but y=4 means 4 neutrons are emitted
thats the point I am making here
In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is -
Majority of the coachings have said the answer to be 1 except AAKASH .
In the below figure this is what AAKASH says (Image has been edited by me to clarify my doubt ) -
-
UP 0 DOWN 0 0 23
23 Answers
Maine toh answer 1 mark kiya tha...seeing "controlled nuclear fission" jisme 3 neutrons absorb kar liye jate hain..
Yaar Pritish unglee na karr ...............Aur ab iskaa answer to wahi batayenge jinhone isse banaya hai !
but neutron toh 4 nikle na reacion mein
and Rohit tumne toh yahi sochke kiya tha na ki 3 neutron nikalte hai Nuclear Reactor mein,jabki usme reaction thodi alag hoti hai aur usme 3 nikalte hain and net would be 2 their
point bas itna hai 4 neutron nikal rahe hain reaction mein,aur nikalne ko English mein emit bolte hain
lol
yaar sach kahun toh jabbu ke naam se koi goiit par bhi attention seeking waale kaam kar raha tha..don't know if this is the same jabbu, but you've been warned :P
.......
Don't you think you all should be studying for whatever comes next instead of dwelling on would-have-beens? :D
it is the only possibility
so 4 neutrons emitted na
that's wat I am saying
Bansal and Prerna have given answer 4 only coz of the same reason
One of the possibilities is -
92U235 + 0n1 → 92U236 → 54Xe142 + 38Sr90 + 4 0n1
Aldol dehydration follows an E1CB mechanism. As per that, the most acidic protons for the base to abstract are the ones next to the C=O group. Hence (A) is the only dehydration product which can be obtained.
Let me explain a little more. The fast step in the E1CB reaction is the abstraction of proton, unlike the other eliminations in which either the leaving group goes first(E1) or the leaving group and the hydrogen go simultaneously(E2).
The most acidic proton is the one abstracted. Then the substrate is of the form
| |
- C- - C - L
| |
where L is a leaving group, the minus charge always forces the leaving group out, whether it is a bad leaver or a good leaver(except Ph), and forms a pi bond. Same case here, OH is a bad leaving group but it has to go. And yeah, the leaving part is the slow step of the reaction, and is the RDS.
check bansal solutions or BT
u will get wat I am saying
neutron 4 niklenge isme koi doubt hai nahi
lekin bas wahi baat hai woh log net bol rahe the ya reaction mein kitne nikal rahe hain
nahi aisa nahi hai
woh same reaction nahi hai
pehle 1 neutron maroge usme toh 4 niklenge
BT also says 4 neutron niklege,but answer given is 3 coz they say 4-1=3
check their solution
uska 3 ya 4 pakka nahi hai kya hoga
nai re...1 proton is used...n 3 are produced.....c NCERT!! c any book..bachpan se padhte aaye hai..nuclear fission/fusion[jo bi..:P i get confused with the name] mei 3 nuetron nikalte hai...if they are not absorbed then wo blast wlast ho jayega... :D....toh its always 3!!!
yaar 1 neutron marte hain na,toh 4 nikalte hain yaar
Bansal walon ne 4 diya hai
check kar le
Woh toh 3 hi hoga , bachpan se padhtaa aaya hoon . Aur usme koi logic bhi toh nahee hai !
nice man jabbu
yaar woh neutron wale ka kya hona chahiye??
3 or 4??
also Aldol is a named reaction,it's product is
(ï¡, ï¢-unsaturated carbonyl compound) always so 1 only