
The stability if $M{e_2}C = C{H_2}$ is more than that of $MeC{H_2}CH = C{H_2}$ due to
A. Inductive effect due to $Me$ group
B. Resonance effect due to $Me$ group
C. Hyperconjugation effect due to $Me$ group
D. Resonance as well inductive effect due to $Me$ group
Answer
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Hint- In order to deal with this question first we will understand the term hyperconjugation further according to its property we will determine which compound has more hyperconjugation amongst the given compounds on the basis of the criterion for comparison.
Complete step-by-step answer:
Hyperconjugation effect is a lasting phenomenon under which the position of $C - H$ bond of an alkyl group immediately attached to an unsaturated cell atom or an unshared p orbital atom exists.
We also note that hyperconjugation helps to balance the carbocation as it helps to distribute positive charge. Therefore we can assume that the greater the number of alkyl groups bound to a positively charged carbon atom, the greater the relationship between hyperconjugation and carbocation stabilization
Now, In $M{e_2}C = C{H_2}$ with respect to Alkene the number of $\alpha - $ Hydrogen atoms are 6
(i.e. the number of hydrogen atoms taking part in hyperconjugation).
While in $MeC{H_2}CH = C{H_2}$ , the number of $\alpha - $ Hydrogen atoms are 2.
So, the extent of hyperconjugation is more in
$M{e_2}C = C{H_2}$ than in $MeC{H_2}CH = C{H_2}$
Thus, $M{e_2}C = C{H_2}$ is more stable than $MeC{H_2}CH = C{H_2}$ .
Hence, the stability if $M{e_2}C = C{H_2}$ is more than that of $MeC{H_2}CH = C{H_2}$ due to Hyperconjugation effect due to $Me$ group.
So, the correct answer is option C.
Note- The resonance effect is the polarity created by the interaction of a lone electron pair with a pi bond in a molecule, or the interaction of two pi bonds in neighboring atoms. It is commonly found in double-bonded conjugated molecules or in at least one lone pair and one double bonded molecule.
The inductive effect is an effect on the transfer of uneven bonding electron sharing through a series of atoms in a molecule, resulting in a permanent dipole in a bond.
Complete step-by-step answer:
Hyperconjugation effect is a lasting phenomenon under which the position of $C - H$ bond of an alkyl group immediately attached to an unsaturated cell atom or an unshared p orbital atom exists.
We also note that hyperconjugation helps to balance the carbocation as it helps to distribute positive charge. Therefore we can assume that the greater the number of alkyl groups bound to a positively charged carbon atom, the greater the relationship between hyperconjugation and carbocation stabilization
Now, In $M{e_2}C = C{H_2}$ with respect to Alkene the number of $\alpha - $ Hydrogen atoms are 6
(i.e. the number of hydrogen atoms taking part in hyperconjugation).
While in $MeC{H_2}CH = C{H_2}$ , the number of $\alpha - $ Hydrogen atoms are 2.
So, the extent of hyperconjugation is more in
$M{e_2}C = C{H_2}$ than in $MeC{H_2}CH = C{H_2}$
Thus, $M{e_2}C = C{H_2}$ is more stable than $MeC{H_2}CH = C{H_2}$ .
Hence, the stability if $M{e_2}C = C{H_2}$ is more than that of $MeC{H_2}CH = C{H_2}$ due to Hyperconjugation effect due to $Me$ group.
So, the correct answer is option C.
Note- The resonance effect is the polarity created by the interaction of a lone electron pair with a pi bond in a molecule, or the interaction of two pi bonds in neighboring atoms. It is commonly found in double-bonded conjugated molecules or in at least one lone pair and one double bonded molecule.
The inductive effect is an effect on the transfer of uneven bonding electron sharing through a series of atoms in a molecule, resulting in a permanent dipole in a bond.
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