
An organic compound A with molecular formula ${C_3}{H_6}O$ is resistant to oxidation but forms compound B ${C_3}{H_8}O$ on reduction, compound B reacts with $HBr$ form C and C reacts with $Mg$ from Grignard reagent. D reacts with A to form a product which on hydrolysis gives E. Identify A, B, C, D, E .
Answer
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Hint: The main difference between aldehydes and ketones is that in an aldehyde a hydrogen is attached to the carbon-oxygen double bond while in Ketones there is no such hydrogen attached to the carbon-oxygen double bond.
Complete step by step answer:
In the given question A has formula ${C_3}{H_6}O$ which applies to the general formula of ketones and aldehydes which is ${C_n}{H_{2n}}O$ where $n$ can be any natural number starting from one for aldehydes while for ketones the minimum value of $n$ must be three so that the first ketone begins with three carbon atom
Now as mentioned in the question A is resistant to oxidation and we all know that ketones does not have any hydrogen attach to the carbon-oxygen double bond making them resistant to oxidation hence the compound A is a ketone having IUPAC name as propan$ - 2 - $one and with common name acetone
Next the compound A on reduction gives B and we all know that ketone give alcohols on reduction hence B having formula ${C_3}{H_8}O$ is isopropanol or propan$ - 2 - $ol
Further $HBr$ is added to B which is an alcohol and this is a type of substitution reaction in which $O - H$is replaced by $Br$ forming C which is an alkyl halide having formula $C{H_3} - CHBr - C{H_3}$
Finally C reacts with $Mg$ obtained from Grignard Reagent to form D which has formula $
C{H_3} - CHMgBr - C{H_3} \\
\\
$
And we all know that D on reaction with A will form a product that on hydrolysis will form the compound E as stated in the following reaction.
Note:
For oxidation of ketones we need to rupture the $C - C$ bond and thus it’s done with the help of strong oxidizing agent like $KMn{O_4}$ or ${K_2}C{r_2}{O_7}$ and will produce an ester which on hydrolysis give rise to an alcohol and carboxylic acid.
Complete step by step answer:
In the given question A has formula ${C_3}{H_6}O$ which applies to the general formula of ketones and aldehydes which is ${C_n}{H_{2n}}O$ where $n$ can be any natural number starting from one for aldehydes while for ketones the minimum value of $n$ must be three so that the first ketone begins with three carbon atom
Now as mentioned in the question A is resistant to oxidation and we all know that ketones does not have any hydrogen attach to the carbon-oxygen double bond making them resistant to oxidation hence the compound A is a ketone having IUPAC name as propan$ - 2 - $one and with common name acetone
Next the compound A on reduction gives B and we all know that ketone give alcohols on reduction hence B having formula ${C_3}{H_8}O$ is isopropanol or propan$ - 2 - $ol
Further $HBr$ is added to B which is an alcohol and this is a type of substitution reaction in which $O - H$is replaced by $Br$ forming C which is an alkyl halide having formula $C{H_3} - CHBr - C{H_3}$
Finally C reacts with $Mg$ obtained from Grignard Reagent to form D which has formula $
C{H_3} - CHMgBr - C{H_3} \\
\\
$
And we all know that D on reaction with A will form a product that on hydrolysis will form the compound E as stated in the following reaction.
Note:
For oxidation of ketones we need to rupture the $C - C$ bond and thus it’s done with the help of strong oxidizing agent like $KMn{O_4}$ or ${K_2}C{r_2}{O_7}$ and will produce an ester which on hydrolysis give rise to an alcohol and carboxylic acid.
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