Isopropyl bromide

Mix 40 g. (51 ml.) of isopropyl alcohol with 460 g. (310 ml.) of constant boiling point hydrobromic acid in a 500 ml. distilling flask, attach a double surface (or long Liebig) condenser and distil slowly (1-2 drops per second) until about half of the liquid has passed over. Separate the lower alkyl bromide layer (70 g.), and redistil the aqueous layer when a further 7 g. of the crude bromide will be obtained (1). Shake the crude bromide in a separatory funnel successively with an equal volume of concentrated hydrochloric acid (2), water, 5 per cent, sodium bicarbonate solution, and water, and dry with anhydrous calcium chloride. Distil from a 100 ml. flask the isopropyl bromide passes over constantly at 59°. The yield is 66 g. 

The residue in the flask may be mixed with the aqueous layer of the first distillate, 40 g. of isopropyl alcohol added, and the slow distillation repeated. The yield of crude isopropyl bromide in the second distillation is only slightly less than that obtained in the original preparation. Subsequently most of the residual hydrobromic acid may be recovered by distillation as the constant boiling point acid (126°). 

Ratio of second order rate constant k for indicated alkyl bromide to k for isopropyl bromide at 25 C... 

Assume that you need to prepare 4 methyl 2 pentyne and discover that the only alkynes on hand are acetylene and propyne You also have available methyl iodide isopropyl bromide and 1 1 dichloro 3 methylbutane Which of these compounds would you choose in order to perform your synthesis and how would you carry it out" ... 

Treat the sodium salt of diethyl acetamidomalonate with isopropyl bromide Remove the amide and ester functions by hydrolysis in aqueous acid then heat to cause (CH3)2CHC(C02H)2... 

A large excess of hydiazine oi bulky alkyl groups favor monosubstitution. For example, a 60—70% yield of monoisopropylbydrazine [2257-52-7J is achieved by reaction of isopropyl bromide and a fivefold excess of hydrazine (56). 

Displacement of the hydroxyl group is exemplified by the production of isopropyl haUdes, eg, isopropyl bromide [75-26-3] by refluxing isopropyl alcohol with a halogen acid, eg, hydrobromic acid [10035-10-6] (12). 

Alkylations of cinnolin-4(lf/)-one (8) with methyl iodide, ethyl iodide, dimethyl and diethyl sulfates, isopropyl bromide, benzyl chloride, etc. take place predominantly at position 2 to give 2-alkyl-4-hydroxycinnolinium anhydro salts (83), together with small amounts of l-methylcinnolin-4-one (84). 

How can we predict whether substitution or elimination will be the principal reaction observed with a particular- combination of reactants The two most important factors are the structure of the alkyl halide and the basicity of the anion. It is useful to approach the question from the premise that the character-istic reaction of alkyl halides with Lewis bases is elimination, and that substitution predominates only under certain special cucurnstances. In a typical reaction, a typical secondary alkyl halide such as isopropyl bromide reacts with a typical Lewis base such as sodium ethoxide mainly by elimination ... 

FIGURE 8.11 When a Lewis base reacts with an alkyl halide, either substitution or elimination can occur. Substitution (Sn2) occurs when the Lewis base acts as a nucleophile and attacks carbon to displace bromide. Elimination (E2) occurs when the Lewis base abstracts a proton from the p carbon. The alkyl halide shown is isopropyl bromide, and elimination (E2) predominates over substitution with alkox-ide bases. 

The real world of Sn reactions is not quite as simple as the discussion has so far suggested. The preceding treatment in terms of two clearly distinct mechanisms, SnI and Sn2, implies that all substitution reactions will follow one or the other of these mechanisms. This is an oversimplification. The strength of the dual mechanism hypothesis and its limitations are revealed by these relative rates of solvolysis of alkyl bromides in 80% ethanol methyl bromide, 2.51 ethyl bromide, 1.00 isopropyl bromide, 1.70 /er/-butyl bromide, 8600. Addition of lyate ions increases the rate for the methyl, ethyl, and isopropyl bromides, whereas the tert-butyl bromide solvolysis rate is unchanged. The reaction with lyate ions is overall second-order for methyl and ethyl, first-order for tert-butyl, and first- or second-order for the isopropyl member, depending upon the concentrations. Similar results are found in other solvents. These data show that the methyl and ethyl bromides solvolyze by the Sn2 mechanism, and tert-butyl bromide by the SnI mech-... 

Replacement of one of the phenyl groups by an alkyl group of similar bulk, on the other hand, alters the biologic activity in this series. Alkylation of phenylacetonitrile with isopropyl bromide affords the substituted nitrile, 136. Treatment of the anion prepared from 136 with strong base with 2-dimethylamino-l-chloropropane gives isoaminile (137). It is of note that alkylation of this halide, isomeric with that used in the early methadone synthesis, is apparently unaccompanied by isomer formation. Isoaminile is an agent with antitussive activity. 

Schiff s base (j ) derived by reaction of p-chloro-anil ine and borohydride followed by acylation with phenylacetyl chloride produces amide 22,. Selective hydrolysis with HBr followed by alkylation with isopropyl bromide completes the synthesis of lorcainide (20). ... 

To a well stirred suspension of 9 g of sodium phenyl acetate and 2.4 g of magnesium turnings in 25 cc of anhydrous ether, a solution of 9.4 cc of isopropyl bromide in 50 cc of anhydrous ether are added. The mixture is refluxed for one hour (during which time propane is evolved) and then 5 cc of cyclopentanone in 25 cc of anhydrous ether are added dropwise. The mixture is then refluxed for one hour and poured over ice water containing some hydrochloric acid. The ether solution is separated and extracted with 200 cc of 5% sodium hydroxide. The alkaline solution on acidification gives the free acid which is filtered off, dried in a desiccator and recrystallized from a mixture of ethylene dichloride and petroleum ether. 

An interesting case arises when different products result from parallel pathways. The hydrolysis of isopropyl bromide in alkaline solutions is instructive. The two prod-... 

Parallel reactions. Show how the three rate constants that characterize the hydrolysis of isopropyl bromide in alkaline solution, Eqs. (3-68)—(3-70), can be obtained from studies of the kinetics and yields determined over a range of [OH" j. 

We shall now examine some applications of these ideas. Experiments were carried out on the reaction between isopropyl bromide and ethoxide ions. This process consists of competing elimination and nucleophilic substitution reactions,19... 

Deuterium isotope effects have been found even where it is certain that the C—H bond does not break at all in the reaction. Such effects are called secondary isotope effectsf" the term primary isotope effect being reserved for the type discussed previously. Secondary isotope effects can be divided into a and P effects. In a P secondary isotope effect, substitution of deuterium for hydrogen p to the position of bond breaking slows the reaction. An example is solvolysis of isopropyl bromide ... 

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