The reason for the lack of optical activity in the 2-chlorobutane after it is examined with a polarimeter is to be explained. Concept introduction: Enantiomers are one of the two molecules that are a mirror image of each other and are non-superimposable. The ability of certain substances to rotate the plane of a plane-polarized light as it passes through a crystal, a liquid, or a solution is known as optical activity. The rotation of the polarized light could be in clockwise or counterclockwise direction. If the rotation of the polarized light will be in clockwise, it is dextrorotatory. If the rotation of the polarized light will be in counterclockwise, it is leavorotatory. If both the enantiomers are present in same amount, then it is optically inactive. A scientific instrument thatis used to measure the angle of rotation caused by passing the polarized light through an optically active substance is known as a polarimeter.

Question

Want to see the full answer?

Answer 1

Question

Chapter 25, Problem 87AP

Interpretation Introduction

Interpretation:

The reason for the lack of optical activity in the 2-chlorobutane after it is examined with a polarimeter is to be explained.

Concept introduction:

Enantiomers are one of the two molecules that are a mirror image of each other and are non-superimposable.

The ability of certain substances to rotate the plane of a plane-polarized light as it passes through a crystal, a liquid, or a solution is known as optical activity.

The rotation of the polarized light could be in clockwise or counterclockwise direction.

If the rotation of the polarized light will be in clockwise, it is dextrorotatory.

If the rotation of the polarized light will be in counterclockwise, it is leavorotatory.

If both the enantiomers are present in same amount, then it is optically inactive.

A scientific instrument thatis used to measure the angle of rotation caused by passing the polarized light through an optically active substance is known as a polarimeter.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³

fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer

34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10