(a)
Interpretation:
Structure of lecithin and the naturally occurring enantiomer need to be drawn and if phosphodiester is in C2 position will there be chirality has to be discussed.
Concept introduction:
Phosphoglycerides are same as triglycerides but the only difference is that in triglycerides there are three fatty acid residues. But in phosphoglyceride one of the fatty acid residue in triglyceride is replaced by a phosphoester. Hydrolysis of phosphoglycerides yield glycerol, fatty acid, phosphate.
To draw: the structure of lecithin
(b)
Interpretation:
Structure of lecithin and the naturally occurring enantiomer need to be drawn and if phosphodiester is in C2 position will there be chirality has to be discussed.
Concept Introduction:
Chirality is the ability of the compound to rotate the plane polarized light when it is passed through them. A chiral center is the one when the carbon atom is attached to a four different groups. If the plane polarized light is rotated clockwise then the compound is said to be dextrorotatory and if it is anti-clockwise then the compound is levorotatory.
(c)
Interpretation:
Structure of lecithin and the naturally occurring enantiomer need to be drawn and if phosphodiester is in C2 position will there be chirality has to be discussed.
Concept Introduction:
Chirality is the ability of the compound to rotate the plane polarized light when it is passed through them. A chiral center is the one when the carbon atom is attached to a four different groups. If the plane polarized light is rotated clockwise then the compound is said to be dextrorotatory and if it is anti-clockwise then the compound is levorotatory. If the carbon is not attached to four different groups means then it is said to be achiral.
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Chapter 26 Solutions
ORGANIC CHEMISTRY- NEXTGEN PACKAGE
- 2' P17E.6 The oxidation of NO to NO 2 2 NO(g) + O2(g) → 2NO2(g), proceeds by the following mechanism: NO + NO → N₂O₂ k₁ N2O2 NO NO K = N2O2 + O2 → NO2 + NO₂ Ко Verify that application of the steady-state approximation to the intermediate N2O2 results in the rate law d[NO₂] _ 2kk₁[NO][O₂] = dt k+k₁₂[O₂]arrow_forwardPLEASE ANSWER BOTH i) and ii) !!!!arrow_forwardE17E.2(a) The following mechanism has been proposed for the decomposition of ozone in the atmosphere: 03 → 0₂+0 k₁ O₁₂+0 → 03 K →> 2 k₁ Show that if the third step is rate limiting, then the rate law for the decomposition of O3 is second-order in O3 and of order −1 in O̟.arrow_forward
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