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Chemistry Question 75 – JEE-MAIN 2025

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Consider the above sequence of reactions. 2-Bromopentane→alcoholic KOHP (Major Product)→Br2Q 151 g of 2-bromopentane is made to react. Yield of major product P is 80\% whereas Q is 100\%. Mass of product Q obtained is _______ g. (Given molar mass in g mol−1 H : 1, C : 12, O : 16, Br : 80)

The first reaction is an elimination reaction (dehydrohalogenation), and the second is an addition reaction (bromination of an alkene).

Step 1: Identify major product P and calculate its actual moles✦ Active

2-Bromopentane undergoes dehydrohalogenation with alcoholic KOH. According to Zaitsev's rule, the major product P is Pent-2-ene (CH3−CH=CH−CH2−CH3). The molar mass of 2-bromopentane (C5H11Br) is 5(12)+11(1)+1(80)=151 g/mol.

Initial moles of 2-bromopentane=151 g151 g/mol=1 mol

Given that the yield of P is 80\%, the actual moles of P obtained are:

Actual moles of P=1 mol×0.80=0.8 mol
Step 2: Identify product Q and calculate its actual moles○ Expand

Pent-2-ene (P) reacts with Br2 via an addition reaction to form 2,3-dibromopentane (Q), which has the formula C5H10Br2. Given that the yield of Q from P is 100\%, the actual moles of Q obtained are:

Actual moles of Q=0.8 mol×1.00=0.8 mol
Step 3: Calculate the mass of product Q obtained○ Expand

The molar mass of Q (C5H10Br2) is calculated as:

Molar mass of Q=5(12)+10(1)+2(80)=60+10+160=230 g/mol

The mass of product Q obtained is:

Mass of Q=Moles of Q×Molar mass of Q=0.8 mol×230 g/mol=184 g
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