Important Question of Chemistry

Q. Write the IUPAC name of the following compound:

Ans 2, 5-Dimethyl hexane-1, 3-diol.

Q. Arrange the following compounds in an increasing order of their acid strengths: (CH₃)₂CHCOOH, CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH

Ans

Q. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.

Ans

Q. Explain as to why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions.

Ans In haloarenes C—X bond acquires a partial double bond character due to resonance. As a result the bond cleavage in haloarenes is difficult than haloalkanes and therefore, they are less reactive towards nucleophilic substitution reaction.

Q State Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention two applications of the law.

Ans It states that at constant temperature the mass of a gas(m) dissolved in a given volume of the liquid is directly proportional to the pressure of the gas () Ppresent in equilibrium with the liquid.

Mathematically, m ∝ P

or m= K_HP

where K_H is the Henry’s law constant.

Applications of Henry’s law are (i) To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure. (ii) To minimize the painful effects accompanying the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas.

Q A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its t_1/2 value.

Ans For a first order reaction

$\displaystyle K=\frac{{2.303}}{t}\log \frac{{\left[ {{{R}_{0}}} \right]}}{{\left[ R \right]}}$

When t = 40 minutes, $\displaystyle \frac{{\left[ {{{R}_{0}}} \right]}}{{\left[ R \right]}}=\frac{{100}}{{100-30}}=\frac{{10}}{7}$

$\displaystyle K=\frac{{2.303}}{{40}}\log \frac{{10}}{7}$

= $\displaystyle \frac{{2.303}}{{40}}\log 1.428$

= $\displaystyle \frac{{2.303}}{{40}}\times 0.1548$

= $\displaystyle 8.91\times {{10}^{{-3}}}{{\min }^{{-1}}}$

$\displaystyle {{t}_{{\frac{1}{2}}}}=\frac{{0.693}}{k}=\frac{{0.693}}{{8.91\times {{{10}}^{{-3}}}}}$

t_½ = 77.78 min.

Q Define the following terms in relation to proteins: (i) Peptide linkage (ii) Denaturation

Ans (i)Peptide linkage: A peptide linkage is an amide linkage (-CONH-) formed between -COOH group of one α-amino acid and NH₂ group of the other amino acid by the elimination of a water molecule.

(ii) Denaturation: When a protein in its native form is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and proteins loses its biological activity. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact, e.g., coagulation of egg white on boiling.

Q List the reactions of glucose which cannot be explained by its open chain structure.

Ans The following reactions of glucose cannot be explained by its open chain structure. (i) Despite having the aldehyde group glucose does not give 2, 4-DNP test, Schiffs test and it does not form the hydrogen sulphite addition product with NaHSO₃. (ii) The pentacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group. (iii) When D-glucose is treated with methanol in the presence of dry hydrogen chloride gas, it gives two isomeric mono methyl derivatives known as α-D glucoside and methyl β-D glucoside. These glucosides do not react with hydrogen cyanide or with hydroxylamine.

Q What is a ligand? Give an example of a bidentate ligand.

Ans The ion, atom or molecule bound to the central atom/ion in the coordination entity is called ligand. A ligand should have lone pair of electrons in their valence orbital which can be donated to central metal atom/ion.

bidentate ligand

Q pK_b for aniline is more than that for methylamine.

Ans In aniline due to resonance the lone pair of electrons on the nitrogen atom are delocalized over the benzene ring. As a result, the electron density on the nitrogen decreases. On the other hand, in methyl amine +I effect of CH₃ increases the electron density on the nitrogen atom. Therefore aniline is a weaker base than methyl amine and hence its pK_b value is higher than that of methyl amine.

Q Aniline does not undergo Friedel-Crafts reaction.

Ans Aniline being a Lewis base, reacts with lewis acid AlCl₃ to form a salt. Due to this N atom of aniline acquires positive charge and hence acts as a strong deactivation group for further reaction.

Q Describe the following reactions: (i) Cannizaro reaction (ii) Cross aldol condensation

Ans Cannizzaro reaction: Aldehydes which do not have an α-hydrogen, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.

2HCHO $\displaystyle \xrightarrow{{conc.KOH}}$ CH3OH + HCOO^–K⁺

(ii) Cross aldol condensation: When aldol condensation is carried out between two different aldehydes and/or ketones, it is called cross aldol condensation.

If both of them contain α-hydrogen atoms. It gives a mixture of four products.

Q Aldehydes are more reactive than ketones towards nucelophiles.

Ans This is due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically two alkyl groups reduce the positivity of the carbonyl carbon more effectively in ketones than in aldehydes.

Q The boiling points of aldehydes and ketones are lower than of the corresponding acids.

Ans This is due to intermolecular hydrogen bonding in carboxylic acids.

Q The aldehydes and ketones undergo a number of addition reactions.

Ans Due to greater electronegativity of oxygen than carbon the C atom of the >C = O bond acquires a partial positive charge in aldehydes and ketones and hence readily undergo nucleophilic addition reactions.

Q What is primary cell? Give an example.

Ans A primary cell is one in which the redox reaction occurs only once and the cell becomes dead after some time and cannot be used again, e.g., dry cell.

Q State Raoult’s law for solutions of volatile liquids. Taking suitable examples explain the meaning of positive and negative deviations from Raoult’s law.

Ans Raoult’s law: It states that for a solution of volatile liquids the partial pressure of each component is directly proportional to its mole fraction.

Mathematically P_A∝ x_A P_B ∝x_B

P_A=P_A⁰x_A P_B=P_B⁰x_B

Positive deviation from Raoult’s law: In this type of deviation the partial pressure of each component of solution is greater than that calculated from Raoult’s law, i.e., P_A>P_A⁰x_A & P_B > P_B⁰x_B .

Example: A solution of water and ethanol.

Negative deviation from Raoult’s: In this type of deviation the partial pressure of each component of solution is less than that expected from Raoult’s law, i.e., P_A< P_A⁰x_A & P_B < P_B⁰x_B .

Example: A solution of acetone and chloroform.

Q Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?

Ans Osmotic pressure (π) defined as the extra pressure that must be applied to the solution side in order to prevent the flow of solvent molecules into it through a semipermeable membrane.

$\displaystyle \pi =\frac{{{{n}_{B}}}}{V}RT=CRT$

where V is the volume of solution in litres containing nB moles of the solute. If W_B grams of the solute whose molecular mass M_B is present in the solution then

$\displaystyle \pi =\frac{{{{W}_{B}}RT}}{{{{M}_{B}}RT}}$

$\displaystyle {{M}_{B}}=\frac{{{{W}_{B}}RT}}{{\pi V}}$

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Important Question of Chemistry

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