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Chemistry 7 Online
OpenStudy (ilovebmth1234):

What is the name of this hydrocarbon? 2–methylpentyne 2–dimethylpentene 2,3–dimethylpentane 3,2–dimethylpentane

OpenStudy (ilovebmth1234):

OpenStudy (ilovebmth1234):

its not A

OpenStudy (aakashtomar):

Okay. Do you know of the lowest sum rule? If not, it is just that when you don't have double bonds, triple bonds or functional groups, and just methyl groups, you decide the direction of locants ( ie, from right to left or left to right) by taking that direction which guarantees you the lowest sum for the locants of the substituents. The structure is actually this : |dw:1471871274566:dw| Please correct me if I'm wrong with this condensed formula structure. So you could either number the methyls 2 and 3 ( left to right ) or 3 and 4 ( right to left ) but since the left to right thing gives us a smaller locant sum ( 5 ) we choose that. And the parent chain can be taken as the long horizontal chain. So it will be 2,3 dimethyl pentane. Right? Sorry, really sorry about the drawing. I 'm bad at it.

OpenStudy (ilovebmth1234):

no your perfectly fine and i guess this would be right i never fully understood this but what little i know it seems to be right

OpenStudy (aakashtomar):

Hmm, right. IUPAC rules are a bit confusing at first. But let me try an analogy. Imagine that every time you number the parent chain, you incur a debt equal to the sum of the numbers you're giving to substituents. Would you like more or less debt? Of course, a smaller debt! So you try to give them as small numbers as possible.

OpenStudy (rushwr):

Hey i hope you don't mind me intruding @AakashTomar :) First what u have to do is fin the longest carbon chain. Try it for me ? @ilovebmth1234

OpenStudy (rushwr):

is to find *

OpenStudy (aakashtomar):

No, no it's perfectly fine. :)

OpenStudy (ilovebmth1234):

okay ummmm

OpenStudy (ilovebmth1234):

how do i do this /.\ im sorry this is defiantly not one of my best subjects

OpenStudy (aakashtomar):

Just try forming a chain which has the highest number of carbons.

OpenStudy (rushwr):

Okai no worries :) I'll teach u how First we gotta find the chain with the highest number of carbon I hope this is clear ? |dw:1471872775055:dw|

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