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25-2Why does the retention onder of peaks 2 and 3 change on the polar embedded column?

Short Answer

Expert verified

Compound 2 has polar functionalities that could interact with the polar embedded groups, increasing retention. Compound 3 cannot interact with these groups

Step by step solution

01

Define interpretation

Interpretation:

The reason why the retention order of compound 2 and compound 3 changes on polar embedded column has to be explained..

Concept Introduction:

Retention of the compound depends upon the functional groups that are present in it and the column that is used for HPLC. If a polar column is used means the polar compounds gets attracted more and vice-versa.

02

Explanation

Compound 2 is 4-butylbenzoic acid and compound 3 is toluene. By looking into the two compounds, we find that compound 2 has a polar acid group, while compound 3 does not have polar groups. Therefore, when using polar embedded column for HPLC, compound 2 interacts more than compound 3 due to the presence of the polar acidic group. Compound 3 does not have any polar groups and this does not interact with the polar embedded column. Hence, the compound 2 has more retention compared to compound 3 . This is the reason why the retention order is changed in the polar embedded column.

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Most popular questions from this chapter

(a) When you try separating an unknown mixture byreversed-phase chromatography with 48%acetonitrile 50%water,the peaks are too close together and are eluted in the range k = 2- 6Should you use a higher or lower concentration of acetonitrile in thenext run?

(b) When you try separating an unknown mixture by normal-phasechromatography with 50%hexane50% methyl t-butyl ether, thepeaks are too close together and are eluted in the range k = 2 - 6Should you use a higher or lower concentration of hexane in thenext run?

Retention factors for three solutes separated on aC8non-polar stationary phase are listed in the table. Eluent was a 70 : 30 (vol/vol) mixture of 50 mM citrate buffer (adjusted to pH withNH3) plus methanol. Draw the dominant species of each compound at each pH in the table and explain the behavior of the retention factors.

A bonded stationary phase for the separation of optical isomers has the structure

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When the mixture is eluted with 20 vol% 2-propanol in hexane, the (R)-enantiomer is eluted before the (S)-enantiomer, with the following chromatographic parameters:

Resolution=Δ³Ùrwav=7.7

Relative retention(α)=4.53

k for (R)-isomer =1.35

tm=1.00min

wherewavis the average width of the two Gaussian peaks at their base.

(a) Findt1,t2andwawith units of minutes.

(b) The width of a peak at half-height isrole="math" localid="1663582749865" w1/2(Figure 23-9). If the plate number for cach peak is the same, findw1/2for each peak.

(c) The area of a Gaussian peak is1.064×peakheight×w1/2. Given that the areas under the two bands should be equal, find the relative peak heights(heightR//heightS).

A mixture of 14compounds was subjected to a reversed-phase gradient separation going from 5%to 100%acetonitrile with

a gradient time of 60min. The sample was injected at t =time. All peaks were eluted between 22and 50min.

(a) Is the mixture more suitable for isocratic or gradient elution?

(b) If the next run is a gradient, select the starting and ending %acetonitrile

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