Question

For making distinction between \[2-\]pentanone and \[3-\]pentanone, the reagent to be employed is:
A.\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}_{2}}S{{O}_{4}}\]
B.$Zn-Hg/HCl$
C.$Se{{O}_{2}}$
D.\[Idoine/NaOH\]

Answer 1

Hint: We know that Ketones can be distinguished by two chemical tests namely iodoform test and \[2,4-\]DNP test. \[2,4\text{ }\]DNP test is done for identifying carbonyl compounds on the whole but iodoform test is performed to distinguish between different ketones as well.

Complete step by step answer:
In iodoform test when iodine reacts with methyl ketone in the presence of base a yellow precipitate is formed which has an antiseptic smell. Iodoform test is valid for some specific secondary alcohols which have methyl groups at alpha position. For iodoform test it is necessary to have a methyl ketone group. Silver-mirror test which is also known as tollen’s test, it helps to differentiate between aldehyde and a ketone. It is used in labs for study of unknown compounds. As per option A it states that Pentan\[-2-\]one will give a silver mirror test. Silver mirror test reacts with aldehyde. It does not react with the ketonic group. Therefore, option A is false, that is that Pentan\[-2-\]one does not give a silver mirror test. As per option B it states that Pentan\[-2-\]one will give an iodoform test.
Ketones having \[C{{H}_{3}}CO-~\] group give yellow coloured iodoform with \[{{I}_{2}}\] ​ and \[NaOH\] . Therefore, \[{{I}_{2}}/NaOH\] is used to distinguish between \[2-\]pentanone and \[3-\]pentanone.

Therefore, the correct answer is option D.

Note:
Remember that Iodoform test is generally used to determine the presence of methyl group on the adjacent carbon of the functional group attached carbon atom. This means an iodoform test is performed to distinguish methyl ketones from all the other ketones.