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DLS
 one year ago
Best ResponseYou've already chosen the best response.0Which carbo cation is least likely to form as as intermediate? dw:1360852772005:dw

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0A) is 3 degree and so is C)

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0B is 2 degree, also has a double bond which adds to the stability

DLS
 one year ago
Best ResponseYou've already chosen the best response.0its definitely not A for sure

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0C is attached to 3 carbons ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0hmm ? well still there are 3 carbons !

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0If i had to put a guess, I'd say D)

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0hmm,, sorry about that then! :

DLS
 one year ago
Best ResponseYou've already chosen the best response.0man I really thought u shud know all that before 2 months of JEE LOL

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0Wish your thinking was true!

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0Well I have never seen a bridged carbocation before.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0are u freaking kidding me ._.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0On a bridge head a positive carbon never occurs ~Wikipedia

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0http://en.wikibooks.org/wiki/Organic_Chemistry/Introduction_to_reactions/Carbocations

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0hmm, that is why it must be least likely to form! hmm, seems legit

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0well since bridged carbocation is not formed, hence it is least likely to be formed here.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0i asked why it isnt form ._. i dont want to accept it as a fact

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0I don't know the fact! I just learnt it!

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0reason of the fact*

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0Here, you may read this, this has the reason http://www.tricity.wsu.edu/Chem540/ar00096a001.pdf

DLS
 one year ago
Best ResponseYou've already chosen the best response.0god i cant read that whole thing :/

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0learn as fact! thats what I just did !

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0In chem, most of times, there aren't reasons, there are rules, !

DLS
 one year ago
Best ResponseYou've already chosen the best response.0http://fc06.deviantart.net/fs70/f/2011/288/3/c/nothing_to_do_here_by_rober_raikd4cxltj.png

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0http://narwhaler.com/img/ys/x/iseeyourpointspearspidermanYsX3vo.jpg

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.0Shuld nt the answer be D............

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0If (A) is benzene, then it's (D). Remember you can't form primary carbocations ever.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0but i want smt more detailed

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Wikipedia says the answer is C?

DLS
 one year ago
Best ResponseYou've already chosen the best response.0On a bridge head a positive carbon never occurs ~Wikipedia

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0The + isn't on the bridge though.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0that is why it cant resonate,the charge is localized.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0so isnt C in a worse condition than D? it cant resonate at all but D still can

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Resonance is not the same as formation of a carbocation

DLS
 one year ago
Best ResponseYou've already chosen the best response.0its about the stability of the carbocation

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Resonance just tells you where electron behaves or forms a more stable structure

DLS
 one year ago
Best ResponseYou've already chosen the best response.0more resonating structures=more stable C h as 0 resonating structures

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Draw them for acetylene.

DLS
 one year ago
Best ResponseYou've already chosen the best response.0dw:1360961235810:dw not sure duh

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0I don't think that would work. Did you move the hydrogen?????

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Resonance is only movement of the electron. Not protons.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Maybe @BluFoot can help right now I'm not on my comp so I can't give detailed answers right now and don't have my notes w/ me. If you're on later I will explain better if no1 else has helped you. I gtg

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1A is very stable because of all the resonance from the phenyl groups. B is kinda stable from the resonance of the double bond. This is called an "allylic carbocation" http://science.uvu.edu/ochem/index.php/alphabetical/ab/allyliccarbocation/ C has no resonance, but it's tertiary at least D is definitely the least stable. Vinylic carbocations (+ on the C=C) are reaalllllllyyyy unstable. The reason is based in orbitals. C=C are sp2, but with a carbocation it becomes sp. sp orbitals are situated close to the nucleus, so the positive charge is very close to the nucleus, which is positive, so it's unstable

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1I've never seen the structure in C btw... do you know what it's called?

DLS
 one year ago
Best ResponseYou've already chosen the best response.0ive been taught its bridgehead carbon (out of the plane)

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Oh right, I haven't learned about those. It's still probably D but don't make my word for it since I know nothing about the bridges.

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1I have no idea, sorry :/

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0It's a bicyclo compound.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Well, "C" would form a radical rather than a carbocation since it's an all hydrocarbon structure. That could be one argument. I'm not entirely sure about this one to be honest. But I can't help but also think the answer could be D since you don't usually ever have primary carbocations. And there's no type of resonance in structure D. If you take out the double bond, you'd have another charge and that would definitely form a very thermodynamically unstable structure. It would definitely prefer to stay.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2If the answer was to be C, then, i imagine that'd it would be because the strain put on it by it's inability to undergo an sp2 conformation (180 degree angles) since it's fixed on the frame and it would make of high energy/unstable
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