Iodo-3-Chloropropane Essay

Reaction 2 - 1.Used a Beral- type pipet, added about 2mL (40 Drops) of 1M hydrochloric acid solution to a small test tube.

The products of the primary alcohol reaction, 1-butanol and HCl, are 1-chlorobutane and water; products of the secondary alcohol, 2-butanol and HCl are 2-chlorobutane and water; products of the tertiary alcohol, 2-methyl-2-propanol are 2-methyl-2-chloropropane and water.

A reaction of the Grignard reagent and carbon dioxide results in an acid, and reaction of a nitrile and a Grignard reagent produce a carbonyl via an imine intermediate. These are show below, respectively.

As detailed in Pavia 's Organic Laboratory techniques the reaction is expected to proceed via the following reaction:

The objective of this laboratory experiment is to study both SN1 and SN2 reactions. The first part of the lab focuses on synthesizing 1-bromobutane from 1-butanol by using an SN2 mechanism. The obtained product will then be analyzed using infrared spectroscopy and refractive index. The second part of the lab concentrates on how different factors influence the rate of SN1 reactions. The factors that will be examined are the leaving group, Br – versus Cl-; the structure of the alkyl group, 3◦ versus 2◦; and the polarity of the solvent, 40 percent 2-propanol versus 60 percent 2-propanol.

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 wet, crude product was placed into the 50 mL Erlenmeyer flask. Small amounts of CaCl2 were added to dry the solution. The flask was sealed and the mixture was swirled and left to settle. Once

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.

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.

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.

However, based on the data obtained in Table 1, Figure 1 shows that all of the chemical substances formed a precipitate, while the tertiary substrate 2-chloro-2-methylpropane formed a precipitate the fastest. This makes sense because as a rule in SN1 reactions, the more stable the carbocation is the faster the reaction will occur. Also, SN1 reactions will prefer tertiary substrates to secondary substrates and secondary substrates to primary substrates. The next substrates to form a precipitate were 2-bromobutane followed by 1-bromobutane. However, it was expected that 2-chlorobutane would form a precipitate before 1-bromobutane because 2-chlorobutane is a secondary substrate, and therefore has a more stable carbocation. The reason that this occurred is because bromine is a better leaving group than chlorine, which allows it to bind easier with the silver ion. The reactions that formed the heaviest precipitate were 2-bromobutane, 1-bromobutane, and 2-chloro-2-methylpropane. This is because these reactions occurred at a faster rate and therefore, generated more of a product than 2-chlorobutane and 1-chlorobutane, which only formed a precipitate upon cooling from the warm water bath.

What is the structure of the complete TCR and BCR complexes? (In addition to text, you may use your own drawings if that is helpful.)

Therefore, the carbocation intermediate must have a bromine atom. 2. A The product must have at least one halogen atom. In step 1 of the electrophilic addition mechanism, the incoming alkene causes the heterolytic fission of the Br-Br bond.

4) Try and propose a mechanism for the reaction using the orders of reaction taking into account the iodine, propanone and sulphuric acid.

Part  C   1) Results  in  table  form   Compound   Water  &  Ethyl  Alcohol   O H O H O H O H H      &       H      &       Cl Cl   H    &     O H    &