Nucleophilic Substitution Reaction Lab Report

A. The reaction between 2-butanol, anhydrous (CH3CH2CHOHCH3) and hydrogen bromine (HBr) is a substitution reaction. Another example of a substitution reaction would be CH3Cl + −OH→ CH3OH + Cl

Determining the identity of unknown compounds in a given solution is an important skill to have. This can be done through a series of steps including extraction, purification, and spectroscopic methods.1

From our results we can conclude that our product is 2-methyl-2-hexanol. We came to this conclusion because our product did not react with the sodium test. Tertiary alcohols are the slowest to react with sodium because the hydroxyl group is very crowded. On the contrary, methanol reacts almost immediately because there are only hydrogen atoms surrounding the hydroxyl group. 2-methyl-2-propanol only reacted after a few minutes and no gas was evolved. It can be assumed that our product is a tertiary alcohol which 2-methyl-2-hexanol is. The next test that was performed was the lucas test. In the lucas test tertiary alcohols react the fastest because a carbocation is made and tertiary carbocations are easiest to make. Tertiary alcohols, such as

Three types of reaction were conducted, unimolecular nucleophilic substitution reaction (SN1), unimolecular elimination reaction (E1), and bimolecular nucleophilic substitution reaction (SN2). Substitution and elimination reactions are important to allow for the transformation of reactants to desired product. This is extremely important when it comes to synthesis of different molecules for research, commercial, and medical purposes. SN1reaction was to create 2-chloro-2-methylbutane from 2-methyl-2-butanol with a reaction yield of 2.28 g (21.4 mmol) and the reaction product was analyzed by infrared spectroscopy (IR) to determine the purity of 2-chloro-2-mehtlybutane. E1 was to create cyclohexene from cyclohexanol with a reaction yield of 0.17

The major reaction that happened was an E1 reaction (otherwise known as a unimolecular elimination reaction). In this reaction the removal of an HX substituent group results in the formation of a double bond via beta hydrogen elimination4. In an E1 reaction the deprotonation of hydrogen in the beta position occurs and from here a carbocation is formed, which results in the formation of an alkene product. This happens when the OH group on the 2-methylcyclohexanol is hydrated by H2SO4, which allows the OH to become an H2O. Since H20 is a better leaving group now it will

The bromide ion is the better nucleophile than the chloride ion because of its larger size and since it is a weaker base. The chloride ion is more electronegative than bromide ion, so it tends to hold electrons in closerE. Since the bromide ion is less electronegative and has more electrons, it is able to share unpaired electrons easier than the chloride ion. In the case of this laboratory experiment, bromide was the better nucleophile in the protic solvent containing hydrogen bonded to oxygen of 1-butanol and 2-methyl-2-butanol as the protic solvent molecules formed strong ion-dipole interactions with the negative-charged nucleophile of the bromide ion, which created a solvent barrier around the nucleophileG. For the electrophile to be attacked

Anti-Markovnikov’s rule expresses the opposite. In an anti-Markovnikov reaction, the halide group attaches at the less substituted carbon, while the hydrogen attaches at the more substituted carbon. The term “less substituted carbon” refers to the carbon with the most surrounding hydrogens, and the term “more substituted carbon” refers to the carbon with the least surrounding hydrogens. The objective of the experiment was to determine which mechanism was used to create the product of either 1-hexanol or 2-hexanol after hydration of

For this experiment, Gas Chromatography (GC) is utilized to observe the streo- and regiochemistry among four elimination products under either basic or acidic conditions. The reagents used in this experiment include primary and secondary alcohols which are in acidic conditions and dehydrobromintion of primary and secondary alkyl halides in basic conditions. When reacted the gas products are collected and undergo further analysis in the GC which will separate the product into its components and will show the different distributions among the product. Theory

The alcohol is attacked by a nucleophile; two hydroxyl groups are then formed as intermediates. After a proton shift, one of the alcohol groups is removed and water and etser are

An electrophilic aromatic substitution reaction occurs when a hydrogen atom from an aromatic ring is replaced with an electrophile. This reaction is of the SN2 type and is a two-step mechanism. The first step is the rate-determining step, where the electrophile forms a sigma-bond to the benzene ring, creating a positively charged intermediate.1 In the second step, the hydrogen atom is removed, resulting in a substituted benzene ring. When a mono-substituted benzene goes through an electrophilic aromatic substitution reaction, there are usually three possible products that can be formed; ortho, para, meta.1 These different products occur because during an electrophilic substitution reaction, certain positions on benzene react faster than the

These compounds are good nucleophiles and are responsible for carbon-carbon formations with other electrophiles.2, 5 This method consists of four steps consisting of protection of your functional group with a dithiane, metalation, reaction with an electrophile, and finally the removal of the protecting group. first, the carbonyl group is protected with 1,3-propanedithiol, an analog to the protecting reagent 1,3-propanediol, in the presence of an acid. This results in the formation of a 1,3-dithiane. In the next step, the central carbon linking the dithiane and -R group is metalated with butylithium replacing the acidic proton with a lithium ion.

( 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.

The reaction for this experiment was the synthesis of an alkyl aryl ether - in this case nerolin - through a Williamson ether synthesis, in which an alcohol (deprotonated 2-naphthol) reacts with an organohalide (iodoethane). First the nucleophile is generated through an acid/base reaction, then ether formation occurs with nucleophilic attack in an Sn2 reaction.

The purpose of this lab was to discover the effects that an alkyl group and solvent have on the rate of SN1 and SN2 reactions. Two separate mechanisms can be used to perform the nucleophilic substitution of alkyl halides: SN1 and SN2. A SN1 reaction, or unimolecular displacement, is a 1st order, nucleophilic substitution that involves two steps. The rate law for this reaction, Rate = k[Rx], doesn’t include the nucleophile in it.1 These two step reactions have a carbocation intermediate. SN1 reactions work best when the central carbon has as many bulky groups surrounding it as possible. These substituents increase the possibility of the carbocation intermediate forming by increasing the steric strain of the molecule. This idea is which causes a carbocation intermediate to develop in reactions of secondary or tertiary alkyl halides, and a primary alkyl halide to not be able to go through a SN1 reaction. The carbocation intermediate that forms is a sp2 carbon that allows the nucleophile to attack from both sides of the molecule. If neither attack is favored, they both occur equally, making it a racemic mixture. The solvent of choice for the SN1 reactions are both polar and protic, because a solvent that will not react with the carbocation intermediate is needed, since this will give an unwanted product. In polar protic solvents you get a hydrogen atom that is very highly polarized since it is attached to an electronegative atom, because of this it can interact

Butanol (ethyl alcohol) is a four-carbon alcohol which has been mainly used as a solvent, chemical intermediate,