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प्रश्न
Which of the following ions is more stable? Use resonance to explain your answer.
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उत्तर
Structure (A) is more stable due to resonance. Structure (B) is non-planar and hence it does not undergo resonance. Double bond is more stable within the ring in comparison to outside the ring.
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संबंधित प्रश्न
Draw the resonance structure for the following compound. Show the electron shift using curved-arrow notation.
C6H5OH
Draw the resonance structure for the following compound. Show the electron shift using curved-arrow notation.
C6H5NO2
Draw the resonance structure for the following compound. Show the electron shift using curved-arrow notation.
CH3CH = CHCHO
Draw the resonance structure for the following compound. Show the electron shift using curved-arrow notation.
C6H5 – CHO
Draw the resonance structure for the following compound. Show the electron shift using curved-arrow notation.
\[\ce{CH3CH = CH\overset{+}{C}H2}\]
In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?
 |
| (A) |
| (i) |  |
| (ii) |  |
| (iii) |  |
| (iv) |  |
What is the hybridisation of each carbon in H2C = C = CH2.
Show the polarisation of carbon-magnesium bond in the following structure.
CH3 – CH2 – CH2 – CH2 – Mg – X
Draw the possible resonance structures for \[\ce{CH3 - \overset{\bullet\bullet}{\underset{\bullet\bullet}{O}} - \overset{+}{C}H2}\] and predict which of the structures is more stable. Give reason for your answer.
Draw the resonance structure of the following compounds;
Draw the resonance structure of the following compounds;
CH2 = CH – CH = CH2
Draw the resonance structure of the following compounds;
\[\begin{array}{cc}
\ce{CH2 = CH - C = O}\\
\phantom{.........}|\\
\phantom{.........}\ce{H}
\end{array}\]
Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.
\[\ce{\underset{I}{CH2 = CH - CH = O} <-> \underset{II}{\overset{⊕}{C}H2 - CH = CH - \overset{Θ}{O}}}\]
