See image — Amines Chemistry Question

💡 Solution & Explanation

Concept: When a primary aliphatic amine undergoes deamination (diazotization followed by loss of N2) with NaNO2/HCl, the diazonium ion formed is unstable and generates a carbocation intermediate. This carbocation can undergo various reactions including elimination to give alkenes, as well as rearrangement. Step 1: Identify the substrate. n-Bu-NH2 is n-butylamine (CH3CH2CH2CH2NH2), a primary amine with the amino group at C1. Step 2: Reaction mechanism. Treatment with NaNO2/HCl generates the diazonium salt CH3CH2CH2CH2-N2+, which rapidly loses N2 to form the primary carbocation CH3CH2CH2CH2+ (1-butyl carbocation). Step 3: Carbocation rearrangement. The primary carbocation at C1 can rearrange via a 1,2-hydride shift to give the more stable secondary carbocation at C2: CH3CH2+CHCH3. Step 4: Elimination from C1 carbocation (1-butyl cation). Loss of a proton from C2 gives 1-butene (CH2=CHCH2CH3). Step 5: Elimination from C2 carbocation (2-butyl cation). Loss of a proton from C1 gives 1-butene (same as above); loss of a proton from C3 gives 2-butene. 2-Butene exists as cis and trans isomers, giving cis-2-butene and trans-2-butene. Step 6: Count the isomeric butenes obtained: 1. 1-butene (CH2=CHCH2CH3) 2. cis-2-butene 3. trans-2-butene These are 3 isomeric butenes accessible from the carbocation intermediates generated during deamination of n-butylamine. Step 7: Why other options fail. - Option (a) 2: Undercounts because it ignores the cis/trans distinction of 2-butene. - Option (c) 4 and (d) 5: Overcount; isobutylene (2-methylpropene) would require a methyl shift to give a branched skeleton, which is less favored under these conditions and is not counted as a butene isomer in the standard treatment of this reaction for this context. Therefore, the correct answer is B.