Linear geometry of a triple bond. Created by Sal Khan.
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- Are the triple bonds what makes the alcohol addicting?(1 vote)
- The addictive property of alcohol is related to the release of endorphins and dopamine in the brain, which are part of the brain's reward and pleasure mechanisms. The release of these chemicals is triggered by alcohol consumption.(27 votes)
- what is the reason for the triple bonds to stretch out in a alcohol(2 votes)
- The carbon atoms in the alykne ARE sp hybridized (not sp2). The 2 sp hybrid orbitals on each carbon in the alkyne were made from one s orbital and one p orbital. Looking at the electron configuration of carbon, if we used up one orbital and one p orbital to make the sp hybrid orbitals, the carbon atom has 2 p orbitals left after hybridization. These 2 unhybridized p orbitals on each carbon (a total of 4 p orbitals) are used to make the 2 pi bonds in the alkyne triple bond. Because the pi bonds are formed from the rigid overlap of 4 p orbitals, the alkyne part of the molecule is essentially locked in a linear position. You can google an image of an sp hybridized molecule to see visually how the pi bonds fix the molecule in a linear position.(6 votes)
- if the triple bonds is drawn that way, how about the double bonds? will it be drawn straight like that?(3 votes)
- it may look like this, >=< (not emozi though)
the one pi bond alongside two sigma bonds between two carbons owns one dimension. thus there are two dimensions left for the groups attached to the carbons. that's why they have to be on a plane
plus, the electrons around them "want" to stay as far as apart from other electrons. thus they make a kind of triangle to have the largest average angle among them
in sum, the double bonded organic compounds(~ene things) look like a "trigonal(triangle) planar(on a plane)" structure(1 vote)
- as per nomenclature rules, double or triple bonds,independent elemental groups and the functional groups are supposed to be given the least number on the longest carbon chain unless the functional group is terminal, right?why did Sal name the triple bond on the 5th carbon and the bromides on the 7th carbon? doesnt this violate the nomenclature rules?(2 votes)
- Because you have to look at priority of functional groups, -OH has higher priority that -Br and triple bond, therefore lowest no. has to go to -OH group.(2 votes)
- 5 and 6 obviously cant rotate individual of each other but can 4 and 7 rotate individual of 5 and 6, while they are all in the same molecule?(2 votes)
- Can we write 7,7-dibromo-oct-5-yn-4-ol as oct-5-yn-4-ol-7,7-dibromo?(1 vote)
- Are alchohol derivatives of only alkanes ?? Not of alkenes??(1 vote)
- Any compound with an -OH attached to a non-carbonyl carbon† is an alcohol, but its name may reflect a higher priority group§.
In the case of alkenes the alcohol is higher priority – e.g. HOCH₂CH=CH-CH₃ is but-2-en-1-ol.
Note however, compounds where the -OH is attached to a carbon that is forming a double bond with another carbon is known as an "enol", which has different properties than a regular alcohol – e.g. HOCH=CHCH₂-CH₃.
If there is a higher priority group then the -OH is called a hydroxyl, so the name will have "hydroxy" in it — e.g. CH₃CHOHCH₂COOH is 3-hydroxybutanoic acid.
†HO- attached to a carbonyl (C=O) is a carboxylic acid group.
§chemical group priorities:
- If there is an double bond then is it goes linear ?(0 votes)
- A carbon involved in one double bond has a trigonal planar, not linear, geometry. A carbon involved in two double bonds (e.g. an allene or a ketene) is linear, as is a carbon involved in a triple bond.(4 votes)
- why sir? why we must straighten out(0 votes)
- I think it may have something to do with the fact that sp hybridized carbons have linear geometry but I'm not sure, if anyone could confirm/correct me that would be much appreciated(1 vote)
- [Instructor] I wanna do a quick clarification on the video on alcohols. In that video, I gave this example of this alkanol right over here, it has a triple bond between the five and six carbons. And I just wanna clarify that in reality, it would not ever be drawn this way. That this was an error to actually draw it this way. The way I should have drawn it is this right over here. And you see, the only difference between these two pictures, starting with the one carbon, two carbon, three carbon, everything looks the same. Four carbon attached to the hydroxyl group, everything looks the same. And then we get to the five carbon, but instead of bending back up, we just keep going straight to the six carbon, that's where we have the triple bond between the five and six carbon, and then we go straight again to the seven carbon that's attached to the two bromo groups. And then we get to the eighth carbon. And the whole point why this right over here, that this is the correct way to draw it, the correct way to draw it, is that triple bonds, this triple bond right here, it forces a linear configuration. So on both sides of that triple bond, you would go straight out. So whenever you see an, actually, whenever you see any type of an alkyne drawn, the triple bonds should essentially straighten out. It should straighten out the molecule.