Reactivity Of Tert-Butyl Chloride Lab Report

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
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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

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
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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

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. An SN1 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

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

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

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 (KOH⁻) base and 2–methyl–1–butene by potassium tert–Butoxide (Kt–
BuO⁻) 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

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

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

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
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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

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

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 ∞–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 | 2° | RT: NO changeAfter heating: Evaporated slightly, made crystals, light yellow |
Slower SN1 reaction due to secondary carbon. | 1–bromoadamantane | 3° | RT: No changeAfter Heating:
Evaporated slightly, clear crystals formed | SN2 |
Sliver nitrate in ethanol reactions: Alkyl Halides | Classification of RX | Observations | Conclusion | 2–
bromobutane | 2° | RT: Precipitate and densely cloudy.No heating required | Slower SN1 reaction due to
secondary carbon. | 2–bromo–2–methylpropane | 3° | RT: ClearAfter heating: No change | Fast SN1 reaction
due to it being a tertiary carbon. | 1–bromobutane | 1° | RT: Precipitate and slightly cloudy.NO heating
required | Heating required due to primary carbon SN2 reaction | ∞–Bromotoluene | 1°Benzylic | RT:
Precipitate and very cloudy.No heating required | Heating required due to primary carbon SN2 reaction |
Bromobenzene | 1°Allylic Vinylic | RT: ClearAfter heating: No change | Heating required due to primary
carbon SN2 reaction | 1–bromoadanabtane | 3° | 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

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

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 the
nucleophile 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

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 ... Show more content on Helpwriting.net ...
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

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

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 more content on Helpwriting.net ...
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 AA B B C
C AA 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.

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 °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.

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

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
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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

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 |
1ᵒ alkane –CH3 | 25.589 | 2ᵒ alkane –CH2 | 34.391 | 2ᵒ 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

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.

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

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 content on
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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

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 Kütüphane 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 Advanced Chemistry – 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,

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 Helpwriting.net ...
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

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 ...
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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.

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 content on Helpwriting.net ...
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

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

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

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

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 (°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 (Rδ–––Mδ+). The partial negative
charge on the carbon allows it to act as a strong nucleophile or base similar to

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

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– ––OH2 R+ + 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

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 (〖AgNO〗_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
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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 80℃. 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

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,

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

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 50°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

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

Reactivity Of Tert-Butyl Chloride Lab Report

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