This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
the points to favor is that both the two groups are same but they are cis and trans so the compound can be chiral. but i am not sure that's why i asked this.
@Gauri shankar Mishra
–
but the cis and trans carbons will rotate the light equally in opposite directions (as all the groups are same) so there will be no effect due to them. my main point is that whether the middle carbon will be chiral or not.
@Neelesh Vij
–
How are you saying the cis and trans groups will rotate light in equal and opposite directions?
I agree with @Gauri shankar Mishra, the compound is chiral.
@Ameya Daigavane
–
Umm. i am not sure about the statement. I thought it similar to R and S configuration. so does that mean geometrical isomers do not rotate light?
@Neelesh Vij
–
I Think that This Compound Is Chiral . The central chiral atom is called pseudo chiral as the groups attached to the carbon can be different due to stereoisomerism.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
In my opinion it's chiral, if one side you take E isomer and the other side the Z isomer the it would be an enantiomer and hence optically active
Hey @neelesh vij This compound is chiral because its mirror image is non-superimposable.
Chiral as it has only one chiral centre.
Log in to reply
The question is - Is that a chiral Centre!
I have already given the answer though.
What are your observations here?
Log in to reply
the points to favor is that both the two groups are same but they are cis and trans so the compound can be chiral. but i am not sure that's why i asked this.
Log in to reply
It is chiral carbon Because the left and right sides when are opposite (cis and trans ) the make the poparised light to rotate
Log in to reply
Log in to reply
@Gauri shankar Mishra, the compound is chiral.
How are you saying the cis and trans groups will rotate light in equal and opposite directions? I agree withLog in to reply
Log in to reply
himanshu pandey qno: (level 2) 22
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
I guess it is not but I am not sure
This is chiral. There is a central carbon bonding with 4 different compounds.
Ciral