Solveeit Logo
Login

Question

Question: An alcohol A on dehydration gives B which on ozonolysis gives acetone and formaldehyde.B decolourise...

An alcohol A on dehydration gives B which on ozonolysis gives acetone and formaldehyde.B decolourises alkaline KMnO4KMn{O_4} but A does not.A and B are:
a) CH3CH2CH2CH2OHC{H_3}C{H_2}C{H_2}C{H_2}OH and CH2CH2CH=CH2C{H_2}C{H_2}CH = C{H_2}
b) CH2CH2−C∣OH−CH2C{H_2}C{H_2} - \mathop C\limits_{\mathop |\limits_{OH} } - C{H_2} and CH2−CH=CH−CH2C{H_2} - CH = CH - C{H_2}
c) (CH3)C−OH(C{H_3})C - OH and (CH3)2C=CH2{(C{H_3})_2}C = C{H_2}
d) (CH3)3CHCH2−OH{(C{H_3})_3}CHC{H_2} - OH and (CH3)3C=CH2{(C{H_3})_3}C = C{H_2}

Explanation

Solution

Alcohols undergo dehydration to form corresponding alkenes.When alkenes are subjected to ozonolysis they tend to produce alcohols,aldehydes,ketones or carboxylic acids.Treatment of organic compounds with alkaline KMnO4KMn{O_4} (Baeyer’s reagent) is a test which is commonly carried out to test unsaturation in organic compounds.

Complete answer:
We will divide the reactions which are occurring in the above question in two parts
Part 1--->
B(Alkene)→OzonolysisCH3−C∣CH3=O(Acetone)+H2C=O(Formaldehyde)B(Alkene)\xrightarrow{{Ozonolysis}}C{H_3} - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = O(Acetone) + {H_2}C = O(Formaldehyde)
The alcohol undergoes dehydration to form corresponding alkene
Part 2--->
B(Alkene)→OzonolysisCH3−C∣CH3=O(Acetone)+H2C=O(Formaldehyde)B(Alkene)\xrightarrow{{Ozonolysis}}C{H_3} - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = O(Acetone) + {H_2}C = O(Formaldehyde)
Ozonolysis can be termed as the addition of ozone across the double bond and subsequent hydrolysis of ozonide thus formed.If we carry out analysis of the reaction occurring in part we will get the product B.In our analysis it’s worth noting that the carbon atoms which are bonded to oxygen will end up losing them and instead bind to each other.
CH3−C∣CH3=O(Acetone)+H2C=O(Formaldehyde)→CH3−C∣CH3=CH2C{H_3} - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = O(Acetone) + {H_2}C = O(Formaldehyde) \to C{H_3} - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = C{H_2}

Hence the alkene which is Isobutylene (B) has been obtained.The alkene undergoes hydration,by following Markovnikov rule
CH3−C∣CH3=CH2→H2O(CH3)C−OHC{H_3} - \mathop C\limits_{\mathop |\limits_{C{H_3}} } = C{H_2}\xrightarrow{{{H_2}O}}(C{H_3})C - OH
Hence we have also obtained the product A.It’s known as tert-Butanol.
From the product A and B which we have obtained in the process we hereby conclude that option (c) is correct.

Note:
Ozonolysis of alkenes offers a valuable method of detection and location of double bonds in complex organic molecules.An alkaline solution of cold Potassium permanganate ( KMnO4KMn{O_4} ) is a known as Baeyer’s reagent.It acts like a strong oxidising agent.