The correct order of stability for the following alkoxides is

 

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mo>(</mo> <mi mathvariant="normal">B</mi> <mo>)</mo> <mo>&gt;</mo> <mo>(</mo> <mi mathvariant="normal">C</mi> <mo>)</mo> <mo>&gt;</mo> <mi mathvariant="normal">A</mi> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mo>(</mo> </math> </span> C) <span class="mathml" contenteditable="false"> <math> <mo>&gt;</mo> <mo>(</mo> </math> </span> A <span class="mathml" contenteditable="false"> <math> <mo>)</mo> <mo>&gt;</mo> <mo>(</mo> <mi>B</mi> <mo>)</mo> </math> </span>&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</p>
Option 3 - <p>&lt;!-- [if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" o:preferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f"> <v:stroke joinstyle="miter"/> <v:formulas> <v:f eqn="if lineDrawn pixelLineWidth 0"/> <v:f eqn="sum @0 1 0"/> <v:f eqn="sum 0 0 @1"/> <v:f eqn="prod @2 1 2"/> <v:f eqn="prod @3 21600 pixelWidth"/> <v:f eqn="prod @3 21600 pixelHeight"/> <v:f eqn="sum @0 0 1"/> <v:f eqn="prod @6 1 2"/> <v:f eqn="prod @7 21600 pixelWidth"/> <v:f eqn="sum @8 21600 0"/> <v:f eqn="prod @7 21600 pixelHeight"/> <v:f eqn="sum @10 21600 0"/> </v:formulas> <v:path o:extrusionok="f" gradientshapeok="t" o:connecttype="rect"/> <o:lock v:ext="edit" aspectratio="t"/> </v:shapetype><v:shape id="_x0000_i1025" type="#_x0000_t75" style='width:4.5pt;\\\\\\\\n height:13pt'> <v:imagedata src="file:///C:/Users/ALOK~1.SIN/AppData/Local/Temp/msohtmlclip1/01/clip_image001.png" o:title="" chromakey="white"/> </v:shape><![endif]--&gt;&lt;!-- [if !vml]--&gt;<img >&lt;!--[endif]--&gt;&lt;!--[endif]--&gt; C) <span class="mathml" contenteditable="false"> <math> <mo>&gt;</mo> <mo>(</mo> <mi>B</mi> <mo>)</mo> <mo>&gt;</mo> <mo>(</mo> <mi>A</mi> <mo>)</mo> </math> </span><br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mo>(</mo> <mi>B</mi> <mo>)</mo> <mo>&gt;</mo> <mo>(</mo> <mi>A</mi> <mo>)</mo> <mo>&gt;</mo> <mo>(</mo> <mi>C</mi> <mo>)</mo> </math> </span></p>
1 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 3
Detailed Solution:
Kindly go through the solution

 

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ΔG° = –RT * 2.303 log K

–nFE° = +RT * 2.303 log K

2 * 96500 * 0.295 = 8.314 * 298 * 2.303 log10 K

10 = log10 K = 1010

It has chiral centre and differently di substituted double bonded carbon atoms.

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten 2025

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