Long Answer Type(a) Complete the following chemical equation
(i) Cu + HNO3 (dilute) --->
(ii) XeF4 + O2F2 -->
(b) Explain the following observation:
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) Oxygen is a gas but sulphur a solid.
(iii) The halogens are coloured. Why?
a) Define the following terms:
(i) Mole fraction
(ii) Ideal solution
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at - 0.34°C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol-1)
(a) Explain the following:
(i) Henry’s law about the dissolution of a gas in a liquid.
(ii) Boiling point elevation constant for a solvent.
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass if glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mol-1)
(a) Write a suitable chemical equation to complete each of the following transformations:
(i) Butan-1-ol to butanoic acid
(ii) 4-methylacetophenone to benzene-1, 4-dicarboxylic acid
(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
(a) Give chemical tests to distinguish between
(i) Propanol and propanone
(ii) Benzaldehyde and acetophenone
(b) Arrange the following compounds in an increasing order of their property as indicated:
(i) Acetaldeyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) Benzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength).
(iii) CH3CH2CH (Br) COOH, CH3CH (Br) CH2COOH, (CH3)2CHCOOH (acid strength)
(a)
(i) Propanone gives iodoform test while propanol does not give this test.
(ii) Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) can be distinguished by iodoform test.
Acetophenone, being a methyl ketone on treatment with I2/NaOH undergoes iodoform reaction to give a yellow ppt. of iodoform. On the other hand, benzaldehyde does not give this test.
(b)
(i) Methyl tert-butyl ketone < Acetone < Acetaldehyde
When HCN reacts with a compound, the attacking species is a nucleophile, CN-.Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same.
(ii) 4-Methoxybenzoic acid < Benzoic acid < 3, 4-Dinitrobenzoic acid
Electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4-ethoxybenzoic acid is a weaker acid than benzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths of acids.
(iii) (CH3)2 CHCOOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH
After losing a proton, carboxylic acids gain a negative charge as shown:
R – COOH ---> R - COO- + H+
Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I effect will decrease the strength of the acids and groups having -I effect will increase the strength of the acids. In the given compounds, -CH3 group has +I effect and Br- group has -I effect. Thus, acids containing Br- are stronger.
The -I effect grows weaker as distance increases. Hence, CH3CH(Br)CH2COOH is a weaker acid than CH3CH2CH(Br)COOH.
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