Bromoethane with nucleophiles

What is the product of the reaction of bromoethane with each of the following nucleophiles a. CH3CH2CH2O- b. CHsC C- c. (CH3)3N d. CH3CH2S-... 

In Chapter 16 (page 218) you learnt how bromoethane undergoes nucleophilic substitution with ammonia to form a mixture of amines. In order to prepare ethylamine (while avoiding the formation of secondary and tertiary amines and ammonium salts) we use excess hot ethanolic ammonia ... 

We call the carbocation, which exists only transiently during the course of the multistep reaction, a reaction intermediate. As soon as the intermediate is formed in the first step by reaction of ethylene with H+, it reacts further with Br in a second step to give the final product, bromoethane. This second step has its own activation energy (AG ), its own transition state, and its own energy change (AG°). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which Br- donates a pair of electrons to the positively charged carbon atom as the new C-Br bond starts to form. 

The nucleophilic hydroxide ion attacks the C atom in the bromoethane from the side opposite to the C-Br bond and begins to form a covalent bond with it. At the same time, the C-Br bond begins to break. A transition state is then reached in which the new 0-C bond is partially formed and the C-Br bond is partially broken. The reaction is completed by the formation of the full 0-C bond and the complete break-up of the C-Br bond. 

Furthermore, allenyl phenyl selenides have been prepared by alkylation of 1-lithioallenyl phenyl selenide, itself produc on metallation of phenyl 1-propynyl selenide or of allenyl phenyl selenide - (Scheme 42). 1-Lithioallenyl phenyl selenide has poorer nucleophilic properties than that of a-seleno-allyllithiums, since the reaction of the former with 2-phenyl-1-bromoethane gives predominantly styrene, whereas the latter leads to products of substitution in good yield. ... 

J Does the fact that bromoethane undergoes substitution faster with the ethoxide anion than with ethanol fit with the order of reactivity of nucleophiles ... 

Bromoethane reacts with ethanol (or with sodium ethoxide) by the Sn2 mechanism. Because ethanol is a poor nucleophile this reaction is very slow. On the other hand. 

Table 2.12. Nucleophilic Strength with Bromoethane as Ordered by the Swain-Scott Equation...

The mechanism for this reaction begins with formation of an oxonium cation. Then, an Sn2 reaction with a bromide ion acting as the nucleophile produces ethanol and bromoethane. Excess HBr reacts with the ethanol produced to form the second molar equivalent of bromoethane. 

Reaction with butyllithium (a strong base) firstly deprotonates the more acidic alcohol, and only then deprotonates the benzylic methyl group to give a resonance stabilised carbanion this nucleophilic carbanion is quenched with bromoethane (5 2 reaction), but only the most reactive benzylic carbon centre reacts with the... 

Steric hindrance. To complete a substitution reaction, the nucleophile must approach the substitution center and begin to form a new covalent bond to it. If we compare the ease of approach by the nucleophile to the substitution center of a 1° haloalkane with that of a 3° haloalkane, we see that the approach is considerably easier in the case of the 1° haloalkane. Two hydrogen atoms and one alkyl group screen the backside of the substitution center of a 1° haloalkane. In contrast, three alkyl groups screen the backside of the substitution center of a 3° haloalkane. This center in bromoethane is easily accessible to a nucleophile, while there is extreme crowding around the substitution center in 2-bromo-2-methylpropane ... 

Draw the product expected when each of the following nucleophiles reacts with bromoethane NaOMe, CHgC CrNa, NaCN, and Nal. Draw the product expected when iodomethane reacts with each of these four nucleophiles. 

The effect of j8-branching in Sj 2 reactions on a primary haloalkane. With bromoethane, attack of the nucleophile is unhindered. V 4th l-bromo-2,2-dimethylpropane, the three /3-branches block approach of the nucleophile to the backside of the C—Br bond, thus drastically reducing the rate of Sj 2 reaction of this compound. 

For example, in protic solvents such as water (H2O) and water-ethanol (ethyl alcohol, CH3CH2OH) mixtures, large ions that are more polarizable and may more readily provide an electron pair for the displacement process are also more poorly solvated than small ions. Thus, with the same substrate (e.g., bromoethane [ethyl bromide, CH3CH2Br]) as shown in Table 7.5, nucleophilicity increases with increasing atomic number in any one group (column) of the periodic table (i.e., T > Br > Cl > F") in such solvents. In this vein, it should be noted that this means that nucleophilicity does not parallel basicity since, as noted earlier, the order of basicity in aqueous solution is Cl > Br > F. 

As mentioned earlier, ammonia can act as a nucleophile through its lone pair of electrons on the nitrogen atom. Ammonia reacts with bromoethane by an S),j2 mechanism to form ethylamine and hydrogen bromide ... 

However, the product ethylamine is also a nucleophile, because the nitrogen atom still has a non-bonding pair of electrons. Thus the reaction can proceed further. The product, ethylamine, can react with more bromoethane to produce diethylamine — a secondary amine ... 

Nucleophilic substitution reactions are important in organic synthesis because the halogen atom on halogenoalkanes can be replaced by other functional groups. The reaction with potassium cyanide is a good illustration of this. The cyanide ion reacts to form a nitrile. For example, bromoethane reacts by an Sj42 mechanism with a solution of potassium cyanide in ethanol to form propanenitrile ... 

Both the nucleophilic substitution reactions we have just considered proceed by an S142 mechanism. Figure 20.22 shows the mechanism involved, with the lone pair on the nitrogen of the ammonia molecule forming a bond to the electron-deficient carbon atom in bromoethane, for instance. 

The first combination will not give the ether product because Sj 2 reactions cannot occur by backside displacement of halogen atoms at sp -hybridi2ed carbon atoms. However, the reaction of the conjugate base of the hydroxyl group at the 2 position of naphthalene with bromoethane occurs readily because bromoethane is an unhindered primary alkyl halide. The nucleophilic oxygen atom of the naphthalene compound is generated by reaction with sodium hydride. 

Hydrogen bromide, HBr, which could react with the ethylamine, is removed by the excess ammonia. It forms ammonium bromide, NH Br. The excess ammonia also reduces the chances of bromoethane being attacked by ethylamine ethylamine is also a nucleophile. 

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