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Hofmann bromamide degradation reaction is shown by:
A: $ArN{H_2}$
B: $ArCON{H_2}$
 C: $ArN{O_2}$
D: $ArC{H_2}N{H_2}$

Last updated date: 28th Feb 2024
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Hint:Degradation reaction is a type of chemical reaction in which reactant decomposes into two or more products. In Hoffmann bromamide reaction amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide.

Complete answer:
In Hoffmann’s bromamide reaction there is the degradation of amide. In this reaction, the amide is treated with bromine in an aqueous solution of the ethanolic solution of sodium hydroxide. This results in the formation of primary amine. Primary amines are those in which the amino group is directly attached to the carbon atom. The primary amine formed will contain one carbon less than the actual compound. In this reaction, we have to find the amide group as in all the options groups are different (Hoffmann bromamide reaction causes degradation of amides). An amide group is a group in which the carbonyl group is linked to the nitrogen atom that is $ArCON{H_2}$. $ArCON{H_2}$ Group is called an amide group. Among the given options, only option B contains this group. This reaction will be as follows:
$ArCON{H_2} + B{r_2} + 4NaOH\xrightarrow[{}]{}ArN{H_2} + N{a_2}C{O_3} + 2NaBr + 2{H_2}O$
Answer to this question is option B that is $ArCON{H_2}$.

Additional information:}
 Pinacol-Pinacolone is an rearrangement for converting $1,2 - $diol to carbonyl compound. This rearrangement takes place in an acidic medium. So this is also not the answer.
In Cannizzaro’s reaction base is sodium hydroxide and then the oxidation product (in which there is the loss of electron) is a carboxylic acid and the reduction product (in which there is the gain of electro) is alcohol. In the tishchenko reaction we use a conjugate base of alcohol so we can consider this as an answer but let’s see what the next option is.
Claisen condensation is a carbon-carbon bond-forming reaction which occurs between two esters or between one ester and one carbonyl compound in presence of a strong base resulting in the formation of $\beta - $keto ester. Therefore this option is also not the answer.

Tischenko reaction is an organic chemical reaction in which there is a disproportionation of an aldehyde in the presence of an alkoxide. Aldol condensation is a reaction in which enol or enolate ion reacts with a carbonyl compound to form $\beta - $hydroxy aldehyde or $\beta - $hydroxy ketone.

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