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 : The pH of 0.289 M solution of lithium acetate at
is 9.1
Explanation :
First we have to calculate the value of
.
As we know that,

where,
= dissociation constant of an acid = 
= dissociation constant of a base = ?
= dissociation constant of water = 
Now put all the given values in the above expression, we get the dissociation constant of a base.


Now we have to calculate the concentration of hydroxide ion.
Formula used :
![[OH^-]=(K_b\times C)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%28K_b%5Ctimes%20C%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
![[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%285.5%5Ctimes%2010%5E%7B-10%7D%5Ctimes%200.289%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[OH^-]=1.3\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.3%5Ctimes%2010%5E%7B-5%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)


Now we have to calculate the pH.

Therefore, the pH of 0.289 M solution of lithium acetate at
is 9.1
Answer:
pH= 12.7
Explanation:
a)Keep an emergency kit and fresh water in your car in the event of _an unexpected event.
b) pOH= -log[OH-]
pOH= -log[5.00×10^-2]
pOH= 1.3
But pH+pOH= 14
pH= 14-pOH= 14-1.3
pH= 12.7