Br(+)

In previous versions of MarvinSketch, in order to represent Br(+), I had to make it a divalent radical; otherwise, Marvin would add two implicit H atoms to the Br atom. Now, when I draw Br and add a +1 charge, MarvinSketch 6.2.1 and 6.3.1 don't add the two H atoms, but they do underline the Br atom in red as if it has an invalid valence, even when I add a divalent radical flag to it. My first problem is, I don't understand why a valence error is being flagged when there is no valence error.

My second problem is that this change in behavior is causing me very serious trouble. A student drew a part of a mechanism as shown below. She did not add a divalent radical flag to the Br(+) atom, because MarvinSketch did not add the two implicit H atoms that it used to do. When ACE calculated the products of the electron-flow arrows in the first mechanistic step, ACE added a divalent radical flag to the calculated Br(+) product. Then, when it compared what it calculated with what the student drew, it failed to match its own Br(+) atom to the student's, because they differed in the presence of the divalent radical flag. 

Only halogen(+) ions are now being made electron-deficient; a N or O with a +1 charge assumes an extra implicit H.

The only way for me to solve this problem myself is to rewrite my electron-flow calculator algorithm to treat halogen(+) differently from other elements. I'd rather not add that complication, and I'd rather you restore the previous behavior. I don't understand why you changed the behavior; I don't see any advantages to the new behavior.

Also the superscripting formatting on this forum sucks.

Here are examples of what I'm talking about. When I put a +1 charge on a heteroatom like N, O, or S, Marvin simply adds one more implicit H atom. However, when I put a +1 charge on a halogen, the atom goes to six electrons, and MarvinSketch incorrectly flags a valence error. When I add a divalent radical flag to the H₂O(+) ion, MarvinSketch correctly reduces the implicit H count by 2, and no valence error is flagged. However, when I add a divalent radical flag to the halogen(+) ion, MarvinSketch continues to incorrectly flag a valence error. Finally, the only way I can generate H₂Br(+) is by drawing out the H atoms explicitly.

No response?

Hi,

We will fix this issue in the near future.

Regards,
Domi

Good to hear. Just to be clear, there are two problems to fix. First, if I put a Br (or any other halogen) on the canvas and then add a +1 charge, it should appear (and be interpreted) as H₂Br⁺, with a full octet. Second, if I add a divalent radical flag to the Br atom in H₂Br⁺ to make it into Br⁺ with only a sextet of electrons, Marvin should not say it has a valence error. 

Hi Bob,

 

Thanks for the clarification.

The fix is ready and it will be available in this week's release.

BRs,

Peter

Which release will that be exactly?

It's in 14.7.14.

You can try it online here.

 

BRs,

Peter