Question

Product(s) formed in the reaction below is(are)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

\(\mathrm{Sn/HCl}\) reduces N-heteroarenes to tetrahydro derivatives via iminium intermediates.
Highly activated aryl tethers can trap iminium ions intramolecularly, giving ring-closed products alongside the simply reduced product.

Answer 1

The Correct Option is A, C

Step 1: Initial reduction under \(\mathrm{Sn/HCl}\).
Quinolines are \emph{chemoselectively reduced} by \(\mathrm{Sn/HCl}\) to the corresponding 1,2,3,4-tetrahydroquinolines via protonation at N followed by stepwise electron transfer. The process generates a \emph{benzylic iminium} at C-2 that can be either \emph{reduced} further or \emph{trapped} intramolecularly. Step 2: Two competing fates of the C-2 iminium.
(i) Direct reduction of the C-2 iminium to the saturated side chain (no C–C bond formation) furnishes the \emph{open} product shown in option (A).
(ii) Because the pendant \(3,4\)-dimethoxybenzyl arene is strongly activated, \emph{intramolecular electrophilic attack} onto the C-2 iminium can also occur. Closure across the side chain forms a medium ring (benzazocine) bearing an \(\mathrm{NH}\) bridge, which corresponds to option (C). Under the strongly acidic, reducing conditions both pathways operate, giving a mixture dominated by (A) with significant (C) from cyclization. \[ \boxed{\text{Products formed: A and C.}} \]