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}](https://tex.z-dn.net/?f=%5Ctext%7BRCOOR%7D%27%20%5Cxrightarrow%5B%5Ctext%7B2.%20H%7D%5E%7B%2B%7D%5D%7B%5Ctext%7B1.%20R%24%5E%7B%5Cprime%20%5Cprime%7D%24MgBr%7D%7D%5Ctext%7BRR%24_%7B2%7D%5E%7B%5Cprime%20%5Cprime%7D%24C-OH%7D)
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.
Using the formula density = mass ÷ volume, I got my answer as 10g/cm^3 or 10g/mL
Answer: The concentration of the acid is 0.01 moles acid/0.040 L = 0.25 moles/L = 0.25 M
Explanation:
Answer:
16.0%.
Explanation:
Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.
V % of ethanol = (volume of ethanol / volume of the solution) x 100.
volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.
∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.