Question
Read the passage given below and answer the following questions: Due to intermolecular hydrogen bonding, the boiling points of alcohols and phenols are much higher than those of corresponding haloalkanes, haloarenes, aliphatic and aromatic hydrocarbons. Among isomeric alcohols, the boiling points follow the order: primary > secondary > tertiary. Boiling points of ethers are much lower than those of isomeric alcohols. The solubility of alcohols in water decreases as the molecular mass of alcohols increases. Amongst isomeric alcohols solubility increases with branching. The solubility of phenols in water is much lower than that of alcohols. Lower ethers such as dimethyl ether and ethyl methyl ether are soluble in water, but the solubility decreases as the molecular mass increases. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Alcohols have higher boiling points than ethers of comparable molecular masses.
Reason: Alcohols and ethers are isomeric in nature.
  1. Assertion: The solubility of phenols in water is much lower than that of alcohols.
Reason: Phenols do not form H-bonds with water.
  1. Assertion: Among n-butane, ethoxyethane, 1-propanol and 2-propanol, the increasing order of boiling points is, 1-butanol < 1-propanol < ethoxyethane < n-butane.
Reason: Boiling point increases with increase in molecular mass.
  1. Assertion: Dimethyl ether and diethylether are soluble in water.
Reason: As the molecular mass increases, solubility of ethers in water decreases.
  1. Assertion: Butan-2-ol has higher boiling point than 2-methylpropan-2-ol.
Reason: Amongst isomeric alcohols, the boiling points decreases with branching.

Answer

  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Explanation:
Due to the presence of intermolecular H-bondi ng in alcohols, they have higher boiling points than isomeric ethers.
  1. (c) Assertion is correct statement but reason is wrong statement.
​​​​​​​Explanation:
Like alcohols, phenols also form H-bond with water. But the solubility of phenols in water is much lower than that of alcohols because of the larger non-polar hydrocarbon part (benzene ring) present in their molecules.
  1. (d) Assertion is wrong statement but reason is correct statement.
​​​​​​​​​​​​​​​​​​​​​Explanation:
Boiling point increases with increase in molecular mass so, 1-butanol has higher boiling point than 1-propanol. Ethers do not form hydrogen bonds thus, they have lower boiling points than the corresponding alcohols.
Due to weak dipole-dipole interactions, the boiling points of lower ethers are only slightly higher than those of the n-alkanes having comparable molecular masses,
Thus, the increasing order of boiling points is n-butane < ethoxyethane < 1-propanol < l-butanol.
  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:
The solubility of lower ethers in water is due to the formation of H-bonds between water and ether molecules. As the molecular mass increases, the solubility of ethers in water decreases due to corresponding increase in the hydrocarbon portion of the molecule.
  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:

Amongst isomeric alcohols,the boiling points decrease with branching due to a corresponding decrease in surface area.

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Read the passage given below and answer the following questions:
At $298 K,$ the vapour pressure of pure benzene $, C_6, H_6$ is $0.256$ bar and the vapour pressure of pure toluene $\ce{C_6 H_5 CH_3}$ is $0.0925$ bar. Two mixtures were prepared as follows:
  1. $7.8g$ of $\ce{C_6 H_6 + 9.2g}$ of toluene
  2. $3.9g$ of $\ce{C_6 H_6 + 13.8g}$ of toluene
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The total vapour pressure $($bar$)$ of solution I is.
  1. $0.128$
  2. $0.174$
  3. $0.198$
  4. $0.258$
  1. Which of the given solutions have higher vapour pressure?
  1. $I$
  2. $II$
  3. Both have equal vapour pressure
  4. Cannot be predicted
  1. Mole fraction of benzene in vapour phase in solution I is.
  1. $0.128$
  2. $0.174$
  3. $0.734$
  4. $0.266$
  1. Which of the following statements is/are correct?
  1. Mole fraction of toluene in vapour phase is more in solution $ I$.
  2. Mole fraction of toluene in vapour phase is less in solution $I.$
  3. Mole fraction of benzene in vapour phase is less in solution $I.$
  1. Only $II$
  2. Only $I$
  3. $I$ and $III$
  4. $II$ and $III$
  1. Solution I is an example of a/an.
  1. Ideal solution.
  2. Non $-$ ideal solution with positive deviation.
  3. Non $-$ ideal solution with negative deviation.
  4. Can't be predicted.
Read the passage given below and answer the following questions : Transition metal oxides are compounds fanned by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic $V_2O_3$ to less basic $V_2O_4$ and to amphoteric $V_2O_5· V_2O_4$ dissolves in acids to give $VO^{2+}$ salts. Transition metal oxides are commonly utilized for their catalytic activity and semi conductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titatnium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants. The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which oxide of vanadium is most likely to be basic and ionic?
  1. $VO$
  2. $V_2O_3$
  3. $VO_2$
  4. $V_2O_5$
  1. Vanadyl ion is:
  1. $\text{VO}^{2+}$
  2. $\text{VO}^{+}_2$
  3. $\text{V}_{2}\text{O}^+$
  4. $\text{VO}^{3-}_4$
  1. Which of the following statements is false?
  1. With fluorine vanadium can form $VF_5.$
  2. With chlorine vanadium can form $VCl^5.$
  3. Vanadium exhibits highest oxidation state in oxohalides $VOCl_3, VOBr_3$ and fluoride $VF_5.$
  4. With iodine vanadium cannot form $Vl_5$ due to oxidising power of $V^{5+}$ and reducing nature of $I^-.$
  1. The oxidation state of vanadium in $V_2O_5$ is:
  1. $\frac{+5}{2}$
  2. $+7$
  3. $+5$
  4. $+6$
  1. Identify the oxidising agent in the following reaction.
$V_2O_{5 }+ 5Ca \rightarrow 2V + 5CaO$
  1. $V_2O_5$
  2. $Ca$
  3. $V$
  4. None of these.
Explain the structure of nucleic acid compounds.
Write detailed note on: Starch
Read the passage given below and answer the following questions:
The amines are basic in nature due to the presence of a lone pair of electron on $N-$atom of the $-NH_2$ group, which it can donate to electron deficient compounds. Aliphatic amines are stronger bases than $NH_3$ because of the $+I$ effect of the alkyl groups. Greater the number of alkyl groups attached to $N-$atom, higher is the electron density on it and more will be the basicity. Thus, the order of basic nature of amines is expected to be $3^\circ > 2^\circ > 1^\circ ,$ however the observed order is $2^\circ > 1^\circ > 3^\circ .$ This is explained on the basis of crowding on $N-$atom of the amine by alkyl groups which hinders the approach and bonding by a proton, consequently, the electron pair which is present on $N$ is unavailable for donation and hence $3^\circ$ amines are the weakest bases.
Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-donating groups such as $-\ce{CH_3, -OCH_3,}$ etc. increase the basicity while electron-withdrawing substitutes such as $\ce{-NO_2, -CN,}$ halogens, etc. decrease the basicity of amines. The effect of these substituents is more at $p$ than at $m-$positions.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which one of the following is the strongest base in aqueous solution?
  1. Methyl amine.
  2. Tri methyl amine.
  3. Aniline.
  4. Dimethyl amine.
  1. Which order ofbasicity is correct?
  1. Aniline $>$ $m-$toluidine $> o-$toluidine
  2. Aniline $> o-$toluidine $> m-$toluidine
  3. $o-$toluidine $>$ aniline $> m-$toluidine
  4. $o-$toluidine $<$ aniline $< m-$toluidine
  1. What is the decreasing order of basicity of primary, secondary and tertiary ethylamines and $NH_3$?
  1. $\ce{NH_3 > C_2H_5NH_2 > (C_2H_5)_2NH > (C_2H_5)_3N}$
  2. $\ce{(C_2H_5)_3N > (C_2H_5)_2NH_{ }> C_2H_5NH_2 > NH_3}$
  3. $\ce{(C_2H_5)_2NH > C_2H_5NH_2> (C_2H_5)_3N > NH_3}$
  4. $\ce{(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3​​​​​​​}$
  1. The order of basic strength among the following amines in benzene solution is:
  1. $\ce{CH_3NH_2 > (CH_3)_3N > (CH_3)_2NH}$
  2. $\ce{(CH_3)_3N > (CH_3)_2NH > CH_3NH_2​​​​​​​}$
  3. $\ce {CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N}$
  4. $\ce{(CH_3)_3N > CH_3NH_2 > (CH_3)_2NH}$
  1. Choose the correct statement.
  1. Methylamine is slightly acidic.
  2. Methylamine is less basic than ammonia.
  3. Methylamine is a stronger base than ammonia.
  4. Methylamine forms salts with alkalies.
Read the passage given below and answer the following questions: Werner, a Swiss chemist in $1892$ prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour. He proposed that, in coordination compounds, metals possess two types of valencies, viz. primary valencies, which are normally ionisable and secondary valencies which are non $-$ ionisable. ln a series of compounds of cobalt $(III)$ chloride with ammonia, it was found that some of the chloride ions could be precipitated as $\text{AgCl}$ on adding excess of $\ce{AgNO_3}$ solution in cold, but some remained in solution. The number ofions furnished by a complex in a solution can be determined by precipitation reactions. The measurement of molar conductance of solutions of coordination compounds helps to estimate the number of ions furnished by the compound in solution. In these questions $(Q$. No. $i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Assertion : The complex $[\ce{Co(NH_3)_3Cl_3]}$ does not give precipitate with silver nitrate solution.
Reason : The given complex is non-ionisable.
  1. Assertion : The complex $\ce{[Co(NH_3)_4Cl_2]Cl}$ gives precipitate corresponding to $2$ mol of $\ce{AgCl}$ with $\ce{AgNO_3}$ solution.
Reason : It ionises as $\ce{[Co(NH_3)_4Cl_2]^+ + Cl}^-.$
  1. Assertion : $\ce{CoCl_3. 4NH_3}$ gives $1$ mol of $\ce{AgCl}$ on reacting with $\ce{AgNO_3},$ its secondary valency is $6$.
Reason : Secondary valency corresponds to coordination number.
  1. Assertion : $1$ mol of $\ce{[CrCl_2(H_2O)_4]Cl· 2H_2O}$ will give $1$ mol of $\ce{AgCl}$ on treating with $\ce{AgNO_3}.$
Reason : $ Cl^-$ ions satisfying secondary valanceis will not be precipitated.
  1. Assertion : $\ce{CoCl_3. 3NH_3}$ is not conducting while $\ce{CoCl_3. 5NH_3}$ is conducting.
Reason : The complex of $\ce{CoCl_3. 3NH_{3 }}$ is $\ce{[CoCl_3(NH_3)_3]}$ while that of $\ce{CoCl_3· 5NH_3}$ is $\ce{[CoCl(NH_3)_5]Cl_3}.$
The progress of the reaction$, \text{A}\rightleftharpoons\text{nB}$ with time is represented in the following figure :

The following questions are multiple choice questions. Choose the most appropriate answer :
  1. What is the value of $n$?
  1. $1$
  2. $2$
  3. $3$
  4. $4$
  1. Find the value of the equilibrium constant.
  1. $0.6M$
  2. $1.2M$
  3. $0.3M$
  4. $2.4M$
  1. The initial rate of conversion of $A$ will be:
  1. $0.1 mol L^{-1}hr^{-1}$
  2. $0.2 mol L^{-1}hr^{-1}$
  3. $0.4 mol L^{-1}hr^{-1}$
  4. $0.8 mol L^{-1}hr^{-1}$
  1. For the reaction, if $\frac{\text{d}[\text{B}]}{\text{dt}}=2\times10^{-4},$ value of $-\frac{\text{d}[\text{A}]}{\text{dt}}$ will be:
  1. $2 \times 10^{-4}$
  2. $10^{-4}$
  3. $4 \times 10^{-4}$
  4. $0.5 \times 10^{-4}$
  1. Which factor has no effect on rate of reaction?
  1. Temperature.
  2. Nature of reactant.
  3. Concentration of reactant.
  4. Molecularity.
Read the passage given below and answer the following questions:
Aldehydes and ketones are reduced to primary and secondary alcohols respectively by $\ce{NaBH}_4$ or $\ce{LiAlH_4}$ as well as catalytic hydrogenation. The carbonyl group of aldehydes and ketones is reduced to  group on treatment with $\ce{Zn-Hg}$ and cone. $\text{HCl} \ ($Clemmensen reduction$)$ or with hydrazine followed by $\ce{NaOH}$ or $\text{KOH}$ in highly boiling solvent such as ethylene glycol $($Wolff $-$ Kishner reduction).Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with $\ce{HNO_3, KMnO_4, K_2Cr_2O_7}$ etc. Even mild oxidising agents mainlyTollens' reagent and Fehling's solution also oxidise aldehydes. Ketones are generally oxidised under vigorous conditions i.e., strong oxidising agents and at elevated temperatures, to give mixture of carboxylic acids having lesser number of $C-$ atoms than the parent ketone.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following cannot be made by reduction of ketone or aldehyde with $\ce{NaBH}_4$ in methanol?
  1. $1-$ Butanol
  2. $2-$ Butanol
  3. $2-$ Methyl $-1-$ propanol
  4. $2-$ Methyl $-2-$ propanol
  1. The carbonyl compound producing an optically active product by reaction with $\ce{LiAlH}_4$ is:
  1. Propanone
  2. Butanone
  3. 3-pentanone
  4. Benzophenone
  1. A substance $\ce{C_4H_{10}O} (X)$ yields on oxidation a compound $\ce{C_4H_8O}$ which gives an oxime and a positive iodoform test.
  2. The substance $X$ on treatment with cone. $\ce{H_2SO}_4$ gives $\ce{C_4H_8}$. The structure of the compound $(X)$ is:
  1. $\ce{CH_3CH_2CH_2CH_2OH}$
  2. $\ce{CH_3CH(OH)CH_2CH_3}$
  3. $\ce{(CH_3)_3COH}$
  4. $\ce{CH_3CH_{2 }- O - CH_2CH_3}$
  1. In the oxidation of  by acidified $\ce{K_2Cr_2O_7},$ the products are:
  1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\ ^\text{14}\text{C}-\text{OH}$ and $\text{CH}_3\text{CH}_2\text{COOH}$
  2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3(\text{CH}_2)_2\text{COOH}-\text{C}-\text{OH}$ and $ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{COOH}$
  3. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}+\text{HCOOH}$
  4. None of these.
  1. The appropriate reagent for the following transformation is:
  1. $\text{Na}_2\text{NH}_2,^-\text{OH}$
  2. $\text{NaBH}_4$
  3. $\frac{\text{H}_2}{\text{Ni}}$
  4. $\text{AICl}_3$
Read the passage given below and answer the following questions: Ligands are atoms or ions which can donate electrons to the central atoms. Ligands can be monodentate, bidentate or polydentate as well. Few ligands can coordinate with the central atom through more than one site, these are called ambidentate ligands. When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelating ligand. In these questions $(Q. No. i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Assertion: Glycinate ion is an example of monodentate ligand.
Reason: Glycinate contains $N $ and $O$ as donor atoms.
  1. Assertion: $EDTA$ forms complex with divalent metals of 3d-series in the ratio of $1 : 1.$
Reason: $EDTA$ has $4 - COOH$ groups.
  1. Assertion: Oxalate ion is a bidentate ligan.
Reason: Oxalate ion has two donor atoms.
  1. Assertion: A chelating ligand must possess two or more lone pairs at such a distance that it may form suitable strain free $5 $ and $6$ membered rings with the metal ion.
Reason: $H_2N - NH_2$ is a chelating ligand.
  1. Assertion: In Zeise's salt coordination number of $Pt$ is five.
Reason: Ethene is a monodentate ligand.
Decrease in concentration of reactant or increase in concentration of product per unit time is called rate of reaction. lt is of two types :
  1. Instantaneous rate of reaction : Rate of change of concentration of reactant or product at a particular time is called instantaneous rate of reaction.
$\text{r}_\text{inst.}=\frac{\text{dC}}{\text{dt}}$
where$, dC =$ infinitely small change in concentration
$dt =$ infinitely small change in time.
  1. Average rate of reaction : Ratio of change in concentration and time required for the change is average rate of reaction.
$\text{r}_\text{av}=\frac{\triangle\text{x}}{\triangle\text{t}}=\frac{\text{Change in concentration}}{\text{Time required for the change}}$
For a reaction of the type$, m_1A + m_2B \rightarrow n_1C + n_2D$
Rate of reaction is given as
$\frac{1}{\text{m}_1}\frac{\text{d[A]}}{\text{dt}}=-\frac{1}{\text{m}_2}\frac{\text{d[B]}}{\text{dt}}=+\frac{1}{\text{n}_1}\frac{\text{d[C]}}{\text{dt}}=+\frac{1}{\text{n}_2}\frac{\text{d[D]}}{\text{dt}}$
In these questions $(Q.$ No. $i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: The kinetics of the reaction, $\text{mA}+\text{nB}+\text{pC}\rightarrow\text{m}'\text{ X}+\text{n}'\text{ Y}+\text{p}'\text{ Z}$ obey the rate expression as $\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^\text{m}[\text{B}]^\text{n}.$
Reason: The rate of the reaction does not depend upon the concentration of $C$.
  1. Assertion : Instantaneous rate of reaction is equal to $\frac{\text{dx}}{\text{dt}}.$
Reason : lt is the rate of reaction at any particular instant of time.
  1. Assertion : For the reaction, $\text{RCl}+\text{NaOH}\rightarrow\text{ROH}+\text{NaCl},$ the rate of reaction is reduced to half on reducing the concentration of $\ce{RCl}$ to half.
Reason : The rate of reaction is represented by $\ce{k[RCl}].$
  1. Assertion : ln rate law, unlike in the expression for equilibrium constants, the exponents for concentrations do not necessarily match the stoichiometric coefficients.
Reason: It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate.
  1. Assertion : ln a reaction$, 2A + B \rightarrow A_2B,$ the reactant $B$ will disappear at twice the rate as $A$ will decrease.
Reason: The rate of disappearance of reactant will be $-\frac{1}{2}\frac{\text{d[A]}}{\text{dt}}=-\frac{\text{d[B]}}{\text{dt}}$