Which of the following is the requirement for the intramolecular Sn2 reaction (intramolecular William Ether Synthesis) to occur
in your reaction? Group of answer choices Anti-periplanar The arrangement of the bromine and hydroxide does not matter. cis Z
1 answer:
Answer:
D. Anti-periplanar
Explanation:
In the <u>second step</u> of the intramolecular William Ether Synthesis mechanism (figure 1) we will have the attack of the negative charge of the oxygen to the carbon bond to the Br. At the same time the Br leaves, so a bond would be broken (the <u>C-Br</u> bond) and a bond would be formed (the <u>C-O</u> bond).
Now, this process can happen only if the <u>attack</u> and the <u>leaving group </u>has an anti configuration (figure 2). In an anti configuration the <u>nucleophile</u> and the <u>leaving group</u> would have <u>opposite directions</u>.
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Answer :
The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g
Given : pH = 6.86
Total concentration of Phosphate buffer = 0.1 M
Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?
Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?
Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :
(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )
<u>Step 1 : To find pka </u>
H₂PO₄⁻ <=> HPO₄²⁻
The above reaction has pka = 7.2 ( from image shown )
<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.
Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :
pH = 6.86 pKa = 7.2
Subtracting both side by 7.2
Removing log
---------------- equation (1)
<u>Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻</u>
Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M
Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M
Assume [H₂PO₄⁻ ] = x
So , [x ] + [ HPO₄²⁻ ] = 0.1 M
[ HPO₄²⁻ ] = 0.1 - x
Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]
[H₂PO₄⁻ ] = x
[ HPO₄²⁻ ] = 0.1 - x
Equation (1) = >
Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :
Cross multiplying
Adding x on both side
Dividing both side by 1.457
x = 0.0686 M
Hence , [H₂PO₄⁻ ] = x = 0.0686 M
[ HPO₄²⁻ ] = 0.1 - x
[ HPO₄²⁻ ] = 0.1 - 0.0686
[ HPO₄²⁻ ] = 0.0314 M
Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .
Molarity is defined as mole of solute per 1 L volume of solution .
Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L
Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L
Since that volume of buffer solution is 1 L , so Molarity = mole
Hence Mole of NaH₂PO₄ = 0.0686 mol
Mole of Na₂HPO₄ = 0.0314 mol
<u>Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄ </u>
Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :
Molar mass of Na₂HPO₄ =
Molar mass of NaH₂PO₄ =
Mass of Na₂HPO₄ = 4.457 g
Mass of NaH₂PO₄ = 8.23 g
