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.
Answer:
a and b are correct
Explanation:
This because both are aqueous solutions,therefore, identity of solvent is same that is water.
And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles
.
Hence answer is that their freezing points and Identity of the solvent shall remain the same.
Answer:
The answer to your question is: Molarity = 0.078
Explanation:
Data
HCl
V = 250 ml
T = 27°C = 300 °K
P = 141 mmHg = 0.185 atm
V2 = 70 ml
NaOH
V = 24.3 ml
Molarity NaOH = ?
Process
1.- Calculate the number of moles of HCl
PV = nRT
n = PV / RT
R = 0.082 atm l / mol K
n = (0.185)(0.25) / (0.082)(300)
n = 0.046 / 24.6
n = 0.0019 moles
2.- Calculate molarity of HCl
Molarity = moles / volume
Molarity = 0.0019 / 0.070
Molarity = 0.027
3.- Write the balanced equation
HCl + NaOH ⇒ H₂O + NaCl
Here, we observe that the proportion HCl to NaOH is 1:1 .
Then 0.0019 moles of HCl reacts with 0.0019 moles of NaOH.
4.- Calculate the molarity of NaOH.
Molarity = 0.0019 / 0.0243
Molarity = 0.078
Answer: (B) a strong acid
Explanation:
Acids turn blue pH paper red
