0
$\begingroup$

In $\ce{CH3CH2-}$ what should be direction of inductive effect?

If going by order mentioned in my book, it is to be from carbon containing negative charge to CH3, but is not the carbon containing the negative charge more electronegative? So should not the direction be towards it?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

The charge on CH2- doesn't say it's electronegative. The direction of inductive effect is from a negatively charged atom to the rest of the uncharged molecule (if there's only one negatively charged atom in a given (canonical) structure) and the effect decreases as we go away from the negative charge. The electronegativity of carbon atoms in organic molecules is more closely related to their hybridization; sp-hybridized carbons are the most electronegative, followed by sp2- and sp3- hybridized carbons. Thus, your textbook is right.

$\endgroup$
0
$\begingroup$

It seems that you use the term electronegativity incorrectly, electronegativity is the tendency of an atom to draw a shared pair of electrons (or electron density) towards itself. (according to wikipedia)

Carbanion is not electronegative (compared to CH3), it tries to get rid of that extra electron making it relatively electropositive so the electron cloud is being pushed away slightly, towards the methyl group.

So your textbook is right, it’s from $\ce{C-}$ to CH3, i.e electropositive to electronegative.

$\endgroup$
2
  • $\begingroup$ So it is pushing the negative charge on it or is it pushing sigma electron away from it because of presence of extra electron? $\endgroup$ Apr 9, 2019 at 18:25
  • 1
    $\begingroup$ It is pushing away the electron electron cloud/ orbital , towards . Also note that I have used pushing term to make it easy to understand though it’s not precise starement $\endgroup$
    – Chemist
    Apr 9, 2019 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.