If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _____%levorotary and ________%dextrorotatory

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Bhavya · Answer

sushmitha raj · Answer

% of one enantiomer =enantiomeric excess /2+50% %dextro rotatory=60/2+50=80% %leavo rotatory=80-60=20%

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If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _%levorotary and ____%dextrorotatory

If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _%levorotary and ____%dextrorotatory

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If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _____%levorotary and ________%dextrorotatory

If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _____%levorotary and ________%dextrorotatory

2 Answers

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If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _%levorotary and ____%dextrorotatory

If a sample of 2 butanol hai an enantiomeric excess of 60% of 1-2-butanol, then per cent of each isomer in the mixture is _%levorotary and ____%dextrorotatory