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What does one mean by, A-H (where A is sp3 hybridised) bond should be in conjugation with polar pi bond and Hydrogen with it should most acidic in the compound ?

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  • $\begingroup$ I'm afraid I don't understand your question … $\endgroup$ – Jan Oct 2 '15 at 17:24
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You can Take conjugation as alternate.

For example, if $\ce{C1}$ has acidic or replacable $\ce{H}$ on it then $\ce{C2}$ will have pi bond.

Most Acidic $H$ will be the one which can be removed most easily. Its order would be like $\ce{sp3}$ > $\ce{sp2}$ > $\ce{sp}$

Just for you to know, the most important difference between resonance and tautomerism is that in resonance there is change in position of the pi bond i.e., pi electrons of purely atomic orbitals but there is no change in the position of $H$ atomic, i.e., it does not attach to other position. But in tautomerism position of $H$ also changes.

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  • $\begingroup$ Can you please explain me about how to find most acidic hydrogen if there like many sp3 hybridised carbons? $\endgroup$ – Dimenein Oct 4 '15 at 5:23
  • $\begingroup$ Look for the $H$ atom which can be removed most easily $\endgroup$ – shaistha Oct 4 '15 at 7:43
  • $\begingroup$ @shaistha: Shouldn't the order of acidity of H be: $sp > sp^2 > sp^3 $ ? $\endgroup$ – chemkatku Jan 13 '16 at 7:17
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The most acidic $H$ is the easiest one to remove. You need check for the electron pushers or pullers attached to that molecule, etc.

Acidic character is directly proprtional to 3 factors, protic character of $H$ atom, polarisation of $H$atom and most important stability of conjugate base. The last one is the most important one.

For example, acidic character

1.$\ce{ phenol > CH3-OH}$ Because phenol conjugate base is resonance stabilised(conjugate base is the ion formed after removal of acidic $H$) 2.$\ce{CCl3-(C=O)-OH > H-(CO)-OH > CH3-CO-OH > CH3-CH2-CO-OH} $ $CCl3$ is electron defficient so acts as $\ce{e-}$ puller, decreasing the strength of $OH$ bond. ( that is acidic $H$ removed)

All alkyl groups i.e., R, are electron pushers i.e., increase $\ce{e-}$ density or concentration, hence incresing the strength of the bond.

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    $\begingroup$ Why did you add this as separate answer? $\endgroup$ – Mithoron Oct 4 '15 at 12:26

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