Question

Aqueous sodium bicarbonate was used to wash the crude n-butyl bromide. a. What was the purpose of this wash? Give equations. b. Why would it be undesirable to wash the crude halide with aqueous sodium hydroxide? 5. Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.

Answer 1

Answer:

See explanation below

Explanation:

a) According to this, this is an excercise that is involving a separation and reaction of different compounds. The n-butyl bromide is formed when, you made the following reaction:

CH2 = CH - CH2 - CH3 + HBr/peroxide -----------> Br - CH2 - CH2 - CH2 - CH3

Now, in this reaction, we still has traces of HBr in the product, so, in order to neutralize these traces, we wash the solution with bicarbonate sodium forming sodium bromide and carbonic acid as follow:

HBr + Na2CO3 ---------> NaBr + Na2CO3

Also, this is a weak base so it will not react with the n-butylbromide to form another product.

b) Basing of what it was stated above, we cannot wash the solution with NaOH because this is a strong base, and not only wil eliminate the traces of HBr, it will also react with the butylbromide causing an elimination and substitution reaction, giving the following products:

BrCH2CH2CH2CH3 + NaOH --->CH2=CHCH2CH3 + OHCH2CH2CH2CH3

That it's why we need to wash this product with a weak base only.

c) The density of 1-chlorobutane is 0.88 g/mL, density of water is 1 g/mL and density of sodium bicarbonate is 2.2 g/cm3, therefore, the one that has a greater density will go at the lower phase.

In this case, after the reflux, it will stay in the lower phase.

after adding water, it will be in the upper phase.

after adding bicarbonate, it will be in the upper phase too.