The correct order of +I strength for the following groups −CH3,−CH2R,−CD3 is :
(Here R is any alkyl group ).
A
−CH2R>−CH3>−CD3
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B
−CH3>−CD3>−CH2R
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C
−CD3>−CH2R>−CH3
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D
−CH2R>−CD3>CH3
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Solution
The correct option is D−CH2R>−CD3>CH3 Since alkyl group is an electron donating group :
More the number of alkyl groups attached more is the +I strength.
So, −CH2R has the highest +I strength.
In case of isotopes, heavier isotopes show more +I effect.
Also, bond length of C−D is less than C−H bond length.
So electron density becomes higher due to shorter bond length. ∴+I effect is more for −CD3
The correct order of decreasing +I strength is : −CH2R>−CD3>CH3