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Bezzdna [24]
3 years ago
9

How would you prepare 2-methyl-2-propanol via a grignard with dimethyl carbonate as your carbonyl source? show all reagents?

Chemistry
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

React it with CH₃MgBr and work up the product with saturated ammonium chloride solution

Explanation:

Grignard reagents convert esters into tertiary alcohols.

The general equation is

\text{RCOOR}' \xrightarrow[\text{2. H}^{+}]{\text{1. R$^{\prime \prime}$MgBr}}\text{RR$_{2}^{\prime \prime}$C-OH}

The Grignard reagent in this synthesis is methylmagnesium bromide. You prepare it by reacting a solution methyl bromide in anhydrous ether with magnesium and a few crystals of iodine.

The reaction consumes 3 mol of CH₃MgBr per mole of dimethyl carbonate, and everything happens in the same pot.

Acid workup of the product usually involves the addition of a saturated aqueous solution of ammonium chloride and extraction with a low-boiling organic solvent.

The mechanism involves:

Step 1. Nucleophilic attack and loss of leaving group

(a) The Grignard reagent attacks the carbonyl of dimethyl carbonate, followed by (b) the loss of a methoxide leaving group.

Step 2. Nucleophilic attack and loss of leaving group

(a) A second mole of the Grignard reagent attacks the carbonyl of methyl acetate, followed by (b) the loss of a methoxide leaving group.

Step 3. Nucleophilic attack and protonation of the adduct.  

(a) A third mole of the Grignard reagent attacks the carbonyl of acetone, followed by (b) protonation of the alkoxide to form 2-methylpropan-2-ol.

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In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign
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Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

3 0
3 years ago
How many moles are found in 133 g of Mg3 N2
konstantin123 [22]
Magnesium nitride weighs 100.95 g/mol

133/100.95 = 1.32 mols
7 0
3 years ago
Read 2 more answers
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