JEE MAIN - Chemistry (2022 - 30th June Morning Shift - No. 22)
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 _____________.
Answer
4
Explanation
Moles of $\mathrm{X}=\frac{10 \mathrm{mg}}{80}=0.125 \mathrm{~m} \mathrm{~mol}$.
moles consumed of $\mathrm{H}_{2}=\frac{8.4}{22.4} 0.375 \mathrm{~m} \mathrm{~mol}$.
$\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{X}}}=\frac{0.375}{0.125}=3$
So, the compound $\mathrm{X}$ have 3 double bond.
Ozonolysis of the compound yield formaldehyde and dialdehyde.
The compound is
$$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} $$
Molecular mass $=(12 \times 6)+1 \times 8=72+8=80 \,\mathrm{amu}$
Ozonolysis form:
_30th_June_Morning_Shift_en_22_1.png)
moles consumed of $\mathrm{H}_{2}=\frac{8.4}{22.4} 0.375 \mathrm{~m} \mathrm{~mol}$.
$\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{X}}}=\frac{0.375}{0.125}=3$
So, the compound $\mathrm{X}$ have 3 double bond.
Ozonolysis of the compound yield formaldehyde and dialdehyde.
The compound is
$$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} $$
Molecular mass $=(12 \times 6)+1 \times 8=72+8=80 \,\mathrm{amu}$
Ozonolysis form:
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