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Question: An organic compound having ‘X’ having molecular formula \({C_5}{H_{10}}O\) yields phenyl hydrazine a...

An organic compound having ‘X’ having molecular formula C5H10O{C_5}{H_{10}}O yields phenyl hydrazine and gives negative response to the iodoform test and tollen’s test. It produces n-pentane on reduction. ’X’ could be:
A.33-pentanone
B.n-amyl alcohol
C.pentanol
D.22-pentanone

Explanation

Solution

By knowing the functional group and the parent chain we can reach to the final answer. As we know that Tollen’s test is given by the aldehyde group. The Iodoform test is given by the methyl-keto group. The molecule yields phenyl hydrazine so we can conclude that carbonyl group is present in the compound.

Complete step by step answer:
First of all we will talk about the test which is given in the question:
Iodoform test: It is used to check the presence of carbonyl compounds with the structure R−CO−CH3R-CO-CH_3 or alcohols with the structure R - CH(OH) - CH3{\text{R - CH}}\left( {{\text{OH}}} \right){\text{ - C}}{{\text{H}}_3} in a given unknown substance. The reaction of iodine, a base and a methyl ketone gives a yellow precipitate along with an antiseptic smell.
Tollen’s test: A chemical reaction used to test for the presence of an aldehyde or a terminal α−\alpha - hydroxy ketone. To perform the test Tollens' reagent (a colorless Ag + {\text{A}}{{\text{g}}^{\text{ + }}}{\text{}}solution) is added to the unknown. If an aldehyde is present Ag+  A{g^ + }\; is reduced to Ag0  A{g^0}\;which precipitates, often as a silver mirror.
According to the question it is given that:
An organic compound 'X' having molecular formula C5H10O{C_5}{H_{10}}O yields phenyl hydrazone. This indicates the presence of a carbonyl group.
Also given that : It gives a negative response to the Iodoform test. It indicates absence of methyl keto group.
Hence, n-amyl alcohol and 22-pentanone are ruled out.
And given that : It gives a negative response to the Tollen's test. It indicates absence of the aldehyde group.
Hence, pentanal is ruled out.
And lastly it is given that : It produces n-pentane on reduction. This indicates the presence of 55-carbon backbone.
Hence 'X' could be 33-pentanone.

Note:
Read the question carefully as what is given in the question.For example: it is given that the molecule gives negative response towards iodoform and tollen’s test and not the positive that means aldehyde group is absent in the molecule and also methyl-keto group is absent.