The correct order of basic strengths of the given amines is:
- A$Me_2NH$ $(2^o)$ > $MeNH_2$ $(1^o)$ > $Me_3N$ $(3^o)$ > $NH_3$ (in polar protic solvent)
- B$Et_2NH$ $(2^o)$ > $Et_3N$ $(3^o)$ > $EtNH_2$ $(1^o)$ > $NH_3$ (in polar protic solvent)
- C$Me_3N > Me_2NH > MeNH_2 > NH_3$ (in gas phase)
- DAll are correct
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Similar Questions
How will you convert:
$(i)$ Ethanoic acid into methanamine
$(ii)$ Hexanenitrile into $1$-aminopentane
$(iii)$ Methanol to ethanoic acid
$(iv)$ Ethanamine into methanamine
$(v)$ Ethanoic acid into propanoic acid
$(vi)$ Methanamine into ethanamine
$(vii)$ Nitromethane into dimethylamine
$(viii)$ Propanoic acid into ethanoic acid?
$(i)$ Ethanoic acid into methanamine
$(ii)$ Hexanenitrile into $1$-aminopentane
$(iii)$ Methanol to ethanoic acid
$(iv)$ Ethanamine into methanamine
$(v)$ Ethanoic acid into propanoic acid
$(vi)$ Methanamine into ethanamine
$(vii)$ Nitromethane into dimethylamine
$(viii)$ Propanoic acid into ethanoic acid?
Easy
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Consider the above sequence of reactions. The number of bromine atom$(s)$ in the final product $(P)$ will be:
DifficultJEE MAIN 2026
View SolutionWhich of the following reactions does not yield an amine?
Medium
View SolutionHow will you convert:
$(i)$ Benzene into aniline
$(ii)$ Benzene into $N,N$-dimethylaniline
$(iii)$ $Cl-(CH_2)_4-Cl$ into hexane-$1,6$-diamine?
$(i)$ Benzene into aniline
$(ii)$ Benzene into $N,N$-dimethylaniline
$(iii)$ $Cl-(CH_2)_4-Cl$ into hexane-$1,6$-diamine?
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