I
iTeachChem•10mo ago
MGK

basic strength

answer is b. someone explain pls
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22 Replies
trinny
trinny•10mo ago
do you know about the molecular orbital theory?
MGK
MGKOP•10mo ago
yes
myst1cboomer
myst1cboomer•10mo ago
you can do it by checking quasiaromaticity
Deleted User
Deleted User•10mo ago
Waht bro waht
myst1cboomer
myst1cboomer•10mo ago
distribute the pi bond and if you attain aromaticity that is 🔥
Deleted User
Deleted User•10mo ago
I don't get it
myst1cboomer
myst1cboomer•10mo ago
No description
myst1cboomer
myst1cboomer•10mo ago
now?
thegamerpopz
thegamerpopz•10mo ago
I’m curious what about a and b
myst1cboomer
myst1cboomer•10mo ago
-I order, more N more -I
nermal
nermal•10mo ago
@ᴘɪᴄᴄʜɪ | 🎧 🎶 i think that bot didnt reply here
ᴘɪᴄᴄʜɪ | 🎧 🎶
uss time issues the abse ho toh btana
nermal
nermal•10mo ago
oh okay :salute:
!Nimay¡
!Nimay¡•10mo ago
One could do it with HOMO LUMO but that would be out of scope for jee purposes where you have to simply check via delocalisation and electronic effects
myst1cboomer
myst1cboomer•10mo ago
so is quasiaromaticity the right way to determine two of them?
!Nimay¡
!Nimay¡•10mo ago
Yes
iTeachChem
iTeachChem•10mo ago
shall we mark this one as solved?
!Nimay¡
!Nimay¡•10mo ago
Ask the dosa eater maybe Wait I’ll sum it up though If u split the double bond heterolytically (giving Nitrogen the negative charge since it is more electronegative), rings a and b will be aromatic, c will have 2 aromatic rings (above one being quasi-aromatic hence even more stable than normal aromatic rings) and d will have 1 aromatic and 1 anti-aromatic ring Hence c is the most basic and d is the least basic Now for a and b you gotta check the inductive effect of nitrogen (the lone pairs won’t be delocalised so you cannot take mesomeric effect into account). In that case since a has more Nitrogens, there is more withdrawal of electron, and hence is less basic than b Final order becomes c > b > a > d Explanation via HOMO LUMO (gives a more concrete idea): For an acid base reaction there should be less HOMO (of base) and LUMO (of acid) gap (HOMO is highest occupied molecular orbital and LUMO is lowest unoccupied molecular orbital) since the electron goes from pi mo of base to the acid. So basicity is given by the energy of HOMO of the bases. More the HOMO energy more the basicity since lesser the HOMO LUMO gap. Now if you understood this, you need to know when is energy of HOMO lowered in a pi system. In this case, Nitrogen is highly electronegative and exerts -I effect, it decreases the energy of HOMO of the base. Since A and D have more Nitrogens, they are comparatively weaker than B and C. Now choosing between A and D, D has more delocalisation so energy of HOMO is lowered. Between B and C, delocalisation of C is higher so B is the better base. Final order: c>b>a>d
iTeachChem
iTeachChem•10mo ago
@MGK we good?
MGK
MGKOP•10mo ago
Yeah all set Sorry I forgot to inform @iTeachChem
iTeachChem
iTeachChem•10mo ago
+solved @!Nimay¡
iTeachChem Helper
iTeachChem Helper•10mo ago
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