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what happens when (+) 2-iodobutane is treated with NaI in acetone?
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After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.
After going through various articles, I am getting reason such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.
And there is a same question where we are treating (R)2-bromobutane with NaI in acetone, but the answer is (S)2-iodobutane.
What could be the reason?
I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.
organic-chemistry halides optical-properties
asked 5 hours ago
kiv
112
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kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.
After going through various articles, I am getting reason such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.
And there is a same question where we are treating (R)2-bromobutane with NaI in acetone, but the answer is (S)2-iodobutane.
What could be the reason?
I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.
organic-chemistry halides optical-properties
asked 5 hours ago
kiv
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
1
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.
After going through various articles, I am getting reason such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.
And there is a same question where we are treating (R)2-bromobutane with NaI in acetone, but the answer is (S)2-iodobutane.
What could be the reason?
I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.
organic-chemistry halides optical-properties
asked 5 hours ago
kiv
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.
After going through various articles, I am getting reason such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.
And there is a same question where we are treating (R)2-bromobutane with NaI in acetone, but the answer is (S)2-iodobutane.
What could be the reason?
I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.
organic-chemistry halides optical-properties
organic-chemistry halides optical-properties
asked 5 hours ago
kiv
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
kiv
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
kiv
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
kiv
112
asked 5 hours ago
kiv
112
112
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
kiv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
1
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago
add a comment |Â
1
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
1
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago
1
1
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
1
1
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction
answered 5 hours ago
Waylander
4,5841619
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction
answered 5 hours ago
Waylander
4,5841619
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
add a comment |Â
up vote
4
down vote
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction
answered 5 hours ago
Waylander
4,5841619
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction
answered 5 hours ago
Waylander
4,5841619
There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction
answered 5 hours ago
Waylander
4,5841619
edited 4 hours ago
answered 5 hours ago
Waylander
4,5841619
answered 5 hours ago
Waylander
4,5841619
answered 5 hours ago
Waylander
4,5841619
4,5841619
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
add a comment |Â
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
But in the first reaction, could the final major product be only (-)2-iodobutane?
– kiv
5 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
No, because the iodides keep exchanging so you get the mixtures.
– Waylander
4 hours ago
add a comment |Â
kiv is a new contributor. Be nice, and check out our Code of Conduct.
kiv is a new contributor. Be nice, and check out our Code of Conduct.
kiv is a new contributor. Be nice, and check out our Code of Conduct.
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1
Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.
– orthocresol♦
4 hours ago
1
If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.
– user55119
1 hour ago