Competing Nucleophiles Lab

Competing Nucleophiles Lab
Title: Competing Nucleophiles (Exp 24, pp 211–221, pp 808–823, pp 836–842)
Purpose:
The purpose of this experiment is to determine the nucleophilic strength of chloride and bromide
ions as it reacts with 1–butanol (n–butyl) and 2–methyl–2–propanol (t–butyl alcohol) under SN1
and SN2 conditions.
Method:
40 g of ice and approximately 30 ml of sulfuric acid is cautiously added to a 100 mL beaker
respectively. Weigh 7.6 g of ammonium chloride and 14.0 g of ammonium bromide and place it in
another beaker, crushing the lumps until a powdery mixture remains. The powdery mixture is then
transferred to a 125 mL Erlenmeyer flask. Add the ammonium salts into the sulfuric acid mixture.
Heat is applied to dissolve the salt. Once the ... Show more content on Helpwriting.net ...
The solution that was performed in this experiment was to use sulfuric acid in order to form a
protonated alcohol, so when the halogen or nucleophile back attacks the compound, water is
displaced. Once the alcohol is protonated, the solution reacts in either an SN1 or SN2 mechanism. A
unimolecular nucleophilic substitution or SN1 is a two–step reaction that occurs with a first order
reaction. The rate–limiting step, which is the first step, forms a carbocation. This would be the
slowest step in the mechanism. The addition of the nucleophile speeds up the reaction and stabilizes
the carbocation. This reaction is more favorable with tertiary and sometimes secondary alkyl halides
under strong basic or acidic conditions with secondary or tertiary alcohols. In this experiment, the t–
butyl halide underwent an SN1 reaction. Nucleophiles do not necessarily effect the reaction because
the nucleophile is considered zero order, (which makes it a first order reaction.) The ion that should
have the strongest effect in an SN1 reaction is the bromide ion. The bromide ion should be stronger
because it has a lower electronegativity than chloride as well as a smaller radius. In a bimolecular
nucleophilic substitution or SN2 reaction, there is only one–step. This occurs because the addition
of the nucleophile and the elimination of the leaving group spontaneously occur at the same time.
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Essay about Rxn of Iodoethane with Sodium Saccharin
Title: Reaction of Iodoethane with Sodium Saccharin– Ambient Nucleophile
Dates Performed: February 21 + 28, 2013
Date Submitted: March 14, 2013
Abstract:
The product ratio of N–ethylsaccharin to O–ethylsaccharin that occurred due to alkylation with
iodoethane at 80 oC was determined to be 81.5% to 18.5%, respectively, based on an analysis of the
1H NMR spectrum that was collected. The melting point range of 87.8–94.7 oC also indicated that
the mixture was largely composed of N–ethylsaccharin. The more prevalent product structure is:
C2H5I
C2H5I
And/Or
And/Or
N–ethylsaccharin product O–ethylsaccharin product
Experimental Procedure:
Sodium saccharin ... Show more content on Helpwriting.net ...
N– saccharin is more stable than O–saccharin, so it should be the major product of the reaction, if
not the only product, if the reaction reaches thermal equilibrium. A reaction involving the oxygen as
the nucleophile should occur faster than one involving nitrogen because the oxygen is more
electronegative. This electronegativity would attract the attack of the nucleophile more so than the
Nitrogen atom would. Based on the 1H NMR spectrum that was collected, a few things can be
determined. Based on deshielding and electronegativity, the peak that occurs around 4.7ppm is
associated with the O–ethylsaccharin product and the peak at 3.8 ppm is associated with the N–
ethylsaccharin product. Based on the height ration, the N–ethylsaccharin product is the more
prevalent result. The solvent that was chosen for the experiment is a polar aprotic solvent that can be
used to stabilize charge. This stabilization can also be used to stabilize an ionic transition state. This
stabilization of the transition state along with the nature of the products based on the chemical
properties would encourage the reaction to proceed toward the N–ethylsaccharin product. This
product is more stable

Taking a Look at Nucleophilic Reactions
Nucleophilic reactions occur when there is an electron pair donor and an electron pair acceptor (2).
There are two types of ways that nucleophilic reactions occur. There is the SN1 reaction and the
SN2 reaction. An SN1 is a two–step reaction that occurs when a molecule first forms a carbocation.
Once the carbocation is formed, the nucleophile comes in and attaches to the molecule (2). Below is
a general reaction scheme of an SN1 reaction:
Below is the mechanistic scheme of SN1:
In an SN2 reaction, it is a one–step reaction and occurs when a nucleophile attacks a molecule and
forces the leaving of a leaving group. Below is the mechanistic scheme of SN2: Reaction kinetics is
the study of the rates of chemical reactions. From these chemical mechanisms, one derives the rate
laws which will show how fast or slow a reaction is occuring and figure out if a first order or a
second order reaction is occuring (1). The first order reaction is an SN1 reaction. A first order
reaction has a rate proportional to the concentration of one reaction. A first order reaction formula
will be :
Rate =k[A] or rate=k[B] A second order reaction is an SN2 reaction. A second order reaction has a
rate proportional to the concentration of both reactants. The formula is as follows:
Rate=k[A][B]
SN1 reactions proceed through a carbocation, the product is a racemic mixture of the substitution
product which is a 50/50 racemic mixture. SN2 reactions do not form carbocation, but require the

Solvolysis Of T-Butyl Brromide Lab Report
Solvolysis of t–Butyl Bromide: The Effect of Solvent Polarity on Reaction Rates
Katherine Lee
Georgetown University
Introduction
The solvolysis of t–butyl bromide is an SN1 reaction, or a first order nucleophilic substitution
reaction. An SN1 reaction involves a nucleophilic attack on an electrophilic substrate. The reaction
is SN1 because there is steric obstruction on the electrophile, bromine is a good leaving group due
to its large size and low electronegativity, a stable tertiary carbocation is formed, and a weak
nucleophile is formed. Since a strong acid, HBr, is formed as a byproduct of this reaction, SN1
dominates over E1. The first step in an SN1 reaction is the formation of a highly reactive
carbocation, in which a leaving group is ejected. The ionization to form a carbocation is the rate
limiting step of an SN1 reaction, as it is highly endothermic and has a large activation energy. The
subsequent nucleophilic attack by solvent and deprotonation is fast and does not contribute to the
rate law for the reaction. The Hammond Postulate predicts that the transition state for any process is
most similar to the higher energy species, and is more affected by changes to the free energy of the
higher energy species. Thus, the reaction rate for the solvolysis of t–butyl bromide is unimolecular
and entirely dependent on the initial concentration of t–butyl bromide. Mechanism 1. The Solvolysis
of t–Butyl Bromide
The reaction shown above is the reaction mechanism for t–butyl

A Study On Molecular Studies
Cyclopropenyl cation (1), the smallest hydrocarbon molecule is a centro–symmetric molecule with a
D_3h symmetry. It's the simplest aromatic species demonstrating aromatic features linked to the
delocalization of the two π electrons. It follows the Hückel's rule, (4n+2) π electron with n = 0.
Aromaticity cannot directly be experimentally measured. However, various methods were
performed such as computational studies from the Schleyer group^1, which address the energy,
structural changes and magnetic properties of cyclic conjugated system.
Computational studies were performed using the Gaussian 94 computational package with ab initio
method and the complete basis set amended by Peterson and co–workers was used for
calculations.^2 The ... Show more content on Helpwriting.net ...
〖〖(CH)〗_3〗^++cyclopropane →cyclopropyl cation (1)
Reaction
Different substituents have different effects on the stability of 1. Effect of these substituents were
studied on the relative stability of 1. Either these substituents have an electron withdrawing effect or
electron donating effect on the ring. Electron withdrawing group such as trialkylsilyl groups
attached directly to the carbenium ion centre have a destabilizing effect. However,
tris(trimethylsilyl)cyclopropenylium hexachloroantimonate (2) could be synthesize from readily
available materials as shown in figure 2 and it's relatively stable.^4
As shown in figure 3, C–C bond lengths for the tris(trimethylsilyl)cyclopropenylium cation are not
equal. The ring is significantly distorted from its ideal D_3h symmetry since one of the carbon
atoms in the ring system is close to the chlorine atoms in the hexachloroantimonate. This distortion
shows that it is caused by the counterion and lattice effects.^4 Using the isodesmic hydride transfer
reaction, the effect of trisilyl and trismethyl substitution was calculated. Three sillyl groups stabilize
1 by 22.4 kcal/mol while three methyl groups has a larger effect and more stable than 1 by 31.4
kcal/mol. Thus, Me > Si >>H was found as the order of the

Elimination Reactions
In this lab, the goal is to explore the idea of elimination reactions and investigate ways to
synthesizes alkynes. The product that will be produced is diphenylacetylene and this will be done by
starting with meso–stilbene dibromide using elimination reactions. The balanced reaction is:
For an E2 reaction to occur, strong bases should be utilized for this lab since E2 reactions favor well
with strong bases. KOH is a strong base that is also acts as a strong nucleophile. This is because it
contains a hydroxyl group and they are both strong bases and nucleophiles.
Heat will also be applied to encourage the chemicals to go through the elimination reaction. There
will be two ways to apply heat so that the reaction can occur. The first way is through a thermal
reaction process using sand baths to heat the reaction. The other method of heating is to microwave
the product so that the elimination reaction can ... Show more content on Helpwriting.net ...
A flea stir bar and 4.0 mL of 95% ethanol were added to the microwave reactor vessel. The solution
was then stirred using a magnetic stirrer until all the KOH dissolved. 1.0 g of meso–stilbene
dibromide was added in the microwave reactor vessel. The cap of the vessel was placed, tightened
and the vessel was placed in a microwave reactor carousel. After the reaction was complete, the
vessel with the mixed reaction was placed in a tap water bath so that it may be cooled. After it
cooled down, 10 mL of DI water was placed in a 50 mL and the reaction mixture was poured into
there to precipitate the product. The product was then collected by vacuum filtration and Hirsch
funnel. The product was rinsed with 3 mL of DI water three times. The product dried on the filter
paper with the vacuum on for 5 minutes after the rinse. The product was then placed in a drawer so
that it may completely dry so that the weight and the melting point may be recorded later. The final
product was a light brown

Synthesis Of T-Pentyl Chloride Lab Report
Introduction:
The purpose of this experiment was to synthesize t–pentyl chloride from the reaction of t–pentyl
alcohol and concentrated HCl. This reaction occurred through an SN1 reaction, a unimolecular
nucleophilic substitution reaction. This was a First Order Rate Reaction where the rate of t–pentyl
chloride was dependent only on the concentration of t–pentyl alcohol. After the reaction was
completed, the products were achieved via 3 liquid–liquid extractions and then after by simple
distillation. In the liquid– liquid extractions a solute was transferred from one solvent to another.
Then in the simple distillation the miscible liquids or the solution, was separated by differences in
boiling points. After this the product was determined through infrared spectroscopy.
Procedure:
Isolation of Crude Product
A mixture of 22 ml ... Show more content on Helpwriting.net ...
In the process, extraction and distillation techniques were used. The theoretical amount of t– pentyl
chloride was 17.358g, while 15.78 g was the actual amount produced which gave a percent yield of
90.9%. An error occurred while performing the experiment, the filtered dried product in the
distillation process was placed in the wrong flask. Due to this that part of the experiment had to be
redone and the new filtered product had some aqueous solution in it, which caused the boiling point
to be under the specified temperature. The boiling point then was at 50 ℃ compared with the
expected range of 79– 84 ℃. The IR spectrum used above was from another group's results. The
experimental IR spectrum has more prominent peaks in the 3000 cm–1 range compared to the
expected IR spectrum. Nonetheless, the experimental IR spectrum resembles the expected IR
spectrum in the sense that the peaks are closely around the same wavenumber range. This is
probably due to the product being distilled at the right boiling

Strong Nucleophilic Substitution Report
Identifying Substitution Reactions of Strong Nucleophilic Compounds and Polar Protic Solvents
with Analytical and Experimental Instruments
Oshi Bonhomme, Courtney Lasher, Olivia Trusty*
Department of Chemistry and Chemical Biology, IUPUI, 402 N. Blackford St., Indianapolis, IN
46202 otrusty@indiana.edu The three reactions investigated through various experimental testing
contain an alcohol leaving group positioned on primary or secondary carbons. The substitution
method (SN1 or SN2) used in the reactions vary and are influenced by various factors. The position
of the alcohol and the number of steps are variables used to identify the substitution method. The
products formed are influenced by the substitution mechanism. The data reported shows ... Show
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In order for SN1 and SN2 reactions to occur, the leaving group must be attached to an alkyne or
alkene (alkyl halides) 3. In nucleophilic substitution, there are two events that occur, development of
a new σ bond to the nucleophile and the σ bond to the leaving group breaks. The timing of these
events determines the type of mechanism2. The main difference between the two mechanisms is that
the SN2 reaction occurs in one step and the SN1 reaction occurs in two steps. The number of steps
in the reactions is influenced by many factors, including the rate law, nucleophile, and solvent. The
rate law (unimolecular or bimolecular) is the rate determining step. For a SN1 reaction has a weak
(neutral) nucleophile and is unimolecular, which means that the rate of the reaction is depending on
the concentration of the substrate. A SN2 reaction has a strong (negatively charged) nucleophile and
is bimolecular, where the rate of the reaction is dependent on the concentration of the substrate and
nucleophile 3. The strength of the nucleophile is important in identifying the type of reactions
because SN1 usually has weak nucleophiles and SN2 usually has strong nucleophiles. The type of
solvent used in the reaction is also an important factor in determining the number of steps because
SN1 reactions favor polar protic solvents (alcohols, carboxylic acids, and water). SN2 reactions tend
to proceed with polar aprotic solvents (DMSO and acetone) 3. Another factor involves the location
of the leaving group. SN2 reactions favor leaving groups in the primary location but can also work
for leaving groups in the secondary location. SN1 reactions favor leaving groups in the tertiary
location and can work for the secondary

Sn1 Lab Essay examples
Experiment 3: SN1 – Report Sheet
Last Name:
Abbassi–Mohadjel | First Name:
Nora | Date: OCT. 26, 2012 | Section Number:
005 (Friday–>1:35–5:35pm) |
Objective: 1 mark
(What is the purpose of this experiment?)
In this lab, spectrochemical technique was used to determine qualitively the product of a reaction.
The purpose was to learn what a SN1 reaction was with combining 2,5–dimethyl–2,5–hexanediol
and HCl and to learn how to use an infrared spectrum for analyzing the reaction. The infrared
spectrum determined the alcohol content present in the final product. If the IR of a molecule with an
alcohol is present, there will be an IR peak between 3400 and 3200 cm^–1. However, if the
experiment was done successfully by ... Show more content on Helpwriting.net ...
Results – Yield and Physical Properties of 2,5–Dichloro–2,5–Dimethylhexane 5 marks | 2,5–
Dichloro–2,5–Dimethylhexane | Yield (g) | 0.250g | Yield (%) | 96.2% | Appearance | White
precipitate, very thick liquid | Melting Point | 62.0ºC – 65.0ºC |
Figure 1. IR Spectrum
Attach IR Spectrum of Your Starting Material (2,5–dimethyl–2,5–hexanediol) and Product (2,5–
Dichloro–2,5–Dimethylhexane ) – label important stretching frequencies.
Discussion Questions: 10 marks
1. Suggest at least one chemical reason why your percent yield is less than 100%. (Incomplete
reactions? Side reactions? Stability of reagents?)
Incomplete reactions could be a possible chemical reason to why the percent yield is less than
100%. During the procedure, the experiment might not have been mixed long enough for the Cl–
ion to fully form.
2. SN1 and E1 reactions both involved the initial formation of a carbocation intermediate. In this
experiment, why is the SN1 product favoured over the E1 product?

The SN1 mechanism leads to substitution products, and the E1 mechanism leads to formation of
alkenes, therefore in this case, it is shown that this mechanism leads to a substitution of products
since the Cl– ion is replacing the OH group by the addition of a strong acid (HCl). When the
nucleophile

Nucleophilic Substitution Lab Report
Lab Report 9: SN1 and SN2 Mechanisms: Nucleophilic Substitution
Raekwon Filmore
CM 244
Section 40
March 27, 2018 Introduction:
For this week's experiment, involved nucleophilic substitution of SN1 and SN2 reactions. On two
separate days, SN2 was done first followed by SN1 being done in the next lab. The substitution of
one group from a saturated sp3 hybridized carbon atom is typically a reaction that is used to
interconvert different functional groups. In this reaction in particular, nucleophiles have the property
of having at least one lone pair of electrons (not shared) having either a neutral or negative charge.
The nonbonding pair of electrons of the nucleophile is donated to an electrophilic carbon. Some
examples of nucleophiles ... Show more content on Helpwriting.net ...
10ml of 2–chloro–2–methylbutane was added to 25ml of concentrated (12M) Hydrochloric Acid in
a separatory funnel. The contents were swirled gently without the stopper and venting was done
frequently to release excess pressure. Shake the funnel until the layers separate and able to
recognize which one is the aqueous layer and which is the organic layer. The organic layer was
washed with 10ml portions of saturated aqueous sodium chloride and cold saturated aqueous
sodium bicarbonate. The organic layer was washed again with 10ml of water and saturated aqueous
sodium chloride. The aqueous layer was carefully removed and the 2–chloro–2–methylbutane was
transferred to a clean erlemmeyer flask and was dried with several tip full of anhydrous sodium
sulfate. After drying, the crude was carefully transferred with a Pasteur pipet to a round bottom flask
and was set up for shortpath distillation. Because finding an accurate boiling point of 2–chloro–2–
methylbutane is difficult, two separate fractions or distillates was taken. The first distillate was
collected below 75°C and the second fraction was collected everything above 75°C. The percent
yield of the reaction was 73%. The refractive index of refraction for both fractions were relatively
almost the same. The index of refraction for the first distillate was 1.404 and the index of refraction
of the second distillate was

Nucleophilic Substitution
Samantha Gutierrez
Introduction:
The purpose of this lab is to investigate how different factors affect the rate of SN1 and SN2
reactions. SN2 reactions proceed via a one step mechanism in which the incoming nucleophile
attacks the electrophilic carbon center from the opposite side of the leaving group. This reaction
mechanism implies that the stereochemistry of a chiral center will be inverted. SN1 reactions
proceed via two steps, slow dissociation of the C–X bonds to form an intermediate carbocation and
a fast second step in which the C–Nucleophile bond is formed. Since the intermediate carbocation is
trigonal planar, the nucleophile can attack with equal probability from ... Show more content on
Helpwriting.net ...
In the other test tube, 0.5mL of 0.2 M 2–chloro–2–methylpropane was added. At the same time,
1mL of 1.0 M AgNO3 in absolute ethanol was added to each test tube. Observations were recorded
and rates of reactions were compared.
1mL of AgNO3 was added to one test tube. 0.5mL of AgNO3 and 0.5mL of ethanol was added to
another test tube. At the same time, 1mL of 1.0 M 2–chloro–2–methylpropane was added to each
test tube. Observations were recorded and rates to the two reactions were compared.
Results:
The reactions that we were specifically looking at in this lab were:
R–X + NaI → R–I + NaX AgNO3
R–X + CH3CH2OH → R–OCH2CH3 + AgX + H+ + NO3–
When 1–bromobutane, 2–bromobutane, and 2–methyl–2–bromobutane were added to 15% NaI
solution, we observed cloudy with precipitants, clear with small precipitants, and a tint of yellow,
respectively. When 1–bromobutane and 1–bromo–2–methylpropane were added to 15% NaI
solution, we observed large precipitants and clear with small precipitants, respectively. When 1–
bromobutane and 1–chlorobutane were added to 15% NaI solution, we observed small precipitants
at first then large precipitants as time went on and a slight tint of yellow, respectively. When 1.0 M
of 1–bromobutane and 2.0 M of 1–bromobutane were added to 15% NaI solution, we observed large
precipitants with a fast reaction and small precipitants with a slower reaction time, respectively.
When 7.5% NaI solution and 15%

2-Methylphenoxyacetic Acid Synthesis Lab Report
The design of this experiment aimed to synthesize 2–methylphenoxyacetic acid from 2–
methylphenol and sodium chloroacetate by means on phenol alkylation. 2–Methylphenoxyacetic
acid has been used to control the growth in plants and exemplifies a product of synthesis as it is a
complex molecule constructed from two simple molecules. This synthesis proceeds by a
nucleophilic substitution reaction, more specifically, via SN2 mechanism (see below). An SN2
mechanism's reactants are a nucleophile and a good leaving group (usually a halide ion). Once the
crude product is obtained, it is purified by recrystallization with water and collected. To start, a 0.32
g (2.9 mmol) sample of 2–methyphenol (the nucleophile) was weighed into a 10 ml Erlenmeyer ...
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The product was then suspended in 2 ml of water with a stir rod in a 50 ml Erlenmeyer flask and
heated to boiling. Water was added in one milliliter increments until all the product was dissolved
(18 ml added total). The saturated solution was allowed to slowly cool, and gradual white crystal
formation was observed. Recrystallized product was collected once more by suction filtration with
the Hirsch funnel once crystallization ceased. Collected product dried on a watch glass for a week,
weighed 0.14 g (1.2 mmol), and the melting point was 139°–141°

Nucleophilic Substitution Reaction Lab Report
Substitution Reactions, Purification, and Identification Introduction (Q:1&2) An unknown
(nucleophile) was subjected to SN2 reaction when combined with benzyl bromide (electrophile) in a
sodium hydroxide (base) solution. The experiment provided a clear understanding in conducting
nucleophilic substitution reaction where heat must be applied in order for the SN2 reaction to move
forward. The goal was to determine the unknown by incorporating knowledge of polarity, the laws
of physics, and the role of energy when forming crystals; "like dissolves like", "matter cannot be
created nor destroyed". (Q3:Techniques) Melting point, recrystallization, and thin layer
chromatography (TLC) mediated the ability to find which unknown nucleophile was provided ...
This was concluded by combining information on melting points and TLC; melting range narrowed
when filtered product was mixed with the standard product. Also, the Rf value of the pure product is
closely related to the Rf value of the standard. TLC of filtrate showed no movement of the substance
in the mixture under 9:1 ratio declaring the substance to be extremely polar. Of the three potential
unknown reactants, 4–methoxyphenol would be the most polar and therefore would travel least up
the TLC plate. (Q14:Yield) 81.2% product yield was collected. "Matter cannot be created nor
destroyed", therefore some product could have filtered through. TLC of filtrate confirmed remnants
of product present. Filtering the filtrate could have increased the yield. (Q15:Recovery) The percent
recovery of the product makes sense because it is the mass of the crystallized product divided by the
crude product: 94.9%. The percentage reflects the mass of pure product (without the presence of
impurities). (Q16:MP) Melting point coincides with the unknown nucleophile being 4–
methoxyphenol because when the standard product was combined with our pure product, the
melting range narrowed. When compared to the melting ranges obtained when mixed with the other
two possible products the melting ranges significantly decreased and widened. This is often an
indication of impurities being present, but because this was a

2-Methylbutane Lab Report
For this experiment, nucleophilic substitution was first observed. The 2–methyl–2–butanol was
converted into 2–chloro–2–methylbutane by using an SN1 reaction. The alcohol on the 2–methyl–
2–butanol was a tertiary alcohol. To form a good leaving group since the hydroxide ion is a bad
leaving group, the tertiary alcohol was protonated with HCl, but since it consists of a fully
substituted carbon atom from the carbon oxygen bond, an SN2 displacement cannot occur.
Therefore, water is the leaving group that disassociates and leaves a relatively stable tertiary
carbocation behind. From HCl, the chloride anion is a good nucleophile, and this adds to the
available carbocation forming a carbon chlorine bond and thus 2–chloro–2–methylbutane is the
resulting ... Show more content on Helpwriting.net ...
The ring stand was used to make sure that the separatory funnel did not tip over and break. Then 2–
methyl–2–butanol was added to the funnel along with concentrated HCl. These two chemicals were
swirled together to combine them. A glass stopper was placed on the top of the separatory funnel. It
was then inverted and vented immediately. Then the stopcock was closed, and the separatory funnel
was vigorously shaken for twenty seconds and then vented. This process was repeated. The
separatory funnel was vented frequently due to hydrochloric acid having a high vapor pressure
which causes there to be a pressurized system. This is dangerous because the buildup of pressure
can cause the glass to break and HCl to get everywhere. HCl is a very strong acid which is
extremely dangerous upon contact. After this, the separatory funnel was placed back into the ring
stand, the glass stopper was removed, and the bottom aqueous layer that separated was drained into
an Erlenmeyer flask. The separatory funnel was placed such that the bottom was resting on the top
of the receiving flask to avoid spills. Then sodium bicarbonate was slowly added, and some gas was
produced. The funnel was swirled until the bubbling ceased and the glass stopper was put on. The
same process from before with inverting and venting was repeated. After this, the separatory funnel
was placed back into the ring stand, the glass stopper was removed, and the bottom aqueous layer
was drained into the same Erlenmeyer flask as before. Some bubbling occurred when combining, so
it was drained slowly. These extracts were waste. A second wash was performed with sodium
chloride solution and the third was performed with distilled water. The organic layer was drained
into a clean Erlenmeyer flask and four spatulas of anhydrous sodium sulfate were added to the
solution. The solution was swirled for about fifteen seconds.

The Diel-Alder Nucleophilic Acyl Substitution
The Diel–Alder/Nucleophilic Acyl Substitution Cascade Reaction Introduction The Diels–Alder
reaction was discovered and named after the Nobel Prize winning scientists Kurt Alder and Otto
Diels in 1928. Such a reaction occurs when a diene with two adjacent double bonds is mixed with a
dienophile consisting of a double bond in order to create a cyclohexene. The diene must be in the s–
cis conformation in order for the electron transfer to engage correctly. If the diene in question is in
s–trans conformation then the access to the molecules is limited, thus, no reaction can occur. The
dienophile we used was maleic anhydride. Maleic anhydride possesses high electron withdrawing
characteristics which caused a very quick reaction. The reaction will

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

In this experiment, a nucleophilic substitution was performed, where a chloride nucleophile
substituted a tertiary hydroxyl group on 2–methyl–2–butanol. In a nucleophilic substitution reaction,
an electron rich nucleophile attacks a positively or partially positively charged electrophile, and
replaces a leaving group. In this reaction, chloride ions are the nucleophile, the tertiary carbon in 2–
methyl–2–butanol is the electrophile, and water is the leaving group.
In the mechanism for this reaction, the oxygen from the hydroxyl group of the 2–methyl–2–butanol
attacks the hydrogen of the HCl, causing heterolytic cleavage of the HCl, resulting in a chloride ion,
and in the oxygen bonding to an extra hydrogen, and becoming positively charged. This changes the
leaving group from a hydroxyl to a water, which is a far more stable leaving group. The leaving
group leaves the hydrocarbon chain through heterolytic cleavage, creating a water molecule, ...
While a technique like gas chromatography could have been used, for this experiment, halide tests
and IR spectroscopy were used to identify the product. Halide tests are an old fashioned test that can
be used to confirm the identity and connectivity of the functional group using various substitution
reactions. In this experiment, NaI and AgNO3 were used, which produce a precipitate if a chloride
functional group is present. The NaI reaction produces a yellow precipitate, whereas the AgNO3
reaction produces a white precipitate. Both reactions involved the chloride group being substituted
by a different nucleophile, in this case, iodide or nitrate. NaI supports an SN2 reaction, which favors
primary carbons, whereas AgNO3 undergoes an SN1 reaction, which favors tertiary carbons. Hence,
depending on which one of these precipitates formed at a faster rate, one could determine if the
carbocation was primary or

Essay on Reaction Iodoethane with Saccharin, an Ambident...
Discussion: In this experiment, we alkylate sodium saccharin to N–ethylsaccharin with iodoethane
in an aprotic solvent N,N dimethylformamide. Nucleophiles in this experiment will react better in an
aprotic solvent. Aprotic solvents have dipoles due to its polar bonds but they do not have H atoms
that can be donated into a H–bond. The anions which are the O– and N– of sodium saccharin are not
solvated therefore are "naked" and the reaction is not inhibited and preceded in an accelerated rate.
The reaction was an SN2 reaction. Since the Oxygen and Nitrogen are more electronegative than the
carbon on which they're attached electrons are pulled towards O– and N– attracting the ethane from
Iodoethane. Iodine being more electronegative ... Show more content on Helpwriting.net ...
The two remaining sp2 hybrid orbitals on oxygen are used to hold oxygen's lone pairs(bruice). O–
ethylsaccharin is then less stable because the Ethyl group is attached to the Oxygen that used to be a
carbonyl group, giving the Carbon an sp3 configuration (joined to two other carbons, the Oxygen
with the Ethyl group and a Hydrogen). This put strain on the ring, and therefore is less stable."
(Richard y.a.). Upon mixing the reactants, sodium saccharin slightly dissolved producing a clear
colorless liquid. When placed at 80°C hot bath, the solution completely dissolved and turned
yellowish–green. Iodoethane is a clear and colorless liquid, but when exposed to light and air the
Iodide dissociates from ethane and gives off a yellow color as a sign of decomposition. The solution
was covered to prevent this from happening. But, as iodide dissociates from CH3CH2 that then gave
off its yellowish color which shows SN2 reaction taking place. SN2 reaction happened fast The
limiting Reagent is Iodoethane , as the alkylating agent; it was not used in excess and dictated how
far the reaction went. The Na+ binds with I–(noted disappearance of the yellow color, as I– binds
with Na+) then the ethane group bonds with either the Oxygen of saccharin or the Nitrogen of
saccharin. The final product after vacuum filtration had some unreacted material, indicated by some
yellow green solid formation.
The actual product was homogenous white powder. Therefore, one cannot tell if two

Preparing A Yield Of Tert Butylchloride Through A...
Objective The objective of the experiment was to prepare a yield of tert–butylchloride through a
unimolecular (SN1) nucleophilic substitution reaction. Procedure Part A– Preparation of tert–Butyl
Chloride A test tube was acquired, and 7 mL of concentrated HCl were added to this test tube
through a graduated cylinder, this was done in the fume hood in order to avoid any noxious fumes
from the sample, which were observed when pouring in the concentrated HCl. After adding the HCl,
a glass pipet was used to add 2.5 mL of tert–butyl alcohol to the test tube. It was observed that the
alcohol and HCl separated into two layers distinct layers. The solution was mixed by using the pipet
to take the bottom layer and squirt it onto the top, this mixing process was performed for 15 minutes
to allow the solutions to mix well. Afterwards, the lower aqueous layer of the test tube was removed
with the pipet; it was imperative to remove as much of the lower layer as possible without removing
too much from the top layer. The solution was then washed with water by pouring 2.5 mL of water
into the test tube, letting it sit for about a minute and then removing the aqueous layer again, in
order to help the mechanism for this process. Additionally, about 2.5 mL of sodium bicarbonate 5%
were also added to he test tube, were allowed to sit for about a minute, and then removed once
again; this was done in order to remove any HCl impurities from the solution. The remaining upper
layer was moved to

A unimolecular nucleophilic substitution of t–amyl alcohol should be done to convert into t–amyl
chloride through extraction and simple distillation. A concentrated hydrochloric acid was added to
the reactant to force the reaction to take place. Infrared spectroscopy was utilized to verify that the
contents of the tert–amyl chloride should be different from the starting material.
In this reaction, a rate–determining step should occur through the ionization between carbon and –
OH bond to form an intermediate.11 This step should be followed by rapid reaction of a nucleophile
to wrap up the substitution.11 For this experiment, hydrochloric acid was used to drive off the
reaction, which contains a chlorine ion, a common nucleophile. (1)Chlorine ion is more effective as
a nucleophile than water; because an ion holds a negative charge and resulting in a faster rate of
reaction, whereas water holds a neutral charge, resulting in a slower rate of reaction with a
carbocation intermediate.13 The starting ... Show more content on Helpwriting.net ...
One supporting evidence is that the resulting infrared spectra in Figure 2 for the product still
contains an OH stretch of 3343 cm–1. In fact, the stretch of OH in the product is broader than the
stretch of the starting material in Figure 1. This could imply that the layers in the separatory funnel
initially did not form, although a lot of time has taken to settle. Adding 10% sodium chloride to the
mixture helped to form layers, but that does not necessarily mean that the –OH group had
completely separated from the carbocation. And so, the resulting product is not pure. Furthermore, it
is expected that the C–Cl stretch (between 850 to 550 cm–1) and C–H wag (1300–1150 cm–1)
should appear on Figure 2, but that did not happen. Moreover, the reaction failed since the amount
of tert–amyl alcohol used was not a lot enough to form an expected

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

Sn2 Reaction Lab Report
Pronin, Reiher, and Shenvi provide a shortcut to produce compounds with tertiary alkylisonitriles or
tertiary alkylamines from tertiary alcohols through a transformation similar to a Sn2 reaction. This
could make it easier to synthesize natural products that have anticancer, antimalarial, or antifungal
properties. In a Sn2 reaction, a nucleophile displaces a leaving group so that the substrate's
stereochemistry is flipped. The structural flip could change a compound's chemical properties. This
cannot take place on tertiary carbons because triply substituted carbon electrophiles are too crowded
for the Sn2 reaction's backside attack. This method transforms tertiary alcohols into tertiary
alkylisonitriles while inverting the stereochemistry. In this method, a special acid catalyst and a
nitrogen–containing molecule, a derivative of cyanide is used. The special acid catalyst is scandium
(III) trifluoromethansulfate. The acid helps detach a fluorous functional group from one side of the
central carbon and then nitrogen forms a new bond on the other side. The substrate alcohol is ...
Figure 2 was unorganized and harder to follow. There were three different sections, a through c, but
they weren't spaced out and it was difficult to decide if the chart was with section b or c. It would
have been easier to read if each section was its own figure (Figure 1, Figure 2, etc.) instead of 2a,
2b, or 2c. Figure 2 was about the Ritter–type reaction, this was not mentioned in the article
straightforward saying this was the Ritter–type but the article provided a useful reference. The other
figures were clearly labeled and were useful in understanding this method. The stereochemistry was
shown in all compounds in the figures. This had to be present in order to show that particular
reactions and products were stereoselective and had stereochemistry. If not the figures would be
incomplete and wouldn't be

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

Nucleophilic Substitution SN1-SN2
The nucleophilic substitution SN1/SN2 typically occur in a competitive regime. There are various
conditions that define the predominant reaction mechanism taking place. Since SN1 leads to the
racemic mixture, SN2 is more popular in asymmetric organic synthesis. So, detailed computational
studies of model SN2 reactions have been carried out during the last three decades[2–6, 9]. The
influence of solvation of the nucleophile with several common solvents on the rate constant of the
reactions F–(Sn) + CH3Cl → CH3F + Cl–(Sn) where S is a solvent molecule and n=0–3, was
studied experimentally (flowing plasma mass spectroscopy) by Bohm and Raksit[2] . The results of
their work are summarized in Table 1:
Table 1. Rate constants measured for reactions of solvated fluoride ions at room temperature in the
gas–phase. Values of kr are given in units of 10–9 cm3mol–1s–1.
F–Sn
kr at different n
0
1
2
3
F–(D2O)n
1.9
0.015
0.0003
0.003
F–(CH3OH)n
1.9
0.0006
0.0003
0.0003
F–(CH3CH2OH)n
1.9
0.0003
0.0003
–
It is clear that the solvation slowers the reaction at least 100 times. This work suggests the existence

of higher barriers on the potential energy surface for the solvated nucleophile. Morokuma[3], using
HF/3–21G level of theory, showed that the solvation in protic polar solvents (such as water or
alcohols) increases the activation energy accordingly to the number of solvent molecules, which
form hydrogen bonds with the nucleophile. (see Figure 1)
Doi et al.[4] studied

Para-Chlorophenoxyacetic Acid Lab Report
The goal of this was to successfully accomplish the synthesis of para–Chlorophenoxyacetic acid. In
this experiment, para–Chlorophenoxyacetic acid was synthesized from 4–chlorophenolate and
chloroacetic acid using an SN2 reaction. The product obtained was determined to be the para isomer
of Chlorophenoxyacetic acid. This was confirmed by the melting point of 157.3–157.9 ◦C. The
percent yield determined at the end of the experiment was 37.83 %. The TLC analysis showed that
P–Chlorophenol was less polar than P–Chlorophenoxyacetic Acid because it had an Rf value of 0.38
in comparison to the value of 0.33 on a 50:50 hexane and ethyl acetate solvent mixture. In the NMR
comparison, it was shown that both the starting material of chloroacetic acid and product contained
a peak of integration two around 4 ppm representing the acidic proton. In the FT–IR comparison, it
was determined that the Chloroacetic acid and the para–Chlorophenoxyacetic acid both had an OH
bond at 3416 cm–1 and 3429.72 cm–1 respectively. The Chloroacetic acid and para–
Chlorophenoxyacetic acid also both had a carbon–oxygen double bond at 1648 cm–1 and 1654.81
cm–1 respectively. The para–Chlorophenoxyacetic acid also contained a peak at 1236.18 cm–1
which represents the C–O–C bond.
Introduction
The goal of this experiment was to the ¬para isomer of Chlorophenoxyacetic acid. Synthesizing
para–Chlorophenoxyacetic acid is an important aspect in the agricultural world. It is commonly used
to promote growth and

Introduction:–. To natural science the substitution responses is the The greater part significant
reactions, particularly Nucleophilic fragrant substitution responses the place nucleophile strike sure
charge alternately incompletely certain accuse Concerning illustration it can so, it replaces An
weaker nucleophile which after that gets to be An abandoning bunch. The remaining sure alternately
incompletely certain particle gets a electrophile. The general type of the response is:. Nuc: + R–LG
→ R–Nuc + LG:. Those electron combine (:) starting with the nucleophile (Nuc :) strike the
substrate (R–LG) framing another covalent bond Nuc–R–LG. The former state about accuse will be
restored The point when the abandoning aggregation (LG) departs for ... Show more content on
Helpwriting.net ...
Need simply one step. Those assault of the reagent and the removal of the abandoning bunch happen
all the while. This system generally brings about reversal from claiming setup. On the substrate that
is under nucleophilic strike is chiral, those response will prompt a reversal from claiming its
stereochemistry called a Walden reversal. To a sample (chloromethane for bromide particle. ( SN2
ambush might happen Assuming that the posterior course of ambush is not sterically hindered by
substituents on the substrate. Subsequently this system generally happens during a unhindered
essential carbon focal point. Assuming that there may be steric swarming on the substrate close to
those abandoning group, for example, such that at a tertiary carbon center, those substitution will
include a SN1 as opposed an SN2 mechanism, (an SN1 might Additionally make less averse in this
the event in light a sufficiently stable carbocation go–between Might a chance to be formed). At the
substrate is a fragrant compound those response kind will be nucleophilic fragrant substitution.
Carboxylic corrosive subsidiaries respond for nucleophiles for nucleophilic acyl substitution. This
sort of response camwood a chance to be advantageous in get ready

Bromopropane And Potassium Hydroxide Lab Report
The purposes of this experiment were to model a bimolecular nucleophilic substitution reaction
between potassium hydroxide (KOH) with 1–bromopropane and determine whether it follows a
second–order rate law mechanism. A rate constant of 0.0684 M–1 min–1 was obtained for this
reaction at 45.1°C, which was determined through equilibrating the reaction and performing
titrations of 0.390 M KOH with 0.1000 M hydrochloric acid (HCl). The activation energy calculated
from class data was 50.188 kJ/mol, which deviated largely from the literature range value of 72.80–
83.76 kJ/mol. It was concluded that the reaction was consistent with the predicted SN2 mechanism,
based on the regression of a trendline.
Many reactions that exist in nature involve a double displacement between ions and reactants with
solvents. A bimolecular nucleophilic substitution, or SN2 reaction, involves a nucleophilic attack on
a substrate and the departure of a leaving group. A nucleophile is a compound or ion that donates
electrons to promote bond formation (Caldwell, 1984). In order for a leaving group in a compound
to leave, it must possess the characteristics of a weak base and be able to occupy electrons. Several
factors affect the rate and favorability of such reaction, such as (Bateman, 1940). In addition, the
substrate that is attacked by the nucleophile is commonly an unhindered primary substrate to allow
the reaction to occur quicker. An SN2 reaction follows the second–order rate law. In this

Aromaticity Of Pyridine Lab Report
Aromaticity and reactions of pyridine Introduction
Pyridine is a nitrogen containing aromatic analogue of benzene. The N in pyridine is sigma bonded
to two atoms and has a lone pair, and is therefore sp2 hybridized.
This leaves one electron in an unhybridized p orbital, which contributes to the system, making a
total of 6, and therefore an aromatic molecule (5 x C–H contribute 5 electrons, N contributes 1, = 6,
4N+2). The lone pair on the N is in an sp2 orbital, which means it is directed away from the ring but
in the same plane.
The lone pair of electrons are not involved in the aromatic system, and stick out away from the
molecule.
Pyridine is aromatic, and displays aromatic characteristics such as high resonance energy
(27kcal/mol), ... Show more content on Helpwriting.net ...
Hellwinkel, D. (1998). Die systematische Nomenklatur der Organischen Chemie (4th ed.). Berlin:
Springer. p. 45. ISBN 3–540–63221–2.
2. Gossauer, A. (2006). Struktur und Reaktivität der Biomoleküle. Weinheim: Wiley–VCH. p. 488.
ISBN 3–906390–29–2.
3. Curvall, Margareta; Enzell, Curt R.; Pettersson, Bertil (1984). "An evaluation of the utility of four
in vitro short term tests for predicting the cytotoxicity of individual compounds derived from
tobacco smoke". Cell Biology and Toxicology. .
4. Aeschbacher, HU; Wolleb, U; Löliger, J; Spadone, JC; Liardon, R (1989). "Contribution of coffee
aroma constituents to the mutagenicity of coffee". Food and Chemical Toxicology. 27 (4): 227–232.
doi:10.1016/0278–6915(89)90160–9. PMID

1-Bromo-3-Phenylpropane Reaction
Reaction 1 involved a primary alcohol (OH), weak leaving group in the starting material and a
reaction with a strong nucleophile (sodium bromide) and a polar protic solvent (sulfuric acid). The
reaction was carried out through reflux and the product had a relatively high yield (75%) (Scheme
1). Scheme 1. Substitution of 3–phenyl–1–propanol to form 1–bromo–3–phenylpropane The
reaction was determined to be SN2 after careful reading of the data obtained and the procedure
followed. Increasing the amount of NaBr used in the reaction did not have an effect on the product
yield. The conditions of the reactions were acidic due to the solvent used. The acidic conditions
prevent an E2 substitution reaction from occurring. In this reaction, the OH leaving group is
replaced with the nucleophile Br. The bromine attacks the first carbon and the OH group leaves in
the same step. This is a concerted process or bimolecular substitution mechanism (SN2). The
location of the leaving group (primary) also indicates the reaction is SN2. The data obtained from
1H NMR was found to confirm the correct product was formed. The data showed the correct
location and coupling of hydrogens in the product. There should be a difference of polarities
between the starting material and the product. The product should travel further on the TLC plate
than the starting material because the starting material should be more polar than the product. The
TLC results confirm the product was less polar than 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

Substitution And Elimination Reaction Lab Report
Substitution and Elimination Reactions Substitution reactions replace a functional group with a new
group. These reactions compete with elimination reactions in which a group is eliminated and a π
(pi) bond is formed. Substitution reactions occur when a nucleophile is added to an
electrophile/substrate. Elimination reactions occur when a base is added to an electrophile/substrate.
The electrophile must contain a leaving group to be considered a substrate. Alkyl Halides Alkyl
Halides are the common substrate used in substitution and elimination reactions. Alkyl Halides are
halogen containing alkane compounds. These compounds are commonly used as the electrophile in
substitution reactions. Naming Alkyl Halides is similar to naming alkanes. Halogen substituents are
named just like alkane substituents, for example if there is a chlorine attached to an ethane the
chlorine substituent will be named using the word chloro–. The other halogens are named using the
words flouro–, bromo–, and iodo– respectively. These possible substituents are listed alphabetically
along with the alkane substituents. When dealing with alkyl halides it is important to note that the
carbons attached to the halogens and the carbons attached to those carbons are involved in the
reaction process, therefore we designate these carbons with special symbols to differentiate them
from the rest of the molecule. The carbon atom that is bonded to the halogen is designated the α
(alpha) carbon. The carbons that are

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

The purpose of this experiment was to perform a nucleophilic substitution reaction to construct a
biologically active compound from two simple parts and then to recrystallize the product collected,
which is a purification technique that purifies solids based on differences in solubility. In order to
accomplish this, other techniques such as heating at reflux, and suction filtration were used. Heating
at reflux is a technique used in lab that allows a solution to be heated for a certain amount of time
once it begins boiling. Suction filtration is a separation technique that is combined with a water
aspirator and was used to collect the product from this experiment, which was 2–
methylphenooxyacetic acid. Reaction Scheme: (Draw reaction scheme here) In this ... Show more
content on Helpwriting.net ...
Upon the addition of water, it was noted that a layer separation occurred and the water layer
remained on top, with the 2–methylphenol layer on the bottom layer. Then, conversion calculations
were performed to determine the appropriate amount of 3M NaOH to be added to the 2–
methylphenol solution. From the calculations, it was determined that 1.08 mL were to be added. 3M
NaOH itself was a cloudy solution in appearance and upon the addition of 3M NaOH to the 2–
methylphenol solution, it was noted a color change occurred and it became a yellow–green solution.
Following this, the same calculations used previously, were used to determine the appropriate
amount of sodium chloroacetate, which was found to be 0.38 g (3.26 mmol). Sodium choloroacetate
was a white, crushed solid that was then combined with 1 mL of water and was swirled until the
sodium chloroacetate completely dissolved. This sodium choloracetate solution was then transferred
to the 2–methylphenol solution by the use of a medicine

Nucleophilic aliphatic substitution is the substitution of one functional group for another functional
group. The substitution is made at a saturated carbon atom. A saturated carbon is a carbon atom that
is sp3–hybridized. For a compound to be considered a nucleophile they must possess certain
properties. Nucleophiles must contain at least one pair of non–bonding electrons and be neutral or
have a negative charge. Through nucleophilic substitution the non–bonding electrons are donated to
the electrophile. The leaving group can also be either neutral or have a negative charge. Said leaving
group must also accept a pair of bonding electrons from the carbon atom that the leaving group was
attached to (1). The electrons are pulled as the bond between the carbon atom and the leaving group
breaks. Some leaving groups are ... Show more content on Helpwriting.net ...
Safety is a priority. Always add acid to water, this reduces some of the heat created and does not
create any acid "spitting" in the hood. If sulfuric acid is spilled and comes in contact with skin, the
area should be washed with cold water and then with a dilute sodium bicarbonate solution. The
original solution once the 1–butanol, hydrobromic acid, and sulfuric acid were added was yellow–
orange in color. After distilling the distillate was clear then became partially cloudy. The calculated
yield of the product was approximately 5.41%. The calculated yield was low due to a fluctuating
temperature during the reaction and possible impurities formed. The presence of tertiary alcohols
creates an increase in the competitive elimination reaction when treated with phosphorus trihalides.
To produce a decent yield of tertiary alkyl halides the tertiary alcohols should be treated with
concentrated hydrogen halides (1). Creating 2–chloro–2–methylbutane from 2–methyl–2–butanol
and hydrochloric acid is an SN1 reaction that demonstrates the transformation. This mechanism can
be shown

The Substitution Mechanisms ( Sn1 And Sn2 ) With Reactions
This paper was written in order to relate nucleophilic substitution mechanisms (SN1 and SN2) with
reactions that involved converting alcohol–containing compounds to alkyl halides. This experiment
was conducted by combining the initial alcohol with reagents and heating under reflux when
necessary. It was determined that mechanistic pathways of substitution depend principally on the
structure of the initial alcohol; that is, substrate is considered primary, secondary, or tertiary.
Synthesis of 1 (46%) was achieved through SN2 mechanism from a primary alcohol, while
syntheses 2 (97%) and 3 (98%) were completed by SN1 mechanisms from secondary and tertiary
alcohols as determined by the presence of rearranged products. Many mechanistic reactions are
utilized and are interrelated in order to produce the desired product in an organic synthesis. These
mechanisms may be generally categorized2 into the following classes: addition, elimination,
substitution, and rearrangement. Of these, substitution reactions are of particular significance in
halogenation reactions of alcohol–containing compounds, in which the alcohol functional group is
replaced by a halide ion (i.e. Cl– or Br–). Alcohols are the most common precursors to alkyl halides
and signify a very substantial reaction in organic synthesis4. Substitution reactions can occur via
two mechanisms2, namely unimolecular nucleophilic substitution (SN1, a stepwise process) and
bimolecular nucleophilic substitution (SN2, a

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 ...
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

Sn2 Nucleophilic Substitution Lab Report
1 and 2) In this experiment, Sn2 nucelophilic substitution between an unknown nucleophile solution
and the known reagents benzyl bromide and NaOH occurred to form a benzyl ether product. 3)
Refluxing, recrystallization, melting point, and TLC were used to purify and identify the products of
the reaction. 4) In order for nucleophilic substitution to take place, an electron pair donor (the
nucleophile) and an electron pair acceptor (the electrophile) with a good leaving group must be
present. In this experiment, Sn2 nucelophilic substitution occurred. In Sn2 nucelophilic substitution,
the nucleophile performs a "backside attack" at the same time the leaving group "leaves", resulting
in an inversion of configuration of the electrophile to form ... Show more content on Helpwriting.net
...
TLC in 9:1 hexanes/ethyl acetate solvent gave Rf Values for given solutions A–F and the filtrate as
.102, .061, .122, .347, .298, .404, .426, .064, and .280, respectively. 13) The product obtained was
4–tert–butylbenzylphenolether. The melting point range of the pure, unknown compound (35–44
°C) was closest to the one measured for 4–tert–butylbenzylphenolether (36–51 °C). Also, the scale–
like appearance of the unknown solid was very similar to the appearance of 4–tert–
butylbenzylphenolether. The NMR spectra solution #33 (the unknown) matched closest to that of 4–
tert–butylbenzylphenolether 14) The yield (.468 g) makes sense because it was lower than the
theoretical yield calculated (1.57 g). Yield may have been lost because the crystals formed too
quickly, causing some of them to be small in size and seep through the filter and into the filtrate. 15)
The percent recovery of 55.7% makes sense. The recovery is lower than 100% because the product
contained impurities before it was recrystallized, which made it heavier in mass. After
recrystallization, these impurities were removed, causing the final pure product to have less mass.
These impurities were confirmed to exist through

Nucleophilic Aromatic Substitution of...
Experiment #1 Nucleophilic Aromatic Substitution of 2,4–dinitrochlorobenze Name: Anouk Deck–
Leger Student I.D: 9380868 Date performed: September 13th, 2010 Due Date: September 20th,
2010 Introduction: The company DNCB produces large amounts of 2,4–dinitrochlorobenzene and
they sell this product to treat against warts and severe and chronic hair loss. It can also be used as an
alternative treatment for HIV. The supervisor notices an excess amount of m–aminobenzoic acid
stored away which is currently not being used for anything. This reactant can be used in certain
reactions to produce valuable solutions for ophthalmologists. This product is going to waste, and our
objective is to see if we can obtain a usable end product when ... Show more content on
Helpwriting.net ...
FT–IR spectrum The major peaks are C=O, O–H, N–H, C=C, and NO2.1 A peak between 1660 and
1820 cm–1 (presence of C–O) and a broad peak between 2400 and 3400cm–1 (presence of O–H)
indicates the presence of carboxylic acid (COOH). In this graph there is a peak at 3248 cm–1 (C–O)
and a peak at 1589 cm–1 (O–H) which confirms the presence of carboxylic acid. To confirm the
presence of amides, a medium absorption near 3400 cm–1 should be observed which is the case in
this graph. Absorptions in the region of 1450 to 1600 cm–1 often designate an aromatic ring and
when consulting the C–H region we can confirm its aromatic since the peak is to the left of the 3000
cm–1. Nitro groups give two strong absorptions at 1530–1600 cm–1 and 1300–1390 cm–1. H NMR
spectrum Nuclei tend to be deshielded by groups which withdraw electron density. Deshielded
nuclei resonate at higher δ values, whereas shielded nuclei resonate at lower δ values.Examples of
electron withdrawing substituents are –OH, –OCOR, –OR, –NO2 and halogens. Discussion: We
began this experiment by weighing out the known values of mass for 1–chloro–2,4–dinitrobenzene
and m–aminobenzoic acid which was 1.012g and 0.686g respectively. These values were calculated
before the experiement based on the mole ratios of the balanced equation in question. Here the
reaction was a 1:1 ratio so we were able to determine the theoretical mass of our product, m–(2,4–
dinitroanilino)

Competing Nucleophiles Lab

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