\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3<\mathrm{CH}_3 \mathrm{OCH}_3<\mathrm{CH}_3 \mathrm{CHO}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\) Stronger the attractive forces, higher is the boiling point. Hydrocarbons are non-polar having weakest attractive forces, ethers are polar and aldehydes have strong dipolar interaction. Alcohols have maximum intermolecular forces due to the presence of H-bonding.

(i) In l-bromo-l-methylcyclohexane, the \(\beta \)-hydrogens on either side of the Br atom are equivalent, therefore, only 1- alkene is formed.

ii) 2-Boro-2-methylbutane has two different sets of equivalent \(\beta \)-hydrogens and hence, in principle, can give two alkenes (I and II). But according to SaytzetT rule, more highly substituted alkene (II), being mor~ stable, is the major product.

(iii) 3-Bromo-2, 2, 3-trimethylpentane has two different sets of \(\beta \)-hydrogens and hence, in principle, can give two alkenes (I and II). But according to SaytzetT rule, more highly substituted alkene (II), being more stable, is the major product.

o-Nitrophenol is steam-volatile due to chelation (intramolecular H-bonding) and hence can be separated by steam distillation from p-nitrophenol which is not steam volatile because of intermolecular H-bonding. 

\(A_m={1000\times k\over M}\)
138.9 S cm² mol^-1=\(1000\times k\over 1.5\)
\(k={138.9\times 1.5\over 1000}={208.05\over 1000}={2.0805}\times 10^{-1}\ S\ cm^{-1}\)

Molar mass of solute,
 \({ M }_{ B }=\frac { { K }_{ f }\times{ W }_{ B }\times1000 }{ { W }_{ A }\times \triangle { T }_{ f } } \)
\( { W }_{ B }=1.0g\)
\( { W }_{ A }=50.0g\)
\( \triangle { T }_{ f }=0.40K\)
\( { K }_{ f }=5.12 \ K \ kg \ { mol }^{ -1 }\)
\(\\ { M }_{ B }=\frac { 5.12\times1.0\times1000 }{ 50\times0.40 } \)
\(=256 \ g/mol\)

(i) Since compound (E) with molecular formula, C₃H₆0 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C₃H₆O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH₃COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)₂ therefore, (B) must be acetic acid (ethanoic acid), CH₃COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH₃CH₂OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH₃COOC₂H₅
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH₃COO)₂O.
(vi) All the reactions involved in this problem can now be explained as follows

\(\underset { n-Butylchloride }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }CI } +KOH(alc.)\overset { \triangle }{ \longrightarrow } \underset { But-l-ene }{ { CH }_{ 3 }{ CH }_{ 2 }CH={ CH }_{ 2 }+KCI+{ H }_{ 2 }O } \)

\((iv)\quad { 2CH }_{ 3 }Br+2Na\overset { Dry\quad ether }{ \underset { \quad Wurt\\ reaction }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }{ CH }_{ 3 } } +2NaBr\)
\(\\ (vi)\quad \underset { Methyl\\ chloride }{ { CH }_{ 3 }CI } +KCN\longrightarrow \underset { Methyl\\ chloride }{ { CH }_{ 3 }CN } +KCI\)

(i) \({ Ca }^{ 2+ }(aq)+2{ e }^{ - }\longrightarrow Ca(s)\)
2 F is required for 40 g of Ca
1 F is required for 20g of Ca
(ii) \({ Al }^{ 3+ }(aq)+3{ e }^{ - }\longrightarrow Al(s)\)
\(Eq.wt.=\frac { 27 }{ 3 } =9,Z=\frac { Eq.wt. }{ 96500 } =\frac { 9 }{ 96500 }\)
\(m=Z\times Q\)
\(50=\frac { 9 }{ 96500 } \times Q\)
\(Q=\frac { 50\times 96500 }{ 9 } =5.3611\times { 10 }^{ 5 }C=\frac { 5.3611\times { 10 }^{ 5 }F }{ 96500 } =5.56F\)

\(\text { No. of moles of solute, } n_{2}=\frac{100}{140}=\frac{5}{7} \text { mole }\)
\(\text { No. of moles of solvent, } n_{1}=\frac{1000}{180}=\frac{50}{9} \text { mole }\)
Mole fraction of solute
\(x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{5 / 7}{5 / 7+50 / 9}=0.114\)
\(\text {Mole fraction of solvent, } x_{1}=\left(1-x_{2}\right)=(1-0 \cdot 114)\)
= 0.886
According to Raoult's law
\( P_{A}=x_{A} P_{A}^{0}=0.114 \times P_{A}^{0} \)
\(P_{B}=x_{B} P_{B}^{\circ}=0.886 \times 500=443 \text { torr } \)
\(P_{\text {Total }}=P_{A}+P_{B}\)
\(475=0.114 P_{A}^{\circ}+443\)
\( P_{A}^{\circ}=\frac{475-443}{0 \cdot 114}=280 \cdot 7 \text { torr } \)
\(\therefore P_{A}=0 \cdot 114 \times 280 \cdot 7=32 \text { torr. }\)

In the presence of sulphuric acid (H₂SO₄), KI produces HI
\(2 \mathrm{KI}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{KHSO}_{4}+2 \mathrm{HI}\) 
Since H₂SO₄ is an oxidizing agent, it oxidizes HI (produced in the reaction to I₂₎.
\(2 \mathrm{HI}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{I}_{2}+\mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) 
As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H₃PO₄ is used.

(i) Kolbes reaction.
It is a decarboxylative dimerisation of two carboxylic acids (or carboxylate ions)  to form carbon carbon bond of alkane.

(ii) Reimer-Tiemann reaction.
It involves the ortho-formylation of phenols. Phenol is converted to salicylaldehyde.

(iii) Williamson ether synthesis.
Sodium alkoxide reacts with alkyl halide to form an ether.
\(\mathbf{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{NaCl}\)
(iv) Unsymmetrical ether.
Two different alkyl groups are attached to O atom. The general formula is R−O−R′. An example is ethyl methyl ether 
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\)

Moles of glucose = 18 g/ 180 g mol^–1 = 0.1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg^-1
For water, change in boiling point
\(\triangle\)T_b = K_b × m = 0.52 K kg mol^–1 x 0.1 mol kg^–1 = 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure, therefore, the
boiling point of solution will be 373.15 + 0.052 = 373.202 K.

\({ { \wedge }^{ ° } }_{ \left( HCOOH \right) }={ { \lambda }^{ ° } }_{ \left( { H }^{ + } \right) }+{ { \lambda }^{ ° } }_{ \left( { HCOO }^{ - } \right) }\)
\(=349.6+54.6\\=404.2S{ cm }^{ 2 }{ mol }^{ -1 }\)
\(\alpha =\frac { { { { \wedge } }^{ m } }_{ c } }{ { { { \wedge } }^{ m } }_{ 0 } } =\frac { 46.1 }{ 404.2 } =0.114\)
\({ K }_{ c }=\frac { c{ \alpha }^{ 2 } }{ 1-\alpha } =\frac { 0.025\times { \left( 0.114 \right) }^{ 2 } }{ 1-0.114 } =3.67\times { 10 }^{ -4 }\)

0.01 M MgCI₂ = 0.03 M particle concentration which is three times in comparison to 0.01 M glucose. Hence, depression will be about three times.

(b) Aquatic species are more comfortable in cold water rather than in warm water because the solubility of gases decreases with increase of temperature.

(b): Presence of acid intensifies the partial positive charge on carbonyl carbon and hence, activates the group. 

Presence of base activates a-methylene component of the carbonyl compounds by converting them in carbanions.
\(\mathrm{RCH}_{2} \mathrm{CHO}+: B^{-} \rightarrow \mathrm{R} \overline{\mathrm{C}} \mathrm{HCHO}+B \mathrm{H}\)

(d): The ease of dehydration of alcohols can be explained on the basis of stability of the intermediate carbocation. Greater the stability of the carbonation formed, greater will be the rate of reaction. The order of stability of carbocation formed is:

This is due to the electron releasing (+I) effect of the alkyl groups. Therefore, the ease of dehydration of alcohols follows the order: Tertiary> Secondary> Primary Dehydration of alcohols proceed through carbocation formation.

(b): Among dichlorobenzenes, the p-isomer being symmetrical, packs closely in the crystal lattice and hence has much higher melting point than o- and m- isomers.

(d) : The electrical resistance of any object is directly proportional to its length I, and inversely proportional to its area of cross-section, A. So, it increases with increase in length of object and decreases with increase in area of cross-section of object.

(i) (c) : Condensation reaction is the reverse of hydrolysis, which splits a chemical entity into two parts through the action of the polar water molecule

(iii) (a)
(iv) (d)

(i) (b) : Moles of C₆H₆ = \(\frac{7.8}{78}=0.1\)
Mole C₆H₅CH₃ = \(\frac{9.2}{92}=0.1\)
Mole fraction of C₆H₆ = \(\frac{0.1}{0.1+0.1}=0.5\)
=> Mole fraction of C₆H₅CH₃ = 0.5
Vapour pressure of toluene = Vapour pressure of pure toluene x mole fraction of toluene
= 0.0925 x 0.5 = 0.04625
Vapour pressure of benzene = 0.256 x 0.5 = 0.128
Total vapour pressure of solution = 0.17425
(ii) (a) : Moles of benzene in solution-II = \(\frac{3.9}{78}=0.05\)
Moles of toluene in solution-II = \(\frac{13.8}{92}=0.15\)
Vapour pressure of solution
= 0.256 x 0.05 + 0.0925 x 0.15
= 0.0128 + 0.013875 = 0.026675
(iii) (c) : Mole fraction of benzene in vapour phase
\(y_{\text {benzene }}=\frac{p_{\text {benzene }}}{P_{\text {total }}}=\frac{0.128}{0.17425}=0.734\)
(iv) (a) : Benzene and toluene form an ideal solution.