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Maslowich
3 years ago
10

Explain why all glassware must be dry and the solvent anhydrous during formation and reaction of a grignard reaction. Further, e

xplain whether acetone would be a viable solvent for generation of a grignard reagent. Include one or more balanced chemical equations in your answer.
Chemistry
1 answer:
mario62 [17]3 years ago
7 0

Answer:

The glassware and solvent for the production of grignard reagent and its reaction should by dry and anhydrous so as to prevent the conversion of the grignard reagent into saturated alkane. Acetone, is not a good solvent for the generation of grignard reagent because it has the potential of forming alcohol.

Explanation:

Grignard reagent is a compound formed by the reaction of halide of alkyl or alkene with magnesium metal. This makes the compound more nucleophilic. Thus the availability of an electrophile around it will evoke a quick reaction. The presence of water in an acidic condition will generate hydroxonium ions which are highly electrophilic. There will therefore be an introduction of hydrogen ions to the grignard reagent and which will displace the MgBr leading to the formation of alkane.

Acetone tends to produce alcohol when they are exposed to grignard reagent, thus they are not appropriate to be used as solvent. This is due to the resultant highly electrophilic nature of the carbonyl carbon on the acetone, thus will react with the nucleophilic carbon on the grignard. The reaction can be represented as follows:

RMgBr + CH3(CO)CH3 + H (with hydrogen ions) >>>RCOH + Mg(OH)Br

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Answer:

No precipitate is formed.

Explanation:

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In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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