1. Molecular formula and degree of functionality

Given: $\mathrm{A} = \mathrm{C_5H_{10}O_5}$

This is the formula of a pentose-type compound. Also, $\mathrm{A}$ gives a tetraacetate with acetic anhydride, $\mathrm{Ac_2O}$.

A tetraacetate means 4 hydroxyl groups are present in $\mathrm{A}$, because each $-OH$ gets acetylated.

So, among 5 oxygens:

• 4 are in $-OH$ groups
• 1 oxygen must be part of a carbonyl group

Hence, $\mathrm{A}$ is a pentahydroxy carbonyl compound with one carbonyl and four alcohol groups, i.e. a pentose sugar in open-chain form.

2. Oxidation with $\mathrm{Br_2/H_2O}$

$\mathrm{Br_2/H_2O}$ is a mild oxidizing agent that oxidizes an aldehyde to a carboxylic acid, but does not oxidize a ketone under these conditions.

Given oxidation product: $\mathrm{C_5H_{10}O_6}$

Compared to $\mathrm{A}$: $\mathrm{C_5H_{10}O_5} \to \mathrm{C_5H_{10}O_6}$

One extra oxygen is added, consistent with: $-CHO \to -COOH$

Therefore, $\mathrm{A}$ must be an aldose, not a ketose.

So structure of $\mathrm{A}$ should be of the form: $\mathrm{CHO-CH(OH)-CH(OH)-CH(OH)-CH_2OH}$

3. Reduction with HI gives isopentane

On prolonged heating with HI, all oxygen-containing groups are removed and the carbon skeleton remains as the corresponding hydrocarbon.

Since the product is isopentane $(2\text{-methylbutane})$, the carbon skeleton of $\mathrm{A}$ must already be that of isopentane.

Now, a normal aldopentose has an unbranched carbon chain and would give n-pentane, not isopentane.

Therefore, $\mathrm{A}$ must be a branched aldopentose whose carbon framework corresponds to $2$-methylbutane.

The suitable open-chain aldehyde with 4 hydroxyl groups is: $\mathrm{CHO-CH(OH)-CH(OH)-CH(CH_2OH)-OH}$

which is equivalently written on the isopentane skeleton as a branched aldopentose.

After exhaustive reduction with HI, this gives: $\mathrm{CHO} \to \mathrm{CH_3}, \quad \mathrm{CH(OH)} \to \mathrm{CH_2}, \quad \mathrm{CH_2OH} \to \mathrm{CH_3}$

leading to isopentane.

4. Conclusion from all data

So $\mathrm{A}$ must satisfy all three:

1. Molecular formula $\mathrm{C_5H_{10}O_5}$
2. Contains 4 alcohol groups and 1 aldehyde group
3. Has a branched 5-carbon skeleton that gives isopentane on HI reduction

Thus, among the given options, the correct structure is the one corresponding to option B.

5. Answer check with stored correct answer

Stored correct answer: B

My derived answer: B

So they agree.