[mass]

Just a quick doubt:

I have been asked to draw an aldo hexoxe, the OH group on the 4th carbon is on the left side compared to all other OH groups which are on the right within the molecule. Now if I was asked to draw an aldotetrose or aldopentose would I have to follow the same mechanism?

[shelanachium]

I presume you mean aldohexOse not aldohexAse, which would be an enzyme acting on an aldohexose.

An aldohexose contains a chain of 6 C atoms, with each bearing an OH group except one of the terminal ones, which is in an aldehyde group, i.e.

HOCH2CH(OH)CH(OH)CH(OH)CH(OH)CHO is an aldohexose.

An aldotetrose is similar with a chain of four C atoms, and an aldopentose with five.

Other sugars are KETOSES, with a ketone carbonyl inside the chain; in natural sugars this is usually the carbon next to the end. Thus

HOCH2CH(OH)CH(OH)COCH2OH is a ketopentose.

The diagrams you have been drawing sound like Fischer diagrams, which show the stereochemistry of each C atom. In these it matters which side you draw the OH groups and H atoms as their stereochemistry depends on which particular sugar you are drawing. For example, glucose, galactose and mannose are all aldohexoses, but differ in the stereochemistry of their -CH(OH)- groups.

This stereochemistry does not matter if you have been asked to draw a generalised aldohexose.