Advertisements
Advertisements
Question
Show the heterolysis of a covalent bond by using curved arrow notation and complete the following equation. Identify the nucelophile.
CH3 – Br + KOH →
Advertisements
Solution
\[\ce{CH3 – Br + KOH →}\]
\[\ce{CH3 – Br + KQH → CH3OH + KBr}\]
Nucleophile is: OH–
APPEARS IN
RELATED QUESTIONS
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?
Which of the group has highest + I effect?
Which of the following species does not exert a resonance 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 ______.
Which of the following represent a set of nucleophiles?
Which of the following species does not acts as a nucleophile?
Write a short note on resonance.
What is electrophiles? Give a suitable example.
Explain the electromeric effect.
