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- IIT JAM
- Chemistry (CY)
18. identify the correct acidic strength order in the following compounds: n o (1) (i) (c)i> ill>ii (a) ii > i> ili (b) ii > ii >i (d) ii > i>ii 19. which among
18. Identify the correct acidic strength order in the following compounds: N O (1) (I) (c)I> Ill>II (a) II > I> IlI (b) II > II >I (d) II > I>II 19. Which among the following compound will give maximum enol content in the solution: (a) CgHs-C-CH C-CH (c) HC-C-CH2-CH-CH3 b) H,C--CH-C-CH; (d) HC-C-CH2-COOCaHs
2 Answer(s)
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Ashutosh singh
In B, there's no cross conjugation + There are two -CO groups completely available to delocalise the negative charge and involve it in resonance which increases the stability of conjugate anion and hence, increases its enol content because of high acidity.
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Ashutosh singh
In C, there's only one -CO, therefore, rule it out very first.
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Ashutosh singh
So, the anion which will be generated after removal of H+ will not be very much stabilised.. because -CO is conjugated with Phenyl in A, and with O in B.
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Ashutosh singh
In Q19, in options A and D, there's cross conjugation with Phenyl and Oxygen respectively.
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Ashutosh singh
Sourav, let me explain this to you.. Nitrogen is more electronegative as compared to Carbon. therefore, when deprotonation occurs in ii, it will leave a minus charge on Nitrogen...which is more stable as compared to negative charge over Carbon. Therefore, ii is most acidic.
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Now, come to I and III. In option 3, there's an Oxygen which is in conjugation with one of the -CO group... therefore after acquiring minus charge at Carbon, that negative charge will not be delocalised effectively over -CO due to this cross conjugation.
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When deprotonation occurs in iii, it will not effectively participate in resonance because, left hand side -CO is already involved in conjugation with O of the ring. Therefore, the anion is not stabilized as much as it can in 1, therefore the conjugate base of the acid is less stable. Hence, iii is less acidic than i.
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Priyanshu kumar Best Answer
option A is correct answer
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in structure 2 after deprotonation the negative charge is there is on more EN nitrogen atom..so it is most stable among these
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In structure 1st negative charge is there on carbon which is less EN than nitrogen
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got sourav?
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how 1 more acidic than 3?
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in structure 3 there is cross conjugation after deprotonation...so CB base formed is least stable
Sourav
https://youtu.be/69AyGv7dL2Q sir can you go through this explanation?