Draw a perspective formula for each of the following: a. (S)-3-chloro-1-pentanol. b. (2R,3R)-2,3-dibromopentane. c. (2S,3R)-3-methyl-2-pentanol. d. (R)-1,2-dibromobutane

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$\fbox{1}$ First write a condensed structure for the compound to determine the asymmetric centers:

$	ext{$\overset{OH\;\;}{\overset{|\;\;\;\;\;}{CH_2}}CH_2 \overset{Cl\;\;}{\overset{|\;\;\;}{CH}}CH_2CH_3$}$

C-3 is the asymmetric center because it is bonded to four different groups.

$\fbox{2}$ Rank the four groups in order of the priority.

$\bullet$ The higher in atomic number, the higher in priority.

Cl has the greatest atomic number. H has the lowest atomic number.

The other two atoms attached directly to the asymmetric center are Cs. So, we will consider the atomic number of the atoms attached to them.

The C in $CH_2CH_3$ is bonded to (H-H-C). The C in $CH_2CH_2OH$ is also bonded to (H-H-C). So, we will consider the atomic number of atoms attached to them to break the tie.

The second C in $CH_2CH_3$ is bonded to (H-H-H). The second C in $CH_2CH_2OH$ is bonded to (H-H-O). O has a greater atomic number than H.

$ightarrow$ the order of priority is $Cl > CH_2CH_2OH > CH_2CH_3 > H$

$\fbox{3}$ Draw the asymmetric center atom bearing the four different groups. The lowest atom in priority(Hydrogen) is bonded by a hatched wedge bond, The highest group in priority($Cl$) is bonded by any other remaining bond. 
Then draw an arrow counterclockwise from ($Cl$) to H to form the S configuration. Draw in the counterclockwise direction the second group in priority which is  $CH_2CH_2OH$ then the third group in priority which is  $CH_2CH_3$.

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Draw a perspective formula for each of the following: a. (S)-3-chloro-1-pentanol. b. (2R,3R)-2,3-dibromopentane. c. (2S,3R)-3-methyl-2-pentanol. d. (R)-1,2-dibromobutane

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