$p$-aminophenol reacts with one equivalent of acetyl chloride in the presence of pyridine to give mainly

What is the end product of the following reaction sequence?
$CH_3CH_2NH_2$ $\xrightarrow[H^+]{KMnO_4}$ $\xrightarrow{SOCl_2}$ $\xrightarrow[\Delta]{NH_3}$ $\xrightarrow{NaOH/Br_2} ?$

$RCONH_2$ is converted into $RNH_2$ by means of Hofmann bromamide degradation. In this reaction,$RCONHBr$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.
$1.$ How can the conversion of $(i)$ to $(ii)$ be brought about?
$(A)$ $KBr$
$(B)$ $KBr + CH_3ONa$
$(C)$ $KBr + KOH$
$(D)$ $Br_2 + KOH$
$2.$ Which is the rate determining step in Hofmann bromamide degradation?
$(A)$ Formation of $(i)$
$(B)$ Formation of $(ii)$
$(C)$ Formation of $(iii)$
$(D)$ Formation of $(iv)$
$3.$ What are the constituent amines formed when the mixture of $(i)$ and $(ii)$ undergoes Hofmann bromamide degradation?
Give the answer for question $1$,$2$,and $3$.

When propionamide reacts with $Br_2$ in the presence of alkali the product is :

Identify the major product from the following reaction sequence.

$Isopropyl$ amine with an excess of acetyl chloride will give: