Which of these would NOT be considered “maintaining homeostasis”?
1 answer:
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i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The structural formula of Y is and Z is
.
iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.
iv. The process of ozonolysis is useful in the field of pharmaceutics.
v. The test of unsaturation can be performed by passing a compound through Bromine water.
<h3>What is ozonolysis?</h3>Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.
i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.
The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The formulas of Y is and Z is
. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.
iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.
iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.
v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.
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