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The concept of nucleophilic substitution reactions, focusing on the preparation of 2-chloro-2-methylpropane from 2-methyl-2-propanol using hydrochloric acid. It discusses the roles of nucleophiles and electrophiles, the importance of leaving groups, and the physical properties and safety considerations of the reactants and products.
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The simplest organic molecules contain only the elements carbon and hydrogen. As these elements have similar electronegativities, the resultant compounds are essentially non-polar. Further, if all the carbon atoms are sp^3 hybridised ( i.e. there are no π-bonds), the molecule will be quite unreactive.
Carbon forms strong bonds with a large number of other elements (called hetero-atoms). In the majority of cases, there is a large difference between the electronegativities of carbon and the hetero-atom, and the resultant bond is therefore polarised. This means that many organic molecules have slightly negatively charged (δ–) and slightly positively charged (δ+) centres. These centres are very important in a vast number of chemical reactions and have been given special names - viz. nucleophile and electrophile.
A nucleophile has an electron rich site and forms a bond by donating an electron pair to an electrophile. (From the Greek for “nucleus loving”.) Examples of nucleophiles are:
An electrophile has an electron-poor site and will react with the electron pair of a nucleophile to form a bond. (From the Greek for “electron loving”.) Examples of electrophiles are:
δ δ
δ
The most common reaction of the functional groups in saturated alkyl compounds is nucleophilic substitution. The nucleophile, or electron rich species, attacks the electrophilic carbon of the alkyl group to give the substituted product. A different nucleophile is generated as a by-product of the reaction.
Table E26-1 on the following page gives a number of examples of nucleophilic substitution reactions. Note the use of “curly arrows”, which indicate the movement of electrons, i.e. the breaking and formation of bonds. The nucleophilic and electrophilic centres in the starting materials are indicated with δ and δ⊕ symbols respectively.
E26-
Table E26-1 Mechanism of some Nucleophilic Substitution Reactions
Substrate Nucleophile Products
δ
δ
δ δ
δ δ
δ
δ
In this experiment, 2-chloro-2-methylpropane is prepared from 2-methyl-2-propanol by the action of concentrated hydrochloric acid. The mechanism of the reaction is shown below.
3
Not all nucleophiles react with all electrophiles. For a nucleophilic substitution reaction to occur, it is necessary that the bond to be broken does so at an appreciable rate. This is highly dependent on the nature of the leaving group (displaced nucleophile). Note that the actual leaving group in this reaction is H 2 O, not OH. Also note the formation of the electron- deficient intermediate (III), which is called a carbocation.
LAB-WORK
Physical properties of 2-chloro-2-methylpropane and 2-methyl-2-propanol
Test the miscibility of these two compounds with water in the following way. Set up a clean semi-micro test-tube containing 0.5 mL of water. Add 3 drops of 2-chloro-2-methylpropane, shake, and observe whether or not the liquids are miscible. If the liquids are immiscible note which is less dense. Repeat using 2-methyl-2-propanol in place of the alkyl halide. Summarise your results below.
The reaction between 2-methyl-2-propanol and hydrochloric acid can be represented as follows.
The full sequence of operations, however, is best summarised using a flow chart. Complete the following flow chart.
solid liquid
pure product
distil
discard down sink
bottom layer (^) shake top layer
discard in bin
add NaHCO 3 solution discard down sink
add CaCl 2 / filter
top layer separate bottom layer
shake vigorously HCl
CH 3 C CH (^3)
CH (^3) OH
hydrated CaCl (^2)
Preparation of 2-chloro-2-methylpropane
In a clean 100 mL conical flask collect 2-methyl-2-propanol (15.0 g = 19.0 mL) from the calibrated dispenser. Support a 250 mL separatory funnel on a retort stand and ring. Ensure that the tap is closed. Transfer the 2-methyl-2-propanol to the separatory funnel and add concentrated hydrochloric acid from a measuring cylinder (60 mL, 10 M, kept in fume hood). Swirl the funnel until the two liquids are thoroughly mixed. Stopper the funnel, invert it and immediately relieve the pressure by opening the tap. Close the tap and shake the funnel gently for 2 - 3 seconds only. Again, relieve the pressure immediately. Repeat the procedure of shaking followed by the regular release of pressure for approximately 15 minutes. Allow the layers to separate. If conversion is complete, the lower acid layer should be clear.
Remove the stopper , run off the lower acid phase and shake the remaining 2-chloro-2- methylpropane with sodium hydrogencarbonate solution (5%, 10 mL) to remove traces of acid. Shake cautiously to begin with, and relieve pressure of evolved CO 2 by inverting the funnel and opening the tap. Remove the lower aqueous layer. Now pour the crude product out of the top of the separatory funnel into a small dry conical flask. Dry the 2-chloro-2-methylpropane over a spatula full of anhydrous calcium chloride. Calcium chloride is used as the desiccant in this experiment as in addition to absorbing water, it also absorbs alcohols (and some other classes of compounds). It therefore removes traces of the starting alcohol as well as water.
Holding a separatory funnel during shaking;
Always point the stem away from others during venting (opening the tap to reduce pressure)
Spectroscopic identification of 2-methyl-2-propanol
POST-WORK
Consider the mechanism of the reaction given in the Introduction and repeated below.
3
What general name is given to the positively charged species, (III)?
Would you expect 2-propanol, (CH 3 ) 2 CHOH, to undergo the same reaction more or less readily? Give a reason for your answer.
No reaction occurs when 2-methyl-2-propanol is shaken with concentrated aqueous potassium chloride. Account in mechanistic terms for the difference in reactivity between HCl and KCl (that is, explain why H⊕^ is necessary in addition to Cl ).
Why does 2-methyl-2-propanol have a higher boiling point than 2-chloro-2-methylpropane, although its molecular weight is lower? (Hint: consider the intermolecular bonding in the two compounds.)
Demonstrator’s Initials