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$C{H_3} - C{H_2} - CO - H\xrightarrow{{\operatorname{Re} dP + HI}}A$
Product A is:
C.Propanoic acid

Last updated date: 13th Jun 2024
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Hint: -At first think about the behaviour of reagent used in the reaction and the properties and conditions of ketones. $HI/\operatorname{Re} dP$ is a strong reducing agent. So in the given reaction the ketone is reduced.

Complete step by step answer:
In the question we have given a ketone in the presence of red phosphorus and hydrogen iodide. We all know that $HI/\operatorname{Re} dP$ is a strong reducing agent as it reduces alcohols, aldehydes and ketones to alkanes. So propanone in the presence of red phosphorus and hydrogen iodide forms propane. The overall reaction is,
$C{H_3} - C{H_2} - CO - H\xrightarrow{{\operatorname{Re} dP + HI}}C{H_3} - C{H_2} - C{H_3}$
From the above reaction we can say that the correct option is A.
Additional Information:-Propane is a three carbon alkane with molecular formula ${C_3}{H_8}$. It is a gas at standard temperature and pressure, but compressible to a transportable liquid. It is a by-product of natural gas processing and petroleum refining and it is commonly used as a fuel. Propane has become a popular choice for barbecues and portable stoves because its low boiling point makes it vaporize as soon as it is released from its pressurized container. It is also used in some locomotive diesel engines to improve combustion. It is a colorless, odorless gas. At normal pressure it liquefies below its boiling point and solidifies below its melting point. In the presence of excess oxygen, propane burns to form water and carbon dioxide.

Reducing agent is an element or compound that loses an electron to an electron recipient in a redox chemical reaction. Reducing agents reduce oxidizing agents. Strong reducing agents are used because they easily lose electrons.