Question

What is the correct order of inductive effect?
(A) $(-{{O}^{_{-}}}>-C{{H}_{3}}>-CM{{e}_{3}})$
(B) $~(-CO{{2}^{-}}>-{{O}^{-}}>-CHM{{e}_{2}})$
(C) $(-{{O}^{-}}>-CHM{{e}_{2}}> -D> -H)$
(D) None

Answer 1

Hint: To solve this question, we should know that inductive effect is the transmission of unequal sharing of the bonding electron through a chain of atoms in a molecule, leading to a permanent dipole in a bond.

Complete step by step solution:

To solve this question, we should first try to understand the inductive effect. So, we should know that inductive effect refers to the phenomenon wherein a permanent dipole arises in a given molecule due to the unequal sharing of the bonding electrons in the molecule. This effect can arise in sigma bonds, whereas the electromeric effect can only arise in pi bonds.
We should also know about the electromeric effect. It is the instantaneous formation of a dipole in the molecule of an organic compound due to the complete transfer of shared pi electron pairs to one of the atoms under the influence of an attacking reagent.
To answer this question, we should know that inductive effect depends on electronegativity, bonding order and charge.
- Electronegativity: We should know that atoms or functional groups that are electronegative relative to hydrogen such as the halogens, oxygen, nitrogen, etc. may have a negative inductive effect (-I). These atoms withdraw electron density from the single bond structure of a compound and can assist in the stabilization of negative charge that may form in reactions. Atoms or functional groups that are electron donating (hydrocarbons, anions) have a positive inductive effect (+I). These groups can help stabilize positive charges in reactions such as protonation of bases.
- Bonding order and charge: Let us understand by taking one example. So, if we take an example of a hydroxyl group, we came to know that oxygen in a hydroxyl group (-OH) is electron withdrawing by induction (-I) because the oxygen atom is relatively electronegative and is uncharged in that bonding arrangement. On the other side we should know that oxygen in an alkoxide $(-{{O}^{-}})$structure is electron donating (+I) by induction because in this bonding order (a single bond to oxygen) it has an excess of electron density.
So, now we will answer this question.
$(-{{O}^{_{-}}}>-C{{H}_{3}}>-CM{{e}_{3}}):$​ In first option, we find that $(-{{O}^{-}})$ is placed on first position. And it is correct because it is more electronegative than the rest of the options. And as we know that more electronegative atoms or functional groups have more inductive effect. So, it is the correct order.
$~(-CO{{2}^{-}}>-{{O}^{-}}>-CHM{{e}_{2}})$: This is incorrect order because as we know that $(-{{O}^{-}})$electronegativity is more than of $(-C{{O}_{2}}^{-}).$ So, $(-{{O}^{-}})$is more inductive. And hence it is incorrect order.
$(-{{O}^{-}}>-CHM{{e}_{2}}> -D> -H):$This is incorrect order because inductive effect of H is more than that of D. Because of one extra neutrons size of D is higher than H as a result this increased size reduces electronegativity of D.
So, from the above discussion we now know that, option A is the correct option.

Note: We should note that when a group displaying the -I effect is bonded to a molecule; the electron density of the resulting molecule effectively reduces, making it more likely to accept electrons and thereby increasing the acidity of the molecule. When a +I group attaches itself to a molecule, there is an increase in the electron density of the molecule. This increases the basicity of the molecule since it is now more capable of donating electrons.