
The arrangement of ${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - ,{\left( {C{H_3}} \right)_2}CH - ,C{H_3}C{H_2} - $ when attached to benzene or an unsaturated group in increasing order of inductive effect is
A.${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \langle {\left( {C{H_3}} \right)_2}CH - \langle C{H_3}C{H_2} - $
B.\[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}} - \langle {\left( {C{H_3}} \right)_2}CH - \langle {\left( {C{H_3}} \right)_3}C - \]
C.${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{CH}} - \langle {\left( {C{H_3}} \right)_4}C - \langle C{H_3}C{H_2} - $
D.${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \langle C{H_3}C{H_2} - \langle {\left( {C{H_3}} \right)_2}CH - $
Answer
558.6k+ views
Hint:
We can see that Inductive effect is based upon the number of alkyl groups attached. So, it must be influenced by the number of carbon atoms.
Complete answer:
Inductive effect can be understood by this, Polarization of a $\sigma $ bond because of the electron withdrawing or electron donating effect of adjacent groups or atoms.
We can see that the alkyl group $\left( { - C{H_3}} \right)$ has +I (inductive effect) and with increasing the number of alkyl groups, inductive effect also increases.
We can see that that the order of the primary, secondary and tertiary alkyl of +I effect is as follow,
$3^\circ \,\rangle \,2^\circ \,\rangle \,1^\circ $
We can also conclude that the greater the number of alkyl groups attached to the positively charged carbon atoms, greater is the inductive effect and also, stabilization of cation.
From above we can see that the trend above molecules follow is more the number of alkyl, more will be an inductive effect. Hence the molecule having greater no. of carbons has a greater inductive effect.
Therefore, option A is correct ${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \langle {\left( {C{H_3}} \right)_2}CH - \langle C{H_3}C{H_2} - $
Therefore, Option B is correct.
Additional Information:
There are two types of inductive effect that is positive inductive effect and negative inductive effect (-I).
Positive inductive effect refers to the electron releasing or donating nature of the groups or atoms and therefore denoted as +I.
Examples of groups in the decreasing order of +I effect,
\[{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \rangle {\left( {C{H_3}} \right)_2}CH - \rangle C{H_3}C{H_2} - \,\rangle - {\rm{C}}{{\rm{H}}_{\rm{3}}}\,\rangle {\rm{H}}\,\]
Negative inductive effect refers to the electron withdrawing nature of the groups or atoms and therefore denoted as -I.
Examples of groups in the decreasing order of -I effect,
\[N{H_3}^ + \,\rangle \,N{O_2}\,\rangle \,CN\,\rangle \,S{O_3}H\,\rangle \,CHO\,\rangle \,F\,\rangle \,Cl\,\rangle \,Br\,\rangle \,I\]
Note: Please note that the increase in no. of alkyl chain contributes to the increase in inductive effect. Tertiary methyl is the most stable and primary is least stable.
We can see that Inductive effect is based upon the number of alkyl groups attached. So, it must be influenced by the number of carbon atoms.
Complete answer:
Inductive effect can be understood by this, Polarization of a $\sigma $ bond because of the electron withdrawing or electron donating effect of adjacent groups or atoms.
We can see that the alkyl group $\left( { - C{H_3}} \right)$ has +I (inductive effect) and with increasing the number of alkyl groups, inductive effect also increases.
We can see that that the order of the primary, secondary and tertiary alkyl of +I effect is as follow,
$3^\circ \,\rangle \,2^\circ \,\rangle \,1^\circ $
We can also conclude that the greater the number of alkyl groups attached to the positively charged carbon atoms, greater is the inductive effect and also, stabilization of cation.
From above we can see that the trend above molecules follow is more the number of alkyl, more will be an inductive effect. Hence the molecule having greater no. of carbons has a greater inductive effect.
Therefore, option A is correct ${\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \langle {\left( {C{H_3}} \right)_2}CH - \langle C{H_3}C{H_2} - $
Therefore, Option B is correct.
Additional Information:
There are two types of inductive effect that is positive inductive effect and negative inductive effect (-I).
Positive inductive effect refers to the electron releasing or donating nature of the groups or atoms and therefore denoted as +I.
Examples of groups in the decreasing order of +I effect,
\[{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{C}} - \rangle {\left( {C{H_3}} \right)_2}CH - \rangle C{H_3}C{H_2} - \,\rangle - {\rm{C}}{{\rm{H}}_{\rm{3}}}\,\rangle {\rm{H}}\,\]
Negative inductive effect refers to the electron withdrawing nature of the groups or atoms and therefore denoted as -I.
Examples of groups in the decreasing order of -I effect,
\[N{H_3}^ + \,\rangle \,N{O_2}\,\rangle \,CN\,\rangle \,S{O_3}H\,\rangle \,CHO\,\rangle \,F\,\rangle \,Cl\,\rangle \,Br\,\rangle \,I\]
Note: Please note that the increase in no. of alkyl chain contributes to the increase in inductive effect. Tertiary methyl is the most stable and primary is least stable.
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