Destroyer GC # 78

[jdent]

---

Members do not see this ad.

I remember a while ago there was much debate and bewilderment and sleepness nights while many SDNers pondered this question. After much though myself the main point of this question is make you realize that its harder for more particles to come together successfully then less. Thats why Choice C is least likely to occur. Many of us thought it was choice d.
Well there really is no basic example of this reaction. However if you were to change it it could be similar like 2X yields Y+Z. Now that would be a canizzaro reaction. But you get the main point of this question!

SEE UPDATE BELOW!

 

[mistero]

Wow, what school did you go to, where did you hear about the canizarro reaction?

 

[orthodocs12]

thank you! that is very helpful! I guess I can sleep now...

 

[jdent]

Even better: The choice X yields X and Y is actually correct- while sleeping last night this following reaction came to mind:
A tertiary radioactive halide- ex. Iodine radioactive, attached to a tertiary carbon. Now form the carbocation and then treat it with Iodine radioactive- the product will yield either R or S . If you started with the R enantiomer, youll get the R and the S!
You see even when you think destroyer doesn't make sense--- It does its just that more awesome!!!!

WOW- DESTROYER I LOVE YOU!!!!!!!!!!!!

 

[klutzy1987]

Even better: The choice X yields X and Y is actually correct- while sleeping last night this following reaction came to mind:
A tertiary radioactive halide- ex. Iodine radioactive, attached to a tertiary carbon. Now form the carbocation and then treat it with Iodine radioactive- the product will yield either R or S . If you started with the R enantiomer, youll get the R and the S!
You see even when you think destroyer doesn't make sense--- It does its just that more awesome!!!!

WOW- DESTROYER I LOVE YOU!!!!!!!!!!!!

Exactly, very clever, the products, X and Y are just the 2 sterioisomers, R and S.