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Amines — Chemistry STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryAmines4 Marks
Question

Read the passage given below and answer the following questions:
$\ce{RCONH_2}$ is converted into $\ce{RNH_2}$ by means of Hoffmann bromamide degradation. During the reaction amide is treated with $\ce{Br_2}$ and alkali to get amine. This reaction is used to descend the series in which carbon atom is removed as carbonate ion $(\text{CO}^{2-}_3)$ Hoffmann bromide degradation reaction can be written as:

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Hoffmann bromamide degradation is used for the preparation of
  1. Primary amines.
  2. Secondary amines.
  3. Tertiary amines.
  4. Secondary aromatic amines.
  1. Which is the rate determining step in Hoffmann bromamide degradation?
  1. Formation of $(i)$
  2. Formation of $(ii)$
  3. Formation of $(iii)$
  4. Formation of $(iv).$
  1. Which of the following are used for the conversion of $(i)$ to $(ii)?$
  1. $\ce{KBr}$
  2. $\ce{KBr + CH_3ONa}$
  3. $\ce{KBr + KOH}$
  4. $\ce{Br_2 + KOH}$
  1. Identify Bin the following reaction.
$\text{R}-\text{C}\equiv\text{N}\xrightarrow[\text{(Partially hydrolysis) }]{\text{Cone. HCI}}\text{A}\xrightarrow{\frac{\text{Br}_2}{\text{KOH}}}\text{B}$
  1. $\ce{RCONH_2}$
  2. $\ce{RNH_2}$
  3. $\ce{RNHBr}$
  4. $\ce{R = N = C = O}$
  1. What are the constituent amines formed when the mixture of $(i)$ and $(ii)$ undergoes Hoffmann bromamide degradation?
 
✓

Answer

  1. $(a)$ Primary amines.
  1. $(d)$ Formation of $(iv).$
The rate determining step is probably loss of $Br^-$ to form is ocyanate as this is the slowest step.
  1. $(d)\  \ce{Br_2 + KOH}$

  1. $(b)\ \ce{RNH_2}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}\equiv\text{N}\xrightarrow[\text{(partially hydrolysis)}]{\text{Cone. HCI}}\text{R}-\text{C}-\text{NH}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \xrightarrow{\frac{\text{br}_2}{\text{KHO}}}\text{R}-\text{NH}_2$
  1. $(b)$ In ethylamine there is no resonance while in acetamide the lone pair of electrons on $N-$atom is delocalised and is less available for protonation.

Since, the overall reaction is intermolecular, hence there will be no effect on product fonnation.
 

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