The cycloalkane ( X ) on bromination consumes one mole of bromine per mole of ( X ) and gives the product ( Y ) in which \(\mathrm{C}: \mathrm{Br}\) ratio is \(3: 1\). The percentage of bromine in the product \((\mathrm{Y})\) is \_\_\_\_ \%. (Nearest integer) (Given : Molar mass in \(\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12\), \(\mathrm{O}: 16, \mathrm{Br}: 80\) )

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Hydrocarbons

The cycloalkane ( X ) on bromination consumes one mole of bromine per mole of ( X ) and gives the product ( Y ) in which

C : Br

ratio is

3 : 1

. The percentage of bromine in the product

(Y)

is
____ %. (Nearest integer)
(Given : Molar mass in

g mol^{- 1} H : 1, C : 12

O : 16, Br : 80

)

Answer: 66

Consuming one mole of bromine adds two bromine atoms to each molecule of X, so the product Y contains two bromine atoms. A carbon-to-bromine ratio of

3 : 1

then fixes six carbon atoms, and the saturated dibromide formed by adding

B r_{2}

across one ring double bond is

C_{6} H_{10} B r_{2}

. Its molar mass is

6 (12) + 10 (1) + 2 (80) = 72 + 10 + 160 = 242 g mo l^{- 1}

, and the mass fraction of bromine is

\frac{160}{242} \times 100 = 66.1%

. To the nearest integer the bromine content of Y is

66%

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer