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If the fraction of glucose present as the alphaanomer is a the fraction present as the ββ anomer is b, and the rotation of the mixture is +52.6 degree, we havea(+112.2degree) + b(18.7) =52.6 degreeThere is a very little of the open chain form present, so the fraction present as the α anomer (a) plus the fraction present as the β anomer (b) should account for all the glucose a+b = 1 or b = 1-a.Thus a(+112.2degree) + (1-a)(18.7) =52.6 degreesolving this equation for (a), we have a = 36% i hope it will help u
If the fraction of glucose present as the alphaanomer is a the fraction present as the ββ anomer is b, and the rotation of the mixture is +52.6 degree, we havea(+112.2degree) + b(18.7) =52.6 degreeThere is a very little of the open chain form present, so the fraction present as the α anomer (a) plus the fraction present as the β anomer (b) should account for all the glucose a+b = 1 or b = 1-a.Thus a(+112.2degree) + (1-a)(18.7) =52.6 degreesolving this equation for (a), we have a = 36% i hope it will help u
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