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Biomolecules — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryBiomolecules4 Marks
Question
What is fructose? Explain the structure of fructose.
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Answer

→Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide sucrose.
→It is natural monosaccharide found in fruits, honey and vegetables. In it's pure form it is used as a sweetner.
Structure of fructose:
→Fructose has the molecular formula $C _6 H _{12} O _6$.
→On the basis of it's reactions it was to contain a ketonic functional group at carbon number 2 and six carbons in straight chain as in the case of glucose.
It belongs to D-series and is a laevorotatory compound. It is appropriately written as D(-) fructose.
→Its open chain structure is as shown.
Image
→It also exists in two cyclic forms which are obtained by the addition ofOH at $C_5$ to theImage group.
→The ring, thus formed is a five membered ring and is named as furanose with analogy to the compound furan.
→Furan is a five membered cyclic compound with one oxygen and four carbon atoms.
Image
→The cyclic structures of two anomers of fructose are represented by Haworth structures as given.
Image

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Similar questions

Read the passage given below and answer the following questions:

When an aldehyde with no et-hydrogen reacts with concentrated aqueous NaOH, half the aldehyde is converted to carboxylic acid salt and other half is converted to an alcohol. In other words, half of the reactant is oxidized and other half is reduced. This reaction is known as Cannizzaro reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives:
  1. Benzyl alcohol and sodium formate.
  2. Sodium benzoate and methyl alcohol.
  3. Sodium benzoate and sodium formate.
  4. Benzyl alcohol and methyl alcohol.
  1. Which of the following compounds will undergo Cannizzaro reaction?
  1. CH3CHO
  2. CH3COCH3
  3. C6H5CHO
  4. C6H5CH2CHO
  1. Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compounds is:
  1. 2, 2, 2-trichloroethanol.
  2. Trichloromethanol.
  3. 2, 2, 2-trichloropropanol.
  4. Chloroform.
  1. In Cannizzaro reaction given below:

$2\text{PhCHO}\xrightarrow{\stackrel{-}{\hbox{ OH}}}\text{PhCH}_2+\text{OH}+\text{PhCO}_2^-$ the slowest step is:

  1. The attack -OH at the carboxyl group.
  2. The transfer of hydride to the carbonyl group.
  3. The abstraction of proton from the carboxylic group.
  4. The deprotonation of PhCH2OH.
  1. Which of the following reaction will not result in the formation of carbon-carbon bonds?
  1. Cannizzaro reaction.
  2. Wurtz reaction.
  3. Reimer-Tiemann reaction.
  4. Friedel-Crafts' acylation.
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Read the passage given below and answer the following questions:

(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. Isomers (A) and (C) give positive Tollen's test whereas isomer (B) does not give Tollen's test but gives positive iodoform test. Isomers (A) and (B) on reduction with $\frac{\text{Zn(Hg)}}{\text{conc.}}.$ HCl give the same product (D).

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Compound A is:

  1. $\text{CH}_3-\text{CH}-\text{CHO}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

  2. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$

  3. $\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_3$

  4. None of these.
  1. Compound (C) is:
  1. Iso-butyraldehyde
  2. Butyraldehyde
  3. Crotonaldehyde
  4. Acrolein
  1. Compound (B) can be obtained by:

  1. $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_2-\text{CH}_3\xrightarrow[333\text{K}]{\text{dil.H}_2\text{SO}_4+\text{HgSO}_4}$

  2. $(\text{CH}_3\text{CH}_2\text{COO})_2\text{Ca}\xrightarrow{\text{Dry distill}}$

  3. $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3\xrightarrow[\frac{\text{H}_2\text{O}_2}{\text{NaOH}}]{\frac{\text{B}_2\text{H}_6}{\text{THF}}}$

  4. $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\xrightarrow[\frac{\text{ZN}}{\text{H}_2\text{O}}]{\text{O}_3}$

  1. Out of (A), (B) and ( C) isomers, which one is least reactive towards addition of HCN?
  1. A
  2. B
  3. C
  4. All are equally reactive.
  1. What will be the product when (B) reacts with ethylene glycol in presence of HCl gas?
  1.  

  1.  

  1.  

  1. None of these.
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Number of molecules which must collide simultaneously to give product is called molecularity. It is equal to sum of coefficients of reactants present in stoichiometric chemical equation.

For reaction, m1A + m2B → Product

Molecularity = [m1 + m2]

ln complex reaction each step has its own molecularity which is equal to the sum of coefficients of reactants present in a particular step. Molecularity is a theoretical property. Its value is any whole number. Number of concentration terms on which rate of reaction depends is called order of reaction or sum of powers of concentration terms present in the rate equation is called order of reaction.

If rate equation ofreaction is: Rate $=\text{k}\cdot\text{C}^{\text{m}_1}_\text{A}\cdot\text{C}^{\text{m}_2}_\text{B}$

Then order of reaction = m1 + m2.

ln simple reaction, order and molecularity are same.

ln complex reaction, order of slowest step is the order ofover all reaction. This step is known as rate determining step. Order is an experimental property. Its value may be zero, fractional or negative.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Higher order(> 3) reactions are rare due to:
  1. Shifting of equilibrium towards reactants due to elastic collisions.
  2. Loss of active species on collision.
  3. Low probability of simultaneous collision of all the reacting species.
  4. Increase in entropy and activation energy as more molecules are involved.
  1. The molecularity of the reaction:

$6\text{FeSO}_4+3\text{H}_2\text{SO}_4+\text{KClO}_3\rightarrow\text{KCl}+3\text{Fe}_2(\text{SO}_4)_3+3\text{H}_2\text{O}$ is:

  1. 6
  2. 10
  3. 3
  4. 7
  1. Which of the following statements is false in the following?
  1. Order of a reaction may be even zero.
  2. Molecularity of a reaction is always a whole number.
  3. Molecularity and order always have same values for a reaction.
  4. Order of a reaction depends upon the mechanism of the reaction.
  1. The rate of reaction, A + 2B → products, is given by the following equation:

$-\frac{\text{d}[\text{A}]}{\text{dt}}=\text{k}[\text{A}][\text{B}]^2$

If B is present in large excess, the order of the reaction is:

  1. Zero
  2. First
  3. Second
  4. Third
  1. The rate of the reaction, A + B + C → products, is given by $\text{r}=\frac{\text{d}[\text{A}]}{\text{dt}}=\text{k}[\text{A}]^\frac{1}{2}[\text{B}]^\frac{1}{3}[\text{C}]^\frac{1}{4}.$ The order of the reaction is:

  1. $\frac{1}{3}$​​​​​​​

  2. $\frac{1}{4}$

  3. $\frac{1}{2}$

  4. $\frac{13}{12}$

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Read the passage given below and answer the following questions:

RCONH2 is converted into RNH2 by means of Hoffmann bromamide degradation. During the reaction amide is treated with Br2 and alkali to get amine. This reaction is used to descend the series in which carbon atom is removed as carbonate ion $(\text{CO}^{2-}_3)$ Hoffmann bromide degradation reaction can be written as:

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Hoffmann bromamide degradation is used for the preparation of
  1. Primary amines.
  2. Secondary amines.
  3. Tertiary amines.
  4. Secondary aromatic amines.
  1. Which is the rate determining step in Hoffmann bromamide degradation?
  1. Formation of (i)
  2. Formation of (ii)
  3. Formation of (iii)
  4. Formation of (iv).
  1. Which of the following are used for the conversion of (i) to (ii)?
  1. KBr
  2. KBr + CH3ONa
  3. KBr + KOH
  4. Br2 + KOH
  1. Identify Bin the following reaction.

$\text{R}-\text{C}\equiv\text{N}\xrightarrow[\text{(Partially hydrolysis) }]{\text{Cone. HCI}}\text{A}\xrightarrow{\frac{\text{Br}_2}{\text{KOH}}}\text{B}$

  1. RCONH2
  2. RNH2
  3. RNHBr
  4. R = N = C = O
  1. What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hoffmann bromamide degradation?

 

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Standard electrode potentials are used for various processes:

  • It is used to measure relative strengths of various oxidants and reductants.
  • It is used to calculate standard cell potential.
  • It is used to predict possible reactions.
A set of half-reactions (in acidic medium) along with their standard reduction potential, Eº (in volt) values are given below:

I2 + 2e- → 2I- ; Eº = 0.54 V

Cl2 + 2e- → 2Cl- ; Eº = 1.36 V

Mn3+ +e- → Mn2+ ; Eº = 1.50 V

Fe3+ + e- → Fe2+ ; Eº = 0.77 V

O2 + 4H+ + 4e- → 2H2O ; Eº = 1.23 V

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following statements is correct?
  1. Cl- is oxidised by O2.
  2. Fe2+ is oxidised by iodine.
  3. I- is oxidised by chlorine.
  4. Mn2+ is oxidised by chlorine.
  1. Mn3+ is not stable in acidic medium, while Fe3+ is stable because:
  1. O2 oxidises Mn2+ to Mn3+
  2. O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+
  3. Fe3+ oxidises H2O to O2
  4. Mn3+ oxidises H2O to O2
  1. The strongest reducing agent in the aqueous solution is:
  1. I-
  2. Cl-
  3. Mn2+
  4. Fe2+
  1. The emf for the following reaction is:

$\text{I}_2+\text{KCl}\rightleftharpoons2\text{KI}+\text{Cl}_2$

  1. -0.82 V
  2. +0.82 V
  3. -0.73 V
  4. +0.73 V
  1. Which of the following statements is correct for the following reaction?

Fe3+ + Mn2+ → Fe2+ + Mn3+

  1. The emf of the cell is positive.
  2. Fe3+ oxidises Mn2+.
  3. The reaction does not occur.
  4. All are correct.
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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 V2O3 to less basic V2O4 and to amphoteric V2O5· V2O4 dissolves in acids to give VO2+ 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. V2O3
  3. VO2
  4. V2O5
  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 VF5.
  2. With chlorine vanadium can form VCl5.
  3. Vanadium exhibits highest oxidation state in oxohalides VOCl3, VOBr3 and fluoride VF5.
  4. With iodine vanadium cannot form Vl5 due to oxidising power of V5+ and reducing nature of I-.
  1. The oxidation state of vanadium in V2O5 is:
  1. $\frac{+5}{2}$

  2. +7
  3. +5
  4. +6
  1. Identify the oxidising agent in the following reaction.

V2O5 + 5Ca → 2V + 5CaO

  1. V2O5
  2. Ca
  3. V
  4. None of these.
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Read the passage given below and answer the following questions:

Ethers are readily cleaved by HI or HBr at 373K to form an alcohol and an alkyl halide.

$\text{R}-\text{O}-\text{R}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{R}-\text{OH}$

$\text{R}-\text{OH}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{H}_2\text{O}$

Mixed ether, containing primary or secondary alkyl group, when heated with hydrogen halide, the lower alkyl group forms halide and higher will form an alcohol. Tertiary alkyl ether when heated with hydrogen halide gives tertiary alkyl halide.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?

  1. $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}- \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 $

  2. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

  3. $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{O}-\text{CH}_3$

  4. $\text{CH}_3-\text{CH}_2-\text{CH}-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

  1. When CH2 = CH - O - CH2 - CH3 reacts with one mole of HI, one of the products formed is:
  1. Ethane.
  2. Ethanol.
  3. Iodoethene.
  4. Ethanal.
  1. (CH3)3COCH3 and CH3OC2H5 are treated with hydroiodic acid. The fragments obtained after reactions are respectively:

  1. (CH3)3CI + CH3OH; CH3I + C2H5OH

  2. (CH3)3CI + CH3OH; CH3OH + C2H5I

  3. (CH3)3COH + CH3I; CH3OH + C2H5I

  4. CH3I + (CH3)3COH; CH3I + C2H5OH

  1. Which of the following ether is unlikely to be cleaved by hot cone. HBr?

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Read the passage given below and answer the following questions:

Williamson's synthesis is used for the preparation of symmetrical as well as unsymmerical ether. It is SN2 reaction mechanism. In Williamson's synthesis, 1º alkyl halide are used for preparation of ethers because 2º and 3º alkyl halide give alkene. Ethers are cleaved by hydrogen halides to alcohol and alkyl halide where alkyl halide is corresponding to that alkyl which has less number of carbon atom (it is because of less steric hindrance). In polar media unsymmetrical ether like tertiary butyl ethyl ether gives ethyl alcohol and tertiary butyl halide as reaction proceeds via carbocation.

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: Rate of reaction of alkyl halide in Williamson's synthesis reaction is 1ºRX > 2ºRX > 3ºRX.

Reason: It is a type of bimolecular substitution reaction (SN2).

  1. Assertion: T-Butyl methyl ether is not prepared by the reaction of t-butyl bromide with sodium methoxide.

Reason: Sodium methoxide is a weak nucleophile.

  1. Assertion: Williamson's synthesis method cannot be used for preparing diphenyl ether.

Reason: Aryl halides do not undergo nucleophilic substitution easily.

  1. Assertion: When isopropyl bromide is treated with sodium isopropoxide, di-isopropyl ether is obtained as a major product.

Reason: With secondary alkyl halides, both substitution and elimination occur.

  1. Assertion: Both symmetrical and unsymmetrical ethers can be prepared by Williamson's synthesis.

Reason: Williamson's synthesis is an example of nucleophilic substitution reaction.

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The electrochemical cell shown below is concentration cell.

M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm-3) |M The emfof the cell depends on the difference in concentrations of M2+ 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 (Ksp' mol3 dm-9) of MX2 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 × 10-15
  2. 4 × 10-15
  3. 3 × 10-12
  4. 1 × 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 foUowing 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 × 1012
  2. 6.5 × 1010
  3. 5.2 × 109
  4. 3.4 × 1012
  1. The solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of the cell

Ag|Ag+ (satd. Ag2CrO4 soln) || Ag+ (0.1 M) | Ag

is 0.164V at 298 K, is:

  1. 3.359 × 10-12 mol3 L-3
  2. 2.287 × 10-12 mol3 L-3
  3. 1.158 × 10-12 mol3 L-3
  4. 4.135 × 10-12 mol3 L-3
  1. To calculate the emf of the cell, which of the foUowing options is correct?
  1. emf = Ecathode - Eanode
  2. emf = Eanode - Ecathode
  3. emf = Eanode + Ecathode
  4. None of these.
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Read the passage given below and answer the following questions:

The addition reaction of enol or enolate to the carbonyl functional group of aldehyde or ketone is known as aldol addition. The $\beta-$ hydroxyaldehyde or $\beta-$hydroxyketone so obtained undergo dehydration in second step to produce a conjugated enone. The first part of reaction is an addition reaction and the second part is an elimination reaction. Carbonyl compound having $\propto-$hydrogen undergoes aldol condensation reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Condensation reaction is the reverse of which of the following reaction?
  1. Lock and key hypothesis.
  2. Oxidation.
  3. Hydrolysis.
  4. Glycogen formation.
  1. Which of the following compounds would be the main product of an aldol condensation of acetaldehyde and acetone?
  1. CH3CH = CHCHO
  2. CH3CH = CHCOCH3
  3. (CH3)2C = CHCHO
  4. (CH3)2C = CHCOCH3
  1. Which combination of carbonyl compounds gives phenyl vinyl ketone by an aldol condensation?

  1. Acetophenone and Formaldehyde.
  2. Acetophenone and acetaldehyde.
  3. Benzaldehyde and acetaldehyde.
  4. Benzaldehyde and acetone.
  1. Which of the following will undergo aldol condensation?
  1. HCHO
  2. CH3CH2OH
  3. C6H5CHO
  4. CH3CH2CHO
  1. Which of the following does not undergo aldol condensation?
  1. CH3CHO
  2. CH3CH2CHO
  3. CH3COCH3
  4. C3H2CHO
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