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Question: A compound has two isomers (A) and (B) of formula \(\,{C_5}{H_{10}}O\,\). Isomer (A) on treating wit...

A compound has two isomers (A) and (B) of formula C5H10O\,{C_5}{H_{10}}O\,. Isomer (A) on treating with NaOH(aq)\,NaO{H_{(aq)}}\, give 2,2\,2,2 - \,dimethylpropane-1-ol and 2,2\,2,2 - \, dimethylpropanoic acid salt. The isomer (B) on treating with NaOH(aq)\,NaO{H_{(aq)}}\, gives 3\,3 - \,hydroxy- 2\,2 - \,propyl heptanal. What are A and B?
A.A: 2,2\,2,2 - dimethylpropanal B:Isobutanal
B.A: Isobutanal B:Pentanal
C.A: 2,2\,2,2 - \, dimethylpropanal B: Pentanal
D.A:Pentanal B:Isobutanal

Explanation

Solution

To solve this question, we have to first analyse each option and their molecular formula to test whether they are the actual isomers with the given formula. Then we have to analyse which of them gives the products on reaction with respective compounds mentioned in the question.

Complete answer:
Let us first understand what are isomers;
Isomers are molecules or polyatomic ions in chemistry with similar molecular formulas, the same number of atoms of each substance, but different atomic configurations in space. Isomerism is the probability or presence of isomers.
Now, let us analyse each option one by one;
A.A: 2,2\,2,2 - \, dimethylpropanal B:Isobutanal

Let us first write the molecular formula of the above compounds;
2,2\,2,2 - \,dimethylpropanalC5H10O\, - {C_5}{H_{10}}O\,
IsobutanalC4H8O\, - {C_4}{H_8}O\,
From this, it is clear that 2,2dimethyl propanal\,2,2 \text{dimethyl propanal} is an isomer of the compound mentioned in the question as it has the same molecular formula also it gives the products mentioned in the reaction with sodium hydroxide;2,2\,2,2 - \,dimethylpropanalNaOH(aq)\xrightarrow{{NaO{H_{(aq)}}}}\, 2,2\,2,2 - \, dimethylpropane 1\, - 1 - \,ol+\, + \, 2,2\,2,2 - \, dimethylpropanoic acid salt
Hence, this is the isomer A, but isobutanal is not an isomer as its molecular formula is different. So, we can reject this option as one of the compounds here is not even the isomer of the mentioned compound.
B.A: Isobutanal B:Pentanal
Let us analyse the molecular formula;
IsobutanalC4H8O\,\, - {C_4}{H_8}O\,\,
PentanalC5H10O\, - {C_5}{H_{10}}O\,
From this it is clear that only Pentanal is the isomer of the compound and iso-butanal is not. However, pentanal is the isomer B as it gives 3\,\,3 - \, hydroxy 2\,2 - \, propyl heptanal on reaction with aqueous sodium hydroxide. However, isomer A is not correct. Hence, this is also not the correct option.
C.A: 2,2\,2,2 - \,dimethylpropanal B: Pentanal
As we already saw above both of them are the isomers of the compound as their molecular formula is the same;
2,2\,2,2 - \, dimethylpropanal C5H10O\,\,\, - {C_5}{H_{10}}O\,\,
PentanalC5H10O\, - {C_5}{H_{10}}O\,\,
Now, we have to check whether they give the given products in the reaction mentioned in the question;
2,2\,2,2 - \, dimethylpropanal NaOH(aq)2,2\xrightarrow{{NaO{H_{(aq)}}}}\,\,\,2,2 - \,\, dimethylpropane 1\, - 1 - \,\,ol +2,2\, + \,\,2,2 - \,\, dimethylpropanoic acid salt
So, we have now found out our compound A
Now, let us see the reaction of pentanal with sodium hydroxide;
Pentanal NaOH(aq)3\,\xrightarrow{{NaO{H_{(aq)}}}}\,3 - \,hydroxy 2\,2 - \, propyl heptanal
So, we have found out isomer B also. Hence, option is correct
D.A:Pentanal B:Isobutanal
Here, we have already seen their chemical formula, from that we know, only pentanal is the isomer of this compound, but it doesn’t give the above mentioned products in the reaction with sodium hydroxide rather it gives 3\,3 - \, hydroxy 2\,2 - \, propylheptanal. Hence, this is the isomer B. And isobutanal has a different molecular formula and it is not even an isomer of the above compound.
Hence, option C is the correct answer for this question.

Note:
Note that related chemical or physical properties are not inherently shared by isomers.
The reaction of 2,2\,2,2 - \, dimethylpropanal with sodium hydroxide is referred to as Cannizzaro reaction. Due to the +I\, + I\, effect of the methyl group and simple transition of H\,{H^ - }\, ion from one molecule to another, it undergoes the Cannizzaro reaction.