How do you name this compound using IUPAC system (including steps)?How would you name this organic compound?Nomenclature with Complex SubstituentsNumber of stereoisomers of 3-ethyl-1-pentene-1,4-diolIUPAC name of this compoundDifficulty in cis, trans isomers in AlkeneWhat is IUPAC name of this compound?How did they assign absolute configuration to these cis and trans 2-methylcyclohexanols?Assigning the IUPAC name of this compoundAssigning locants when both unsaturation and substituents are present in a hydrocarbonHow do you name this compound?
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How do you name this compound using IUPAC system (including steps)?
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How do you name this compound using IUPAC system (including steps)?
How would you name this organic compound?Nomenclature with Complex SubstituentsNumber of stereoisomers of 3-ethyl-1-pentene-1,4-diolIUPAC name of this compoundDifficulty in cis, trans isomers in AlkeneWhat is IUPAC name of this compound?How did they assign absolute configuration to these cis and trans 2-methylcyclohexanols?Assigning the IUPAC name of this compoundAssigning locants when both unsaturation and substituents are present in a hydrocarbonHow do you name this compound?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
And including stereochemistry (cis trans or R S )
I was told I should demonstrate some effort to explain my knowledge of underlying concepts (which are concepts that I already know?)
So here it is : I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question,so now it should be 3-bromo-3-methyl-1-hexene, my quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)
I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.
trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?
organic-chemistry nomenclature
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
$endgroup$
add a comment |
$begingroup$
And including stereochemistry (cis trans or R S )
I was told I should demonstrate some effort to explain my knowledge of underlying concepts (which are concepts that I already know?)
So here it is : I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question,so now it should be 3-bromo-3-methyl-1-hexene, my quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)
I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.
trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?
organic-chemistry nomenclature
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
$endgroup$
1
$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago
add a comment |
$begingroup$
And including stereochemistry (cis trans or R S )
I was told I should demonstrate some effort to explain my knowledge of underlying concepts (which are concepts that I already know?)
So here it is : I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question,so now it should be 3-bromo-3-methyl-1-hexene, my quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)
I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.
trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?
organic-chemistry nomenclature
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
$endgroup$
And including stereochemistry (cis trans or R S )
I was told I should demonstrate some effort to explain my knowledge of underlying concepts (which are concepts that I already know?)
So here it is : I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question,so now it should be 3-bromo-3-methyl-1-hexene, my quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)
I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.
trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?
organic-chemistry nomenclature
organic-chemistry nomenclature
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
edited 7 hours ago
ALPHAz CoC
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
asked 8 hours ago
ALPHAz CoCALPHAz CoC
294 bronze badges
294 bronze badges
1
$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago
add a comment |
1
$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago
1
1
$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago
$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.
(3R)-3-bromo-3-methylhex-1-ene.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
add a comment |
$begingroup$
When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:
- Solid lines are on plane of your paper;
- Wedges are above the paper; and
- Dashed lines are below the paper.
That means, $ce-Br$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $ce-Br$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.
(3R)-3-bromo-3-methylhex-1-ene.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
add a comment |
$begingroup$
Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.
(3R)-3-bromo-3-methylhex-1-ene.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
add a comment |
$begingroup$
Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.
(3R)-3-bromo-3-methylhex-1-ene.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.
(3R)-3-bromo-3-methylhex-1-ene.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
answered 7 hours ago
Michael GreenMichael Green
917 bronze badges
917 bronze badges
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Michael Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
add a comment |
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
3
3
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
(1) R should be italicised; (2) current IUPAC recommendations also state that the locant "3" should be associated with the stereodescriptor "R", i.e. (3R)- instead of (R)-; (3) no hyphen between methyl and hex. (This may be pedantic, but nomenclature itself is arguably an exercise in pedantry, so it makes no sense to go half the distance.)
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
$begingroup$
This is why I'm a materials chemist haha
$endgroup$
– Michael Green
7 hours ago
3
3
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
$begingroup$
To be fair, few people care about the actual rules (organic chemists will just use ChemDraw to generate names), and the way it's taught is often inconsistent with the actual rules...
$endgroup$
– orthocresol♦
7 hours ago
add a comment |
$begingroup$
When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:
- Solid lines are on plane of your paper;
- Wedges are above the paper; and
- Dashed lines are below the paper.
That means, $ce-Br$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $ce-Br$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
$endgroup$
add a comment |
$begingroup$
When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:
- Solid lines are on plane of your paper;
- Wedges are above the paper; and
- Dashed lines are below the paper.
That means, $ce-Br$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $ce-Br$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
$endgroup$
add a comment |
$begingroup$
When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:
- Solid lines are on plane of your paper;
- Wedges are above the paper; and
- Dashed lines are below the paper.
That means, $ce-Br$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $ce-Br$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
$endgroup$
When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:
- Solid lines are on plane of your paper;
- Wedges are above the paper; and
- Dashed lines are below the paper.
That means, $ce-Br$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $ce-Br$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
answered 6 hours ago
Mathew MahindaratneMathew Mahindaratne
10.4k1 gold badge12 silver badges37 bronze badges
10.4k1 gold badge12 silver badges37 bronze badges
add a comment |
add a comment |
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$begingroup$
Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.
$endgroup$
– user55119
7 hours ago