Given below are two statements:
Statement I: Hyper conjugation is permanent effect.
Statement II: Hyper conjugation in ethyl cation (CH₃—CH₂⁺) involves the overlapping of Csp² – H₁ₛ bond with empty 2p orbital of other carbon.
Choose the correct option:
Given below are two statements:
Statement I: Hyper conjugation is permanent effect.
Statement II: Hyper conjugation in ethyl cation (CH₃—CH₂⁺) involves the overlapping of Csp² – H₁ₛ bond with empty 2p orbital of other carbon.
Choose the correct option:
Option 1 - <p>Statement I is incorrect but Statement II is true.<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>Statement I is correct but Statement II is false.<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>Both Statement I and Statement II are false.<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>Both Statement I and Statement II are true.</p>
2 Views|Posted 5 months ago
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1 Answer
A
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5 months ago
Correct Option - 1
Detailed Solution:
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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