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प्रश्न
Show the heterolysis of a covalent bond by using curved arrow notation and complete the following equation. Identify the nucelophile.
CH3 – Br + KOH →
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उत्तर
\[\ce{CH3 – Br + KOH →}\]
\[\ce{CH3 – Br + KQH → CH3OH + KBr}\]
Nucleophile is: OH–
APPEARS IN
संबंधित प्रश्न
For the following reactions
- \[\ce{CH3CH2CH2Br + KOH → CH3 – CH = CH2 + KBr + H2O}\]
- \[\ce{(CH3)3CBr + KOH → (CH3)3COH + KBr}\]
Which of the following statement is correct?
Decreasing order of nucleophilicity is ______.
Homolytic fission of covalent bond leads to the formation of ______.
Which of the group has highest + I effect?
Assertion: Tertiary Carbocations are generally formed more easily than primary Carbocations ions.
Reason: Hyper conjucation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
Heterolytic fission of C–C bond results in the formation of ______.
Write a short note on resonance.
Write a short note on hyperconjucation.
What is nucleophiles? Give a suitable example.
Explain inductive effect with suitable example.
