Question

Describe the appearance of:

Silver nitrate test

Commercial 1- bromobutane: unheated? and heated?

Commercial t-butly chloride: unheated? and heated?

Sodium iodide test

Commercial 1- bromobutane: unheated? and heated?

Commercial t-butly chloride: unheated? and heated?

Answer 1

Silver nitrate test

The overall reaction is shown in the following equation:

                                                           RX + AgNO₃ =====> AgX + RONO₂

Such a reaction will be of the S_N1 type. Tertiary halides are more reactive in an S_N1 reaction than secondary halides, which are in turn more reactive than primary halides.

The order of reactivity is follows, primary < secondary < tertiary.

Here the least reactive are primary alkyl halides and the most reactive is tertiary alkyl halides.

a).But primary alkyl bromides and iodides are reactive compared to chlorides. So here 1-bromobutane reacts with silver nitrate upon heating.

b) Since it is an SN1 reaction tertiary butyl chloride has high reactivity.( because tertiary butyl cations more stable).

Thus tertiary butylchloride reacts with AgNO3 at room temperature itself forming a precipitate of silver chloride.

Sodium iodide test

The reactions that occur are S_N2 substitutions in which iodide ion is the nucleophile; the order of reactivity is primary > secondary > tertiary.

So the highest reactivity is for primary alkylhalides.

                    

                                      acetone

                     RBr + NaI ------------> RI + NaBr   

            {With the reagent, primary bromides give a precipitate of sodium bromide in about 3 min at room temperature, whereas the primary and secondary chlorides must be heated to about 50oC before reaction occurs. Secondary and tertiary bromides react at 50oC, but the tertiary chlorides fail to react in a reasonable time. It should be noted that this test is necessarily limited to bromides and chlorides.}

a). So here 1-bromobutane which is a primary alky halide which reacts faster with out heating forming a precipitate of sodium bromide [ SN2 Reaction]

CH3CH2CH2CH2Br + NaI === > NaBr + CH3CH2CH2CH2I ( no heating required)

b). But tertiary butylchloride is unreactive with sodium iodide

(CH3)3CCl + NaI == = > No reaction