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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:

1)0.2

2)0.72

3)0.01

Explanation:

Formula is 1)454÷2270

2)0.6×1.2

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2 years ago
A certain chemical reaction releases 36.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of re
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Answer:

0.038 g of reactant

Explanation:

Data given:

Heat release for each gram of reactant consumption = 36.2 kJ/g

mass of reactant that release 1360 J of heat = ?

Solution:

As  36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J

As we know

1 KJ = 1000 J

So

36.2 kJ = 36.2 x 1000 = 36200 J

So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release

Apply unity formula

                 36200 J of heat release ≅ 1 gram of reactant

                 1360 J of heat release ≅ X gram of reactant

Do cross multiplication

              X gram of reactant = 1 g x 1360 J / 36200 J

              X gram of reactant = 0.038 g

So 0.038 g of reactant will produce 1360 J of heat.

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
NeX [460]

Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

[SO^{2-}_{3}] = [Na_{2}SO_{3}] = 0.58 M

The sulphite anion will act as a base and react with H_{2}O to form HSO^{-}_{3} and OH^{-}.

As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

                x = 0.0003 M

So,   x = [HSO^{-}_{3}] = [OH^{-}] = 0.0003 M

[SO^{2-}_{3}] = 0.58 - 0.0003

                     = 0.579 M

Now, we will use [HSO^{-}_{3}] = 0.0003 M

The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              K_{b} = \frac{x^{2}}{0.0003 - x}

        K_{b} = \frac{K_{w}}{K_{a_{1}}}

                      = \frac{10^{-14}}{1.4 \times 10^{-2}}

                      = 7.14 \times 10^{-13}

Therefore,  7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

                              = 0.00214 \times 10^{-13}

                     x = 0.0146 \times 10^{-6}

x = [OH^{-}] = [H_{2}SO_{3}] = 1.46 \times 10^{-8}

    [H^{+}] = \frac{10^{-14}}{[OH^{-}]}

                = \frac{10^{-14}}{0.0003}

                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

5 0
3 years ago
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