- AThe nature of the electrolyte added.
- ✓The nature of the electrode used.
- CConcentration of the electrolyte.
- DThe nature of solvent used.
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In this reaction, $\mathrm{RCONHBr}$ 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?
$(A)$ $\mathrm{KBr}$ $(B)$ $\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$ $(C)$ $\mathrm{KBr}+\mathrm{KOH}$ $(D)$ $\mathrm{Br}_2+\mathrm{KOH}$
$2.$ 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?
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give the answer question $1$, $2$, and $3.$
How many ${H^ \ominus }$ shifts are involved in above rearrangement :
$3 \mathrm{PbCl}_2+2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NH}_4 \mathrm{Cl}$
If $72 \mathrm{mmol}$ of $\mathrm{PbCl}_2$ is mixed with $50 \mathrm{mmol}$ of$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$, then amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is......... mmol. (nearest integer)