C4 H(11) N on reaction with HNO2 forms t...

`C_4 H_(11) N` on reaction with `HNO_2` forms tertiary alcohol. Thus, `C_4 H_(11) N` is

Primary amine

Secondary amine

Tertiary amine

Quaternary ammonium salt

Text Solution

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The correct Answer is:

`C_4 H_(11) N overset(HNO_2)to 3^@` alcohol. Thus , `C_4 H_(11) N` has- `NH_2` group . Thus , it is `1^@` amime . Since product is `3^@` alcohol , hence `C_4 H_(11) N` is `3^@` butyl amine `underset( 1^@ "amine")((CH_3)_3 C- NH_2) overset(HNO_2)to underset(3^@ "alcohol")((CH_3)_3 C-OH)`

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