Question

a) Illustrate the following name reactions giving suitable example in each case:
i) Clemmensen reduction
ii) Hell-Volhard-Zelinsky reaction

b) How are the following conversions carried out?
(i) Ethyl cyanide to ethanoic acid
(ii) Butan-1-ol to butanoic acid
(iii)Benzoic acid to m-bromobenzoic acid

Answer 1

Hint: As we know clemmensen reduction is based on the reduction of ketones, or aldehydes, and the reaction Hell-Volhard-Zelinsky reaction is related to the alpha position addition in the carboxylic acid. If we talk about the conversions, then these conversions will be done with the help of bromine, or potassium permanganate.

Complete step by step answer:
Now, we will talk about all the reactions step by step.
First, we will discuss part (a).
The first (i) reaction is Clemmensen reduction.
As mentioned this reaction includes the reduction of aldehydes or ketones using hydrochloric acid, and zinc amalgam, and it is reduced to the alkanes. So, let us take an example of acetophenone.
The chemical reaction is

Here, we can see acetophenone is reduced to the Ethyl benzene, or 1-Phenyl Ethane
Now, the second (ii) reaction is Hell- Volhard-Zelinsky, it involves the halogenation of carboxylic acids at the alpha carbon position. This reaction is carried out with the catalytic amount of phosphorus tribromide, and the addition of diatomic bromine.
The chemical reaction is

Here we can see the general carboxylic addition forms alpha-bromo carboxylic acid.
Now, we will discuss the part (b).
The first (i) is the conversion of Ethyl cyanide to ethanoic acid.
The conversion will take place thru hydrolysis, then diatomic bromine, and potassium permanganate. The chemical reaction is

Now, the second (ii) is the conversion of butan -1- ol to butanoic acid. It is done by the use of potassium permanganate. The chemical reaction is

The third (iii) is the conversion of benzoic acid to m-bromobenzoic acid. It takes place through the bromination of benzoic acid. The chemical reaction is

Now, in the end we can conclude that in part (a) (i) the product is alkane, and in (ii) the product is alpha – bromo carboxylic acid. In the part (b) we have shown the conversion reactions.

Note: It is important to know that in the Hell-Volhard- Zelinsky reaction, there is no fluorination, and iodination of carboxylic acids. So, we just consider the bromination of carboxylic acids.