Question
Read the passage given below and answer the following questions: Carbohydrates are polyhydroxy aldehydes and ketones and those compounds which on hydrolysis give such compounds are also carbohydrates. The carbohydrates which are not hydrolysed are called monosaccharides. Monosaccharides with aldehydic group are called aldose and those which free ketonic groups are called ketose. Carbohydrates are optically active. Number of optical isomers $= 2^n$ Where $n =$ numberofasymmetric carbons. Carbohydrates are mainlysynthesised by plants during photosynthesis. The monosaccharides give the characteristic reactions of alcohols and carbonyl group $($aldehydes and ketones$).$ It has been found that these monosaccharides exist in the form of cyclic structures. In cyctization, the $-OH$ groups $($generally $C_5$ or $C_4$ in aldohexoses and $C_5$ or $C_6$ in ketohexoses$)$ combine with the aldehyde or keto group. As a result, cyclic structures of five or six membered rings containing one oxygen atom are formed, e.g., glucose forms a ring structure. Glucose contains one aldehyde group, one $IO$ alcoholic group and four $2^\circ$ alcoholic groups in its open chain structure. The following questions are multiple choice questions. Choose the most appropriate answer:
  1. First member of ketos sugar is:
  1. Ketotriose.
  2. Ketotetrose.
  3. Ketopentose.
  4. Ketohexose.
  1. In $\ce{CH_2OHCHOHCHOHCHOHCHOHCHO,}$ the number of optical isomers will be:
  1. $16$
  2. $8$
  3. $32$
  4. $4$
  1. Some statements are given below:
  1. Glucose is aldohexose.
  2. Naturally occurring glucose is dextrorotatory.
  3. Glucose contains three chiral centres.
  4. Glucose contains one $1^\circ $ alcoholic group and four $2^\circ$ alcoholic groups.
Among the above, correct statements are:
  1. $1$ and $2$ only
  2. $3$ and $4$ only
  3. $1, 2$ and $4$ only
  4. $1, 2, 3$ and $4$
  1. Two hexoses fonn the same osazone, find the correct statement about these hexoses.
  1. Both of them must be aldoses.
  2. They are epimers at $C-3.$
  3. The carbon atoms $I$ and $2$ in both have the same configuration.
  4. The carbon atoms $3, 4$ and $5$ in both have the same configuration.
  1. Which of the following reactions of glucose can be explained only by its cyclic structure?
  1. Glucose forms cyanohydrin with $\text{HCN}.$
  2. Glucose reacts with hydroxylamine to form an oxime.
  3. Pentaacetate of glucose does not react with hydroxylamine.
  4. Glucose is oxidised by nitric acid to gluconic acid.

Answer

  1. $(a)$ Ketotriose.
  1. $(a)\ 16$
  1. $(c)\ 1, 2$ and $4$ only
Glucose contains four chiral centres.
  1. $(d)$ The carbon atoms $3, 4$ and $5$ in both have the same configuration.
In the formation of osazone, $C-1$ and $C-2$ react with phenylhydrazine to form phenylhydrazone. If $C-3, C-4, C-5$ have same configuration they will form same osazone even if they differ in configuration at $C-1$ or $C-2.$
  1. $(c)$ Pentaacetate of glucose does not react with hydroxylamine.
Pentacetate of glucose does not react with hydroxylamine showing absence of free $-\text{CHO}$ group. This cannot be explained by open structure of glucose.

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A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. For a first order reaction of the type $A \rightarrow$ Products, the rate of the reaction is given as: rate $= k[A]$. The differential rate law is given as : $\frac{\text{dA}}{\text{dt}}=-\text{k}[\text{A}].$ The integrated rate law is : In $\frac{[\text{A}]}{[\text{A}]_0}=-\text{kt}, [A]$ is the concentration of reactant left at time $t$ and $[A]_0$ is the initial concentration of the reactant$, k$ is the rate constant.
The following questions are multiple choice questions. Choose the most appropriate answer :
  1. The unit of rate constant for a first order reaction is:
  1. $S^{-1}$
  2. $mol\ L^{-1} s^{-1}$
  3. $L\ mol^{-1} s^{-1}$
  4. $L^2\ mol^{-2} s^{-1}$
  1. Half$-$life period of a first order reaction is $10$ min. Starting with initial concentration $12M,$ the rate after $20$ min is:
  1. $0.693 \times 3M\ min^{-1}$
  2. $0.0693 \times 4M\ min^{-1}$
  3. $0.0693 \times M\ min^{-1}$
  4. $0.0693 \times 3M\ min^{-1}$
  1. $50\%$ of a first order reaction is complete in $23$ minutes. Calculate the ti me required to complete $90\%$ of the reaction.
  1. $70.4$ minutes.
  2. $76.4$ minutes.
  3. $38.7$ minutes.
  4. $35.2$ minutes.
  1. For a first order reaction$, (A) \rightarrow$ products, the concentration of $A$ changes from $0.1M$ to $0.025M$ in $40$ minutes. The rate of reaction when the concentration of $A$ is $0.01M,$ is:
  1. $3.47 \times 10^{-4} M/ min$
  2. $3.47 \times 10^{-5} M/ min$
  3. $1.73 \times 10^{-4} M/ min$
  4. $1.73 \times 10^{-5} M/ min$
  1. The half$-$life period ofa $1^{st}$ order reaction is $60$ minutes. What percentage will be left over after $240$ minutes?
  1. $6.25\%$
  2. $4.25\%$
  3. $5\%$
  4. $6\%$
Read the passage given below and answer the following questions:
Haloarenes are less reactive than haloalkanes. The low reactivity of haloarenes can be attributed to:
  • Resonance effect.
  • $sp^2$ hybridisation of $C - X$ bond.
  • Polarity of $C - X$ bond
  • Instability of phenyl cation $($formed by self$-$ionisation of haloarene$).$
  • Repulsion between the electron rich attacking nucleophiles and electron rich arenes.
Reactivity of haloarenes can be increased or decreased by the presence of certain groups at certain positions for example, nitro $(-NO_{^2})$ group at o/ p positions increases the reactivity of haloarenes towards nucleophilc substitution reactions.
The following questions are multiple choice questions Choose the most appropriate answer:
  1. Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to
  1. The formation of less stable carbonium ion.
  2. Resonance stabilisation.
  3. Larger carbon$-$halogen bond.
  4. Inductive effect.
  1. Which of the following aryl halides is the most reactive towards nucleophilic substitution?
  1. Which one of the following will react fastest with aqueous $\text{NaOH}$?
  1. Which chloro derivative of benzene among the followings would undergo hydrolysis most readily with aqueous sodium hydroxide to furnish the corresponding hydroxy derivative?
  1. $\ce{C_6H_5Cl}$
  1. The reactivity of the compounds $\ce{(i)\ MeBr, (ii)\ PhCH_2Br, (iii)\ MeCI, (iv)\ p-MeOC_6H_4Br}$ decreases as:
  1. $\ce{(i) > (ii) > (iii) > (iv)}$
  2. $\ce{(iv) > (ii) > (i) > (iii)}$
  3. $\ce{(iv) > (iii) > (i) > (ii)}$
  4. $\ce{(ii) > (i) > (iii) > (iv)}$
Read the passage given below and answer the following questions:
An organic compound $(A)$ having molecular formula $C_6H_6O$ gives a characteristic colour with aqueous $FeCl_3$ solution. $(A)$ on treatment with $CO_2$ and $\text{NaOH}$ at $400K$ under pressure gives $(B),$ which on acidification gives a compound $(C).$ The compound $(C)$ reacts with acetyl chloride to give $(D)$ which is a popular pain killer.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Compound $(A)$ is:
  1. $2-$Hexanol.
  2. Dimethyl ether.
  3. Phenol.
  4. $2-$Methyl pentanol.
  1. Compound $(C)$ is:
  1. Salicylic acid.
  2. Salicyladehyde.
  3. Benzoic acid.
  4. Benzaldehyde.
  1. Number of carbon atoms in compound $(D)$ is:
  1. $7$
  2. $6$
  3. $8$
  4. $9$
  1. The conversion of compound $(A)$ to $(C)$ is known as:
  1. Reimer$-$Tiemann reaction.
  2. Kolbe's reaction.
  3. Schimdt reaction.
  4. Swarts reaction.
  1. Compound $(A)$ on heating with compound $(C)$ in presence of $POCl_3$ gives a compound $(D)$ which is used:
  1. In perfumery as a ftavouring agent
  2. As an antipyretic
  3. As an analgesic
  4. As an intestinal antiseptic.
Read the passage given below and answer the following questions :
Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal$-$ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five $d-$orbitals to split$-$up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy $(\Delta_0)$ depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of $\Delta_0$ and $P\ ($pairing energy$)$.
If $\Delta_0<\text{P},$ then complex will be high spin.
If $\Delta_0>\text{P},$ then complex will be low spin
The following questions are multiple choice questions. Choose the most appropriate answer :
  1. Which of the following ligand has lowest $\Delta_0$ value?
  1. $CN^-$
  2. $CO$
  3. $F^-$
  4. $NH_3$
  1. The crystal field splitting energy for octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complex is related as :
  1. $\Delta_\text{t}=\frac{1}{2}\Delta_0$
  2. $\Delta_\text{t}=\frac{4}{9}\Delta_0$
  3. $\Delta_\text{t}=\frac{3}{5}\Delta_0$
  4. $\Delta_\text{t}=\frac{2}{5}\Delta_0$
  1. On the basis of crystal field theory, the electronic configuration of $d_4$ in two situations : $(i)\ t.0 > P$ and $(ii)\ t.0$
  $(i)$ $(ii)$
$(a)$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
$(b)$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
$(c)$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
$(d)$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
  1. Using crystal field theory, calculate magnetic moment of central metal ion of $[FeF_6]^{4-}.$
  1. $1.79B.M.$
  2. $2.83B.M.$
  3. $3.85B.M.$
  4. $4.9B.M.$
  1. Electronic configuration of $d-$orbitals in $[Ti(H_2O)_6]^{3+}$ ion in an octahedral crystal field is:
  1. $\text{t}^1_{2\text{g}}\text{e}^0_\text{g}$
  2. $\text{t}^2_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{t}^0_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{t}^1_{2\text{g}}\text{e}^1_\text{g}$
The electrochemical cell shown below is concentration cell. $M|M^{2+}\ ($saturated solution of a sparingly soluble salt$, MX_2) || M^{2+} (0.001\ mol\ dm^{-3})\ |M$ The emfof the cell depends on the difference in concentrations of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 K$ is $0.059V$. The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The solubility product $(K_{sp'}\ mol^3\ dm^{-9})$ of $MX_2$ at $298 K$ based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$
  1. $2 \times 10^{-15}$
  2. $4 \times 10^{-15}$
  3. $3 \times 10^{-12}$
  4. $1 \times 10^{-12}$
  1. The value of $\triangle\text{G}\ ($in $kJ\ mol^{-1})$ for the given cell is $($take $1 F = 96500 C\ mol^{-1})$
  1. $3.7$
  2. $-3.7$
  3. $10.5$
  4. $-11.4$
  1. The equilibrium constant for the following reaction is:
$\text{Fe}^{2+}+\text{Ce}^{4+}\rightleftharpoons\text{Ce}^{3+}+\text{Fe}^{3+}$
(Given, $\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}=1.44\text{V}$ and $\text{E}^\circ_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}=0.68\text{V})$
  1. $7.6 \times 10^{12}$
  2. $6.5 \times 10^{10}$
  3. $5.2 \times 10^9$
  4. $3.4 \times 10^{12}$
  1. The solubility product of a saturated solution of $\ce{Ag2CrO4}$ in water at $298 K$ if the emf of the cell
$Ag|Ag^+ ($satd. $\ce{Ag2CrO4}$ soln$) || Ag^+ (0.1 M) | Ag$
is $0.164V$ at $298 K,$ is :
  1. $3.359 \times 10^{-12}\ mol^3\ L^{-3}$
  2. $2.287 \times 10^{-12}\ mol^3\ L^{-3}$
  3. $1.158 \times 10^{-12}\ mol^3\ L^{-3}$
  4. $4.135 \times 10^{-12}\ mol^3\ L^{-3}$
  1. To calculate the emf of the cell, which of the following options is correct?
  1. $emf = E_{cathode }- E_{anode}$
  2. $emf = E_{anode }- E_{cathode}$
  3. $emf = E_{anode} + E_{cathode}$
  4. None of these.
Read the passage given below and answer the following questions:
The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared, for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at 20° C is 17.5 mm of Hg)
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Relative lowering of vapour pressure for the given solution is.
  1. 0.00348
  2. 0.061
  3. 0.122
  4. 1.75
  1. The vapour pressure (mm of Hg) of solution will be.
  1. 17.5
  2. 0.61
  3. 17.439
  4. 0.00348
  1. Mole fraction of sugar in the solution is.
  1. 0.00348
  2. 0.9965
  3. 0.061
  4. 1.75
  1. If weight of sugar taken is 5g in 108g of water, then molar mass of sugar will be.
  1. 358
  2. 120
  3. 240
  4. 400
  1. The vapour pressure (mm of Hg) of water at 293K when 25g of glucose is dissolved in 450g of water is.
  1. 17.2
  2. 17.4
  3. 17.120
  4. 17.02
Read the passage given below and answer the following questions: Lucas test is a test to differentiate between primary, secondary and tertiary alcohols. This test consists of treating an alcohol with Lucas' reagent, and turbidity, due to the formation of insoluble alkyl chloride, is observed. Lucas test is based on the difference in reacting of three classes of alcohols with hydrogen chloride via $S_N1$ reaction. The different reactivity reflects the differing ease of formation of the corresponding carbocations. 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: Equimolar mixture of cone. $\ce{HCI}$ and anhydrous $\ce{ZnCl_2}$ is called Lucas' reagent.
Reason: Lucas' reagent can be used to distinguish between methanol and ethanol.
  1. Assertion: $2-$Methyl$-2-$butanol gives no turbidity with Lucas' reagent at room temperature.
Reason: It is a $3^\circ$ alcohol.
  1. Assertion: Tertiary alcohols react fastest with Lucas' reagent by $S_N1$ mechanism.
Reason: $3^\circ$ carbocation is most stable.
  1. Assertion: Amongst the compounds, $\ce{H2C = CHCH_2OH (I), C_6H_5OH (II), CH_3CH_2CH_2OH (III)}$ and $\ce{(CH_3)_3COH (IV),}$ only $\ce{(IV)}$ reacts with Lucas' reagent at room temperature.
Reason: Tertiary alcohol gives turbidity immediately with Lucas' reagent.
  1. Assertion: Lucas test can be used to distinguish between $1-$propanol and $2-$propanol.
Reason: Lucas test is based upon the difference in reactivity of primary, secondary and tertiary alcohols with cone. $\ce{HCI}$ and anhyd. $\ce{ZnCl_2.}$
Read the passage given below and answer the following questions: Arrangement of ligands in order of their ability to cause splitting $(\Delta)$ is called spectrochemical series. Ligands which cause large splitting $($large $\Delta)$ are called strong field ligands while those which cause small splitting $($small $\Delta)$ are called weak field ligands. When strong field ligands approach metal atom/ ion, the value of $\Delta_0$ is large, so that electrons are forced to get paired up in lower energy $t_{2g}$ orbitals. Hence, a low$-$spin complex is resulted from strong field ligand. When weak field ligands approach metal atom/ ion, the value of $\Delta_0$ is small, so that electrons enter high energy egorbitals rather than pairing in low energy $t_{2g}$ orbitals. Hence, a high$-$spin complex is resulted from weak field ligands. Strong field ligands have tendency to form inner orbital complexes by forcing the electrons to pair up. Whereas weak field ligands have tendency to form outer orbital complex because inner electrons generally do not pair up. 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: In tetrahedral coordination entity formation, the $d$ orbital splitting is inverted and is smaller as compared to the octahedral field splitting.
Reason : Spectrochemical series is based on the absorption of light by complexes with different ligands.
  1. Assertion: In high spin situation, configuration of $d^5$ ions will be $\text{t}^3_{2\text{g}}\text{e}^2_\text{g}.$
Reason : In high spin situation, pairing energy is less than crystal field energy.
  1. Assertion: $F^-$ ion is a weak field ligand and fonns outer orbital complex.
Reason : $F^-$ ion cannot force the electrons of $d_{z^2}$ and $d_{x^2-y^2}$ orbitals of the inner shell to occupy $d_{xy}, d_{yz}$ and $d_{zx} $orbitals of the same shell.
  1. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.
Reason: In spectrochemical series, ligands are arranged in a series of increasing field strength.
  1. Assertion: $NF_3$ is a weaker ligand than $N(CH_3)_3.$
Reason: $NF_3$ ionizes to give $F^-$ ions in aqueous solution.
Read the passage given below and answer the following questions:
Amines are basic in nature. The basic strength of amines can be expressed by their dissociation constant, $K_b$ or $pK_b.$
$\text{RNH}_2+\text{H}_2\text{O}\rightleftharpoons\text{RNH}^+_3+\text{OH}^-$
$\text{k}_\text{b}=\frac{[\text{RNH}^+_3][\text{OH}^-]}{[\text{RNH}_2]}\text{and}\text{ pk}_\text{b}=-\log\text{k}_\text{b}$
Greater the $K_b$ value or smaller the $pK_b$ value, more is the basic strength of a mine. Aryl amines such as aniline are less basic than aliphatic amines due to the involvement of lone pair of electrons on $N-$atom with the resonance in benzene. In derivatives of aniline, the electron releasing groups increase the basic strength while electron withdrawing groups decrease the basic strength. The base weakening effect of electron withdrawing group and base strengthening effect of electron releasing group is more marked at $p-$position than at $m-$position. $a-$Substituted aniline is less basic than aniline due to ortho effect and is probable due to combination of electronic and steric effect.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following has lowest $pK_b$ value?
  1. The strongest base among the following is:
  1. $\ce{C_6H_5NH_2}$
  2. $\ce{p-NO_2 - C_6H_4NH_2}$
  3. $\ce{m-NO_2 - C_6H_4NH_2}$
  4. $\ce{C_6H_5CH_2NH_2}$
  1. Maximum $pK_b$ value of:
  1.  
  1.  
  1. $\ce{(CH_3CH_2)_2NH}$
  2. $\ce{(CH_3)_2NH}$
  1.  The order of basic strength among the following amines in benzene solution is:
  1. Methylamine is more basic than $NH_3.$
  2. Amines form hydrogen bonds.
  3. Ethylamine has higher boiling point than propane.
  4. Dimethylamine is less basic than methylamine.
  1. $\ce{CH_3CH_2NH_2}$ contains a basic $-NH_2$ group, but $\ce{CH_3CONH_2}$ does not because:
  1. Acetamide is amphoteric in character.
  2. In ethylamine the electron pair on $N-$atom is delocalised by resonance.
  3. In ethylamine there is no resonance while in acetamide the lone pair of electrons on $N-$ atom is delocalised and is less available for protonation.
  4. None of these.
For the reaction : $2\text{NO}_\text{(g)}+\text{Cl}_{2\text{(g)}}\rightarrow2\text{NOCl}_\text{(g)},$ the following data were collected. All the measurements were taken at $263K.$
Experiment No.
Initial $\ce{[NO] (M)}$
Initial $\ce{[Cl_2] (M)}$
Initial rate of disapp. of $\ce{Cl_2 (M/ min)}$
$1.$ $0.15$ $0.15$ $0.60$
$2.$ $0.15$ $0.30$ $1.20$
$3.$ $0.30$ $0.15$ $2.40$
$4.$ $0.25$ $0.25$ $?$
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The molecularity of the reaction is:
  1. $1$
  2. $2$
  3. $3$
  4. $4$
  1. The expression for rate law is:
  1. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]$
  2. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]$
  3. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]^2$
  4. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]^2$
  1. The overall order of the reaction is:
  1. $2$
  2. $0$
  3. $1$
  4. $3$
  1. The value of rate constant is:
  1. $150.32\ M^{-2}\ min^{-1}$
  2. $200.08\ M^{-1}\ min^{-1}$
  3. $177.77\ M^{-2}\ min^{-1}$
  4. $155.75\ M^{-1}\ min^{-1}$
  1. The initial rate of disappearance of $Cl_2$ in experiment $4$ is:
  1. $1.75\ M\ min^{-1}$
  2. $3.23\ M\ min^{-1}$
  3. $2.25\ M\ min^{-1}$
  4. $2.77\ M\ min^{-1}$