Compound A, C$$_5$$H$$_{10}$$O$$_5$$, given a tetraacetate with AC$$_2$$O and oxidation of A with Br$$_2$$ - H$$_2$$O gives an acid, C$$_5$$H$$_{10}$$O$$_6$$. Reduction of A with HI gives isopentane. The possible structure of A is:

The molecular formula of compound $$A$$ is $$C_5H_{10}O_5$$.

Step 1 : Information from the acetylation
With $$Ac_2O$$, $$A$$ gives a tetra-acetate, therefore $$A$$ contains four free $$-OH$$ groups (each converts into an acetate). One oxygen atom is already present in some other functional group, so the total number of oxygens (5) is accounted for as

$$4\;(\text{from } -OH)\;+\;1\;(\text{from a carbonyl})=5$$

Step 2 : Information from the mild oxidation
Bromine water $$(Br_2\!-\!H_2O)$$ oxidises an aldehyde to a monocarboxylic acid but does not attack ketones or alcohols. Because $$A$$ is converted into an acid of the same carbon count, $$A$$ must contain one aldehydic $$-CHO$$ group and no other oxidisable carbon-carbon bonds.

Step 3 : Information from reduction with HI/red P
Hot $$HI$$ in the presence of red phosphorous replaces every $$-OH$$ by $$-H$$ and reduces the aldehyde to $$-CH_3$$ without altering the carbon skeleton. For $$A$$ the product is isopentane ($$2\!-\!methylbutane$$):

$$\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}$$

This hydrocarbon shows that the carbon skeleton of $$A$$ is not a straight chain of five carbons; it is a four-carbon chain carrying one methyl substituent at the second carbon.

Step 4 : Constructing the structure of $$A$$
Let the main (unbranched) chain be $$\mathrm{C_1-C_2-C_3-C_4}$$ with the aldehyde at $$C_1$$. Because the HI-reduction product has a methyl branch on the carbon that is second from one end, the branch must be attached to $$C_2$$ in the original molecule. To retain an overall total of five oxygens and form four acetates, the four hydroxyl groups are placed on:

$$C_2,\;C_3,\;C_4$$ and on the branching carbon itself.

The only arrangement satisfying all these points is

$$\mathrm{HO\!-\!CH_2\!-\!CH(OH)\!-\!CH(OH)\!-\!CHO}$$
                  $$|$$
                $$HO\!-\!CH$$

i.e. $$2\!-\!(\!hydroxymethyl)\!-\!2,3,4\!-\!trihydroxybutanal$$.

Verification
1. Four $$-OH$$ groups → tetra-acetate ✔
2. One aldehyde carbon → oxidised to $$C_5H_{10}O_6$$ by $$Br_2/H_2O$$ ✔
3. After complete hydrogenolysis (HI/red P)     $$CHO\rightarrow CH_3$$, every $$OH\rightarrow H$$ → $$CH_3-CH(CH_3)-CH_2-CH_3$$ (isopentane) ✔

Thus the only structure fulfilling all the conditions is the one shown in Option C.

Final answer: Option C - $$2\!-\!(\!hydroxymethyl)\!-\!2,3,4\!-\!trihydroxybutanal$$ (open-chain form).