the solvent has the ability to stabilize the carbocation. A polar protic
solvent is favored as it will form hydrogen bonds to further stabilize the carbocation intermediate. Overall, SN1 reactions are widely dependent on the
formation of a stable carbocation intermediate. Since SN1 reactions are dependent on the formation of the carbocation, and that step occurs separately
from the nucleophilic addition, the rate of the reaction is not typically affected by the nature or concentration of the nucleophile. The mechanism of an
SN1 reaction is consistent. First, the leaving group leaves forming the carbocation. The intermediate carbocation is planar and achiral. The nucleophile
is then free to attack from either side of the carbocation, which forms the substituted product in either a racemic mixture or with some form of

2-Methylbutane Lab
In this lab, concentrated HCl was added to 2–methyl–2–butanol in a separatory funnel. The reactants were shaken to help the reaction occur quickly.
The separatory funnel was vented several times, to reduce pressure that resulted from the reaction which was due to the high vapor pressure of HCl.
The Hydrochloric acid will dissociate, and the H+ will react with the OH–C to form H2O–C. The H2O is a good leaving group. As the LG is on a
tertiary carbon, an Sn1 reaction should occur. In the rate determining step, the water will leave, to produce a tertiary carbocation. In the second (fast)
step, the Cl– nucleophile from the dissociated HCl will attack the carbocation and 2–chloro–2–mehtylbutane forms. In this reaction, some E1 also
occurs; however, HCl will react with the product of E1 to produce the 2–chloro–2–methylbutane product. In the lab, after the reaction occurred, the
product needed to be purified as excess HCl was used and still remained. To purify the product, a series of reactions were completed.... Show more
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The percent recovery was calculated by finding the moles of the initial and final products. The moles of the initial product were found by multiplying
the volume used by the density of 2–methyl–2–butanol, and then dividing by the molar mass. The percent recovery was good, showing few errors
were made and that the lab went as expected. To improve the recovery, the extractions could be done even more carefully. To analyze the purity of the
product, an IR test was run. 2–methyl–2–butanol has a large OH peak; 2–chloro–2–butane does not, but otherwise looks quite similar. The main peak
on 2–chloro–2–butane is a C–Cl peak; however, this peak is not revealed with this IR machine, so it could not be analyzed. The IR chart for the
experimental product did not have the OH peak and looked like the theoretical IR chart, meaning it was dry and that the desired product had been

Reactivities Of Some Alkyl Halides
Karen Dennie
Professor Tjandra
Chemistry 75A
Due: November 18, 2013
Experiment 20: Reactivities of Some Alkyl Halides
Purpose: The purpose of this experiment is to examine the reactivities of various alkyl halides under both SN2 and SN1 reaction conditions. The alkyl
halides will be examined based on the substrate types and solvent the reaction takes place in.
Procedure: For the first part of this experiment, six dry test tubes were obtained and labeled accordingly to test the following halides: 2–chlorobutane,
2–bromobutane, 1–chlorobutane, 1–bromobutane, 2–chloro–2–methylpropane, and bromobenzene. To each of the six test tubes 2ml of 15% sodium
iodide in acetone was added. 4 drops of the appropriate halide was added to the test tube labeled for that specific halide. After adding the halide, the
test tube was then shaken to mix thoroughly. If a precipitate formed the time it took was recorded. Since none of the solutions formed a precipitate at
room temperature after five minutes, the test tubes were placed inside of a hot bath at about 50В°C. After one minute, the test tubes were taken out
of the hot bath and allowed to cool. If any test tubes formed a precipitate, the time it took was recorded on a table. For the second part of the
experiment, again six dry test tubes were obtained and labeled accordingly to test the following halides: 2–chlorobutane, 2–bromobutane,
1–chlorobutane, 1–bromobutane, 2–chloro–2–methylpropane, and bromobenzene. To each of the six

Alkyl Halides
The synthesis of alkyl halides from alcohol is the basis for this experiment, providing reactions with interesting contrast in mechanisms. Not only
synthesizing, but extracting is another important procedure that involves quick actions and judgement, when removing unnecessary layers in a
separatory funnel. This allows us to learn and grasp more of an understanding between organic compounds in the laboratory.
Experimental To begin the experiment, a 125 mL separatory funnel is needed and a gathered amount of t–pentyl chloride at 10.0 mL should be inserted
into the funnel. 20.0 mL of concentrated HCl (Hydrochloric acid) was gathered and also inserted into the separatory funnel with the 10.0 mL of t–pentyl
chloride. A diagram of separatory funnel and its indicated parts will be shown in the following:
[Figure 1]
Once added, a reaction should occur immediately and a large amount of gas should be formed and visible. Begin to swirl in order for the mixture to
be fully mixed and cause all the immediate gas to be formed and released with the stopper taken off. After a minute or so, ... Show more content on
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Analytically, both graphs have a peak forming at ~less than 3000 cm–1, which indicates an Sp3 C–H stretch in both tested compounds. Another peak
both IR spectrum graphs share is a peak formed at ~800–700 cm–1, which indicates a C–Cl bond in both tested compounds. The last peak both IR
spectrums graph show is a peak formed at ~1500–1450 cm–1, which indicate both an –CH3 bend and –CH2– bend. Based off this analysis, this
concludes that the pure t–pentyl chloride gathered through our experiment is indeed t–pentyl chloride, based off comparing the IR spectrums of the
actual compound, t–pentyl chloride, shown in the

Sn2 Lab Report
As we knew from the background that these reactions are depend on several factors such as, the substrate which is the most priority, the solvent,
and the nucleophile. For the substrate, the SN1 is favor tertiary alkyl halide, while the SN2 is favor the primary alkyl halide. For the solvent, it is an
important factor. For example, in this experiment, we used AgNO3 in ethanol and NaI in acetone solvent. We used AgNO3 in ethanol in SN1
because the ethanol is protic polar and will able to stabilize the intermediate state of the SN1. For SN2, we used NaI. The nucleophilic specie in this
reaction is the Iodine. The acetone will act to not allow for the intermediate state to form in SN2. Depending on the observation of the reaction, some
compounds are reacting in SN2 and... Show more content on Helpwriting.net ...
This compound is not very reactive in SN2. The reason the reaction happens is because the Iodine is strong nucleophilic. viii. Bromocyclohexane:
In SN1, it started to form participation after 5 second and its complete that participation in 7 minutes in room temperature. This compound is
reactive in SN1. The reason is the bromine which is a good leaving group, and the nitrate which a weak nucleophilic. In SN2, Bromocyclohexane
didn't form a participation. That's mean this compound is not reactive in SN2. There reason is the ring which cause steric hindrance. ix. Benzyl
Chloride: In SN1, it started to form participation after 12 second and its complete that participation in 2 min and 18 sec in room temperature. This
compound is reactive in SN1. The reason is the chlorine which is a good leaving group, and the nitrate which a weak nucleophilic. Also, the
carbocation will be more stable due to resonance. In SN2, it formed a participation after 10 sec and complete formation after 1 minute and 5 sec.
That's mean this compound is reactive in SN2. There reason it is a primary alkyl halide. The carbocation will be a primary

Chemical Properties Of Unknown White Compound
Abstract Unknown white compound (823U) was discovered in the lab. In order to dispose of it correctly, the substance and its physical and chemical
properties had to be identified. The unknown white compound was one of a list of 15 compounds. 5g of the unknown compound were given in order
to correctly identify and discover its physical and chemical properties. In order to do so, a solubility test, a flame test, and ion tests were conducted.
From the results of these initial tests and the given list of compounds, the unknown white compound was thought to be composed of sodium and a
halide (I–, Br–, or Cl–). Of the list, NaCl was the appropriate compound, however NaC2H3O3 was also tested out of skepticism. To verify the identity
of the substance, the solubility and flame tests were performed again along with a pH test. The pH tests of NaCl and NaC2H3O2 did not match that of
the unknown white compound. The list of compounds had been entirely ruled out. The identity of the unknown white compound was revealed to be
calcium chloride. To synthesize at least a gram (calculated to produce 1.2g) of CaCl2, the following reaction was completed.
2 HCl + CaCO3+heat –––> CaCl2 + CO2 + H2O
To verify that CaCl2 was the compound that had been synthesized, the compound was tested with the flame test, solubility test, halide ion test, and pH
test. CaCl2 was confirmed to be the synthesized compound as the results of these tests matched the results of the unknown white compound.Introduction
When

Grignard Reagent From An Unknown Aryl Halide
7) Discussion:
In this experiment, the goal is to prepare a Grignard reagent from an unknown aryl halide and identify the identity of the aryl halide by converting it to
a carboxylic acid to determine its melting point and molar mass (determined by titration). The experiment began by dissolving 0.25g of magnesium
powder in a 25mL round–bottom with 5mL of anhydrous ether and stirring with a stir bar. Then the round–bottom flask was set up for reflux using a
Claisen adapter where the vertical part is covered with a septum to prevent air from mixed with the solution. The septum is very important because the
Grignard product can react with oxygen to produces a carboxylic acid, which is not wanted. Also, the choice of anhydrous solvent is important because
the Grignard product can react with water to produce an alkane. With the reflux set up, the next step was to add the halide. 1.2mL of the unknown
bromoarene mixed with 2.5mL ether was slowly added dropwise through the septum using the needle and syringe. The bromoarene had to be added
slowly because there Grignard product would undergo another unwanted side reaction by reacting with the unknown bromoarene. The product with be
a new carbon–carbon bond between the unknown bromoarene and the Grignard product. If the bromoarene is added slowly, the chances of the
Grignard product reacting with the bromoarene over the magnesium is low because magnesium exists in larger concentration in the solution. Once all
the unknown bromoarene

Introduction Of A Primary Alcohol
Substitution
5. Introduction In this experiment, a primary alcohol was converted into a primary bromoalkane using hydrobromic acid. The reaction was done under
reflux and then distilled to obtain a product of higher purity. The degree of the alkyl halide obtained from the experiment was tested with silver nitrate
and sodium iodide. An infrared (IR) spectra and the weight of the product were obtained for further analysis. The IR gave information on the present
functional groups and product weight was used to calculate the percent yield.
6. Data and Results The product obtained after reflux was a yellow liquid. With the halide tests, the product reacted faster with sodium iodide to form a
precipitate than with silver nitrate. This indicates that the degree of the product was primary and that bromine did replace the alcohol group. The IR of
the product showed four peaks at 3326.36, 2932.86, 1465.34, and 1028.73. These peaks indicate O–H, C–H, C–H3, and C–O bonds, respectively,
within the product. 2.032 grams of product were recovered from the reaction. With this, a percent yield of 27.14% was calculated. 7. Discussion and
Conclusion Nucleophilic aliphatic substitution is the replacement of one group for another at a saturated, sp3–hybridized carbon atom. This process is
often used to interconvert functional groups, such as in the preparation of alkyl halides. In these reactions, nucleophiles attack the carbon atom–which
the electronegative leaving group breaks its bond

Experiment 6 Alkyl Halides
In day 2 of the experiment in step 1, two millitliters of 2% silver nitrate solution in ethanol were placed insix test tubes. Two drops of the six alkyl
halides were added to the test tubes. The alkyl halides included, n–butyl chloride, sec–butyl chloride, t–butyl chloride, n–butyl bromide, allyl chloride,
and chlorobenzene. Once the alkyl halides were added, the contents were swirled, and left alone to react. The t–butyl chloride, n–butyl bromide, allyl
chloride, and chlorobenzene all reacted and precipitate was formed. It was observed that it took the t–butyl chloride 30 seconds to react and form a
precipitate. It was observed that it took the n–butyl bromide four minutes to react, the allyl chloride 1 minute to react, and the chlorobenzene

Chemistry Of Film
Chemistry of Film Exposure and Development
Film is composed of an acetate base coated with a photographic emulsion layer. The base supports the emulsion, which consists of small silver halide
crystals suspended in gelatin. The crystals are Silver–Iodobromide (silver bromide with added silver iodide). (Earle L. Kitts) When the emulsion is
struck by light, the silver halide is converted to solid silver. (Janice Clifton) A crystal absorbs a photon of light, with energy equal to Planck's constant
times the frequency of the light, liberating an electron from the ion. This leads to the formation of a photoelectron and neutral bromine atom. (Earle L.
Kitts) The photoelectron combines with a silver ion to form a neutral metallic silver atom. (Earle ... Show more content on Helpwriting.net ...
First, the unprocessed negatives are treated with developer, an agent that turns silver halide to solid silver. (Janice Clifton) The most common
developing agents are phenidone and hydroquinone. The active agents have a high pH, causing the gelatin to swell and allowing for a more rapid
penetration throughout the emulsion. (Earle L. Kitts) The bromide in the gelatin acts as a development restrainer, preventing the development of
unexposed silver bromide grains that would cause fog. (Earle L. Kitts) Development is a chemical reaction, so time and temperature must be
regulated. (Janice Clifton) After development, some photosensitive silver halide crystals remain. (Janice Clifton) A stop bath of Acetic acid
neutralizes the developer with no effect on the silver halide. The developed negatives are treated with a fixer solution of high concentration sodium
hyposulfite to dissolve the unused silver halide.(Earle L. Kitts)(Janice Clifton) Once the film is removed from the fixer, it can be exposed to regular
light since only solid silver negative remains. A hypo–cleaning agent is then used to neutralize fixer on the film. This step is optional however the
alternative is to run the film under water for an hour as opposed to 6 minutes. Photoflo, a wetting agent, is then added to break down the surface
tension of water droplets so the film dries spot–free, similarly to dishwasher detergent with glass. (Janice

Alkyl Halides
In this lab, the experimenters will determine how the structure of an alkyl halide (i.e. methyl, primary, secondary, or tertiary), steric effects, nature of a
leaving group, and solvent polarity affect the relative rates of SN1 and SN2 reactions. In addition, comparing the rates of reactions under varying
concentrations of alkyl halides and nucleophiles will help determine the rate laws for both types of nucleophilic substitution reactions. During SN2
reactions, a good nucleophile, like iodide, will displace the leaving group on an alkyl halide in a single step that results in an inversion of
configuration. Methyl, primary, and some secondary halides undergo this bimolecular substitution reaction. Minimal steric effects combined with a
good leaving group and a polar, aprotic solvent ... Show more content on Helpwriting.net ...
The more stable the resulting carbocation, the quicker this step occurs. After this rate–limiting step, rapid reaction with a weak nucleophile (e.g.
ethanol) attacking from either side of the carbocation completes the substitution reaction. A racemic mixture forms with a slight favoring of the
inverted molecule because of ion pair formation. A good leaving group capable of delocalizing a negative charge aids in the formation of the initial
carbocation. Additionally, polar, protic solvents capable of solvating and stabilizing a carbocation support the SN1 mechanism. Tertiary alkyl halides
provide the ideal substrate for the formation of a stable tertiary carbocation that favors the SN1 mechanisms. When testing the factors affecting the SN2
reaction below, the experimenters will achieve the fastest reaction when using a primary alkyl halide with minimal branching and the best leaving
group (i.e. 1–bromobutane). Varying the concentrations of the alkyl halide or nucleophile while keeping the other constant will change the rate of
reaction for the SN2 reaction. The fastest SN1 reaction will occur for the tertiary alkyl halide with the strongest leaving group reacted in pure

Conversions of Alcohols to Alkyl Halides
Conversions of Alcohols to Alkyl Halides: 1–Propanol and 2–Pentanol Introduction One way scientist gets alkyl halides is by using the manipulation of
an alcohol. When alcohols are treated with HBr or HCl; they can undergo a nucleophile substation reaction to generate an alkyl halide and water2.
Using the structure of the alcohols they are able to use SN1 or SN2 mechanisms. For both these mechanisms though, the –OH group must be pronated
shown in Figure 1. R–OH + H–Br+ R–OH2 +Br– Figure 1. This figure is the protonation of the –OH group from the alcohol SN1 is a limiting
mechanism that is a unimolecular nucleophic substation reaction. In this mechanism it involves two steps, one in which the leaving group leaves and
then forms a carbocation intermediate, shown in figure 2. Then it is able to break bonds between carbon and making it able for carbon to leave the
group before the bond forming with nucleophile begins1. R– ––OH2R+ + H2O Br– Figure 2.In this figure it shows the cation intermediate that is
forming and the attack of Br– resulting in the production of RBr While the mechanisms for SN2 is a biomolecular reaction. These two species are
involved in the rate–determine step3. Once the nucleophile attacks, though it is allowed to form a nucleophile, that then forms a transition state and
then results in the final product, as shown in Figure 3. + R– ––OH2 RBr +H2O Br– Figure 3. Figure 3 shows the attacking of Br– and water

An Unknown Grignard Reagent Experiment Essay
Grignard
5. Introduction In this experiment, an unknown Grignard reagent was prepared from an aryl halide. The unknown reagent was then reacted with
carbon dioxide to form a carboxylic acid. The solid acid was then isolated and recrystallized before the melting point was taken. The precipitate was then
dissolved in water and titrated to determine the molecular weight. The melting point and molecular weight were then used to determine the unknown
acid obtained from the experiment.
6. Data and Results The product obtained had a melting point of approximately 107 В°C and a weight of .324 grams. Some of the product would not
dissolve in water and so was removed through vacuum filtration, which left .141 g not dissolved in solution. It took 13.2 mL of sodium hydroxide to
turn the solution of the product dissolved in water pink. A molecular weight of 138.63 g/mol was calculated from the data. These results indicate that
the product was 2–methylbenzoic acid, the Grignard reagent was 2–methylphenylmagnesium bromide, and the unknown bromide solution was
2–methylbromobenzene. Calculations showed that the limiting reagent of the Grignard preparation was magnesium and that the experiment had a 23.13
% yield. 7. Discussion and Conclusion Organometallic compounds, such as Grignard reagents, are molecules containing carbon–metal bonds and are
often used to create new carbon–carbon bonds. Grignard reagents–or organomagnesims– are specifically those that have a carbon–magnesium bond.

Alcohol and Ir Spectrum
Conversion of Alcohols to Alkyl Halides
Title: Conversion of Alcohols to Alkyl Halides
Abstract: In this experiment the conversion of alcohols to alkyl halides are investigated through reflux and simple distillation. These are common
procedures used to separate substances. After the reflux and distillation is complete 13C NMR and IR spectrum is used to identify the product or
products for each reaction: 1a, 1b, and 2. Every individual in the group was assigned either 1a (1–propanol) or 1b (2–pentanol), and 2
(1,4–dimethyl–3–pentanol). The purpose of this experiment was to understand and become familiar with the reaction mechanisms and be able to
observe and compare the product or products for each of the reactions using 13C NMR and IR. ... Show more content on Helpwriting.net ...
To prevent the escape of organic vapors, the reaction mixture is cooled with an ice bath before removing the condenser. The next technique used in
this experiment was simple distillation. This is a physical separation of the components of the mixture. This technique is accomplished once the drip
rate of the product into the collection vessel diminishes considerably. After the reflux and distillation is complete 13C NMR and IR is used to
identify the product or products for each reaction. 13C NMR is used to observe the carbon skeleton of an organic molecule. Analysis of this spectrum
allows certain stretches to be observed. An IR spectrum has energy measured as frequency recorded on a horizontal axis and intensity of the absorption
on the vertical axis. Analysis of the IR allows us to differentiate between certain characteristics and functional groups in organic chemistry.
Results:
1–propanol–1a
Mass of product=1.39
Theoretical yield=5.079g
% Yield=mass of product/theoretical yield x 100
=1.395g/5.079 x 100
% Yield = 27.5%
13C NMR and IR spectrum results for 1a (1–propanol) 13C NMR peaks| 13C NMR descriptions| 11.993| 1бµ’ alkane –CH3| 25.589| 2бµ’ alkane
–CH2| 34.391| 2бµ’ alkane –CH2|
IR peaks| IR descriptions | 2921.38| C–H stretch| 2360.59 & 2341.06 | N/A| 667.94 & 571.62 & 559.56 &565.37| Fingerprint area|
* The final result and

Formal Report
Experiment 6: Synthesis of an Alkyl Halide
Maria Alexandria Buraga Ammuyutan
Institute of Chemistry, University of the Philippines, Diliman, Quezon City 1101 Philippines
–––––––––––––––––––––––––––––––––––––––––––––––––
Department of Food and Science Nutrition, College of Home Economics, Univeristy of the Philippines, Diliman, Quezon City 1101 Philippines
–––––––––––––––––––––––––––––––––––––––––––––––––
ABSTRACT
–––––––––––––––––––––––––––––––––––––––––––––––––
Alcohols react with hydrogen halides (HCl is used in this experiment) to yield the resultant alkyl halides and water. The insolubility of the alkyl halide
in water allows the separation of it from the aqueous layer using a separatory funnel. The alkyl halide, then, were purified ... Show more content on
Helpwriting.net ...
The experiment was accomplished.
References
[1] Chapter 4 : Alkyl Halides. (n.d.). TED Ankara Koleji KГјtГјphane ve Bilgi Merkezi. Retrieved January 1, 2012, from http://library.tedankara.k12.tr
/carey/ch4–2.html [2] What Is The Use Of Boiling Chips In The Distillation Process?. (n.d.). Ask Questions, Get Free Answers– Blurtit. Retrieved
January 2, 2012, from http://www.blurtit.com/q803268.html [3] Nuffield AdvancedChemistry – In the synthesis of an alkyl halide experiment (n.d.).
Nuffield Advanced Chemistry – Home page. Retrieved January 2, 2012, from http://www.chemistry–react.org/go/Faq/Faq_29604.html [4] Organic
Chemistry Laboratory Manual.University of the Philippines Diliman, QuezonCity. 2008 ed.
Appendices 1. Why is it necessary to use cold concentrated HCl? Why is it added in excess?
It is important to use cold concentrated HCl because keeping the temperature of the acid low will increase the fraction of the SN1 product so to
increase the yield. Keeping it cold will help prevent loss of the volatile product after it is formed.
2. Why is solid NaHCO3 used instead of aqueous NaHCO3?
It is the primary standard for acid–base titrations because it is solid and air–stable,

Cyclohexane Reaction Lab Report
Aim: The experiment aims to compare the reactivity of halogens by studying their displacement by one another.
Prediction (P1): The more reactive halogens will displace the more reactive halides from their ion solutions. As you go down the group the reactivity of
the halogens decrease. Chlorine will displace bromide and iodide ions, Bromine will displace iodide ions but not chloride ions, and Iodine will not
displace either bromide or chloride ions from the solution due to the order of their reactivity.
Materials: (A list of all the materials you will use)
Sodium Hypochlorite
Potassium iodide solution (1 mol/litre)
Bromine water (1 mol/litre)
Cyclohexane Iodine–potassium iodide solution (dissolve 20g of KI and ... Show more content on Helpwriting.net ...
The aqueous/denser layer settles at the bottom. Leaving the less dense cyclohexane on the top layer.
2.What was the purpose of cyclohexane in this practical?
Halogens are not very soluble in water because they are non–polar, but are soluble in cyclohexane. Using solvent extraction with cyclohexane, halogens
can be separated from the halide ions allowing them to be easily identified.
Conclusions The general reactivity of halogens and their ability to form ionic compounds decreases down the group. This is influenced by increased
size, electronegativity, and nuclear charge; therefore the ease with which electrons are accepted to form an ion also decreases. As observed from the
results of the experiment, the reactivity orders of the Halogens are; F2>Cl2 > Br2 > I2
References
1. Alyea, H.Gorman, R. (1969). Group VII. The Halogens. E. Iodine. J. Chem. Educ., 46 (6), p.A451. American Chemical Society (ACS) [Online].

Available at: doi:10.1021/ed046pa451.1.
2. Anon. (2016). BBC – GCSE Bitesize Science – Trends within the periodic table : Revision, Page 6. [Online]. Available at: http://www.bbc.co.uk
/schools/gcsebitesize/science/triple_aqa/periodic_table/trends_periodic_table/revision/6/ [Accessed: 8 May

Grignard Reagent Lab Report
The purpose of this experiment is to prepare a Grignard reagent by reacting with alkyl or aryl halide and to ultimately react the Grignard reagent with
carbon dioxide in order to produce carboxylate. The formed carboxylate is then protonated with an acid to produce carboxylic acid that could be used
with liquid–liquid extraction to isolate the unknown acid from the other products from side reactions. The final unknown product is identified by
measuring the melting point and calculating the molecular weight obtained from titration.
Organometallic compounds are compounds that contain carbon–metal bonds (C–M bonds), in which carbon bears a partial negative charge because
metal is less electronegative than the carbon. This partial negative charge of the carbon atom allows it to be a good nucleophile that attacks the
electrophile to make a new carbon–carbon bond. There are several examples of organometallic compounds, such as organolithium, Gilman reagents,
and Grignard reagents (organomagnesium reagent). In this experiment, Grignard reagents are prepared and reacted with other electrophilic carbon to
form a new carbon–carbon bond. Victor Grignard discovered Grignard reagent around 1890s and received a Noble Prize in 1912 with his discoveries.
In this experiment, an alkyl halide or aryl halide is reacted with magnesium metal to prepare a Grignard reagent (R–MgX). In this Grignard reagent
preparation reaction, halide is typically used with bromine (sometimes with chlorine, not

Grignard Essay
Introduction:
Grignard was the child of a sail producer. In the wake of concentrating on arithmetic at Lyon he exchanged to science and found the manufactured
response bearing his name (the Grignard response) in 1900. In 1909, he assumed responsibility of the Department of Organic Chemistry at Nancy and
was granted the Nobel Prize in Chemistry in 1912. At the beginning of the First World War, he studied chemical warfare agents, especially the produce
of phosgene and the identification of mustard gas. (2) The Grignard reagent is exceptionally responsive and responds with most natural mixes. It
likewise responds with water, carbon ... Show more content on Helpwriting.net ...
In the principle, you get an augmentation of the Grignard reagent to the carbon dioxide. (1)
Dry carbon dioxide is bubbled through a reply of the Grignard reagent in ethoxyethane, made as depicted already. (1)
For instance: The item is then hydrolyzed (responded with water) inside observing a weaken dangerous. By and large, you would join weaken
sulphuric ruinous or cripple hydrochloric damaging to the game–plan encompassed by the response with the CO2. (1) A carboxylic acid is conveyed
with one more carbon than the main Grignard reagent.
The usually quoted equation is: All sources cite the arrangement of a basic halide, for instance, Mg (OH) Br as the other result of the reaction. That is
truly tricky in light of the way that these blends react with dilute acids. What you wind up with would be a blend of a mixture of ordinary hydrated
magnesium particles, halide particles and sulfate or chloride ions – depending upon which weaken acid you added. (1)
–Reaction with carbonyl compounds:–
The responses between the various sorts of carbonyl blends and Grignard reagents can look uncommonly caught, However truth be told they all
respond similarly – every one of that movements are the get–together connected with the carbon–oxygen twofold bond.

Free-Radical Chain Reaction Lab Report
Free–radical chain reactions involve the formation of halides and alkyl halides by reacting diatomic halogens with reactive hydrogens attached to
hydrocarbons. In this experiment, diatomic bromine was reacted with various arenes to produce hydrobromic acid and alkyl halides. The mechanism
behind this reaction can be characterized by three distinct phases: initiation, propagation, and termination. During the initiation step, bromine radicals
are produced via thermal or photochemical homolysis. These bromine radicals then react with hydrogens attached to a hydrocarbon in the propagation
step to produce hydrobromic acid and a carbon radical. The chain reaction continues since the carbon radical formed can react with another diatomic
bromine molecule, producing a carbon–bromine bond and regenerating a bromine radical. The termination step ends the reaction by reacting two
bromine radicals with each other, lowering the concentration of highly reactive bromine radicals in solution.... Show more content on Helpwriting.net ...
For example, when propane and diatomic bromine are reacted together, a 3:97 ratio of primary alkyl halide to secondary alkyl halide is produced. In
contrast, when propane and diatomic chlorine are reacted together, a theoretical 3:1 ratio of primary alkyl halide to secondary alkyl halide is produced;
experimentally, the ratio is closer to 45:55. Thus, it was determined that diatomic bromine was better suited to examine the reaction rates of the various

S. N. 1 Lab Report
As discussed in lecture, the determining factors of a substitution reaction is likely to occur in either S_N 1 or S_N 2 by: substrates, leaving group,
strength of the nucleophile, and the type of solvent.
When conducting the S_N 1 portion of the experiment, the solvent used was 1% silver nitrate in ethanol solution (гЂ–AgNOгЂ—_3). This solvent is
polar protic. This is the best type of solvent needed for an S_N 1 reaction because the hydrogen atom is positively charged while the nucleophile is
negatively charged achieving a hydrogen bond. Two factors that contributed to the rate of reaction for S_N 1 are stability of the carbocation and the
leaving group departure. The more stable the carbocation, the faster the reaction generates and the better the leaving group, it creates a carbocation
faster, which in turn leads to a ... Show more content on Helpwriting.net ...
The following six of the nine alkyl halides produced precipitates at room temperature: 2–bromobutane, t–butyl chloride, crotyl chloride,
bromocyclohexane, bromocyclopentane, and benzyl chloride. All of these alkyl halides reacted in room temperature because they were either
secondary or tertiary carbons except for crotyl chloride. These types of structures (tertiary) are favored in the case of S_N 1 while secondary can
undergo either S_N 1 or S_N 2. The only primary carbon that reacted quickly was crotyl chloride because it is an allylic substrate, which can undergo
S_N 1. On the other hand the three remaining alkyl halides did not react at room temperature: 1–chlorobutane, 2–chlorobutane, and bromobenzene.
These three alkyl halides were then heated at 80в„ѓ. 1–chlorobutane's alpha carbon is primary, so heat is needed in order to create more collision
resulting in a reaction. 2–chlorobutane and 2–bromobutane have similar structures, but the latter reacted in room temperature because Br is a better
leaving group than Cl. Lastly, bromobenzene did not react even after adding heat

Taking a Look at Photographic Film
Photographic film A film can be defined as a material which is chemically reactive and when exposed to a light source records a fixed or still
image. In other words it helps in capturing the image of a the photographed object, formed from the light reflecting back from its surface.
Technically it is a photographic material consisting of a celluloid base that is covered with a photographic emulsion that is later used to make the
negatives or the transparencies that maybe contained inside a roll, or maybe a cassette or a cartridge. A film, when developed will produce
transparent negatives. A negative film is a photographic image that is formed on a narrow transparent strip of sheet whose surface is generally
made up of plastic or glass. It is a type of photographic image that shows the bright areas of the subject photographed as dark and the dark areas of the
subject as light. A negative film can be found in all formats, and can be black and white or colour. A negative can produce a positive image when light
is made to pass through it. A positive image is actually a normal image while a negative image is a total inversion were the light areas of the picture
becomes dark and the dark as light. Hence a positive is the final image. A positive can also be made directly from a film, like transparencies ( were a
film is processed to give a positive image directly rather than to give a negative ) what is it made up of film is actually a light sensitive emulsion on top
of a

Organometallic Reactions : Identification Of An Unknown...
LAB 7: ORGANOMETALLIC REACTIONS: IDENTIFICATION OF AN UNKNOWN BROMIDE (Preparative) Introduction The first purpose of the
lab was to prepare an unknown organomagnesium bromide, an organometallic reagent, reacting an unknown aryl bromide with magnesium in
anhydrous ether. The unknown was chosen from a predetermined list of benzoic acid derivatives with varying molecular weights and melting points
(see Supplement C). The second purpose of this lab was to prepare an unknown carboxylic acid by reacting the unknown aryl–magnesium bromide
with carbon dioxide and diethyl ether then protonating.The third purpose of this lab was to determine the neutralization equivalence point of the
unknown carboxylic acid by titrating with sodium hydroxide. The fourth purpose of this lab was to ascertain the identity of the unknown carboxylic
acid, and thus the original unknown aryl bromide, using its molecular weight determined from neutralization and melting point. Data and Results
Compound Molecular Weight (g/mol) Melting Point (В°C) Unknown Carboxylic Acid (R––COOH) 121.18 116–119 Discussion Organometallic
reagents are compounds with carbon–metal (R––M) bond. In the carbon–metal bond, carbon is more electronegative than the metal atom which creates
a dipole moment where carbon possesses a partial negative charge and the metal atom possessing a partial positive charge (RОґ–––MОґ+). The partial
negative charge on the carbon allows it to act as a strong nucleophile or base similar to

Alkyl Halide Lab
Elimination Reactions of Alkyl Halides Purpose The purpose of this lab is to understand the process of eliminating an alkyl halide to form an alkene.
The experiment is carried out by first converting the alcohol, 2–methy–2–butanol, into the alkyl halide of 2–chloro–2–methylbutane that will then be
put through dehydrohalogenation that favors elimination reaction (E2) to create a mixture of 2–methyl–2–butene by alcoholic potassium hydroxide
(KOHвЃ») base and 2–methyl–1–butene by potassium tert–Butoxide (Kt–BuOвЃ») base. A fractional distillation will be taken to purify the mixture
and an additional gas chromatography will be done to further analyze the mixture composition. A bromide test will be done to determine the production
of an alkene in the... Show more content on Helpwriting.net ...
The group decide to carry out the elimination reaction with alcoholic potassium hydroxide. A dry 100 milliliter round–bottom flask was filled with 5
grams of potassium hydroxide and 50 milliliters of absolute ethanol. The flask was stoppered and swirled for no more than 10 minutes to have the
potassium hydroxide mostly dissolved. Then 5.3 grams of the 2–chloro–2–methylbutane was added along with a stir bar. A fractional distillation was
set up to run the reaction. A Hempel column served first as a reflux condenser while the elimination reaction was carried out. The Hempel column was
attached with the round–bottom flask while being lubricated with silicone grease to not allow the joints to freeze. Water hoses were attached to the
Hempel column and water was circulated from bottom to top during the reflux period. An additional rubber hoses were connected to the vacuum
adapter that was also attached to the receiving flask (50 milliliter round–bottom flask). The reaction was heated for 30 minutes for a gentle reflux. A tan
Variac was set to 55 (rate of heating) to allow for the vapor condensation front to be about half–way up the Hempel column. After the reflux period,
then distillation was started by connecting the water hose on the vacuum adapter to the water to circulate during the fractional distillation. The
distillation was run until

Relative Reactivity Of Alkyl Halides
Relative Reactivity of Alkyl Halides in Nucleophilic Substitution Reactions
Charlie Doyle
Madison McGough
Annie Chang
Introduction
Both Sn1 and Sn2 reactions are nucleophilic substitution reactions, though they are slightly different. Sn2 reactions have bimolecular displacement and
are also concerted, meaning the bond making and the bond breaking processes happen in one step.1 Sn1 reactions require two steps and have
unimolecular displacement. This difference can be seen when comparing Figure 1 and Figure 2 below. The strength of thenucleophile does not effect
Sn1 reactions, while the strongest nucleophile is required for Sn2 substitution reactions.2 Other important considerations include the effect of ... Show
more content on Helpwriting.net ...
For Part A: five drops of 2–bromo–2–methylpropane were pipetted into test tube 1; five drops of 2–bromobutane were pipetted into test tube two; five
drops of 1–bromobutane were pipetted into test tube three; and five drops of 1–chlorobutane were pipetted into test tube four and each were labeled
accordingly. Then, twenty drops of a 15% solution of sodium iodide (NaI) in acetone (for Sn2 reactions) was added to all four test tubes, shaking
each once to mix contents, and the time from when the first drop hit to when cloudiness or a precipitant formed was recorded. The solutions were
then disposed of in the appropriate waste container. For Part B, the test tubes were cleaned and allowed to dry. The process of pipetting the alkyl
halides into the individual four test tubes was repeated. 20 drops of a 1% solution of silver nitrate (AgNO3) in ethanol (for Sn1 reactions) was added to
each tube, shaking each once to mix contents, and again recording the

Reactivity Of Tert-Butyl Chloride Lab Report
Courtney Copeland
Alexis Madrazo
TA: Katrinah Tirado
October 10, 2017
Synthesis and Reactivity of tert–butyl Chloride via an Sn1 Reaction
Copeland 1
Introduction/Background
Substitution reactions are important chemical processes that contribute to the production of new compounds. Simplistically, these reactions take place
through a series of steps in which one functional group is replaced by another (March). There are two types of nucleophilic substitution reactions,
first–order and second–order, but this experiment only involves the Sn1 first–order reactions. Sn1 reactions are considered unimolecular meaning that
only one molecule is involved in the rate determining step, the slowest step of the reaction which determines the overall speed of the reaction. In
contrast, Sn2 reactions are considered bimolecular, and complete the substitution reaction in one step. The main components of these reactions are the
nucleophile and the leaving group; a nucleophile replaces the leaving group by donating its electrons to form a new bond to the carbon (Weldegirma).
For this experiment, the nucleophiles and leaving groups of the Sn1 reactions are alkyl halides and alcohols respectively. When a hydrogen atom is
replaced by halogen in an alkane, the resulting compound is referred to as a alkyl halide. There are certain factors that affect both Sn1 and Sn2
reactions which include the structure of the substrate, and the concentration and reactivity of the nucleophile (Sn2 only). If a

Synthesis Tert-Butyl Chloride Sn2 Report
In order to synthesis tert–butyl chloride, HCl is used in a substitution reaction to displace an OH molecule that is connected to the tert–butyl molecule.
Substitution reactions for alkyl halides can go one of two ways; an Sn1 reaction or an Sn2 reaction. An Sn1 reaction is unimolecular (only depends on
the substrate), and requires a very weak nucleophile, a polar–protic solvent and favors tertiary alkyl halides as the electrophiles. An Sn2 reaction is the
opposite and is bimolecular (depends on the substrate and nucleophile), requires a very strong nucleophile, a polar–aprotic solvents and favors primary
alkyl halides as the electrophile. There are a number of factors that can affect the efficiency of these reactions. An Sn2 reaction is affected

Rates of Reaction of the Halogenoalkanes Essay
Rates of reaction of the halogenoalkanes
scientific knowledge and understanding:
Halogenoalkanes are classified as either primary, secondary or tertiary. In primary halogenoalkanes the halogen atom is covalently bonded to a carbon
atom which is bonded to one other carbon, to two other carbons in secondary and three others in tertiary.
Apparatus for a reflux
Increasing the size of the halogen atom in the halogenoalkane decreases the bond strength, increasing the reactivity of the molecule, as the bond is
easier to break.
BOND
BOND ENTHALPY (kJ mol–1)
carbon–fluorine 467 carbon–chlorine 340 carbon–bromine 280 carbon–iodine 240
It can clearly be seen ... Show more content on Helpwriting.net ...
R–Hal + H2O ROH + H+ + Hal–
The ethanol in the aqueous solution will dissolve the halogenoalkanes so they are able to mix with water.
Bonds in the water molecules will then break giving OH and H ions.
The OH ions will behave as a nucleophile because they have a lone pair of electrons, which when accepted by the carbon, will form a stronger bond
than the halogen does.

In each reaction the halogen atom will be eliminated, forming a negative halogen ion, which will then bond with a positive silver ion to form a visible
precipitate.
The rate, at which the silver halide precipitates form, is the same rate at which the halogen atoms are being eliminated by nucleophilic substitution.
C2H5Cl + :OH– C2H5OH + Cl–
C2H5Br + :OH– C2H5OH + Br=
C2H5I + :OH– C2H5OH + I–
The mechanism for this reaction is:
The free halide ions will then bond with the positive silver ions. The precipitation of the halides will occur as follows:
Ag+(aq) + Cl–(aq) AgCl(s)
Ag+(aq) + Br–(aq) AgBr(s)
Ag+(aq) + I–(aq) AgI(s)
aim:
By carrying out this experiment I hope to investigate how the rate of displacement of the halogenoalkanes differs in relation to the C–X
(carbon–halogen) bond.
prediction:
Based on my 'scientific knowledge and understanding' I predict the rate of hydrolysis of the halogenoalkanes

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