CHAPTER 1: Question 6

2D Explanations

a) First rotate the structure so that the C-CH₃ and C-CO₂H bonds are vertical and pointing away from you with the acid group at the top, whilst the C-H and C-NH₂ bonds are horizontal and pointing towards you. It is important that the acid group is at the top of the structure, as the carbonyl carbon would be numbered 1 in organic nomenclature. This is the difficult part of the process, and a molecular model may be very useful. Next redraw the wedges and hashes as normal bonds, then delete bonds to hydrogen atoms to give the Fischer projection. The process is shown below.

b) First rotate the structure so that the C-CH₂OH and C-CO₂H bonds are vertical and pointing away from you with the acid group at the top, whilst the C-H and C-NH₂ bonds are horizontal and pointing towards you. It is important that the acid group is at the top of the structure, as the carbonyl carbon would be numbered 1 in organic nomenclature. This is the difficult part of the process, and a molecular model may be very useful. Next redraw the wedges and hashes as normal bonds, then delete bonds to hydrogen atoms to give the Fischer projection. The process is shown below, though note that the orientation of the OH group attached C3 is not important as C3 is not a stereocentre.

c) First rotate the structure by 180^o about the midpoint of the C2-C3 bond so that the acid group is at the top of the diagram. Next, rotate C3 about the C2-C3 bond so that it is pointing in the same direction as the acid group. This is the difficult part of the process, and a molecular model may be very useful. View the molecule from the side opposite to the acid and methyl groups which will give a projection in which the C1-C4 bonds are vertical and directed away from you, whilst the other bonds are horizontal and directed towards you. Next redraw the wedges and hashes as normal bonds, then delete bonds to hydrogen atoms to give the Fischer projection. The process is shown below.

d) First rotate the structure anti-clockwise by 90^o, so that the aldehyde is at the top. If the resulting structure is viewed from the left side, then the bonds around C2 and C4 are correctly oriented to allow a Fischer projection to be drawn, whilst if the molecule is viewed from the right side then the bonds around C3 are correctly located to allow a Fischer projection to be drawn. Finally, redraw the wedges and hashes as normal bonds, then delete bonds to hydrogen atoms to give the Fischer projection. The process is shown below.

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