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The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due to
  • a)
    inductive effect of the Me group                        
  • b)
     resonance effect of the Me group
  • c)
    hyperconjugative effect of the Me group              
  • d)
    resonance as well as inductive effect of the Me group
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due toa)i...
Me2C = CH2 is   more stable than MeCH2CH = CHdue to hyperconjugative effect.
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The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due toa)i...
Explanation:

Introduction:
The stability of a molecule can be influenced by various factors, such as the presence of functional groups, the nature of the bonding, and the distribution of electron density. In the case of Me2C=CH2 and MeCH2CH=CH2, the stability difference can be attributed to the presence of the Me group and its effects on the molecule.

Hyperconjugation:
Hyperconjugation is the phenomenon in which the overlap of a filled bonding orbital with an adjacent empty p-orbital or an unhybridized orbital leads to the stabilization of the molecule. In the case of Me2C=CH2, the presence of the Me groups adjacent to the double bond allows for hyperconjugation to occur. The filled bonding orbitals of the Me group can overlap with the empty p-orbital of the adjacent carbon, leading to the stabilization of the molecule. This hyperconjugative effect increases the stability of Me2C=CH2 compared to MeCH2CH=CH2.

Inductive Effect:
The inductive effect is the polarization of electron density along a sigma bond due to the electronegativity difference between atoms. In the case of Me2C=CH2 and MeCH2CH=CH2, the Me group has a +I (electron-donating) effect due to its lower electronegativity compared to carbon. This inductive effect can stabilize the molecule by dispersing the electron density away from the double bond. However, the inductive effect alone is not sufficient to explain the difference in stability between Me2C=CH2 and MeCH2CH=CH2.

Resonance Effect:
The resonance effect is the delocalization of pi electrons through a conjugated system. In the case of Me2C=CH2, the presence of the Me groups adjacent to the double bond allows for resonance to occur. The pi electrons can be delocalized between the carbon atoms and the Me groups, leading to the stabilization of the molecule. This resonance effect increases the stability of Me2C=CH2 compared to MeCH2CH=CH2.

Conclusion:
The stability of Me2C=CH2 is more than that of MeCH2CH=CH2 due to the hyperconjugative effect of the Me group, which allows for the stabilization of the molecule through the overlap of bonding orbitals. The inductive effect and resonance effect also contribute to the stability, but they are not the primary factors in this case.
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The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due toa)inductive effect of the Me group b)resonance effect of the Me groupc)hyperconjugative effect of the Me group d)resonance as well as inductive effect of the Me groupCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due toa)inductive effect of the Me group b)resonance effect of the Me groupc)hyperconjugative effect of the Me group d)resonance as well as inductive effect of the Me groupCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The stability of Me2C=CH2 is more than that of MeCH2CH = CH2 due toa)inductive effect of the Me group b)resonance effect of the Me groupc)hyperconjugative effect of the Me group d)resonance as well as inductive effect of the Me groupCorrect answer is option 'C'. Can you explain this answer?.
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