1how....
well... beta position pe keto group rehne se...
CO2 jaate jaate jo minus charge chhor jata hai [3] woh stabilised jo jata hai..
not only keto group...
koi electron withdrwing wala grp chaiye faster decarboxylation ke liye..
30 Answers
1
3let it be aldol condensation i am geting a product in which the carbon has two double bonds so it will be unstable cyclic structure.
if the oh- abstracts the hydrogen at the terminal carbon then it leads to the stable aromatic cpd benzene
from the formation of the products it can be concluded dat the abstraction of oh- at the terminal carbon is energetically favourd more
3the option b and a will be rite if the keto gp can be easily tautomerise to the enol form
3i dont know y it is b
but it can be said dat the allylic carbon is some wat got partial negative charge due to resonance we know dat the allylic carbocation is stablised by resonance .in decarboxylation rxn the c-c bond will break but in this case c-c bond is compartively stronger than other ones so it can be easily decarboxylated.
please reply if i blabbered
1naam kyun change karte ho yaar?? !#@$#^%%^ (abhigyan -> abhinav)
well, someone comment on q.2.
1@abhigyan,
.... jo next wala question hai naa ... [woh tumhara doubt nahi hai shayad] .. par i have a doubt ...
i dun think it will be both A and B.
it should onlly be B.
because ... in B only, the carbanion formed during decarboxylation is stabilised by the -R effect of C=0 group.
1it will be 'b' only ..
it wud have been c if there were no double bonds beside -OH .
1yaar in aldol....H is extracted from α(alpha) carbon not the one having aldehydic group....
1prateek
its aldol condensation (self)
here one h atom frm last carbon and the o,h atom frm 1st carbon eliminate as water leaving behind benzene
1i think it shud be c...
don't know why the ans given is b...
@sagar can u tell the mechanism of getting b..as the answer...
