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Published on: 09/10/2019
Fundamentals of Organic Chemistry
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1.
Which among the following hybrid orbitals are highly electronegative? sp, sp2 , sp3 .
2.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous in the compound.
3.
0.40 g of an iodo-substituted organic compound gave 0.235 g of AgI by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127)
4.
0.284 g of an organic substance gave 0.287 g AgCl in a carius method for the estimation of halogen. Find the Percentage of Cl in the compound.
5.
0.16 g of an organic compound was heated in a carius tube and H2SO4 acid formed was precipitated with BaCl2. The mass of BaSO4 was 0.35g. Find the percentage of sulphur [30.04]
6.
In an estimation of sulphur by carius method, 0.2175 g of the substance gave 0.5825 g of BaSO4 calculate the percentage composition of S in the compound.
7.
Write all the possible isomers of molecular formula C4H10O and identify the isomerisms found in them.
8.
Write structural formula for the following compounds
(i) Cyclohexa-1, 4-diene
(ii) Ethynyl cyclohexane.
9.
Write structural formula for the following compounds
m - dinitrobenzene
10.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg2P2O7 by the analysis. Calculate the percentage of P in the compound.
1.
s-character & electronegativity,
\(\therefore\) sp hybrid orbitals are more electronegative among the three.
2.
Weight of an organic compound = 0.24 g
Weight of Mg2P2O7 = 0.66 g = y
\(
\% \mathrm{P} =\frac{62}{222} \times \frac{\mathrm{y}}{\mathrm{w}} \times 100
\)
\(=\frac{62}{222} \times \frac{0.66}{0.24} \times 100=76.80 \%\)
3.
(w) = 0.33 g
(c) = 0.397 g
\(\% I=\frac{127}{235} \times \frac{c}{w} \times 100=\frac{127}{235} \times \frac{0.235}{0.40} \times 100=31.75 \%\)
4.
Weight of the organic substance = 0.284 g
Weight of AgCl is = 0.287 g = a
\( \% \mathrm{Cl} =\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100 \)
\(=\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100=24.98 \% \)
5.
(w) = 0.16 g
(x) = 0.35 g
\(\% S=\frac{32}{233} \times \frac{x}{\mathrm{w}} \times 100=\frac{32}{233} \times \frac{0.35}{0.16} \times 100=30.04 \%\)
6.
w = 0.2175 g
x = 0.5825 g
\(
\% \mathrm{~S} =\frac{32}{233} \times \frac{x}{y} \times 100
\)
\(=\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100
\)
= 36.78%
7.

i) Butan.- I - ol & Butan - 2 -ol are position isomers. - Position isomerism
ii) Butan -I -ol & 2-methyl propan - I - ol are chain isomers - Chainisomerism
iii) Butan -I - ol and I - methoxy propane are functional isomers - Functional isomerism
iv) 1 - methoxy Propane and ethoxy ethane and 2 - methoxy Propane are metamers - Metamerism
8.

9.

10.
(w) = 0.33 g
y = 0.397 g
\(\% \mathrm{P}=\frac{62}{222} \times \frac{\mathrm{y}}{\mathrm{w}} \times 100=\frac{62}{222} \times \frac{0.397}{0.33} \times 100=33.60 \%\)
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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