11Its the result of thinking out of box [3][3]
Happy Birthday.[1]
organic chemistry! :(
1) Arrange according to Acidity..
a) benzoic acid, b) para-methylketo-benzoic acid, c) ortho-methylketo-benzoic acid [where methylketo means -COCH3]
Answer given: b>c>a (reason: H-bonding in c makes it less acidic than b!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!)
Now i will give detailed explanation what i thought and reached my answer of c>a>b
First of all first let us consider resonance effect and inductive effect of the methylketo group.
the group has C=O group and as a result, must have a high d+ charge on carbonyl carbon. Now, this will make C show -I effect, i.e. decrease the electron density by pulling the sigma(single bond) electrons towards it.
But, due to continuous conjugation from O=C-C=C-C=C.. (the underlined part is from benzene) the methyl keto group would release electron towards the ring making electron accumulation especially at ortho and para positions.
Now, the resonance effect is more effective than inductive effect (as is generally).
So, we would generally expect that the conjugate base of benzoic acid with methyl keto substitution at ortho or meta position, to be destabilised highly due to resonance effect whilst it is only partially stabilised by inductive effect...so as a whole, the conjugate base is destabilised..
Thus at first glance we would feel that a>(b or c)..
But then let us look into the matter with a bit more thought, understanding and concentration.
Since both -COOH and -COCH3 are bulky groups, thus there would be the well known ortho effect..making the resonance effect non existent in ortho isomer...thus, the only existing effect of the methyl keto group will be the inductive effect which stabilises the conjugate base.
Thus the (c) acid is more acidic than normal benzoic acid (a).
(b) as already proved (:P) is less acidic than (a)..
thus, order of acidity... (c)>(a)>(b)...that was my answer..
Now there is always the question prevalent .. what is this ortho effect?
So, to understand the logicing for such explanation let us take this example..
Formic acid and benzoic acid. If we look into the pKa values of the 2 acids, we will get
pKa for formic acid is 3.17 and pKa for benzoic acid is 4.17. Thus, benzoic acid is weaker acid than formic acid.
Though we would expect -I effect from Ph, still we get less acidity..why??
Reason: in formic acid and also in benzoic acid, the conjugate base is stabilised by the resonance between the two O atoms, the resonance occurs between two equivalent resonating structures and so is very important and highly stabilising.
so far so OK but.. in benzoic acid there is another resonance taking place..that from the benzene ring. As a result there is interference in the resonance of -COO- which is highly stabilising and thus 100% of this resonance is not possible, a few percent (very few) is taken by the resonance in the benzene ring)
This causes destabilisation of benzoate ion as compared to formate ion..thus formic acid is more acidic..
But if somehow we can stop the resonance partcipation of benzene then we would get as much a strong acid as formic acid..(or might be even more also!!!)
So we have ortho effect..
in it, when bulky group is present ortho to -COOH (any group other than -H, -OH or -F), there is steric replsion in the molecule.
Thus in the benzoate ion there are 2 destabilisations 1)due to resonance interference from benzene ring and 2)Vanderwaal's repulsion due to crowding.
the compound will get rid of both destabilisations by a simple trick.
the -COO- will rotate a bit so that it comes out of the plane of the benzene ring.
This eases the molecule of the steric repulsion
also there is inhibition of resonance effect (this is called steric inhibition of resonance) and so again the resonance in -COO- becomes 100% dominant and molecule is stabilised.
Only effect experienced by the carboxylate ion is any inductive effect of the ortho group which weakly stabilises or destabilises the ion..
-
UP 0 DOWN 0 0 10
10 Answers
492) We have to compare acidity between para-methylketo-phenol and phenol.
I logic in p-me-keto-phenol, there is +R and -I
+R stronger than -I thus, CB less stable in p-me-keto-phenol.
thus, p-me-keto-phenol less acidic.
but book provides just opposite answer.
also in the next question the book mentions that p-hydroxytoluene is more acidic than p-me-keto-phenol..
but resonance destabilised p-me-keto-phenoxide must have been more unstable than hyperconjugation followed by resonance destabilisation in p-hydroxy toluene..isn't it??!! :(
again opposite answer!! :'(
493) We have to compare acidity between phenyl mercaptan (Ph-S-H) and phenol (Ph-O-H)
I logic, the 3p-1s overlap of S-H is not so good as compared to 2p-1s overlap in O-H and as a result, S-H bond is weaker and can be easily broken.
Also, S being larger in size, negative charge developed on losing H+ is reduced by resonance and then rest small charge is further diffused all over the surface of the sulphur atom.
So, Mercaptan must have been more acidic...but then book gives opposite answer! :'(
13) Oxygen being more electronegative than sulphur is more effective in localising negative charge on its surface after losing the proton. Resonance stabilisation in phenoxide ion is greater because of the effective pπ-pπ bonding, whereas in the case of thiophenoxide ion it is pπ-dπ bonding which is not so effective.
49thanks for correcting that bit of concept of mine!! :)
baki do k liye??
393) The book seems to be wrong, subho. I did a little backtrack and checked the pKas of both compounds.
Phenyl mercaptan : 6.62
Phenol : 9.95
I think your explanation is bang on the money. When I was asked this question by Govind(I think), I had the same answer. The size of the sulfur atom allows it to delocalise the charge better, overriding the pπ-dπ bonding problem.
1) I think the book has disregarded that the ortho effect can also be SIR and not just chelation. However if chelation does occur it can make release of -H difficult. Considering all the factors involved from your explanation, we arrive at the conclusion that -COO- has lost its resonance connection due to the bulkiness of -COCH3, and is being stabilized by -I effects. The power of chelation is underlined by how large the equilibrium constant is. A six-membered transition state(the chelate) would be more stable than the conjugate base being stabilized by -I effects, and the equilibrium constant would obviously be large for such an equilibrium. Hence the -H is hard to dislodge and acidity is lowered.
So the book's order may be correct here.
2) Note how differently -O- and -COO- behave as conjugate bases. Losing the resonance connection helps carboxylate but not the phenoxide's oxygen, obviously due to oxygen's direct connection to the ring, helping in delocalisation of charge.
The -COCH3 group only helps matters further. Its -I effect stabilizes the conjugate base more as it resonates across the ring. Hence the book's order is correct.
However p-hydroxytoluene ought to be less acidic than p-methylketophenol. The +H (positive hyperconjugation) effect of -CH3 destabilizes the ring when the negative charge of O is trying to delocalise via resonance...
1@Pritish- How is it that sulphur being a larger atom accomodates negative charge more easily?
I read somewhere that atoms with higher electronegativity reside negative charge more effectively.
23"But, due to continuous conjugation from O=C-C=C-C=C.. (the underlined part is from benzene) the methyl keto group would release electron towards the ring making electron accumulation especially at ortho and para positions."
this is wrong , the ketonic grp will actually show - M , i dont understand wat kind of +M it will show ,
the -M of ketonic grp will decrease e- density on COO- in anion and hence making it a weaker base and hence the conjugate acid is more acidic .
this is completely true for para substituted
now for ortho ,there r 2 factors,
1] the h bonding
2] in the anion , there wont be the delocalisation of charge as in the para substituted anion bcz due to steric effetcs COO- would tend to go out of plane of the ring and hence -M of ketonic grp would be weakened thus the conjugate acid will be more acidic than unsubstituted benzoic acid but less acidic than para substituted
232]subhomoy i think u shud draw the resonating structures then u will understand where u r doing the mistake
COCH3 wil show -M not +M !!
in para substituted HO-Ph-COCH3
-M of methyl keto grp will increase the delocalisation of chanrge on O- , thus decreasing its acidity and hence conjugate acid becomes more acidic
3] "also in the next question the book mentions that p-hydroxytoluene is more acidic than p-me-keto-phenol..
but resonance destabilised p-me-keto-phenoxide must have been more unstable than hyperconjugation followed by resonance destabilisation in p-hydroxy toluene..isn't it??!! :(
again opposite answer!! :'("
obviously keto substituted phenol is more acidic than phydroxytoluene
49sorry sorry....i just realised that...when i came to correct it...with apologies i saw that u have already pointer it out!!
anyways sorry for wasting your valuable time!! [2]