Question

Arrange the following in increasing order of $+I$ effect :
$1^{\circ }\text{Alkyl Group,}{2}^{\circ }\text{Alkyl Group,}{3}^{\circ }\text{Alkyl Group}\text{and Methyl Group}$

• A. $1^{\circ }\text{Alkyl Group}<2^{\circ }\text{Alkyl Group}<3^{\circ }\text{Alkyl Group}<\text{Methyl Group}$
• B. $\text{Methyl Group}<1^{\circ }\text{Alkyl Group}<2^{\circ }\text{Alkyl Group}<3^{\circ }\text{Alkyl Group}$
• C. $1^{\circ }\text{Alkyl Group}<3^{\circ }\text{Alkyl Group}<2^{\circ }\text{Alkyl Group}<\text{Methyl Group}$
• D. $\text{Methyl Group}<3^{\circ }\text{Alkyl Group}<1^{\circ }\text{Alkyl Group}<1^{\circ }\text{Alkyl Group}$

Answer 1

The correct option is B $\text{Methyl Group}<1^{\circ }\text{Alkyl Group}<2^{\circ }\text{Alkyl Group}<3^{\circ }\text{Alkyl Group}$

Alkyl groups are electron donating groups
More the number of alkyl groups attached, higher the strength of $+I$ effect.
In other words, as the degree of ‘C’ increases, the strength of $+I$ effect increases.
So, the order of electron donating tendency or $+I$ effect order will be
$\text{Methyl Group}<1^{\circ }\text{Alkyl Group}<2^{\circ }\text{Alkyl Group}<3^{\circ }\text{Alkyl Group}$

$\text{or}$

$-C{H}_3<-C{H}_{2}C{H}_3<-CH(C{H}_3{)}_2<-C(C{H}_3{)}_3$