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QJEEM22Chcd25: JEE Main 2022 Chemistry - Class 12 JEEM22Chcd25 Solution: A hydrocarbon 'X' is found to have molar mass of 80. A 10.0 mg…

A hydrocarbon 'X' is found to have molar mass of 80. A 10.0 mg of compound 'X' on hydrogenation consumed 8.40 mL of H2 gas (measured at STP). Ozonolysis of compound 'X' yields only formaldehyde and dialdehyde. The total number of fragments/molecules produced from the ozonolysis of compound 'X' is _____________.

Step-by-Step Verified Solution

Moles of X=10mg80=0.125 m mol\mathrm{X}=\frac{10 \mathrm{mg}}{80}=0.125 \mathrm{~m} \mathrm{~mol} .

moles consumed of H2=8.422.40.375 m mol\mathrm{H}_{2}=\frac{8.4}{22.4} 0.375 \mathrm{~m} \mathrm{~mol} .

nH2nX=0.3750.125=3\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{X}}}=\frac{0.375}{0.125}=3

So, the compound X\mathrm{X} have 3 double bond.

Ozonolysis of the compound yield formaldehyde and dialdehyde.

The compound is

H2C=CH−CH=CH−CH=CH2\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}

Molecular mass =(12×6)+1×8=72+8=80 amu=(12 \times 6)+1 \times 8=72+8=80 \,\mathrm{amu}

Ozonolysis form:

JEE Main 2022 (Online) 30th June Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 91 English Explanation

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