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$CH_3-CHCl-CH_2-CH_3$ has a chiral centre. Which one of the following represents its R-configuration?
(Each option is a Fischer projection, described by the groups at the top, left, right and bottom of the central carbon.)
(Each option is a Fischer projection, described by the groups at the top, left, right and bottom of the central carbon.)
A
Top $C_2H_5$, left $H$, right $CH_3$, bottom $Cl$
B
Top $C_2H_5$, left $Cl$, right $CH_3$, bottom $H$
C
Top $CH_3$, left $H$, right $Cl$, bottom $C_2H_5$
D
Top $C_2H_5$, left $H_3C$, right $Cl$, bottom $H$
Detailed Solution
Step 1: assign priorities to the four groups on the chiral carbon by atomic number: $Cl$ (1) > $C_2H_5$ (2) > $CH_3$ (3) > $H$ (4).
($C_2H_5$ ranks above $CH_3$ because its first carbon is attached to (C, H, H) while the methyl carbon is attached to (H, H, H).)
Step 2: in a Fischer projection the vertical bonds point away from the viewer. If the lowest priority group (H) is on a vertical bond, the direction 1 to 2 to 3 seen on paper gives the configuration directly; if H is on a horizontal bond, the direction seen must be reversed.
For the projection with $C_2H_5$ on top, $Cl$ on the left, $CH_3$ on the right and $H$ at the bottom: H is on a vertical bond.
Going from $Cl$ (left) to $C_2H_5$ (top) to $CH_3$ (right) is a clockwise path.
Clockwise with H pointing away means the R-configuration.
In the projection with $H_3C$ on the left and $Cl$ on the right (H at the bottom), the path $Cl \rightarrow C_2H_5 \rightarrow CH_3$ is anticlockwise, so it is S.
In the two projections with H on the left (a horizontal bond), the path seen is clockwise, which on reversal gives S.
So the R-configuration is the projection with $C_2H_5$ at the top, $Cl$ on the left, $CH_3$ on the right and $H$ at the bottom.
($C_2H_5$ ranks above $CH_3$ because its first carbon is attached to (C, H, H) while the methyl carbon is attached to (H, H, H).)
Step 2: in a Fischer projection the vertical bonds point away from the viewer. If the lowest priority group (H) is on a vertical bond, the direction 1 to 2 to 3 seen on paper gives the configuration directly; if H is on a horizontal bond, the direction seen must be reversed.
For the projection with $C_2H_5$ on top, $Cl$ on the left, $CH_3$ on the right and $H$ at the bottom: H is on a vertical bond.
Going from $Cl$ (left) to $C_2H_5$ (top) to $CH_3$ (right) is a clockwise path.
Clockwise with H pointing away means the R-configuration.
In the projection with $H_3C$ on the left and $Cl$ on the right (H at the bottom), the path $Cl \rightarrow C_2H_5 \rightarrow CH_3$ is anticlockwise, so it is S.
In the two projections with H on the left (a horizontal bond), the path seen is clockwise, which on reversal gives S.
So the R-configuration is the projection with $C_2H_5$ at the top, $Cl$ on the left, $CH_3$ on the right and $H$ at the bottom.