Q. \( \mathrm{C}_{15}\mathrm{H}_{31}\mathrm{COO} \)_3 \mathrm{C}_3\mathrm{H}_5 + \mathrm{Br}_2 \)

We need to interpret the expression and determine what it represents in chemical terms. The given problem is:

\(\text{(c15h31coo)3c3h5 + br2}\)

This looks like a triglyceride fragment (or triglyceride-like structure) plus bromine. The key ambiguity is that the terms are not written in standard chemical notation (for example, parentheses indicate grouping, and the “\(\text{c15h31coo}\)” is not written as a fully expanded structure).

So, the most reasonable chemistry interpretation is a reaction where bromine adds to an unsaturated bond. However, we must check the structural clue:

\(\text{(c15h31coo)3}\) suggests three identical “\(\text{C}_{15}\text{H}_{31}\text{COO}\)” groups attached to something named “\(\text{c3h5}\)”. In triglycerides, a common core is \(\text{C}_3\text{H}_5\) (from glycerol after esterification), and three fatty acid chains form esters. That would describe \(\text{(RCOO)}_3\text{C}_3\text{H}_5\).

Then “\(+ \text{Br}_2\)” suggests bromination (often addition of bromine across carbon–carbon double bonds) or bromine substitution, depending on whether the substrate is unsaturated.

The “fatty acid” unit here is written as \(\text{C}_{15}\text{H}_{31}\text{COO}\). That chain corresponds to

\(\text{C}_{15}\text{H}_{31}\text{COO-}\)

which is consistent with a saturated alkyl chain ending at \(\text{C}_{15}\text{H}_{31}\). Saturated chains have no \(\text{C=C}\) double bonds.

Therefore, if the intended triglyceride is fully saturated, bromine will not do an addition reaction across double bonds because there are none. In typical high-school/general organic chemistry contexts, bromine will only add across \(\text{C=C}\) bonds.

So the correct conclusion in the standard reaction-interpretation framework is:

There are no carbon–carbon double bonds in the given structure, so bromine \(\text{Br}_2\) has no add-on product under normal bromination conditions (no reaction of the “addition across double bonds” type).

Final answer: No net reaction (no bromine addition product can be formed from a saturated triglyceride of the form \(\text{(}\text{C}_{15}\text{H}_{31}\text{COO}\text{)}_3\text{C}_3\text{H}_5\) and \(\text{Br}_2\) by the usual bromine addition mechanism).