00:01
So we have to draw the structures for following incorrectly named compound.
00:06
And the first one is 2 ethyl 3, methyl, 5's isopropyl, hexane.
00:54
Okay, so let me first draw the correct structure for this compound.
01:01
Here we have hexane, so this is the main carbon chain.
01:04
So this means hags is the prefix and for hags prefix we have six carbon atoms.
01:10
So 1, 2, 3, 4, 5 and 6.
01:16
And n is the suffix.
01:18
So this means that this is an elegant and these all carbon items are being connected by single bond.
01:24
And if i do numbering then 1, 2, 3, 4, 5 and 6.
01:31
So it says that on second carbon we have ethyl group.
01:37
So this is ch2, ch3.
01:40
So this is we call ithyl group.
01:42
And then on third carbon atom we have methyl group.
01:46
So now this will be a methyl group here.
01:56
And then it says on fifth carbon atom we have isopropile group.
02:00
So what do we mean by isopropile? this substituent containing three carbon atoms in which the central carbon atom containing the two same substituent two same group that is methyl group and here so this is the isopropile group and now completing the valencies with the help of hydrogen atom here we need three hydrogens carbon is tetravalent here we need one hydrogen here we need one hydrogen here we need two hydrogen here we need one hydrogen, here we need three hydrogens.
02:40
So this is the complete structure we have.
02:43
So now the correct iepac name for this compound.
02:47
So the basic rule for the iepac naming is select the longest carbon chain possible.
02:52
So here the longest carbon chain possible is 1 and then 2, 3, 4, 5, 6, 7, 8.
03:07
So we have 8 carbon atoms.
03:10
This is the longest carbon chain possible.
03:13
And now we should always make sure that in this carbon chain, if the carbon containing groups have some substituent, then they should contain the least position possible.
03:24
So here i should do the numbering such that the substituents get the least position possible.
03:29
So the correct method for positioning is.
03:41
So from here, 1, 2, 2, 3, 4, 5, 6, 7, 8.
03:53
So now we can see that on the second carbon atom, we have a methyl group.
03:59
On third carbon atom also we have a methyl group.
04:02
On fifth carbon atom also we have a methyl group.
04:07
And on sixth carbon atom also we have a methyl group.
04:10
So we have total of four methyl groups here as a substituent.
04:15
So whenever the same substituent is being repeated for more than one time then we use prefixes and when they are repeated for four times then the prefix is tetra.
04:25
So this will be iupac name.
04:42
2, 3, 5, 6, tetra methyl and now the longest carbon chain containing of 8 carbon atoms for 8 carbon atoms the prefix.
05:14
Is oct and the carbon items are being connected by single bond so the suffix is an so it will be octane so this will be the correct ibc name now moving on to the next part so in the next part we have two ethyl four turt butyl pantane so accordingly drawing the structure then pantine means we have 5 carbon atoms 1 2 3 4 5 and these all are connected by single bond since this is the alkenchant for suffix aen and prefix is spent so we have 5 carbon atoms and now it says that on second carbon atom i have ethyl group so this will be phdh2 c h3 and then it says that 4 third butile so this means that on 4th carbon atom we have a tertiary butyl substrate.
06:31
So what does this mean? this means that i have a carbon atom which consists of i have a substitute which consists of total of four carbon atoms.
06:46
First that the central carbon is tertiary in nature.
06:51
So now completing the valency with the help of hydrogen bonds, hydrogen atoms.
06:57
So this will be being carbon tetravalent.
07:01
We have three hydrogen atoms...