organic
can someone check if the answers are right the book doesnt provide ans to subjective ques
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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.@Opt this one
Yeah I saw it
besides the second are the others correct
?
Look through 43(i) and 44. You've made mistakes
wait which part lemme see once
Just look at them slowly.
You'll figure it out
okk
Oh I figured out that reaction in 42
Wait
ohh hbr will be replacing
no double bond
wait
its correct
no im not gtting whts wrong in 43 i
ohh the double bond part
tht was stewpid
it will be ch3cbr=chbr
,rotate
No?
wait why
both will be formed the cynaide and ether after h3o+?
A is correct. It's propene
B is also propene because NaNH2 just removes HBr. There's no difference between it and alc.KOH if you have a single halogen.
C is Anti-Markovnikov product, 1-bromopropane
Oh crap H3O(+) hydrolyses cyanide to carboxylic acid
I forgot to add that
but didnt we form alkyne so how will it give alkane
There's no alkyne anywhere
wait
in b part alkyne wont be formed?
doesnt nanh2 act as atrong alkyne former?
That's only if you have a vicinal di halide.
If you have a single halogen, it's the same as alc.KOH
oh okk
noted
,rotate
I'm assuming 1 mol of methanamine because it's not mentioned.
wait im on the step where cyanide is getting attcked
cant we apply this in here or is this the case in only haloalkane ?
@Opt
,rotate
It's specifically for haloalkanes
ohh thts why
Cyanide is a really really bad leaving group.
i applied same here
so in aromatic cn turns to what?
It didn't turn to anything. It stayed there until it was hydrolysed
so what happens to it when it is put in sodium alcohol
What happened was that there is a strong base (ethoxide) which abstracted a proton from the carbon next to CN.
The carbanion is stabilized by resonance and a strong withdrawing group.
That carbanion executes nucleophilic attack on the aldehyde.
This is something you learn later
Don't wrorry
*worry
later chpter?
You can do this after you're done with Aldehyde and Ketones
ohh
but here in the step where uh wait a sec
where did this second compund come from the right side one
It's in the second reaction.
Below the reaction arrow
It's another reagent
wait a sec
oh my god
but why will o come out in the step?
after heating?
It's not O, it's OH
Again, you'll get it after you learn the aldol condensation
The O(-) picks up a H(+) to become an algohol
*alcohol
Heating causes dehydration
na naa the step right after this
Yeah, that's what I explained
where o left forming double bond
O didn't leave?
wait so it will turn to oh and heating will mke dehydrogenation?
Only the π bond shifted , making an O(-)
Dehydration
Same as alcohol → alkene
yeah yeahh dehydration
Oh you haven't learnt that either
ahh ok yeah i hvnt done alcohols
This is all mostly aldehydes and ketones stuff
i think i will revisit this que whn i am done with tht chap adn then tru
gona leave it for now
thks opt
got half of it
+solved @Opt
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