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@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.the triple bond up there is converted to ethane using nanh2 ig...
Yo dude change your titles
Like doubt titles
Make it related to the topic of the question instead
Or maybe the type of reaction in this case
the 2 equivalents of NaNH2 is quite clever
NaNH2 is also a base apart from the Birch reduction reagent
and terminal alkynes are acidic
so NaNH2 abstracts the terminal alkyne proton, making -C≡C(-)
i wanna say Birch reduction but i don't think it can happen with -C≡C(-)
so maybe the second equivalent of NaNH2 abstracts the proton from the -OH, making O(-)
uhh
is it (B) or (D)? if so, which one of those
4 rxn hai ques mei lol kya hi title dalu
B
isnt birch reduction done using na and liq nh3
First is NaNh2, excellent base as an entropically favourable mechanism
Hence the 2 H+ can poof , first the sp carbon and then the oxygen
Next is an alkyl halide with a good leaving group , so the sp hybridized carbonion can attack now
Next is another alkyl hallidE wow , so prol just the oxide ion can do it's job
Now it's just hydrogenation ( rosemund I belive? )
So prol will turn the tripple bond to double , as the hydrogenation is done in the presence of a poison ( pd-baso4)
Making the answer b?
A take away from this question is carbanion even tho being sp hybridized, having less s character than the oxide ( sp3 ) still doesn't do enough to overcome the electronegativity of oxygen in general
Na + liq NH3 forms NaNH2 but anyway birch reduction is not happening here because NaNH2 would much prefer to be a strong base to a terminal alkyne (faster reaction)
my problem is
how would you know which alkyl chain goes where after you get -C≡C(-) and -O(-)
that's exactly why B and D are given as a separate options
So like this
having the same doubt
my guess is triple bond carbon is less electronegative than O ( as pointed above) so -ve charge will be unstable on it , to stabilise ethyl grp goes there
Strong base, 2eq → Abstracts two protons, one from OH, one from terminal alkyne
Ethyl iodide self ionises into ethyl carbocation + iodide. Electrophile carbocation is generated. Carbon nucleophile is better than oxygen nucleophile so ethyl gets connected to the sp-hybridised carbon.
Methyl carbocation in the next step connects to oxygen nucleophile.
Partial reduction by lindlar to cis alkene.
Option B?
Yup, this is the point. Carbon nucleophile is better than O nucleophile due to EN.
got it thanx
@Nimboi can i close
ah, ok
this confuses me fundamentally
how can a triple bonded carbon simply have more EN when EN is supposed to be the property of an atom
ill just look it up actually i need to revise atomic structure and chemical bonding
yeah sure
ab toh adv ho gya na tumhara?
didnt give adv, doing bitsat
that's on 30th for me
and also in june 2nd attempt
not always ig
its some orbital reason
that's why we say "more s character more EN"
we measure EN using s character too
yeahh
alr yeh feel free to close it
this is a different question now anyway
+solved @Opt
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