Text Book Back Questions and Answers - Chapter 11 Fundamentals of Organic Chemistry 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Fundamentals of Organic Chemistry
TextualEvaluation Solved
Question 1.
Select the molecule which has only one ir bond.
(a) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
(b) $\mathrm{CH} 3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$
(c) $\mathrm{CH} 3-\mathrm{CH}=\mathrm{CH}-\mathrm{COOH}$
(d) All of these
Answer:
(a) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
Question 2.
In the hydrocarbon

the state of hybridisation of carbon 1,2,3,4 and 7 are in the following sequence.
(a) $\mathrm{sp}, \mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}^3$
(b) $\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}^3$
(c) $\mathrm{sp}, \mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3$
(d) none of these
Answer:
(a) $s p, s p, \mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}^3$
Question 3.
The general formula for alkadiene is
(a) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$
(b) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-1}$
(c) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$
(d) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{\mathrm{n}-2}$
Answer:
(c) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$
Question 4.
Structure of the compound whose IUPAC name is 5,6 - dimethylhept-2-ene is

(d) None of these 

Answer:

Question 5.
The IUPAC name of the compound is ................

(a) 2,3-Diemethyiheptane
(b) 3-Methyl-4-ethyloctane
(c) 5-ethyl-6-methyloctanc
(d) 4-Ethyl -3-methyloctane.
Answer:
(d) 4-Ethyl-3-methyloctane.
Question 6.
Which one of the following names does not fit a real name?
(a) 3-Methyl-3-hexanone
(b) 4-Methyl-3-hexanone
(c) 3-Methyl-3-hexanol
(d) 2 - Methyl cyclo hexanone.
Answer:
(a) 3-Methyl-3-hexanone
Question 7.
The TUPAC name of the compound $\mathrm{CH} 3-\mathrm{CII}=\mathrm{CH}-\mathrm{C} \mathrm{CH}$ is
(a) Pent $-4-y n-2-$ ene
(b) Pent -3-en - 1- yne
(c) pent $-2-$ en $-4-$ yne
(d) Pent $-1-y n-3-$ ene
Answer:
(b) Pent -3-en - 1- yne
Question 8.

IUPAC name of  is .................
(a) 3,4,4-Trimethylheptane
(b) 2-Ethyl-3,3-dimethyl heptane
(c) 3,4,4-Trimethyloctane
(d) 2 Butyl-2 -methyl-3-ethyl-butane.
Answer:
(c) 3,4,4-Trimethyloctane
Question 9.

IUPAC name of  is .................

(a) 2 -Hydroxypropionic acid
(b) 2, 4. 4-Trimethylpent-3-ene
(c) Propan - 2-ol 1-oie acid
(d) 2, 2,4-Trimethylpent-2-ene
Answer:
(b) 2 -Hydroxy Propanoic acid
Question 10 .

The IUPAC name of the compound  is ................

(a) 3-Ethyl-2-hexene
(b) 3 - Propyl-3. hexene
(c) 4-Ethyl-4-hexene
(d) 3-Propyl-2-hexenc
Answer:
(a) 3-Ethyl-2-hexene
Question 11.
The IUPAC name of the compound  is ...............
(a) 2 -Hydroxypropionic acid
(b) 2 -Hydroxy Propanoic acid
(c) Propan 2-ol-1-oic acid
(d) 1 - Carboxyethanol.
Answer:
(b) 2 -Hydroxy Propanoic acid
Question 12 .

The IUPAC name of  is..............

(a) 2 - Bromo-3-methylbutanoic acid
(b) 2 -methyl-3-bromobutanoic acid
(c) 3-Bromo-2-methylbutanoic acid
(d) 3-Bromo-2. 3-dimethyl propanoic acid.
Answer:
(c) 3-Bromo-2-methylbutanoic acid
Question 13.
The structure of isobutyl group in an organic compound is ........

Answer:

Question 14.
The number of stereoisomers of 1,2-dihydroxycyclopentane is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Question 15 .
Which of the following is optically active?
(a) 3 -Chioropentane
(b) 2 -Chioropropane
(c) Meso - tat-tat-ic acid
(d) Glucose
Answer:
(d) Glucose
Question 16.
The isomer of ethanol is
(a) acetaldehyde
(b) dimethylether
(c) acetone
(d) methyl carbinol
Answer:
(b) dimethylether
Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula $\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}$ ?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9
Question 18 .
Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) $\mathrm{CH}_2 \mathrm{Cl}_2$
Answer:
(c) ethanol
Question 19.

 is ................

(a) resonating structure
(b) taulomers
(c) optical isomers
(d) conformers
Answer:
(b) tautomers
Question 20.
Nitrogen detection in an organic compound is earned out by Lassaigne's test. The blue colour formed is due to the formation of
(a) $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
(b) $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
(c) $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
(d) $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
Answer:
(b) $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
Question 21.
Lassaigne's test for the detection of nitrogen fails in
(a) $\mathrm{H}_2 \mathrm{~N}-\mathrm{CO}-\mathrm{NH} . \mathrm{NH}_2 . \mathrm{HCl}$
(b) $\mathrm{NH}_2-\mathrm{NH}_2 . \mathrm{HCl}$
(c) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{NH}_2 . \mathrm{HCl}$
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2$
Answer:
(c) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{NH}_2 . \mathrm{HCl}$
Question 22.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne's test is separately done.
(a) $\mathrm{NH}_2 \mathrm{NH}_2 \mathrm{HCl}$ and $\mathrm{CICH}_2-\mathrm{CHO}$
(b) $\mathrm{NH}_2 \mathrm{CS} \mathrm{NH}_2$ and $\mathrm{CH}_3-\mathrm{CH} 2 \mathrm{Cl}$
(c) $\mathrm{NH}_2 \mathrm{CH}_2 \mathrm{COOH}$ and $\mathrm{NH}_2 \mathrm{CONH}_2$
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ and $\mathrm{ClCH}_2-\mathrm{CHO}$
Answer:
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ and $\mathrm{ClCH}_2-\mathrm{CHO}$

Question 23.
Sodium nitropruside reacts with suiphide ion to give a purple colour due to the formation of
(a) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{~N} 0\right]^{3-}$
(b) $\left[\mathrm{Fe}(\mathrm{NO})_5 \mathrm{CN}\right]^{+}$
(c) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$
(d) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{3-}$
Answer:
(c) $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$
Question 24.
An organic compound weighing $0.15 \mathrm{~g}$ gave on carius estimation, $0.12 \mathrm{~g}$ of silver bromide. The percentage of bromine in the compound will be close to
(a) $46 \%$
(b) $34 \%$
(c) $3.4 \%$
(d) $4.6 \%$
Answer:
(b) $34 \%$
Question 25.
A sample of $0.5 \mathrm{~g}$ of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in $50 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$. The remaining acid after neutralisation by ammonia consumed $80 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{NaOH}$, The percentage of nitrogen in the organic compound is
(a) $14 \%$
(b) $28 \%$
(c) $42 \%$
(d) $56 \%$
Answer:
(b) $28 \%$
Question 26.
In an organm compound, phosphorus is estimated as
(a) $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$
(b) $\mathrm{Mg}_3\left(\mathrm{PO}_4\right)_2$
(c) $\mathrm{H}_3 \mathrm{PO}_4$

(d) $\mathrm{P}_2 \mathrm{O}_5$
Answer:
(a) $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$
Question 27.
Ortho and para-nitro phenol can be separated by
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer:
(c) steam distillation
Question 28.
The purity of an organic compound is determined by
(a) Chromatography
(b) Crystallisation
(c) melting or boiling point
(d) both (a) and (c)
Answer:
(d) both (a) and (c)
Question 29.
A liquid which decomposes at its boiling point can be purified by
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer:
(b) distillation under reduced pressure
Question 30 .
Assertion: is 3-carbethoxy-2-butenoicacid.

$\mathrm{COOC}_2 \mathrm{H}_5$
gets lowest number followed by double bond (or) triple bond.
Reason: The principal functional group geth the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d both the assertion and reason are false.
Answer:
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
Question 31.
Give the general characteristics of organic compounds.
Answer:
- All organic compounds are covalent compounds of carbon and are insoluble in water and soluble in organic solvents.
- They are inflammable (except $\mathrm{CCl}_4$ ).
- They possess low boiling and melting points due to their covalent nature.
- They are characterised by functional groups.
- They exhibit isomerism.
Question 32.
Describe the classification of organic compounds based on their structure.
classification of organic compounds based on the structure

Answer:

Question 33.
Write a note on homologous series.
Answer:
- A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a $\mathrm{CH}_2$ group is called homologous series.
- e.g., Alkanes, Methane $\left(\mathrm{CH}_4\right)$ ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$, Propane $\left(\mathrm{C}_3 \mathrm{H}_8\right)$ etc.
- Compounds of the homologous series are represented by a general formula. e.g., Alkanes: $\mathrm{C}_2 \mathrm{H}_{2 n} \mathrm{Alkene}$ : $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$
- They can be prepared by general methods.
- They show regular gradation in physical properties but have almost similar chemical properties.
Question 34.
What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(c) dimethyl ether
(d) methylamine
Answer:
1. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way,
irrespective of the organic molecule in which it is present.

Question 35.
Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol $-\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1}+\mathrm{OH}$
(b) Aliphatic ketones $-\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}} \mathrm{O}$
(c) Aliphatic amines $-\mathrm{C}_2 \mathrm{H}_{2 \mathrm{n}+1} \mathrm{NH}_2$
Question 36.
Write the molecular formula of the first six members of homologous series of nitro - alkanes. Nitroalkanes:
Answer:
- $\mathrm{CH}_2 \mathrm{NO}_2$ Nitromethane
- $\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{NO}_2$ Nitroethane
- $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{NO}_2$ 1- nitropropane
- $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$ 1- nitrobutane
- $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2 1$ - nitropentane
- $\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$ 1- nitrohexane
Question 37.
Write the molecular and possible structural formula of the first four members of homologous series of carhoxylic acids.

Question 38.
Give the IUPAC names of the following compounds.
Answer:

Give the IUPAC names of the following compounds.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

(xvi)

(i) 2,3,5-tnmethylhexane
(ii) 2-bromo-3-methylbutane
(iii) methoxymethane
(iv) 2-hydroxybutanal
(v) buta-1,3-diene
(vi) 4-chioropent-2-yne
(vii) 1 -bromobut-2-ene
(viii) 5-oxohexanoic acid
(ix) 3-ethyl-4-ethenylheptane
(x) 2, 4, 4-trimethylpent-2-ene
(xi) 2-methyl-I -phenyipropan- I -amine
(xii) 2,2-dimethyl-4-oxopentanenitrile
(xiii) 2-ethoxypropane
(xiv) I -fluoro-4-methyl-2-nitrobenzene
(xv) 3-bromo-2-methylpentanal
Question 39.
Give the structure for the following compound.
Answer:
(i) 3 - ethyl-2 methyl -1 - pentene
(ii) 1,3,5-Tnmethyl cyclohex-1 - ene
(iii) tertiary butyl iodide
(iv) 3 -Chlorobutanal
(y) 3 -Chlorobutanol.
(vi) 2 -Chloro-2-methyl propane
(vii) 2,2-dimethyl- 1 - chioropropane
(viii) 3 - methylbut - 1 - ene
(ix) Butan -2,2 - diol
(x) Octane $-1,3$ - diene
(xi) 1,3-Dimethylcyclohexane
(xii) 2 -Chlorobut-3-ene
(xiii) 3 - methylbutan - $1-$ ol
(xiv) acetaldehyde

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

Question 40 .
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
Detection of Nitrogen:
The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method. Estimation of halogens:
Answer:
carius method:
1. A known mass of the organic compound is heated with fuming $\mathrm{HNO}_3$ and $\mathrm{AgNO}_3$.
2. $\mathrm{C}, \mathrm{H}$ and $\mathrm{S}$ gets oxidised to $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{SO}_2$ and halogen combines with $\mathrm{AgNO}_3$ to form a precipitate of silver halide

3. The precipitate $\mathrm{AgX}$ is filtered, washed, dried and weighed.
4. From the mass of $\mathrm{AgX}$ and the mass of organic compound taken, the percentage of halogens are calculated.
$
\begin{aligned}
& \text { 5. } \% \text { of chlorine }=\frac{35.5}{143.5} \times \frac{\text { wt. } \% \text { of chlorine }}{\text { wt. of organic compound }} \times 100 \\
& \% \text { of Bromine }=\frac{80}{188} \times \frac{\text { wt. } \% \text { of bromine }}{\text { wt. of organic compound }} \times 100 \\
& \% \text { of Iodine }=\frac{127}{235} \times \frac{\text { wt. } \% \text { of Iodine }}{\text { wt. organic compound }} \times 100
\end{aligned}
$

Question 42.
Give a brief description of the principles of:
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.
2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.
(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.
(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.
(d) In column chromatography, the above principle is carried out in a long glass column.
Question 43.
Explain paper chromatography.
Answer:
Paper chromatography:
1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.
2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.
3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.
Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:

Constitutional isomers:
These isomers having the same molecular formula but differ in their bonding sequence. It is classified into 6 types:
1. Chain (or) nuclear (or) skeletal isomerism:
The phenomenon in which the isomers have similar molecular formula but differ in the nature of carbon skeleton (i.e., straight (or)
branched)
e.g., $\mathrm{C}_5 \mathrm{H}_{12}$ :

2. Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
e.g., $\mathrm{C}_5 \mathrm{H}_{10}$ :

3. Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
$
\text { e.g., } \mathrm{C}_3 \mathrm{H}_6 \mathrm{O} \text { : }
$

4. Metamerism:
This isomerism anses due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to either side of the same functional group and having same molecular formula.
e.g., $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ :

5. Tautomerism:
It is an isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus generally hydrogen.
$
\text { e.g., } \mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \text { : }
$

6. Ring chain isomerism:
It is an isomerism in which compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures.
e.g., $\mathrm{C}_2 \mathrm{H}_6$ :

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.
2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.
3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign $(+)$.
4. The optical isomer which rotates the plane of plane polarised light to the left or in anti-clockwise direction is said to be laveo rotatory and is denoted by the sign $(-)$.
5. Dextrorotatory compounds are represented as ' $d$ ' (or) by $(+)$ sign and leave rotatory compounds are represented as 1 (or) by $(-)$ sign.
6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.
Question 46.
Briefly explain geometrical isomerism in alkenes by considering 2-butene as an example.
Answer:
2-butene: Geometrical isomerism: $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
1. Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
2. In 2-butene, the carbon-carbon double bond is $\mathrm{sp} 2$ hybridised. The carbon-carbon double bond consists of a a bond and $\mathrm{a}$ it bond. The presence of it bond lock the molecule in one position. Hence, rotation around $\mathrm{C}=\mathrm{C}$ bond is not possible.

3.

4. These two compounds are termed as geometrical isomers and are termed as cis and trans form.
5. The cis-isomer is the one in which two similar groups arc on the same side otthe double bond. The transisomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-Irans isomerism.
Question 47.
$0.30 \mathrm{~g}$ of a substance gives $0.88 \mathrm{~g}$ of carbon dioxide and $0.54 \mathrm{~g}$ of water calculate the percentage of carbon and hydrogen in it.
Answer:
Weight of organic compound $=0.30 \mathrm{~g}$
Weight of carbon-dioxide $=0.88 \mathrm{~g}$
Weight of water $=0.54 \mathrm{~g}$
Percentage of hydrogen:
$18 \mathrm{~g}$ of water contains $2 \mathrm{~g}$ of hydrogen
$0.54 \mathrm{~g}$ of water contain $=\frac{2}{18} \times 0.54 \mathrm{~g}$ of hydrogen
$\%$ of hydrogen $=\frac{2}{18} \times \frac{0.54}{0.30} \times 100=\frac{2}{18} \times \frac{54}{0.3}$
$\%$ of $\mathrm{H}=0.111 \times 180=19.888 \approx 20 \%$
Percentage of carbon:
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ contains $12 \mathrm{~g}$ of carbon
0.88 g of $\mathrm{CO}_2$ contains $=\frac{12}{44} \times 0.88 \mathrm{~g}$ of carbon
$\%$ of carbon $=\frac{12}{44} \times \frac{0.88}{0.30} \times 100=\frac{12}{44} \times \frac{88}{0.3}=\frac{24}{0.3}$
$\%$ of carbon $=80 \%$
Question 48.
The ammonia evolved form $0.20 \mathrm{~g}$ of an organic compound by kjeldahl method neutralised $15 \mathrm{~m} 1$ of $\mathrm{N} / 20$ Sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound $=0.20 \mathrm{~g}$
Volume of sulphuric acid taken $=15 \mathrm{ml}$
Strength of sulphuric acid taken $=\frac{N}{20}=0.05 \mathrm{~N}$
Percentage of nitrogen $=\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100$

$
\begin{aligned}
& =\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100 \\
& =\frac{1050}{200}=5.25 \\
& \% \text { of nitrogen }=5.25 \%
\end{aligned}
$
Question 49.
$0.32 \mathrm{~g}$ of an organic compound. after heating with fuming nitric acid and barium nitrate crystals is a scaled tube gave $0.466 \mathrm{~g}$ of barium sulphate. Determine the percentage of sulphur in the compound.
Answer:
Weight of organic compound $=0.32 \mathrm{~g}$
Weight of $\mathrm{BaSO}_4$ formed $=0.466 \mathrm{~g}$
$233 \mathrm{~g}$ of $\mathrm{BaSO}_4$ contains $=32 \mathrm{~g}$ of sulphate
$0.466 \mathrm{~g}$ of $13 \mathrm{aSO}_4$ contain $=\frac{32}{233} \times \frac{0.466}{2.32} \times 100$
$=\frac{32}{233} \times \frac{46.6}{0.32}=19.999 \mathrm{~g}$ of sulphur
$\%$ of sulphur $=20 \%$
Question 50.
$024 \mathrm{~g}$ of an organic compound gave $0.287 \mathrm{~g}$ of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer:
Weight of organic compound $=0.24 \mathrm{~g}$
Weight of silver chloride $=0.287 \mathrm{~g}$
$143.5 \mathrm{~g}$ of $\mathrm{AgCl}$ contains $=35.5 \mathrm{~g}$ of $\mathrm{Cl}$
$0.287 \mathrm{~g}$ of $\mathrm{AgCl}$ contains $=\frac{35.5}{143.5} \times 0.287 \mathrm{~g}$ of $\mathrm{Cl}$
$\%$ of chlorine $=\frac{35.5}{143.5} \times \frac{0.287}{0.24} \times 100=29.58 \%$
Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method $0.35 \mathrm{~g}$ yielded $20.7 \mathrm{~mL}$ of nitrogen at $15^{\circ} \mathrm{C}$ and $760 \mathrm{~mm} \mathrm{Hg}$ pressure. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound $=0.35 \mathrm{~g}$
Volume of moist nitrogen $\left(\mathrm{V}_1\right)=20.7 \mathrm{ml}=20.7 \times 10^{-3} \mathrm{~L}$
Temperature $=\mathrm{T}_1=15^{\circ} \mathrm{C}=273+15^{\circ} \mathrm{C}=288 \mathrm{~K}$
Pressure of moist nitrogen $P_1=760 \mathrm{mmHg}$
$
\begin{aligned}
& \frac{P_1 V_1}{T_1}=\frac{P_0 V_0}{T_0} \\
& V_o=\frac{P_1 V_1}{T_1} \times \frac{T_0}{P_0}=\frac{760 \times 20.7 \times 10^{-3}}{288} \times \frac{273 \mathrm{~K}}{760} \\
& V_0=19.62 \times 10^{-3} \mathrm{~L}
\end{aligned}
$
Percentage of nitrogen $=\frac{28}{22.4} \times \frac{\mathrm{V}_0}{\mathrm{~W}} \times 100$
$
\begin{aligned}
& =\frac{28}{22.4} \times \frac{19.62 \times 10^{-3}}{0.35} \times 100 \\
& =\frac{28}{22.4} \times \frac{19.62}{0.35} \times 10^{-1} \\
& =56.05 \times 10^{-3} \times 100=7.007 \%
\end{aligned}
$

Percentage of nitrogen $=7.007 \%$
Question 1.
Give two examples for each of the following type of organic compounds.
1. non-benzonoid aromatic
2. aromatic heterocyclic
3. alicycic
4. aliphatic open chain.
Answer:
1. Non benzenoid aromatic compounds

2. Aromatic heterocyclic compounds 

3. Alicyclic compounds 

4. Aliphatic open chain compounds
- $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ n-pentane
- $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}$ 1-propanol
Question 2.
Write structural formula for the following compounds
1. Cyclohexa-1, 4-diene
2. Ethynykyclohexane

Question 3.
Write structural formula for the following compounds
1. $\mathrm{m}$ - dinitrobenzene
2. p-dichlorobenzene
3. 1, 3, S- Trimethytbeuzene
Answer:

Question 4 .
Write all the possible isomers of molecular formula $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ and identify the isomerisms found in them. Answer:
$\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ isomers:

Question 5.
$0.2346 \mathrm{~g}$ of an organic compound containing C, $\mathrm{H} \& \mathrm{O}$, o comhution giweb $0.2754 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ and $0.4488 \mathrm{~g}$ $\mathrm{CO}_2$. Calculate the \% composition of $\mathrm{C}, \mathrm{H} \& \mathrm{O}$ in the organic compound.
Answer:
Weight of organic substance $(w)=0.2346 \mathrm{~g}$
Weight of water $(\mathrm{x})=0.2754 \mathrm{~g}$
Weight of $\mathrm{CO}_2(\mathrm{y})=0.4488 \mathrm{~g}$
Percentage of carbon $=\frac{12}{44} \times \frac{y}{w} \times 100$
$
=\frac{12}{44}-\frac{0.4488}{0.2346} \times 100=52.17 \%
$
Percentage of hydrogen $=\frac{2}{18} \times \frac{y}{w} \times 100$
$
=\frac{2}{18} \times \frac{0.2754}{0.2346} \times 100=13.04 \%
$
Percentage of oxygen $=[100-(52.17+13.04)]=100-65.21=34.79 \%$
Question 6.
$0.16 \mathrm{~g}$ of an organic compound was heated in a carlus tube and $\mathrm{H}_2 \mathrm{SO}_4$ acid formed was precipitated with $\mathrm{BaCl}_4$. The mass of $\mathrm{BaSO}_4$ was $0.35 \mathrm{~g}$. Find the percentage of sulphur.
Answer:
Weight of organic substance (w) $=0.16 \mathrm{~g}$
Weight of Barium sulphate $(x)=0.35 \mathrm{~g}$
Percentage of Sulphur $=\frac{32}{233} \times \frac{x}{w} \times 100$
$
=\frac{32}{233} \times \frac{0.35}{0.16} \times 100=30.04 \%
$
Question 7.
$0.185 \mathrm{~g}$ of an organic compound when treated with Conc. $\mathrm{HNO}_3$ and silver nitrate gave $0.320 \mathrm{~g}$ of silver bromide. Calculate the $\%$ of bromine in the compound.
Answer:
Weight of organic substance (w) $0.185 \mathrm{~g}$;
Weight of silver bromide $(\mathrm{x})=0.320 \mathrm{~g}$
Percentage of bromine $=\frac{80}{188} \times \frac{x}{w} \times 100=\frac{80}{188} \times \frac{0.32}{0.185} \times 100=73.6 \%$

Question 8.
$0.40 \mathrm{~g}$ of an iodo-substituted organic compound gave $0.235 \mathrm{~g}$ of Agi by carius method. Calculate the percentage of iodine in the compound. ( $\mathrm{Ag}=108, \mathrm{I}=127)$.
Answer:
Weight of organic substance $(\mathrm{w})=0.40 \mathrm{~g}$
Weight of silver iodide $(\mathrm{x})=0.235 \mathrm{~g}$
$127 \times 1270.235$
Percentage of iodine $=\frac{127}{235} \times \frac{x}{w} \times 100$
$=\frac{x}{w} \times \frac{0.235}{0.40} \times 100=31.75 \%$
Question 9.
$0.33 \mathrm{~g}$ of an organic compound containing phosphorous gave $0.397 \mathrm{~g}$ of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$ by the analysis. Calculate the percentage of $P$ in the compound.
Answer:
Weight of organic compound $=0.33 \mathrm{~g}$;
Weight of $\mathrm{Mg}, \mathrm{P}, 07=0.397 \mathrm{~g}$
$222 \mathrm{~g}$ of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$ contains $62 \mathrm{~g}$ of phosphorous.
$0.397 \mathrm{~g}$ of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$ will contain $\frac{62}{222} \times 0.397 \mathrm{~g}$ of $\mathrm{P}$.
$0.33 \mathrm{~g}$ of organic compound contains $\frac{62}{222} \times 0.397 \mathrm{~g}$ of $\mathrm{P}$
$100 \mathrm{~g}$ of organic compound will contain $\frac{62}{222} \times \frac{0.397}{0.33} \times 100$
$=\frac{2,461.4}{73.26}=33.59 \%$
Percentage of phosphorous $=33.59 \%$
Question 10.
$0.3 \mathrm{~g}$ of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise $30 \mathrm{~mL}$ of $0.1 \mathrm{~N}$ $\mathrm{H}_2 \mathrm{SO}_4$. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound $(w)=0.3 \mathrm{~g}$
Strength of sulphuric acid used $(\mathrm{N})=0.1 \mathrm{~N}$
Volume of sulphuric acid used $(\mathrm{V})=30 \mathrm{~mL}$
$30 \mathrm{ml}$ of $0.1 \mathrm{~N}$ sulphuric acid $30 \mathrm{ml}$ of $0.1 \mathrm{~N}$ ammonia
Percentage of nitrogen $=\left(\frac{14 \times \mathrm{NV}}{1000 \times w}\right) \times 100$
$=\left(\frac{14 \times 0.1 \times 30}{1000 \times 0.3}\right) \times 100=14 \%$
Question 1.
Classify the following compounds based on the structure

1. $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}$
2. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$

3.

4.

Answer:
1. Unsaturated open chain compound
2. Saturated open chain compound
3. Aromatic benzenoid compound
4. Alicyclic compound
Question 2.
$0.26 \mathrm{~g}$ of an organic compound gave $0.039 \mathrm{~g}$ of water and $0.245 \mathrm{~g}$ of carbon dioxide on combustion. Calculate the percentage of $\mathrm{C} \& \mathrm{H}$.
Answer:
Weight of organic compound $=0.26 \mathrm{~g}$
Weight of water $=0.039 \mathrm{~g}$
Weight of $\mathrm{CO}_2=0.245 \mathrm{~g}$
Percentage of hydrogen:
$18 \mathrm{~g}$ of water contains $2 \mathrm{~g}$ of hydrogen
$0.039 \mathrm{~g}$ of water contains $=\frac{2}{18} \times \frac{0.039}{0.26}$ of $\mathrm{H}$
$\%$ of hydrogen $=\frac{0.039}{0.26} \times \frac{2}{18} \times 100=1.66 \%$
Percentage of carbon:
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ contains $12 \mathrm{~g}$ of $\mathrm{C}$
$0.245 \mathrm{~g}$ of $\mathrm{CO}_2$ contains $=\frac{12}{44} \times \frac{0.245}{0.26} \mathrm{~g}$ of $\mathrm{C}$
$\%$ of Carbon $=\frac{12}{44} \times \frac{0.245}{0.26} \times 100=25.69 \%$
Question 3.
In an estimation of sulphur by Carius method, $0.2175 \mathrm{~g}$ of the substance gave $0.5825 \mathrm{~g}$ of $\mathrm{BaSO}_4$, calculate the percentage composition of $S$ in the compound.
Answer:
Weight of organic compound $=0.2175 \mathrm{~g}$
Weight of $\mathrm{BaSO}_4=0.5825 \mathrm{~g}$
$233 \mathrm{~g}$ of $\mathrm{BaSO}_4$ contains $=32 \mathrm{~g}$ of $\mathrm{S}$
$0.5825 \mathrm{~g}$ of $\mathrm{BaSO}_4$ contains $=\frac{32}{233} \mathrm{x} \frac{0.5825}{0.2175} \mathrm{~g}$ of $\mathrm{S}$
Percentage of $\mathrm{S}=\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100=36.78 \%$
Question 4.
$0.284 \mathrm{~g}$ of an organic substance gave $0.287 \mathrm{~g} \mathrm{AgCl}$ in Carius method for the estimation of halogen. Find the percentage of $\mathrm{Cl}$ in the compound.
Answer:
Weight of the organic substance $=0.284 \mathrm{~g}$

Weight of $\mathrm{AgCl}=0.287 \mathrm{~g}$
$143.5 \mathrm{~g}$ of $\mathrm{AgCl}$ contains $35.5 \mathrm{~g}$ of chlorine
$0.287 \mathrm{~g}$ of $\mathrm{AgCl}$ Contains $=\frac{35.5}{143.5} \times \frac{0.287}{0.284}$
$\%$ of chlorine $=\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100=24.98 \%$
Question 5.
$0.24 \mathrm{~g}$ of organic compound containing phosphorous gave $0.66 \mathrm{~g}$ of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$ by the usual analysis. Calculate the percentage of phosphorous in the compound
Answer:
Weight of an organic compound $=0.24 \mathrm{~g}$
Weight of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7=0.66 \mathrm{~g}$
$222 \mathrm{~g}$ of $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$ contains $=62 \mathrm{~g}$ of $\mathrm{P}$
o. $66 \mathrm{~g}$ contains $=\frac{62}{222} \times 0.66 \mathrm{~g}$ of $\mathrm{P}$
Percentage of $\mathrm{P}=\frac{62}{222} \times \frac{0.66}{0.24} \times 100=76.80 \%$
Question 6.
$0.1688 \mathrm{~g}$ when analysed by the Dumas method yield $31.7 \mathrm{~mL}$ of moist nitrogen measured at $14^{\circ} \mathrm{C}$ and $758 \mathrm{~mm}$ mercury pressure. Determine the $\%$ of $\mathrm{N}$ in the substance (Aqueous tension at $14^{\circ} \mathrm{C}=12 \mathrm{~mm}$ of $\mathrm{Hg}$ ).
Answer:
Weight of Organic compound $=0.168 \mathrm{~g}$
Volume of moist nitrogen $\left(\mathrm{V}_1\right)=31.7 \mathrm{ml}=31.7 \times 10^{-3} \mathrm{~L}$
Temperature $\left(\mathrm{T}_1\right)=14^{\circ} \mathrm{C}=14+273=287 \mathrm{~K}$
Pressure of Moist nitrogen $(\mathrm{P})=758 \mathrm{~mm} \mathrm{Hg}$
Aqueous tension at $14^{\circ} \mathrm{C}=12 \mathrm{~mm}$ of $\mathrm{Hg}$
$
\begin{aligned}
& \frac{R V_1}{T_1}=\frac{P_0 V_0}{T_0} \\
& \mathrm{~V}_0=\frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760} \\
& \mathrm{~V}_0=29.58 \times 10^{-3} \mathrm{~L}
\end{aligned}
$
Percentage of nitrogen:
$
\begin{aligned}
& =\left(\frac{28}{22.4} \times \frac{V_0}{\mathrm{~W}}\right) \times 100 \\
& =\frac{28}{22.4} \times \frac{29.58 \times 10^{-5}}{0.1688} \times 100 \\
& =21.90 \%
\end{aligned}
$
Question 7.
$0.6 \mathrm{~g}$ of an organic compound was Kjeldhalised and $\mathrm{NH}_3$ evolved was absorbed into $50 \mathrm{~mL}$ of semi-normal solution of $\mathrm{H}_2 \mathrm{SO}_4$. The residual acid solution ws diluted with distilled water and the volume made up to 150 mL. $20 \mathrm{~mL}$ of this diluted solution required $35 \mathrm{~mL}$ of $\frac{N}{2} \mathrm{NaOH}$ solution for complete neutralisation. Calculate the $\%$ of $\mathrm{N}$ in the compound.
Answer:
Weight of Organic compound $=0.6 \mathrm{~g}$
Volume of sulphuric acid taken $=50 \mathrm{~mL}$
Strength of sulphuric acid taken $=0.5 \mathrm{~N}$

$20 \mathrm{ml}$ of diluted solution of unreacted sulphuric acid was neutralised by $35 \mathrm{~mL}$ of $0.05 \mathrm{~N}$ Sodium hydroxide Strength of the diluted sulphuric acid $=\frac{35 x 0.05}{20}=0.0875 \mathrm{~N}$
Volume of the sulphuric acid remaining after reaction with ammonia $=\mathrm{V}_1 \mathrm{~mL}$
Strength of $\mathrm{H}_2 \mathrm{SO}_4=0.5 \mathrm{~N}$
Volume of the diluted $\mathrm{H}_2 \mathrm{SO}_4=150 \mathrm{~mL}$
Strength of the diluted sulphuric acid $=0.0875 \mathrm{~N}$ $\mathrm{V}_1=\frac{150 x 0.087}{0.5}=26.25 \mathrm{~mL}$
Volume of $\mathrm{H}_2 \mathrm{SO}_2$ consumed by ammonia $=50-26.25=23.75 \mathrm{~mL}$
$23.75 \mathrm{~mL}$ of $0.5 \mathrm{~N} \mathrm{H}_2 \mathrm{SO}_4=23.75 \mathrm{~mL}$ of $0.5 \mathrm{~N} \mathrm{NH}_3$
The amount of Nitrogen present in the $0.6 \mathrm{~g}$ of organic compound $=\frac{14 \mathrm{~g}}{1000 \mathrm{~mL} \times 1 \mathrm{~N}} \times 23.75 \times 0.5 \mathrm{~N}=0.166 \mathrm{~g}$
Percentage of Nitrogen $\frac{0.166}{0.6} \times 100=27.66 \%$