4
$\begingroup$

I would like to address the compound below.

enter image description here

This compound on elaborate elongation would result in this structure.

enter image description here

But is the below given numbering correct? I am just wondering that what if the same compound is given numbering, starting from that carbon which is assigned 7th number in the first elaborate diagram (second diagram), by doing so, triple bonded substituent will also get the least number, as shown in the below given diagram.

enter image description here

I have a similar problem with this compound:

enter image description here

What if you just reverse the numbering? By reversal I mean, the carbon assigned 1st number should be given the last number, am I violating any rule? If I do that, by numbering the compound in reverse order, we get least number for the methyl chain substituent, so is it necessary that methyl group substituted in the side chain should get the highest possible number?

In short, I mean, is the below given numbering valid or not?

enter image description here

|improve this question
$\endgroup$
  • 2
    $\begingroup$ I think your numbering is correct. Your choice is between including the 6-alkene or 6-alkyne in the main chain. The rule is "tie goes to the alkene". This link may provide useful examples starting around p. 9 csupomona.edu/~psbeauchamp/pdf/314_supp_8_nomen_2.pdf $\endgroup$ – ron Sep 15 '14 at 21:32
  • $\begingroup$ @ron yes, i agree, but what if i start the numbering from the number seventh carbon?, is that valid too? $\endgroup$ – agha rehan abbas Sep 16 '14 at 0:50
  • 1
    $\begingroup$ No, see page 5, item 7, "middle" example in the link I supplied. P.S. Good to see you again, hope things are going well. $\endgroup$ – ron Sep 16 '14 at 0:59
  • 1
    $\begingroup$ Note how in the example I mentioned above (6-ethyl-6-methylocta-trans-2-cis-4,7-triene) they do not start numbering from the side with the terminal double bond, but rather from the side with the lowest overall double bond numbers. $\endgroup$ – ron Sep 16 '14 at 1:20
  • 1
    $\begingroup$ the first one, it has the lower substituent numbering (1,4 vs. 2,5) $\endgroup$ – ron Sep 16 '14 at 1:34
2
$\begingroup$

For your first structure, I think your numbering is correct. Your choice is between including the 6-alkene or 6-alkyne in the main chain. The rule is "tie goes to the alkene". This link may provide useful examples starting around p. 9. Also see page 5, item 7, "middle" example in the above link for another relevant example. Note how in this example (6-ethyl-6-methylocta-trans-2-cis-4,7-triene) they do not start numbering from the side with the terminal double bond, but rather from the side with the lowest overall double bond numbers.

In your second compound, your first numbering scheme has the lower substituent numbering (1,4 vs. 2,5).

|improve this answer
$\endgroup$
  • 3
    $\begingroup$ Please note that your link contains several mistakes. In particular, the numbering in the following examples is wrong: 4-methyl-1-ethylidenecyclohexane should be 1-ethylidene-4-methylcyclohexane, 6-methyl-3-methylenecyclohex-1-ene should be 3-methyl-6-methylidenecyclohex-1-ene, 1-methyl-3-(4-methylpent-3-ynyl)cyclohexa-1,3-diene should be 1-methyl-3-(4-methylpent-2-yn-1-yl)cyclohexa-1,3-diene, 4-(2,2-dimethylcycloprop-1-enyl)-6-isopropylcyclooct-1-yne should be 4-(3,3-dimethylcycloprop-1-en-1-yl)-6-(propan-2-yl)cyclooct-1-yne, 4,4-dimethylnon-8-en-1-yne should be 6,6-dimethylnon-1-en-8-yne, … $\endgroup$ – Loong♦ Aug 29 '16 at 19:05
  • 1
    $\begingroup$ … 5-(but-2E-enyl)nona-5Z,7E-dien-1-yne should be (2E,4Z,7E)-5-(but-3-yn-1-yl)nona-2,4,7-triene, and 2-(1-methylpropyl)pent-1-en-3-yne should be 5-methyl-4-methylidenehept-2-yne. $\endgroup$ – Loong♦ Aug 29 '16 at 19:05
2
$\begingroup$

Your first compound should be (3E)-5-ethynylhepta-1,3,6-triene. The 2 conjugated double bonds at the end determine the first carbon.

The second compound should be (4-methylpentyl)cyclobutane. The aliphatic group has less precedence than the cyclobutane group.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.