When 1-pentyne (A) is treated with 4 N alcoholic KOH at 175^@c , it is convertedslowly into an equilibrium mixture of 1.3 T 1-pentyne (A) , 95.2% 2-pentyne (B) and 3.5 % of 1,2-pentadiene C . The equilibrium was maintained at 175^@C. Calculate DeltaG^@ for the followingequilibria . {:(BhArrA,DeltaG_1^@=?),(BhArrC,DeltaG_2^@=?):} From the calculated value of DeltaG_1^@ and DeltaG_2^@ indicate the order of stabilityof (A) , (B) and (C ) . write a reasonable reaction mechanism showing all intermediates leading to (A),(B) and (C ) .

When 1-pentyne (A) is treated with 4 N alcoholic KOH at 175^@c , it is

1.	When 1-pentyne (A) is treated with 4 N alcoholic KOH at 175^@c , it is convertedslowly into an equilibrium mixture of 1.3 T 1-pentyne (A) , 95.2% 2-pentyne (B) and 3.5 % of 1,2-pentadiene C . The equilibrium was maintained at 175^@C. Calculate DeltaG^@ for the followingequilibria . {:(BhArrA,DeltaG_1^@=?),(BhArrC,DeltaG_2^@=?):} From the calculated value of DeltaG_1^@ and DeltaG_2^@ indicate the order of stabilityof (A) , (B) and (C ) . write a reasonable reaction mechanism showing all intermediates leading to (A),(B) and (C ) .
Answer» Solution :`{:("","PENTYNE",overset(KOH)HARR,2-"Pentyne"+1","2,-"Pentadiene"),("",(A),"",(B),(C )),("At eqm"% ,1.3,"",95.5,3.5):}` `K_c=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31` For eqm. `B hArr A` `K_1=([A])/([B])` From Eqs. (i) and (ii) ,`K_1=([C])/(K_c)` `=(3.5)/(256.31)=0.013` `DeltaG^2=-2.303 RTlog_(10)K` `=-2.303 xx 8.314xx448 log _(10)0.013` =161787 =16.178 kJ STABILITY order for A and B is `B gt A` Similarly `B hArr C` `K_2=([C])/([B])` `=(K_cxx[A])/([B]^2)` `=(256.31xx1.3)/(95.2xx95.2)` `therefore DeltaG_2^@=-2.303 RT log _(10)k` `=-2.303 xx 8.314 xx448 log _(10) 0.037` =12282 J =12.282 kJ Thus stability order for B and C is `B gt C` Total ordeer of stability is `B gt C gt A`