haloalkane
how to like get the realtion of reactivity in these type of ques where different functional groups are there and when the double bond is realtivly in differnt posititons to the halogen atom
38 Replies
@Dexter
Note for OP
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Ok, first of all, let me give you a general idea of how to approach these.
oka
ok
1. You should know the mechanism of course. What is the step that determines the rate of the reaction? (That's what reactivity is talking about all the time)
2. What sort of thing needs to happen for that step to become faster or slower? (The product should be stable)
3. What sort of electron movement is going to make that happen?
4. What group/species can effectively cause that type of electron movement?
R and I effect
uhhuhh okk and like the stability of carbocation in solution also if we talk Sn1 and E1
right?
1. It's SN1, so RDS is carbocation formation.
2. Carbocation should be stable.
3. Electron movement should be towards the plus charge
4. +R, +H, +I groups in ortho/para positions can do this.
yeah ig
uhuh okk and like in the ques like the b part and c part how do we do there?
same
concept
uh okk lemme try
Same four steps. What's left is figuring out which groups do what.
The first step is easy here since it's given beforehand.
But if instead of SN1, it was instead given that it happened in a mildly acidic medium, it would have involved some more thinking.
yeah some ques had tht
also opt can u check another organic doubt?
Yeah, the second reaction is tripping me up
b part?
wdym more thinking
like just knowing the sn1 or 2
No, the benzonitrile with ethoxide/ethanol step
Yes. It's not clear sometimes
i see
aha yess thts why i wanna confirm if i did was right or not
so this is where i ended
@Opt
I went to sleep, sorry.
Ok,
23(i) is correct.
23(ii), you've made a mistake. Compound 3 has a carbonyl group, which is withdrawing by resonance, so it'll be an unstable carbocation.
23(iii) why is it 4>3>1>2 and not 4>1>3>2? 1 gives an allylic primary carbocation, which is stabilised by resonance with double bond.
oh no issues
coumpund 3 in ii is ketone na?
then
Yes.
C=O withdraws by resonance
oh wait
oh shi
wait leeme freshn up then i will try this again
Sure lol
wait so in 23 (ii) will the order 4>2>1>3 be correct ?
i realised 4 is secondary carbon attchment
2,4,1,3
Resonance is better than 2°-alkyl
uhuh ok lemme right down these rules
and if its 3 degree thn? will still resonance be more?
Uhhh that's where it gets iffy
Tertiary butyl is more stable than benzyl primary, but I'm pretty sure that's the only exception
It's, in general, resonance > hyperconjugation
uh huhh okk
but is it like for onlu butyl or any 3 degree?
Tertbutyl is the simplest 3° alkyl, so it must hold for higher ones too
ohh ok thn
+solved @Opt
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