Question

An organic compound \[{\text{C}}{{\text{H}}_3}{\text{CH(OH)C}}{{\text{H}}_3}\] on treatment with acidified \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] gives a compound \[{\text{Y}}\], which reacts with \[{{\text{I}}_2}\] and sodium carbonate to form triiodomethane. The compound \[{\text{Y}}\] is:
A.\[{\text{C}}{{\text{H}}_3}{\text{OH}}\]
B.\[{\text{C}}{{\text{H}}_3}{\text{COC}}{{\text{H}}_3}\]
C.\[{\text{C}}{{\text{H}}_3}{\text{CHO}}\]
D.\[{\text{C}}{{\text{H}}_3}{\text{CH(OH)C}}{{\text{H}}_3}\]

Answer 1

Hint: \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] is an oxidizing agent. The product \[{\text{Y}}\] gives iodoform test and the precipitate form has an antiseptic smell.

Complete step by step answer:
Since we have been given the initial reactant and products, let us start by writing reactions one by one and form compounds.
We have been given secondary alcohol, secondary alcohols react with potassium dichromate to ketone. Acidified potassium dichromate is a very good oxidizing agent and hence it oxidizes the alcohol to ketone by removing hydrogen from alcohol, we can write the reaction as follow:
\[{\text{C}}{{\text{H}}_3}{\text{CH(OH)C}}{{\text{H}}_3} + {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} \to {\text{C}}{{\text{H}}_3}{\text{COC}}{{\text{H}}_3}\]
Hence the compound \[{\text{Y is C}}{{\text{H}}_3}{\text{COC}}{{\text{H}}_3}\]. That is called propanone or dimethyl ketone. It is commonly named as acetone and is the first member of ketone series.
The next reagents that are given to us are the iodoform test. It is used to detect the presence of carbonyl groups, in the presence of base iodine reacts with ketone to form triiodo or iodoform product. The appearance of yellow colour precipitates confirms the presence of carbonyl groups. The reaction is as follow:
\[{\text{C}}{{\text{H}}_3}{\text{COC}}{{\text{H}}_3} + {{\text{I}}_2} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{C}}{{\text{H}}_3}{\text{COONa + CH}}{{\text{I}}_3}\]

Hence the correct option is B.

Note:
The iodoform test is used in laboratories to distinguish between various species. For example, it is used to distinguish primary alcohol and a ketone. Compounds that show positive test for iodoform test are acetaldehyde, methyl ketones and secondary alcohol that contain methyl group in alpha position. This test is also used in identification of unknown compounds.