a) CH 3 – CH = CH – CH 3
a) sp, sp, sp 3, sp 2, sp 3
c) C n H 2n – 2
d) 4 – Ethyl – 3 methyloctane
a) 3 – Methyl – 3- hexanone
b) Pent – 3- en – 1- yne
c) 3, 4, 4 – Trimethyloctane
c) 2,2,4 – Trimethylpent – 3 – ene
a) 3 – Ethyl – 2 – hexene
b) 2 – Hydroxy Propanoic acid
c) 3 – Bromo – 2 – methylbutanoic acid
c) 3
d) Glucose
b) dimethyl ether
c) 9
c) ethanol
b) tautomers
b) Fe 4 [ Fe(CN) 6 ] 3
c) C 6 H 5 – NH – NH 2.HCl
d) C 6 H 5 NH 2 and ClCH 2 – CHO
c) [Fe (CN) 5 NOS] 4-
b) 34 %
b) 28 %
a) Mg 2 P 2 O 7
c) steam distillation
d) both (a) and (c)
b) distillation under reduced pressure
a) both the assertion and reason are true and the reason is the correct explanation of assertion.
II. Write brief answers to the following questions:
All organic compounds have the following characteristic properties.
* They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc…
* Many of the organic compounds are inflammable (except CCl 4 ). They possess low boiling and melting points due to their covalent nature.
* Organic compounds are characterized by functional groups. 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. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.
Homologous series:
A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH 2 group is called homologous series.
Example:
Alkanes:
Methane (CH 4 ), Ethane (C 2 H 6 ), Propane (C 3 H 8 ) etc..
Alcohols:
Methanol (CH 3 OH), Ethanol (C 2 H 5 OH) Propanol (C 3 H 7 OH) etc…
Functional group:
An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties.
a) acetaldehyde → – CHO
b) oxalic acid → – COOH
c) di methyl ether → – O –
d) methyiamine → – NH 2
a) Aliphatic monohydric alcohol → C n H 2n + 2 O
b) Aliphatic ketones → C n H 2n O
c) Aliphatic amines → C n H 3n + 2 N
- CH 3 NO 2
- C 2 H 5 NO 2
- C 3 H 7 NO 2
- C 4 H 9 NO 2
- C 5 H 11 NO 2
- C 6 H 13 NO 2
(i) 2, 3, 5 – trimethyihexane
(ii) 2 – bromo – 3 – methylbutane
(iii) methoxymethane
(iv) 2 – hydroxybutanal
(y) buta – 1,3 – diene
(vi) 4 – chloropent – 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 -1 – phenyipropan – 1 -amine
(xii) 2, 2 – dimethyl – 4oxopentanenitrile
(xiii) 2 – ethoxypropane
(xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – bromo – 2 – methylpentanal
(xvi) Acetophenone
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
This method involves the conversion of covalently bonded N, S or halogen present in the organic compounds to corresponding water soluble ions in the form of sodium salts For this purpose a freshly cut piece of Na of the size of a paper, dried it by pressing between the folds of a filter paper is taken in a fusion tube in an iron tand clamping it just near the upper end and it is gently heated.
When it melts to a shiñing globule, put a pinch of the organic compound on it. Heat the tube with the tip of the flame till all reaction ceases and it becomes red hot. Now plunges it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striding against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for qualitative analysis of nitrogen, sulfur and halogens present in organic compounds.
Test for Nitrogen:
If nitrogen is present it gets converted to sodium cyanide which on reaction with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl gives a Prussian blue color or green color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the excess of NaOH on FeSO 4 which would otherwise mark the Prussian blue precipitate. The following reaction takes part in the formation of Prussian blue.
Estimation of halogens (Carius method):
A known mass of the organic compound is heated with fuming HNO 3 along with AgNO 3. C, H & S gets oxidized to CO 2, H 2 O, SO 2 and halogen combines with AgNO 3 to form a precipitate of silver halide.
The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated.
A known mass of the substance is taken along with fuming HNO 3 and AgNO 3 is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made.
Calculation:
Weight of the organic compound: w g
Weight of AgCl precipitate = a g
143. 5 g of AgCl contains 35.5 g of Cl
∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a
w g Organic compound gives a g AgCl
Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\)
Let Weight of silver Bromide be ‘b’ g
188 g of AgBr contains 80 g of Br
∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br
w g Organic compound gives b g AgBr
Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
Let Weight of silver Iodide be ‘c’ g
235 g of AgI contains 127 g of I
∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I
w g Organic compound gives c g AgI
percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
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.
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of ‘ paper acts as an adsorbent. This method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.
A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase. The solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots. on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.
Structural isomerism:
This type of isomers have same molecular formula but differ in their bonding sequence.
(a) Chain or nuclear or skeletal isomerism:
These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched).
(b) 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.
(c) Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
Molecular formula C 3 H 6 O
CH 3 – CH 2 – CHO
propanal (aldehyde group)
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 l (or) by (-) sign.
6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.
In alkenes, the carbon-carbon double bond is sp 2 hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The π bond is formed by the head on overlap of sp 2 hybrid orbitals. The n bond is formed by the side-wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible. This restriction of rotation about C – C double bond is responsible for geometrical isomerism in alkenes.
These two compounds are termed as geometrical isomers and are distinguished from each other by the terms c is and trans. The c is isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis- trans isomerism.
The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the n bond so that rotation about a bond becomes possible. Upon cooling, the reformation of the n bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butene.
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of carbon:
44 g of carbondioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \(\frac{12 \times 0.88}{44}\) g
0.30 g substance contains,
carbon = \(\frac{12 \times 0.88}{44}\) g
100 g substance Contains \(\frac{12 \times 0.88}{44}\) × \(\frac{100}{0.30}\) = 80 g of carbon
Percentage of carbon = 80 %
Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \(\frac{2 \times 0.54}{18}\) g
0.30 g of substance contains hydrogen = \(\frac{2 \times 0.54}{18 \times 0.30}\) g
100 g of substance contains = \(\frac{2 \times 0.54}{18 \times 0.30}\) × 100 g = 20 g of hydrogen
Percentage of hydrogen = 20 %
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%
Mass of the substance taken = 0.32 g
Mass of BaSO 4 formed = 0.466 g
Molecular mass of BaSO 4 = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of BaSO 4 = \(\frac{0.466 \times 32}{233}\)
Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143. 5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) × 0.287
Percentage of chlorine = \(\frac{35.5}{143.5} \times \frac{0.287}{0.24}\) × 100 = 29.58 %
Volume of N 2 at NTP = \(\frac{V \times P}{t+273} \times \frac{273}{760}\)
= V 0 ml
Substituting the various values in the above equation,
V 0 = \(\frac{20.7 \times 760}{288} \times \frac{273}{760}\) = 19.62 ml
weight of 19.62 ml of Nitrogen = \(\frac{28}{22400}\) × 19.62 g
∴ Percentage of Nitrogen = \(\frac{28}{22400}\) × 19.62 × \(\frac{100}{0.35}\)
= 4.9 %