ESTIMATION
Problem 12.20
On complete combustion, 0.246g of an organic compound gave 0.198gm
of carbon dioxide and 0.1014 gm of water. Determine the percentage
composition of carbon and hydrogen in the compound.
Solution
Percentage of carbon = 12 × 0.198 × 100
18 × 0.246
= 21.95%
Percentage of hydrogen = 2 × 0.1014 × 100
44 × 0.246
=4.58%
Problem 12.21
In Dumas method for estimation of nitrogen,0.3g of an organic
compound gave 50ml of nitrogen collected at 300K temperature
and 715mm pressure. Calculate the percentage composition of
nitrogen in the compound. Aqueous tension at 300K = 15mm
Solution
Volume of nitrogen collected at 300k and 715mm pressure is
50ml.
Actual pressure = 715-15 = 700mm
Volume of nitrogen at STP=273 × 700 × 50 / 300 × 760 = 41.9ml
22.400ml of N2 at STP weighs = 28g
41.9ml of N2 weighs = 28 × 41.9g / 22400.
Percentage of N2 = 28 × 41.9 ×100 / 22400 × 0.3
= 17.46%.
Problem 12.22
During estimation of nitrogen present in an organic compound by
Kjeldahl’s method., the ammonia evolved from 0.5g of the
compound in Kjedahl’s estimation of nitrogen, neutralized 10mL
of 1m H2SO4. Find out the percentage of nitrogen in the
compound
Solution
1m of 10mL H2SO4 =1m of 20ml NH3
1000 mL of 1M of ammonia contains 14g nitrogen.
20mL of 1M ammonia contains 14 × 20 / 1000 g nitrogen
Percentage of nitrogen= 14 × 20 × 100 / 1000 × 0.5 = 56.0%.
Problem 12.23
In carius method of estimation of halogen.0.15 gm of an
organic compound, gave 0.12g of AgBr. Find out the
percentage of bromine in the compound.
Solution
Molar mass of AgBr = 108 + 80=188g mol-1
1.188g AgBr contains 80g bromine.
0.12g AgBr contains 80 × 0.12 / 188 g bromine.
Percentage of bromine =80 × 0.12 × 100 / 188 × 0.15
= 34.04%.
Problem 12.24
In sulphur estimation, 0.157 g of an organic compound
gave 0.4813 g of barium sulphate. What is the
percentage of sulphur in the compound?
Solution
Molecular mass of BaSO4 = 137 + 32 + 64 = 233 g
233 g BaSO4 contains 32 g Sulphur
0.4813 g BaSO4 contains 32 × 0.4813 / 233 g Sulphur
Percentage of Sulphur = 32 × 0.4813 × 100 / 233 × 0.157
= 42.10%
Chapter 13
HYDROCARBONS
Problem 13.1
Write structures of different chain isomers of alkanes corresponding to the
molecular formula C6H14. Also write their IUPAC names
Solution
CH3-CH2-CH2-CH2-CH2-CH3
n-Hexane
CH3-CH-CH2-CH2-CH3
2-Methylpentane
CH3
CH3-CH2-CH-CH2-CH3
3-Methylpentane
CH3
CH3-CH-CH-CH3
2,3-Dimethylbutane
CH3 CH3
CH3
CH3-C-CH2-CH3
CH3
2,2-Dimethylbutane
Problem 13.2
Write structures of different isomeric alkyl groups corresponding to the molecular
form C5H11. write IUPAC names of alcohols obtained by attachment of-OH groups
at different carbons of the chains.
Solution
Corresponding alcohols
Structures of C5H11 groups
1.CH3-CH2-CH2-CH2-CH22.CH3-CH-CH2-CH2-CH3
3.CH3-CH2-CH-CH2-CH2
CH3
4.CH3-CH-CH2-CH2-
CH3-CH2-CH2-CH2-CH2-
Pentan-1-ol
CH3-CH-CH2-CH2-CH3
Pentan-2-ol
CH3-CH2-CH-CH2-CH2
CH3
CH3-CH-CH2-CH2-
CH3
6.CH3-C-CH2-CH3
CH3
CH3-CH2-CH-CH2-
2-Methylbutan-1-ol
CH3
CH3-C-CH2-CH3
2-Methylbutan-2-ol
CH3
CH3-C-CH27.CHCH
3-C-CH
23
3-Methylbutan-1-ol
CH3
CH3
5.CH3-CH2-CH-CH2-
Pentan-3-ol
CH3
2,2-Dimethyl propan-1-ol
Problem 13.3
Write IUPAC names of the following compounds
1.
(CH3)3CCH2C(CH3)3
2.
(CH3)2C(C2H5)2
3.
Tetra-tert-butyl methane
Solution



2,2,4,4- Tetramethylpentane
3,3- Dimethylpentane
3,3-Di-tert-butyl-2,2,4,4- Tetramehylpentane
Problem 13.4
Write structures of the following compounds:
i. 3,4,4,5-Tetramethylheptane
ii. 2,5-Dimethyhexane
Solution
CH3
CH3-CH2-CH-C-CH-CH-CH3
CH3 CH3 CH3
CH3
CH3
CH3-CH-CH2-CH2-CH-CH3
Problem 13.5
Write structures for each the following compounds. Why are the given
names incorrect? Write correct IUPAC names.
i.
ii.
2-Ethylpentane
5-Ethyl-3-methylheptane
Solution
Longest Chain is of 6 C atoms and not
that of 5. hence, correct name is 3Methylhexane.
CH3-CH-CH2-CH2-CH3
C2H5
7
6
5
4
3
2
1
CH3-CH2-CH-CH2-CH-CH2-CH3
CH3
C2H5
Numbering is to be started from the end
which gives lower number to ethyl group.
Hence, correct name is 3-ethyl-5methylheptane
Problem 13.6
Sodium salt of which acid will be needed for the preparation of propane? Write
chemical equation for the reaction.
Solution
Butanoic Acid
- +
CH3CH2CH2COONa + NaOH
CaO
CH3CH2CH3 + Na2CO3
Problem 13.7
Write IUPAC names of the following compounds.
i.
(CH3)2CH-CH=CH-CH2-CH
CH3-CH-CH
Problem 13.8
Calculate no. of σ and π bonds in the
above structures:-
C2H5
Solution
ii.
iii. CH2=C(CH2CH2CH3)2
σ bonds
iv. (CH3)CH2CH2CH2 CH2CH3
CH3-CHCH=C-CH2-CHCH3
CH3
Solution
i. 2,8- Dimethyl-3,6-decadiene
ii. 1,3,5,7 Octatetraene
iii. 2-n-Proplypent-1-ene
iv. 4-Ethyl-2,6-Dimethyl-dec-4-ene
π bonds
i.
33
2
ii.
7
4
iii.
23
1
iv.
41
1
Problem 13.9
Write structures & IUPAC names of different structural isomers of alkenes
corresponding to the C5H10.
Solution
CH2=CH-CH2-CH2-CH3
Pent 1-ene
CH3-CH=CH-CH2-CH3
Pent 2-ene
CH3-C=CH-CH3
2-Methylbut-2-ene
CH3
CH3-CH-CH=CH2
3-Methylbut-1-ene
CH3
CH2=C-CH2-CH3
CH3
2-Methylbut-1-ene
Problem 13.10
Draw cis and trans isomers of the following compounds. Write their IUPAC names.
i. CHCl=CHCl
ii. C2H5CCH3=CCH3C2H5
Solution
1.
H C=C
Cl
H
H
Cl
Cl
C=C
Cl
H
2.
CH3 C=C CH3
C2H5
C2H5
CH3 C=C C2H5
C2H5
CH3
Problem 13.11
Which of the following compounds show cis-trans isomerism?
i.
ii.
iii.
iv.
(CH3)2C = CH - C2H5
CH2 = CBr2
C6H5CH = CH - CH3
CH3CH = CClCH3
Solution
3RD AND 4TH
In 1st and 2nd structures, two identical groups are attached to one of the doubly
bonded carbon atoms
Problem 13.12
Write IUPAC names of the products obtaine dby addition reactions of
HBr to Hex-1-ene in absence of peroxide and In presence of peroxide
Solution
CH2=CH-CH2-CH2-CH2-CH3 + H-Br
No Peroxide
CH3-CH-CH2-CH2-CH2-CH2-CH3
Hex-1-ene
Br
2-Bromohexane
Peroxide
CH3-CH-CH2-CH2-CH2-CH3 + H-Br
CH2-CH2-CH2-CH2-CH2-CH3
1-Bromohexane
Problem 13.13
Write structures of different isomers corresponding to the 5th member of alkyne series.
Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different
pairs of isomers ?
Solution
5th member of alkyne has the molecular formula C6H10, the possible isomers are:
A)HC
≡ C-CH2-CH2-CH2-CH3
Hex-1-yne
E)HC ≡ C-CH2-CH-CH3
B)CH3-C ≡ C-CH2-CH2-CH3
Hex-2-yne
CH3
4-methylpent -1-yne
C)CH3 CH2-C ≡ C-CH2–CH3
Hex- 3yne
D)HC ≡ C–CH-CH2-CH3
3–Methylpent-1-yne
CH3
F)CH3 –C C –CH –CH3
CH3
4-methylpent -2-yne
CH3
G)HC C-C-CH3
Position and chain isomerism shown
by different parts.
CH3
3,3-dimethylbut- 1-yne
Problem 13.14
How will you convert ethanoic acid into benzene ?
Solution
CH3COOH
C2H5CL
CH2=CH2
NaOH(aq)
Cl2
Br2
CH3COONa
C2H6
CH2Br
CH2Br
Sodalime
Na/ Dry Ether
Wutz reaction
Alc. KOH
CH4
CH3Cl
CH2=CHBr
CH=CH