The absolute configuration of

is:

The second carbon (C2) is attached to $$-OH$$, $$-CO_2H$$, $$H$$, and the C3 group.

(1) $$-OH$$: Oxygen (atomic number 8) is the highest.

(2) $$-C3$$ group: This carbon is attached to $$(Cl, C, H)$$. Since Chlorine (17) has a higher atomic number than Oxygen (8), this group beats the carboxylic acid.

(3) $$-CO_2H$$: This carbon is attached to $$(O, O, O)$$.

(4) $$-H$$: Hydrogen is always the lowest.

The sequence $$1 \rightarrow 2 \rightarrow 3$$ ($$OH \rightarrow C3 \rightarrow CO_2H$$) moves clockwise. In a Fischer projection, if the lowest priority group ($$H$$) is on a horizontal bond, we reverse the result. Clockwise becomes S.

The third carbon (C3) is attached to $$-Cl$$, the C2 group, $$-CH_3$$, and $$H$$.

(1) $$-Cl$$: Chlorine (17) is the highest.

(2) $$-C2$$ group: This carbon is attached to $$(O, C, H)$$. Oxygen beats the hydrogens of the methyl group.

(3) $$-CH_3$$: Carbon attached to $$(H, H, H)$$.

(4) $$-H$$: Lowest priority.

The sequence $$1 \rightarrow 2 \rightarrow 3$$ ($$Cl \rightarrow C2 \rightarrow CH_3$$) moves counter-clockwise. Since $$H$$ is on a horizontal bond, we reverse the result. Counter-clockwise becomes R.