Chemistry Ii Chapter 11 Alcohols, Phenols And Ethers
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    NCERT Solution For Class 12 Chemistry Chemistry Ii

    Alcohols, Phenols And Ethers Here is the CBSE Chemistry Chapter 11 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Chemistry Alcohols, Phenols And Ethers Chapter 11 NCERT Solutions for Class 12 Chemistry Alcohols, Phenols And Ethers Chapter 11 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 12 Chemistry.

    Question 1
    CBSEENCH12008013
    Question 7
    CBSEENCH12008019

    Identify allylic alcohols in the above examples.

    Solution

    Allylic alcohols are:

    Question 8
    CBSEENCH12008020

    Name the following compounds according to IUPAC system.


    Solution

    3-Chloromethyl-2-isopropyI pentan-1 -ol.
    Question 9
    CBSEENCH12008021

    Name the following compounds according to IUPAC system.




    Solution

    2, 5-Dimethylhexane-1,3-diol

    Question 10
    CBSEENCH12008022

    Name the following compounds according to IUPAC system.





    Solution

    3-Bromocyclohexanol

    Question 11
    CBSEENCH12008023
    Question 12
    CBSEENCH12008024

    Name the following compounds according to IUPAC system.







    Solution
    2-Bromo-3-methylbut-2-en-1-ol

    Question 15
    CBSEENCH12008027

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    Question 18
    CBSEENCH12008030

    Give structures of the products you would expect when each of the following alcohol reacts with,
    (a) HCl—ZnCl2 (b) HBr and (c) SOCl2.

    (i)  Butan-1-ol      (ii) 2-Methyl butan-2-ol.

    Solution

    (HCl—ZnCl2) is lucas reagent, lucas reagent react with alcohol however it does not react with primary alcohol but readily gives turbidity with tertiary alcohols.
    a) Reaction with (HCl—ZnCl2):
    Butan-1-ol is primary alcohol thus no reaction occur.
    where as 2-methyl butan-2-ol is tertiary alcohol, it forms 2-chloro-2-methylbutane.


    b) Reaction with HBr:

    c) Reaction with SOCl2:
    Question 20
    CBSEENCH12008032

    Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of corresponding phenoxide ions.

    Solution
    The electron withdrawing groups are more effective in increasing the acidic strength at the para-position relative to the ortho position because of greater dispersal of charge on oxygen atom.
    Resonance structure of phenoxide ion:

    Resonance structure of paranitrophenol:


    Resonance structures of ortho nitro phenol:


    Thus, presence of nitro group at ortho and para position incerease the acidic character.

    Question 21
    CBSEENCH12008033
    Question 22
    CBSEENCH12008034

    Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methyl pentan-2-ol.

    Solution

    It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.


    Question 23
    CBSEENCH12008035

    Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitro-benzene and why?

    Solution

    In structure (a), group sodium methoxide (CH3ONa) is a strong nucleophile also it is a strong base. Hence, an elimination reaction predominates over a substitution reaction.

    Question 24
    CBSEENCH12008036
    Question 25
    CBSEENCH12008037
    Question 26
    CBSEENCH12008038
    Question 27
    CBSEENCH12008039
    Question 28
    CBSEENCH12008040

    Write IUPAC name of the following compounds:


    Solution
    2, 5-Dimethyl phenol.
    Question 29
    CBSEENCH12008041

    Write IUPAC name of the following compounds:




    Solution
    2-Methoxy propane.
    Question 30
    CBSEENCH12008042

    Write IUPAC name of:

    Solution
    3-Methyl butaoxy benzene.
    Question 31
    CBSEENCH12008043

    Write IUPAC name of:
    CH3O—CH2—CH—OCH3

    Solution
    1, 2-Dimethoxy ethane.
    Question 32
    CBSEENCH12008044

    Name of the following compounds in IUPAC system:




    Solution
    2-chloro-4-methyl phenol.
    Question 33
    CBSEENCH12008045

    Name of the following compounds in IUPAC system:




    Solution
    2, 4, 6-Trinitrophenol.
    Question 34
    CBSEENCH12008046

    Give the structure of 2-nitro-4-acetyl phenol.

    Solution

    2-Nitro-4 acetyl phenol

    Question 35
    CBSEENCH12008047

    Give the structure of 2-methoxy phenol.

    Solution

    2-methoxy phenol

    Question 36
    CBSEENCH12008048

    Give IUPAC name of the following compound:




    Solution
    Its IUPAC name is 2, 2, 3-trimethyl pentan-1-ol.
    Question 37
    CBSEENCH12008049

    Write the structure and IUPAC name of isobutyl alcohol.

    Solution
    IUPAC name is 2-methyl propan-1-ol.

    Question 38
    CBSEENCH12008050

    What is the common name of the following alcohol?
    (CH3)2CHCH2OH.

    Solution
    Common name of 2 methyl -1- propanol is Isobutyl alcohol .
    Question 39
    CBSEENCH12008051

    What is the common name of the following alcohol?
    (CH3)3COH

    Solution
    The common name of 2-methylpropan-2-ol is tert-Butyl alcohol.

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    Question 40
    CBSEENCH12008052
    Question 42
    CBSEENCH12008054

    Write the structure of compound A in the following reactions:
    left parenthesis CH subscript 3 right parenthesis subscript 2 CHCH subscript 2 CHO space rightwards arrow with LiAlH subscript 4 on top space straight A

    Solution
    A is 3-methyl butanol (CH3)2CHCH2CH2OH.
    Question 43
    CBSEENCH12008055

    Write the structure of compound A and B in the following reactions:



    Solution

    When benzaldehyde react with sodium bromide Benzyl alcohol is formed .
     

    Question 44
    CBSEENCH12008056

    Why are alcohols soluble in water?

     

    Solution
    Alcohols are soluble in water because alcohol molecules engage hydrogen bonds with water molecules.

    Question 45
    CBSEENCH12008057

    Why phenol is stronger acid than ethanol?

    Solution

    Acidity of phenols: The reactions of phenol with metals (e.g., sodium, aluminium) and sodium hydroxide indicate its acidic nature. The hydroxyl group, in phenol is directly attached to the sp2 hybridised carbon of benzene ring which acts as an
    electron withdrawing group. Due to this, the charge distribution in phenol molecule, as depicted in its resonance structures, causes the oxygen of –OH group to be positive.

    In case of ethanol, no such resonace structure is shown as phenol. Hence,  phenol is stronger acid than ethanol.

    Question 46
    CBSEENCH12008058

    The boiling points of phenols are higher than the corresponding arenes and haloarenes. Give reason.

    Solution

    The high boiling points of alcohols are mainly due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

    Question 47
    CBSEENCH12008059

    Phenols are sparingly soluble in water. Why?

    Solution

    Phenol is soluble in water to some extent. It is due to its ability to form hydrogen bonding with water molecules. However the large part of phenol molecule is phenyl group that is non polar and hence its solubility if limited in water.

    Question 48
    CBSEENCH12008060

    What happens when 3-methyl butanone is reduced with sodium tetrahydrido borate(III)?

    Solution
    When  3-methyl butanone is reduced with sodium tetrahydrido borate(III) 3-methyl butan-2-ol is formed.

    Question 49
    CBSEENCH12008061
    Question 50
    CBSEENCH12008062
    Question 51
    CBSEENCH12008063
    Question 54
    CBSEENCH12008066

    Compare the relative boiling points of:


    Solution

    Phenol has a higher boiling point than thio phenol although thio phenol has higher molecular mass. This is due to the presence of intermolecular hydrogen bonding in phenol.

    Question 55
    CBSEENCH12008067

    Give the equation of reaction of preparation of phenol from cumene.

    Solution

    Cumene (isopropylbenzene) is oxidised in the presence of air to cumene hydroperoxide. It is converted to phenol and acetone by treating it with dilute acid.


    Question 56
    CBSEENCH12008068

    Compare the relative boiling point of toluene and phenol.

    Solution
    The molecular mass of toluene and phenol are approximately the same such as mass of toluene is 92.14g and mass of phenol is 94.112g. But the phenol has a higher boiling point than toluene because phenol have ability to form intermolecular hydrogen bonding but toluene does not form intermolecular hydrogen bonding.

    Question 57
    CBSEENCH12008069

    You are given benzene, conc.  straight H subscript 2 SO subscript 4 and NaOH. Write the equations for the preparation of phenol using these reagents.

    Solution

    First benzene react with sulphuric acid to give benzensulfonic acid and it react further with sodium hydroxide and sodium benzosulfonate is formed. when benzosulfonate is fused  by sodium hydroxide phenol is obtaine.

    Question 58
    CBSEENCH12008070

    Convert aniline into phenol.

    Solution
    First convert aniline to diazonium salt using NaNO2/HCl at 0-5 degree C, Then diazonium ion reacts with the water at 433K to produced phenol

    Question 60
    CBSEENCH12008072

    Which is more acidic and why? p-chlorophenol or p-nitrophenol?

    Solution

    Both the chlorine and the nitro group at the para position makes the phenol group acidic because they are both electron withdrawing groups that stabilize the negative charge  of phenol. However, the fact that the nitro groups is electron withdrawing goes resonance, allows better stabilize the negative charge of the oxygen then chlorine alone. Thus, p-nitrophenol is more acidic than p-chlorophenol.

    Question 61
    CBSEENCH12008073

    Arrange the following compounds in increasing order of their acid strength.
    Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitro-phenol, 3, 5-dinitrophenol, phenol, 4-methyl phenol.

    Solution

    When any phenol loses its acidic proton, it forms an oxygen with three lone pairs. In phenol, these lone pairs can be dislocated over the aromatic ring, providing resonance stability. This added stability allows for the easier dissociation of phenols. Thus, order of acidity is ,

    Propan-1-ol < 4-methyl phenol, phenol <3-nitrophenol < 3, 5-dinitrophenol < 2, 4, 6-tri nitro-phenol.
    Question 62
    CBSEENCH12008074

    Why has tertiary alcohol higher reactivity towards hydrogen halide compared to secondary and primary alcohols?

    Solution

    Tertiary alcohol has higher reactivity towards hydrogen halide as compared to secondary because carbonium ion formed from tertiary alcohol is the most stable and reaction  go via SN1 reaction. Therefore, order of reaction is,
    Tertiary alcohol > Secondary alcohol > Primary alcohol.

    Question 63
    CBSEENCH12008075

    State what happens when isopropyl chloride is boiled with alcoholic potassium hydroxide solution.

    Solution
    when isopropyl chloride is boiled with alcoholic potassium hydroxide solution, 1-propene is formed


    Question 64
    CBSEENCH12008076

    Give reason: nitration of phenol gives ortho and para products only.

    Solution
    The -OH group present to the benzene ring direct the incoming group to ortho and para position in the ring as these position become electron rich due to the resonance eggect caused by -OH group. Hence,On nitration phenol gives ortho and para product.

    Question 66
    CBSEENCH12008078

    Write the equation for the industrial synthesis of methanol.

    Solution

    Industrial synthesis of methanol is given by:
    CO plus 2 straight H subscript 2 space rightwards arrow from 400 degree straight C comma space 200 space atm to ZnO divided by CrO subscript 3 of CH subscript 2 OH

    Question 67
    CBSEENCH12008079

    Write the structure of the major product and also label the product as 1°, 2° or 3° alcohol.

    Solution

    The product is 1 methylcyclopentanol

    It is 3° alcohol.

    Question 68
    CBSEENCH12008080

    Name the enzymes which convert sucrose into glucose and fructose and finally into ethanol.

    Solution

    (i) Invertase converts sucrose into glucose and fructose.
    (ii) Zymase converts glucose into ethanol.

    Question 69
    CBSEENCH12008081

    What is the expected product from the reaction of



    with
    (i) LiAlH4,    (ii) H2/Pt?

    Solution

    Lithium aluminium hydride is used as a reducing agent. It convert ketones into the corresponding alcohols. Whereas H2/Pt convert alkene to alkane. Thus product are;

    Question 70
    CBSEENCH12008082

    Arrange the following compounds in order of increasing boiling points:
    Penta-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

    Solution

    Boiling point of alcohol increase with increase in the
    number of carbon atoms (increase in van der Waals forces). Thus, order of increasing boiling point,

    Methanol < ethanol < propan-1 -ol < butan-2-ol < butan-1-ol < pentan-1-ol.
    Question 71
    CBSEENCH12008083

    Arrange the following compounds in increasing order of boiling point and account for the order:
    (a) 1-pentanol, (b) 2-methyl-2-butanol, (c) 3-methyl-butanol-2.

    Solution

    In alcohols, the boiling points decrease with increase of branching in carbon chain, because of decrease in van der Waals forces with decrease in surface area. Thus the decreasing order of boiling point is A > C > B

    Question 72
    CBSEENCH12008084

    Why is ethanol less acidic than water?

    Solution

    Ethanol is less acidic than water because of the electron donating nature of the alkyl group attached to the oxygen which destabilizes the hydroxide ion and make it tougher for the oxygen to donate the proton. Hence, water is more acidic than ethanol.

    Question 73
    CBSEENCH12008085

    Compare the acidity of Cl—CH2 CH2OH and CH3CH2OH. Explain.

    Solution
    Cl—CH2CH2OH is a stronger acid than CH3CH2OH because the chlorine is electronegtive element, therefore, chlorine attract all electron from the compound and O-H bond become weaker thus easily available for reaction. Hence it is more acidic than ethanol.

    Question 74
    CBSEENCH12008086

    Arrange the following in order of decreasing basic strength:
    (CH3)3(CO, CH3O, (CH3)2CHO

    Solution
    The correct order is
    CH3O<  (CH3)3CO- < (CH3)2CHO 
    Question 75
    CBSEENCH12008087

    Compare and explain the relative Bronsted basicities of 1°, 2° and 3° alcohols in the liquid state.

    Solution
    Bronsted base is H+ ion acceptor, therefore, more of the electron releasing alkyl groups at α-carbon, more the electron density on oxygen atom and more the tendency to accept H+ ion. Hence, the basicity of alcohol is in order of,
    3° > 2° > 1°
    Question 77
    CBSEENCH12008089

    What happens when ethanol reacts with Na metal? Name the products formed.

    Solution

    When ethanol reacts with Na metal sodium ethoixde.
    2 straight C subscript 2 straight H subscript 5 OH plus space 2 Na rightwards arrow space space stack stack 2 straight C subscript 2 straight H subscript 5 straight O with minus on top Na with plus on top with Sodium space ethoxide below plus straight H subscript 2 with space below

    Question 78
    CBSEENCH12008090

    Arrange the following alcohols in order of increasing reactivity towards Na metal:

    Solution
    The reactivity of alcohol towards Na increases as the acidic strength of alcohol increases. The correct order is

    Question 79
    CBSEENCH12008091

    Explain why sodium (Na) may be used to remove the last trates of H2O from benzene but not from ethanol?

    Solution
    Ethanol has an acidic hydrogen and reacts with Na and give sodium ethoxide ion, whereas benzene has not any acidic hydrogen thus it does not react with benzene, thus when it react with benzene sodium react with water forming sodium hydroxide along with hydrogen. Hence used to remove the last trates of water.
    straight C subscript 2 straight H subscript 5 OH plus Na space rightwards arrow space space space straight C subscript 2 straight H subscript 5 straight O to the power of minus Na to the power of plus space plus space 1 half straight H subscript 2

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    Question 80
    CBSEENCH12008092

    Give the major product of the reaction of ethanol with:
    (i) Ethanoic anhydride (CH3CO)2O.
    (ii) Ethanol chloride in the presence of pyridine.

    Solution

    i) Ethanol react with ethanoic anhydride to form esters.
    CH3CH2OH+  (CH3CO)2O → CH3OCOC2H5

    ii) When ethanol react with ethanol chloride in the presence of pyridine, it also form esters.
    CH3CH2OH +CH3COCl → CH3OCOC2H5

    Question 81
    CBSEENCH12008093

    Convert-propanol into 1-bromo propane.

    Solution
    CH3CH2CH2OH + PBr3  → CH2CH2CH2Br + H3PO3
    Question 82
    CBSEENCH12008094

    What is the decreasing order of reactivity of hydrogen halides in the conversion of ROH into RX?

    Solution
    When alcohols react with a hydrogen halide, a substitution occurs, producing an alkyl halide and water.
    R-OH +H-X → R-X + H-OH
    The decreasing order of reactivity of hydrogen halide is HI > HBr > HCl
    Question 83
    CBSEENCH12008095

    Convert methyl bromide into ethyl alcohol.

    Solution

    When methyl bromide react with magnesium in presence ether Grignard regent is form, further when it is treated with formaldehyde in presence of hydronium ion gives ethyl alchol.

    Question 84
    CBSEENCH12008096

    How will you prepare propan-2-ol from bromomethane?

    Solution

    When bromomethane is treated with magnesium in presence of ether Grignard regent is form, further reaction with ethanal in presence of hydronium ion, propan-2-ol formed.

    Question 85
    CBSEENCH12008097

    Arrange the following alcohols in decreasing order of reactivity towards HBr.

    Solution

    As the number of electron donating R groups at α-carbon increases, the reactivity towards HX increases. The correct order is,
    (A) > (C) > (D) > (B) .

    Question 86
    CBSEENCH12008098
    Question 87
    CBSEENCH12008099

    Write the name and sturcture of the product formed when propene dissolves in cold conc. H2SO4.

    Solution

    when propene dissolves in cold conc. H2SO4 
    isopropyl hydrogen - sulphate.

    Question 88
    CBSEENCH12008100

    Arrange the following in order of decreasing ease of dihydration:

    (CH3)3COH, (CH3)2CHOH, CH3CH2OH

    Solution

    The relative ease of dehyration of alcohols follows the following order:
    Tertiary > Secondary > Primary

    Question 89
    CBSEENCH12008101

    Convert t-butyl alcohol into isobutyl bromide.

    Solution

    Conversion of tert-butyl alcohol to isobutyl bromide.
    .

    Question 90
    CBSEENCH12008102

    Prepare propan-2-ol from propan-1-ol

    Solution

    When propan-2-ol is treated with conc. sulphuric acid propene is formed further reaction with Diborane in presence of water propan-1-ol is obtained

    Question 91
    CBSEENCH12008103
    Question 93
    CBSEENCH12008105

    Write the structure of the oxidation product of the following reactions:







    Solution

    (ii) No reaction, there is no H atom at straight alpha-carbon.
    Question 94
    CBSEENCH12008106

    Give a chemical test to distinguish methanol and ethanol.

    Solution
    Methanol and ethanol can be distinguished by using iodoform test. Ethanol gives a yellow precipitate when treated with I2/NaOH while methanol does not give.

    Question 95
    CBSEENCH12008107

    How can we distinguish between, n-butanol and t-butanol using acidified potassium dichromate?

    Solution

    Acidified K2Cr2O7 is orange in colour. When it oxidises alcohol, its colour changes to bluish green.
    n-butanol is 1° alcohol and gives a bluish-green colour while butanol is 3° alcohol, does not react with K2Cr2O7.

    Question 97
    CBSEENCH12008109

    What happens to 2-methyl propane-2-ol, when it is heated at 573 K with copper?

    Solution
    2-methyl propan-2-ol undergo dehydration with Cu at 573 K resulting in 2- methyl propene.

    Question 98
    CBSEENCH12008110

    Two test tubes A and B containing 2-methyl-2-propanol and 2-methyl propanol-1 respectively. In which test tube the turbidity appears when lucas reagent is added?

    Solution
    The turbidity will appear only in test tube A which contains a tertiary alcohol. Test tube B does not produce any turbidity as it contains primary alcohol.
    Question 99
    CBSEENCH12008111
    Question 100
    CBSEENCH12008112

    Give the structure and IUPAC name of the product from the reaction of phenol with CHCl3 in NaOH followed by acidification.

    Solution

    When phenol react with trichloromethane in presence of sodium hydroxide, it forms 2- hydroxy benaldehyde.
    C6H5OH +CHCl3 +NaOH ---> 
    IUPAC name is 2-hydroxybenzaldehyde.

    Question 101
    CBSEENCH12008113

    What happens when dilute HNO3 is added to phenol? Name the method to separate the isomers obtained.

    Solution

    With dilute nitric acid at low temperature (298 K),
    phenol yields a mixture of ortho and para nitrophenols.


    The isomer can be separated either by steam distillation or fractional crystallization.
    Question 102
    CBSEENCH12008114

    Give the structure of the products resulting from the reaction of sodium phenoxide with CO2 at 4 atm and 400 K followed by addition of aqueous acid.

    Solution

    When sodium phenoxide react with CO2 at 4 atm and 400K in aqueous acid, salicylic acid is obtain.

    Question 103
    CBSEENCH12008115

    What happens when phenol is treated with chloroform in presence of NaOH at 343 K followed by hydrolysis?

    Solution

    When phenol is treated with chloroform in presence of NaOH at 343 K, salicyaldehyde is formed.

    Question 104
    CBSEENCH12008116

    Which of the following compound is more soluble in water and why? Ortho-nitrophenol or para nitrophenol?

    Solution

    p-nitrophenol is more soluble than ortho and para nitrophenol because P-nitrophenol has Inter-molecular Hydrogen Bonding  while o-nitrophenol has Intra-molecular Hydrogen Bonding.
    Now due to the Hydrogen Bonding more energy needs to be supplied to vaporize the compound as there are extra bonds other than that present bonds that hold the molecule.
    There are bonds between molecules in p-nitrophenol which leads to higher boiling point as more energy need to be supplied to separate the molecules. Whereas in o-nitrophenol the bonds exist in the molecule itself due to which there is virtually no change in the boiling point.

    Question 105
    CBSEENCH12008117

    While separating a mixture of ortho and para nitrophenols by stream distillation, name the isomer which is steam volatile. Give reason.

    Solution
    Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding. On other hand p- nitro phenol is less volatile due to having intermolecular hydrogen bonding which cause association of molecules.

    Question 106
    CBSEENCH12008118

    Complete the following:

    Solution

    i)

    ii)

    Question 107
    CBSEENCH12008119

    Name the reagent used in the conversion of phenol into 2, 4, 6-tribromophenol. Also write the equation.

    Solution

    When phenol is treated with bromine water,
    2,4,6-tribromophenol is formed as white precipitate.




    Question 108
    CBSEENCH12008120

    Convert phenol into chlorobenzene.

    Solution

    Phenol is converted to benzene on heating with zinc dust. Further reaction with chlorine followed by FeCl3 gives chlorobenzene.

    Question 109
    CBSEENCH12008121

    Name the electrophile in the Reimer Tiemann reaction and how is it formed?

    Solution
    The electrophile is dichloro carbon, CCl2. It is formed by the reaction of CHCl3 with NaOH.
    CHCl3 + NaOH  → NaCl + H2O + CCl2
    Question 110
    CBSEENCH12008122

    Complete the equation:

    Solution
    Question 111
    CBSEENCH12008123

    Discuss the structural relationship between water, alcohols and ether.

    Solution
    Alcohols and ethers are derivatives of water,

    Replacement of one H of H2O by R (alkyl) gives alcohols,

    The replacement of both hydrogen by R gives ether.
    Question 112
    CBSEENCH12008124

    Write the products of the following reaction:

    Solution
    The products are (CH3)2CHI and CH3OH.

    Question 113
    CBSEENCH12008125
    Question 114
    CBSEENCH12008126

    Name the ether used as a heat transfer medium.

    Solution
    Diphenyl ether used as a heat transfer medium.
    Question 115
    CBSEENCH12008127

    Do ethers have dipole moments? Explain.

    Solution
    In ethers (R-O-R) there are two lone pair of electrons on the oxygen atom which is responsible for ethers to have bent geometry.
    Yes, they are weakly polar. Ethers have a bent structure with a C—O—C bond angle of ~112° and the two C—O bond dipole moments do not cancel.



    Question 116
    CBSEENCH12008128

    Compare the solubility of diethyl ether and n-butane in water.

    Solution
    Solubility of substance in water depend on the polar nature of substance. Diethyl ether being weakly polar is capable of forming intermolecular hydrogen bonding with water. On other hand n-butane is hydrocarbon and it is not polar in nature. Hence, diethyl ether is soluble in water while n-butane is not.
    Question 117
    CBSEENCH12008129

    Give the product(s) from the reaction of one mole diethyl ether with
    (i) one mole of conc. HI and
    (ii) excess of HI.

    Solution
    i) when diethyl ether react with one mole of conc. HI, it form C2H5OH and C2H5I.
    CH3-CH2-O-CH2-CH3 +(conc.)HI ---> C2H5OH +C2H5I
    ii) When diethyl ether react with excess of HI, it form 2C5H5I.
    CH3-CH2-O-CH2-CH3 +(excess)2HI ---> 2C5H5I +H2O

    Question 118
    CBSEENCH12008130

    Why bromination of anisole takes place even in the absence of FeBr3 as a catalyst?

    Solution
    Anisole undergoes bromination with Br2 even in the absence of FeBr3 due to the activation of benzene ring by the methoxy group.


    Question 119
    CBSEENCH12008131

    Name the main product formed in Williamson’s synthesis.

    Solution

    Ether formed as main product in williamson's synthesis.
    R-X + R-O-Na  ---> R-O-R + NaX

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    Question 120
    CBSEENCH12008132

    Name one reagent that converts sodium phenoxide into anisole.

    Solution
    Dimethyl sulphate (CH3)2SO4.
    Question 121
    CBSEENCH12008133

    Why have ethers low boiling points?

    Solution
    Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points.
    Question 122
    CBSEENCH12008134
    Question 123
    CBSEENCH12008135
    Question 124
    CBSEENCH12008136

    How will you obtain diether ether from ethanol?

    Solution
    When ethanol is heated with conc. H2SO4 in a molar ratio of 2 : 1 at 140°C, diethyl ether is obtained.
    2 straight C subscript 2 straight H subscript 5 OH space rightwards arrow with straight H subscript 2 SO subscript 4 on top space stack straight C subscript 2 straight H subscript 5 OC subscript 2 straight H subscript 5 with diethyl space ether below plus space straight H subscript 2 straight O
    Question 125
    CBSEENCH12008137

    Does peroxide formation occur more rapidly with n-propyl ether or di-isopropyl ether?

    Solution
    Formation of peroxide occur more faster in di-isopropyl ether than n-propyl ether because secondary alkyl radical is more stable than primary alkyl radical.
    Question 126
    CBSEENCH12008138

    Why no flame should be allowed in the lab, while working with ether?

    Solution
    Ether is a highly volatile liquid. Its vapour forms explosive mixtures with air which catches fire readily.
    Question 127
    CBSEENCH12008139

    Why is  O—CH3, ortho and para directing in aromatic electrophilic substitution?

    Solution
    Oxygen of —OCH3 forms a partial double bond with carbon of benzene nucleus to which it is attached. –OCH3 is an electron donating group and activates the benzene nucleus and increases electron density at ortho and para position.
    Question 128
    CBSEENCH12008140

    Diethyl ether does not react with sodium. Explain.

    Solution
    Diethyl ether does not react with sodium because diethyl ether does not have acidic hydrogen.
    Question 129
    CBSEENCH12008141

    Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

    Solution
    The secondary and tertiary alcohol being highly crowded under go dehydration in acid presence to form alkene.

    Hence, preparing of ethers by acid dehydration of secondary or tertiary is not a suitable method.
    Question 130
    CBSEENCH12008142

    How is 1-propoxypropane synthesised from propan-1-ol?

    Solution
    By dehydration using H2SO4 at 413 K.

    Question 131
    CBSEENCH12008143

    Which of the following can easily undergo Williamson's synthesis to prepare ether?

    Solution
    Ethers containing substituted alkyl groups such as secondary and tertiary may also prepared by this method. But better results are obtained in primary alkyl halide. Thus, CH3CH2CH2Br  undergoes Williamson’s synthesis because it is 1° alkyl halide, straight alpha-carbon is less hindered while in (CH3)3CCH2Br (although it is 1°), straight alpha-carbon is highly crowded and difficult for alkoxide ion to attack.
    Question 133
    CBSEENCH12008145

    Name the ether used as inhalation an aesthetic. What are the disadvantages of using it?

    Solution
    Ethoxy ethane(CH3CH2OCH2CH3) is an aesthetic which is used as inhalation.
    Disadvantage:
    It has very slow effect and unpleasant recovery period.
    Question 134
    CBSEENCH12008146

    How would you differentiate between phenol and carboxylic acid by a simple chemcial test?

    Solution
    Phenol gives violet coloured complex with aqueous FeCl3, while carboxylic acid does not give such colour.
    Question 135
    CBSEENCH12008147

    Explain why is propanol has higher boiling point than that of hydrocarbon, butane?

    Solution
    Propanol has higher boiling point than butane because of wide spread H-bonding among propanol molecules. A lot of energy is required to break inter molecular H-bonding among propanol molecules. There is no such bonding in butane.

    Question 136
    CBSEENCH12008148

    Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

    Solution

    Alcohols can form H- bonds with due to presence of (-OH) group and break the H-bonds already existing between water molecules. Hence they are soluble in water.

    In contrast, hydrocarbon cannot form H-bonds with water and hence are insoluble in water.

     

    Question 137
    CBSEENCH12008149

    Of the two hydroxy organic compounds ROH and R'OH, the first one is basic and the other is acidic in behaviour. How is R different from R'?

    Solution
    R is an alkyl group (electron repelling). it shows basic behaviour whereas R' is an aryl group. It shows acidic behaviour (electron attracting).

    Question 138
    CBSEENCH12008150

    Why does phenol get coloured on long standing?

    Solution
    On long standing in contact with air, phenol gets slowly oxidised. Hence, it get coloured.
    Question 139
    CBSEENCH12008151

    How is the C—OH bond in phenol stabilised?

    Solution
    C—O bond acquires double bond character and gets stabilised by resonance in phenol molecule.

    Question 140
    CBSEENCH12008152

    Phenol is acidic but does not decompose NaHCO3 solution while carbonic acids like CH3COOH decompose NaHCO3 solution. Why?

    Solution
    Phenol is weaker acid than carbonic acids, Ka value of phenol is very less than that of carbonic acids like CH3COOH, C6H5COOH etc. Therefore phenol does not decompose NaHCO3 solution.
    Question 141
    CBSEENCH12008153

    What is wood alcohol? Why is it so called?

    Solution
    Methanol (CH3OH), is known as wood alcohol. It is so called as it was earlier prepared by heating wood in the absence of air.
    Question 142
    CBSEENCH12008154

    How is methanol (or wood spirit) obtained from carbon monoxide?

    Solution
    By catalytic hydrogenation.

    Question 143
    CBSEENCH12008155

    Write the main products obtained after the cleavage of methyl tert. butyl ether with HI.

    Solution
    The main product are methanol and tert. butyl iodide.
    Question 144
    CBSEENCH12008156

    How is t-butyl alcohol obtained from acetone?

    Solution

    When acetone react with methylmagnesiumbromide, addition product is formed which on reacting with water gives t-butyl alcohol.

    Question 145
    CBSEENCH12008157

    Phenols do not undergo substitutions at the carbon oxygen bond. Why?

    Solution
    Phenol do not undergo substitutions at carbon oxygen bond because carbon to which oxygen is attached-is sp2 hybridised.
    Question 146
    CBSEENCH12008158

    Of benzene and phenol which is more easily nitrated and why?

    Solution
    Phenol is more easily nitrated because of presence of -OH group on the ring activates it towards electrophillic reaction.
    Question 147
    CBSEENCH12008159

    Which bond of alcohol is cleaved during its reaction with (a) carboxylic acid (b) HCl.

    Solution
    (a) O—H  bond is cleaved with carboxylic aicd
    (b)  C—O bond is cleaved with HCl
    Question 148
    CBSEENCH12008160

    C—O bond is much shorter in phenol than in ethanol. Why?

    Solution
    The ionisation of an alcohol and a phenol takes place as follows.

    Due to resonance, C-bond in phenol acquires a partial double bond character. In Ethanol resonance is not possible.

    Question 149
    CBSEENCH12008161

    Why are ethers very little reactive chemically?

    Solution
    The C-O bonds in ehters are polar and thus, ethers have a net dipole moment. Therefore ethers are very little reactive.
    Question 150
    CBSEENCH12008162

    Alkoxy groups attached to benzene ring is ortho and para directing, Justify.

    Solution
    The alkoxy group (-OR) is ortho, para directing  and activates the aromatic ring towards  electrophillic  substitution.

    Question 151
    CBSEENCH12008163

    Sodium metal can be used for drying diethyl ether and benzene but not ethanol. Why? 

    Solution
    Diethyl ether and benzene does not have any replaceable alpha hydrogen. On other hand in ethanol have replaceable hydrogen atom therefore, reacts with sodium.
    Question 152
    CBSEENCH12008164
    Question 153
    CBSEENCH12008165

    How will you distinguish between 1-phenyl ethanol and 2-phenyl ethanol?

    Solution
    1-phenyl ethanol gives iodoform test, where as 2-phenyl ethanol does not give this test.

    Question 154
    CBSEENCH12008166

    How is phenol obtained from aniline?

    Solution

    A diazonium salt is formed by treating an aromatic primary amine with nitrous acid (NaNO2 + HCl) at 273-278 K. Diazonium salts are hydrolysed to phenols by warming with water or by treating with
    dilute acids.

    Question 155
    CBSEENCH12008167

    Explain giving reasons, why Phenol has a smaller dipole moment than methanol.

    Solution
    In phenol, there is double bond character between C—O bond and there is +ve charge on oxygen due to electron withdrawing benzene ring, therefore, dipole moment is less whereas in methanol there is no such positive charge, it has more dipole moment than phenol.

    Question 156
    CBSEENCH12008168

    If an alcohol is treated with excess of sulphuric acid, no esterification takes place. Why?

    Solution
    When alcohol is treated with excess of sulphuric acid there occurs dehydration of alcohol. Hence, no esterification takes place.
    straight C subscript 2 straight H subscript 5 OH space rightwards arrow with straight H subscript 2 SO subscript 4 space space space space on top space CH subscript 2 space equals CH subscript 2 space plus straight H subscript 2 straight O
    Question 157
    CBSEENCH12008169

    How many sigma bonds are present in 3-methyl phenol?

    Solution

    Structure of 3-methyl phenol

    6 (benzene ring) + 4(C—H of benzene ring) +1 (C of benzene ring and CH3 group) +3 (of CH3 group) + 1(C of benzene and O of OH) +1 (O and H of OH group) = 16.

    Question 158
    CBSEENCH12008170

    Give the equation of the reaction of bromine in carbon disulphide with phenol.

    Solution

    When the reaction is carried in the presence of CS2 at low temperature monobromophenols are formed.

    Question 159
    CBSEENCH12008171

    Ethanol and chloro ethane both are polar in nature yet ethanol is miscible in water while chloro ethane is immiscible. Explain.

    Solution
    Ethanol is miscible in water due to its ability to form hydrogen bonds with water.

    Chloroethane does not form hydrogen bonds with water, a result it is immiscible.
    Question 160
    CBSEENCH12008172

    Give IUPAC names of the following compounds:



    Solution
    4-chloro -2, 3-dimethyl pentan-1-ol.
    Question 161
    CBSEENCH12008173

    Give IUPAC names of the following compounds:





    Solution
    2-Ethoxy propane.
    Question 162
    CBSEENCH12008174

    Give IUPAC names of the following compounds:






    Solution
    2, 6-Dimethyl phenol.
    Question 163
    CBSEENCH12008175

    Give IUPAC names of the following compounds:








    Solution
    1-Ethoxy -2-nitrocyclohexane.
    Question 164
    CBSEENCH12008176

    Write IUPAC names of the following compounds:




    Solution
    2, 2, 4-Trimethyl pentan-3-ol.
    Question 165
    CBSEENCH12008177

    Write IUPAC names of the following compounds:


    Solution
    4-Ethyl hexane-1, 3-diol.
    Question 166
    CBSEENCH12008178

    Write IUPAC names of the following compounds:



    Solution
    Butane-2, 3-diol.
    Question 167
    CBSEENCH12008179

    Write IUPAC names of the following compounds:

    HOCH subscript 2 space minus space CH left parenthesis OH right parenthesis space minus space CH subscript 2 OH


    Solution
    Propane-1, 2, 3-triol.
    Question 168
    CBSEENCH12008180
    Question 169
    CBSEENCH12008181
    Question 170
    CBSEENCH12008182
    Question 172
    CBSEENCH12008184
    Question 173
    CBSEENCH12008185

    Give the IUPAC name for the following compound and also label alcohol as 1°, 2° and 3°.




    Solution

    The IUPAC name of the given compound is
    Cyclo 1- pentene methanol. It is primary alcohol.

    Question 174
    CBSEENCH12008186

    Give the IUPAC name for the following compound and also label alcohol as 1°, 2° and 3°.




    Solution

    The IUPAC name of given compound is 2- cyclopropylethanol.
    It is secondary alcohol.

    Question 175
    CBSEENCH12008187

    Write IUPAC names of the following compounds:



    Solution
    2-Methyl phenol.
    Question 176
    CBSEENCH12008188

    Write IUPAC names of the following compounds:







    Solution
    2, 5-Dimethyl phenol.
    Question 177
    CBSEENCH12008189

    Write IUPAC names of the following compounds:







    Solution
    1 -Methoxy-2-methyl propane.
    Question 178
    CBSEENCH12008190

    Write IUPAC names of the following compounds:

    straight C subscript 6 straight H subscript 5 OC subscript 7 straight H subscript 15






    Solution
    Phenoxy heptane
    Question 179
    CBSEENCH12008191

    Write IUPAC names of the following compounds:









    Solution
    2-Ethoxy butane.
    Question 180
    CBSEENCH12008192
    Question 184
    CBSEENCH12008196

    Give the structures and IUPAC names of the products expected from the following reactions:
    (a)    Catalytic reduction of butanal.
    (b)    Hydration of propene in the presence of dilute sulphuric acid.
    (c)    Reaction of propanone with methyl magnesiumbromide followed by hydrolysis.

    Solution

    (a) Catalytic reduction of butanal gives,
    CH3–CH2–CH2–CH2–OH
     Butan-1-ol

    (b) When propene is react with dilute sulphuric acid, it gives


    (c) When propanone react with methylmagnesiumbromide followed by hydrolysis, it gives,

    Question 185
    CBSEENCH12008197

    Arrange the following sets of compounds in order of their increasing boiling points:
    Pentan-1 -ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

    Solution
    The boiling points of alcohols increase with increase in the number of the carbon atoms. Also, the boiling points decrease with increase of branching in carbon chain. Therefore, order of their increasing boiling point.
    Methanol <ethanol<  propan-1-ol< butan-2-ol< butan-1-ol< pentan-1-ol.

     

    Question 186
    CBSEENCH12008198

    Arrange the following sets of compounds in order of their increasing boiling points:
    Penta-1-ol, n-butane, pentanal, ethoxyethane.

    Solution
    Pentan-1-ol molecules are associated with extensive hydrogen bonding, therefore, the boiling point of pentan-1-ol would be highest. Pentanal is more polar than ethoxyethane. Therefore, the intermolecular dipole -dipole attraction is stronger in the former. n-butane is have only weak vander waals forces. Hence increasing order of boiling points of the given compounds is as follows:
    n-Butane< ethoxyethane< pentanal< penta-1-ol.
    Question 188
    CBSEENCH12008200

    Write equations for preparation of propan-2-ol from a Grignard reagent.

    Solution

    Preparation of propan-2-ol from a Grignard reagent.

    Question 189
    CBSEENCH12008201
    Question 190
    CBSEENCH12008202

    Convert Chloro methane into n-butane. 

    Solution

    Conversion of methane into n-butane.

    Question 194
    CBSEENCH12008206
    Question 196
    CBSEENCH12008208

    Show the steps in the industrial preparation of phenol starting with Cumene.

    Solution

    Industrial prepartion of phenol from cumene:

    Question 197
    CBSEENCH12008209

    Show the steps in the industrial preparation of phenol starting with chlorobenzene.

    Solution
    Industrial prepartion of phenol form Chlorobenzene is also known as (Dow’s process):

    Question 199
    CBSEENCH12008211

    Explain why phenol has a much lower pKa value than ethanol.

    Solution

    In phenol, the lone pair of oxygen participates into resonance with the benzene ring.
    As a result, oxygen acquires a partial positive charge. The electron density of O—H bond then shifts towards oxygen decreases around H-atom. H-atom, therefore, can easily be removed as H+ ion. While in ethanol, ethyl group has a + I effect and increases the electron density around H of O—H group making it difficult to remove H as H+. Hence, phenol is more acidic than ethanol and has lower pKa value than ethanol.

    Question 200
    CBSEENCH12008212

    Compare the acidic strength:
    (i) 2, 4-dinitrophenol and 2,4,6-trinitro phenol.
    (ii) O-amino phenol and m-amino phenol. 

    Solution
    (i) 2, 4, 6 trinitro phenol (picric acid) is stronger acid than 2, 4-dinitrophenol as it contains three electron withdrawing NO2 groups.



    (ii) NH3 group is electron donating by resonance when present at ortho and para position while it has -I effect when present at m-position (lone pair can not participate into resonance with benzene from m-position. In m-amino phenol, therefore, the electron density around H of O—H group decreases and H can easily be removed as H+ ion. Hence, m-amino phenol is more acidic than o-amino phenol.
    Question 201
    CBSEENCH12008213

    Explain why is ortho-nitrophenol more acidic than ortho-methoxy phenol?

    Solution

    Ortho nitrophenol is more acidic than ortho-methoxyphenol. This is because the nitro-group is an electron-withdrawing group. The presence of this group in the ortho position in ortho nitro phenol decreases the electron density in the O−H bond. As a result, it is easier to lose a proton.Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.
    In the case of ortho methoxyphenol; methoxy group is an electron-releasing group. Thus, it increases the electron density in the O−H bond and hence, the proton cannot be given out easily. Hence, ortho-nitrophenol is more acidic than ortho-methoxyphenol.


    Question 204
    CBSEENCH12008216
    Question 205
    CBSEENCH12008217

    Give the structures of the compounds A:

    CH subscript 3 COCH subscript 2 CHO space rightwards arrow with NaBH subscript 4 on top

    Solution
    space CH subscript 3 COCH subscript 2 CHO space rightwards arrow with NaBH subscript 4 on top   
    Question 206
    CBSEENCH12008218
    Question 207
    CBSEENCH12008219
    Question 208
    CBSEENCH12008220

    How are the following conversions carried out?
    Propene  → Propane-2-ol.

    Solution

    Conversion of propene to propane-2-ol
    .

    Question 211
    CBSEENCH12008223

    How are the following conversions carried out?
    Methyl magnesium bromide → 2-Methylpropan-2-ol.

    Solution

    Conversion of 2-methylpropan-2-ol from methyl magnesium.

    Question 214
    CBSEENCH12008226
    Question 216
    CBSEENCH12008228

    Explain Oxymercuration and  demercuration.

    Solution
    Oxymercuration-demercuration: In this method alkene reacts with mercuric acetate in presence of water to form hydroxy mercurial compound, which on reaction with NaBH4 yield alcohol. The alcohol is formed according to Markovnikoff’s rule.

    First step is oxymercuration and second step is demercuration.
    Question 217
    CBSEENCH12008229

    Explain Hydroboration.

    Solution
    Hydroboration: It this method, alkene reacts with diborane, B2H6 (an electrophile) to form alkyl boranes which on oxidation with alkaline hydrogen peroxide yields alcohol.

    Question 218
    CBSEENCH12008230

    The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.



    (i) What would be the major product of this reaction?
    (ii) Write a suitable reaction for the preparation of t-butyl ethyl ether.

    Solution
    (i) The major product of the given reaction is 2-methylprop-1 -ene. It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

    Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.
    Question 222
    CBSEENCH12008234

    Write IUPAC names of the following compounds:

    Solution
    5-Ethyl heptane-2, 4-diol.
    Question 223
    CBSEENCH12008235

    Write IUPAC names of the following compounds:

    Solution
    2-Ethoxy butane.
    Question 224
    CBSEENCH12008236

    Write IUPAC names of the following compounds:

    straight C subscript 6 straight H subscript 5 minus straight O minus straight C subscript 7 straight H subscript 15 left parenthesis straight n minus right parenthesis

    Solution
    1-Phenoxy heptane.
    Question 225
    CBSEENCH12008237

    Write IUPAC names of the following compounds:
    straight C subscript 6 straight H subscript 5 space minus straight O minus straight C subscript 7 straight H subscript 15 left parenthesis straight n minus right parenthesis

    Solution
    1-phenoxy heptane
    Question 226
    CBSEENCH12008238

    Write IUPAC names of the following compounds:




    Solution
    4-Methyl phenol.
    Question 237
    CBSEENCH12008249

    (a) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
    (b) Classify the isomers of alcohols in the above part of question as primary, secondary and tertiary alcohols.   

    Solution

    (a) Eight isomeric alcohols are possible:



    b)  Primary alcohol: pentan-1-ol; methylbutan-1-ol; 3-methyl-1-ol; 2,2Dimethylpropan-1-ol.

    Secondary alcohol: Pentan-2-ol; 3-methylbutan-2-ol; pentan-3-ol

    Tertiary alcohol: 2-methylbutan-2-ol

     
    Question 238
    CBSEENCH12008250

    What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

    Solution

    The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction.

    In the first step, alkene reacts with diborane (B2H6) as boron hydride (BH3) to form an alkyl borane. In the next step, the alkyl borane is oxidised by alkaline H2O2 to form an alcohol. The indirect hydration proceeds according to Anti Markovnikoff’s rule. For example,
    bold 3 bold CH subscript bold 3 bold CH bold space bold equals bold space bold CH subscript bold 2 bold plus bold BH subscript bold 3 bold space bold rightwards arrow bold space bold space bold space stack bold left parenthesis bold CH subscript bold 3 bold CH subscript bold 2 bold CH subscript bold 2 bold right parenthesis subscript bold 3 bold B with bold Tri bold minus bold n bold minus bold propyl bold space bold borate below
bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards arrow from bold H subscript bold 2 bold O to bold H subscript bold 2 bold O subscript bold 2 bold divided by bold OH to the power of bold minus of bold space stack bold 3 bold CH subscript bold 3 bold CH subscript bold 2 bold CH subscript bold 2 bold OH with bold Propan bold minus bold 1 bold minus bold ol below bold plus stack bold H subscript bold 3 bold BO subscript bold 3 with bold Boric bold space bold acid below
    Question 239
    CBSEENCH12008251

    Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O. 

    Solution
    Three monohydric phenols are possible which have molecular formula C7H8O. These are:

    Question 240
    CBSEENCH12008252

    Give the equations of reaction for the preparation of phenol from cumene.

    Solution
    Cumene is oxidised into cumene hydro peroxide by air at high pressure which on decomposition by aqueous acid gives phenol and acetone.


    The cumene required in the above reaction is prepared from benzene and propene by Friedel-crafts reaction.

    Question 241
    CBSEENCH12008253

    Write chemical reaction of preparation of phenol from chlorobenzene.

    Solution
    Chlorobenzene is heated with 6 to 8% solution of NaOH to about 623 K under a pressure of 300 atmospheres to form sodium phenate which upon acidification with dilute HCl gives phenol as follows:

    Question 242
    CBSEENCH12008254

    Write the mechanism of hydration of ethene to yield ethanol.

    Solution

    The mechanism  of the reaction involves the following three step:
    Step 1: Proptonation of ethene to form carbocation by electrophilic attack of H3O+.


    Step 2: Nucleophilic attack of water on carbocation.

    Step 3: Deprotonation to form an ethanol.



     
    Question 243
    CBSEENCH12008255

    You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol. Write the equations for the preparation of phenol using these reagents.

    Solution
    Benzene is heated with conc. H2SO4 for a long time and then treated with NaOH. Sodium benzene sulphate, C6H5SO3Na formed is again strongly heated with NaOH at 575 K to give phenol. The reactions are,



    Question 244
    CBSEENCH12008256

    Show how will you synthesize:

    1-phenylethanol from a suitable alkene.

    Solution

    Synthesis of 1-phenyl ethanol from phenylethene.

    Question 245
    CBSEENCH12008257

    Show how will you synthesize:
    cyclohexylethanol using an alkyl halide by an SN2 reaction.

    Solution

     Synthesis of Cyclohexylethanol from chloromethyl cyclohexane.

    Question 246
    CBSEENCH12008258

    Show how will you synthesize:
    Pentan-1-ol using a suitable alkyl halide?

    Solution

    Synthesis of pentan-1-ol by using of 1-chloropentane.
    bold left parenthesis bold CH subscript bold 3 bold right parenthesis bold thin space bold left parenthesis bold CH subscript bold 2 bold right parenthesis subscript bold 4 bold Cl bold plus bold NaOH bold space bold rightwards arrow bold space bold space bold space stack bold CH subscript bold 3 bold left parenthesis bold CH subscript bold 2 bold right parenthesis subscript bold 4 bold OH with bold Pentan bold minus bold 1 bold minus bold ol below bold plus bold NaCl

    Question 247
    CBSEENCH12008259

    Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of control?

    Solution
    (i) Reaction with active metals: Phenol reacts with active metals like Na, K, etc. and H2 is evolved.

    (ii) Reaction with alkalies: Phenol neutralises the caustic alkalies such as NaOH or KOH to form salt and water.

    Phenol shows a considerable acidic character as compared to that of ethanol. It is because of structural difference in C6H5 and C2H5 parts of the two molecules.

    Oxygen connected to sp3 hybridized C atom which is less electronegative and has electron repelling nature. This makes —O—H bond less polar and release of H becomes more difficult. Alkoxide ion formed is not stabilized by resonance.


    Oxygen atom connected to sp2 hybridized C atom of benzene ring which is more electro negative than sp3 hybridized carbon atom. This makes O—H bond more polar and H is released easily. Penoxide ion formed is stabilised by resonance.
    Question 248
    CBSEENCH12008260

    Explain how does the —OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

    Solution

    The -OH group attached to the benzene ring activates it towards electrophillic substitution reaction. It direct the incoming group to ortho and para position in the ring as these positions become electron rich due to the resonance effect the caused by -OH group.

    Since there is –ve charge at o and p position. Therefore, —OH group activates benzene ring towards electrophilic substitution reaction.

    Question 250
    CBSEENCH12008262

    Give equations of the following reactions:

    Bromine in CS2 with phenol.

    Solution
    A mixture of o-bormo phenol and p-bromo phenol is formed.

    Question 251
    CBSEENCH12008263

    Give equations of the following reactions:
    Dilute HNO3 with phenol.

    Solution
    A mixture of o-nitrophenol and p-nitrophenol is formed.

    Question 253
    CBSEENCH12008265

    Explain the following with an example:
    Kolbe’s reaction.

    Solution
    Kolbe reaction: When sodium phenoxide is treated with carbon dioxide under pressure (4 to 7. atom) and at 400 K, sodium salicylate is formed, which on further reaction gives salicyclic acid.


    This reaction is known as Kolbe’s reaction.
    Question 254
    CBSEENCH12008266

    Explain the following with an example:
    Reimer-Tiemann reaction.

    Solution
    Reimer-Tiemann reaction: When phenol is refluxed with chloroform and sodium hydroxide at 340 K followed by hydrolysis, gives o-hydroxy benzaldehyde (main product) and p-hydroxy benzaldehyde (minor product). This reaction is known as Reimer and Tiemann reaction.

    Question 255
    CBSEENCH12008267

    Explain the following with an example:
    Williamson ether synthesis.

    Solution
    Williamson’s synthesis: The reaction of alkyl halides with sodium alkoxide or sodium phenoxide to form ethers is called Williamson’s synthesis. For example,




    This is one of the best methods for the preparation of both simple and mixed ethers.
    Question 256
    CBSEENCH12008268

    Explain the following with an example:
    Unsymmetrical ether.

    Solution
    Unsymmetrical ether: Ethers in which the groups R and R’ are different are called unsymmetrical ethers.
    Examples:

    Williamson synthesis method is used for the preparation of symmetrical and unsymmetrical ethers.
    Question 257
    CBSEENCH12008269

    Write mechanism of acid dehydration of ethanol to yield ethene.

    Solution

    Mechanism: The formation of ethylene and diethyl ether from ethanol and conc. H2SO4 may be explained as follows:
    (a) Formation of ethene or ethylene:
    Step 1. Due to the presence of two lone pairs of electrons on oxygen, alcohols act as weak bases. Therefore, they react with strong mineral acids (HCl, H2SO4 etc.) to form oxonium salts.


    Step 2.
    The presence of a positive charge on the highly electronegative oxygen atom weakens the C—O bond. Thus, the protonated ethanol readily eliminates a molecule of H2O to form ethyl carbocation.

    This step is slow and hence is the rate-determining step of the reaction.
    Step 3. The ethyl carbocation formed in step 2 being a reactive chemical species, readily loses a proton to form the ethene molecule.

    Question 258
    CBSEENCH12008270

    How are the following conversions carried out?
    Propene → Propan-2-ol.

    Solution

    Propene is react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

    Question 259
    CBSEENCH12008271

    How are the following conversions carried out?
    Benzyl chloride → Benzyl alcohol

    Solution

    Conversion of benzyl alcohol from benzyl chloride.

    Question 260
    CBSEENCH12008272

    How are the following conversions carried out?
    Ethyl magnesium chloride → propane-1-ol.

    Solution

    Conversion of propane-1-ol from ethyl manesium chloride.

    Question 261
    CBSEENCH12008273

    How are the following conversions carried out?
    Methyl magnesium bromide → 2-methyl propan-2-ol.

    Solution

    Conversion of 2-methyl propan -2-ol from methyl magnesium bromide.

    Question 262
    CBSEENCH12008274
    Question 263
    CBSEENCH12008275
    Question 266
    CBSEENCH12008278
    Question 267
    CBSEENCH12008279
    Question 268
    CBSEENCH12008280

    Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

    Solution
    Ethanol molecules are associated with extensive intermolecular bonding, therefore, the boiling point of ethanol would be the higest with methoxymethane.
    Question 269
    CBSEENCH12008281

    Give IUPAC names of the following ethers:

    Solution
    1-Ethoxy-2-methyl propane.
    Question 270
    CBSEENCH12008282

    Give IUPAC names of the following ethers:

    CH subscript 3 OCH subscript 2 CH subscript 2 Cl

    Solution
    2-Chloro-1-methoxy ethane.
    Question 271
    CBSEENCH12008283

    Give IUPAC names of the following ethers:

    straight O subscript 2 straight N space minus space straight C subscript 6 straight H subscript 4 space minus space OCH subscript 3 left parenthesis straight p right parenthesis


    Solution
    4-Nitro anisole.
    Question 272
    CBSEENCH12008284

    Give IUPAC names of the following ethers:

    CH subscript 3 CH subscript 2 CH subscript 2 OCH subscript 3



    Solution
    1-Methoxy propane.
    Question 273
    CBSEENCH12008285

    Give IUPAC names of the following ethers:






    Solution
    1-Ethoxy-4,4-dimethyl cyclohexane.
    Question 274
    CBSEENCH12008286

    Give IUPAC names of the following ethers:









    Solution
    Ethoxy benzene.
    Question 279
    CBSEENCH12008291

    Illustrate with examples the limitations of Williamson synthesis for the preparation of certain type of ethers.

    Solution
    Williamson synthesisinvolves SN2 attack may an alkoxide ion on primary alkyl halide. Better result are obtained if the alkyl halides is primary. In case of secondary and tertiary alkyl halide, elimination competes over substitution.
     (i) If the alkyl halide is tertiary, then the major product is an alkene and alkoxide then acts as a strong base not as a nucelophile.
    wiredfaculty.com

    (ii) Chlorobenzene does not give Williamson’s synthesis due to partial double bond character of C—Cl bond. So, aromatic ether are prepared from sodium phenoxide.

    wiredfaculty.com
    Question 280
    CBSEENCH12008292

    How is 1-propoxypropane synthesised from propan-1-ol ? Write mechanism of this reaction. 

    Solution
    1-propoxy propane can be prepared either by dehydration of propan-1-ol by heating it with conc. H2SO4 or by first converting a part of it into 1-bromo propane and other part into its sodium salt and then reducing them together i.e., by Williamson synthesis.

    Mechanism:

    Step 1 :Protonation


    Step 2 :Nucleophilic attack

    Step 3: Deprotonation 


     
    Question 281
    CBSEENCH12008293

    Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

    Solution

    The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) involving the attack of alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.


    Question 282
    CBSEENCH12008294

    Write the equation of the reaction of hydrogen iodide with 1-proxypropane.

    Solution
    If HI is present in limited amount, a mixture of 1-iodo propane and 1-propanol is formed. If excess of HI is present, then 1-propanol formed further reacts with HI and 1-iodo propane only is formed.


    Question 283
    CBSEENCH12008295

    Write the equation of the reaction of hydrogen iodide with methoxybenzene.

    Solution

    O—CH3 bond is weaker than —OC6H5. Therefore, O—CH3 bond is broken easily to form phenol and iodo methane and no CH3OH is formed. It is also due to steric factors during protonation at methoxy benzene.


    Question 284
    CBSEENCH12008296

    Write the equation of the reaction of hydrogen iodide with benzyl ethyl ether. 

    Solution
    Ethyl iodide and benzyl alcohol are formed. When benzyl ethyl ether react with HI.


    Question 285
    CBSEENCH12008297

    Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.

    Solution
    Alkyl aryl ethers are the resonating hybrids of the following structures:


    The alkoxy group (—OR) is ortho and para directing group and it activates the aromatic ring in ortho and para positions due to conjugation with pi electrons of the benzene ring. However, ethers are some what less reactive than phenol in these substitution reactions. Activation of benzene ring takes place as shown above.
    The following reactions show that alloxy group directs the incoming substituents to ortho and para positions in benzene ring.


    Similar products are formed on sulphonation of anisole.
    Question 286
    CBSEENCH12008298

    Write the mechanism of the reaction of HI with methoxymethane.

    Solution

    Mechanism: The cleavage of ethers by halogen acids occur by the following mechanism.
    Step 1. Ethers being Lewis basis, undergo protonation to form oxonium salts:

    Step 2. The protonated ether thus formed undergoes nucleophilic attack by the halide ion to form a molecule of an alkyl halide and a molecule of an alcohol.

    Question 289
    CBSEENCH12008301

    Write equations of the following reactions:
    Bromination of anisole in ethanoic acid medium.

    Solution

    Anisole undergoes bromination with bromine in the ethanoic acid even in the absence of iron(III)bromide catalyst. Due to the activation of benzene ring by the methoxy group. Para isomers is obtained in 90% yield.

    Question 291
    CBSEENCH12008303

    Use simple chemical test to differentiate between the following pairs of organic compounds:
    (i) 2, 4, 6-trinitro phenol and 2, 4, 6-tri methyl phenol.

    (ii) Anisole and benzyl alcohol.

    Solution
    (i) Addition of NaHCO3 gives brisk effervescence with more acidic 2,4.6-trinitro phenol and not with much less acidic 2, 4, 6-tri methyl phenol.
    (ii) Benzyl alcohol C6H5CH3OH is a primary alcohol. Addition of acidic-warm KMnOwill oxidise the benzyl alcohol and the purple colour will disappear. The anisole, an ether resists this oxidation.
    Question 292
    CBSEENCH12008304

    Show with the help of chemical equations what happens when ethanol is treated with phosphorus tribromide.

    Solution

    When ethanol is treated with phosphrous tribromide it gives bromoethane.

    Question 293
    CBSEENCH12008305

    Show with the help of chemical equations what happens when cumene hydroperoxide is treated with dilute H2SO4.

    Solution

    When cumene hydroperoxide is treated with dilute sulphuric acid, it gives phenol and acetone.

    Question 294
    CBSEENCH12008306

    Give a chemical test to distinguish between: 2-pentanol and 3-pentanol.

    Solution
    2-pentanol on oxidation gives methylketone.

    This compound has a CH3C = O group.
    Therefore, it will form iodoform (CHI3) with iodine and alkali on warming. Iodoform is a yellow crystalline substance.

    3-pentanol does not show this test.

    Question 295
    CBSEENCH12008307

    Write the resonance structures of phenol (which is similar to that of chlorobenzene) and explain why it does not form chlorobenzene on reaction with PCI5 Explain whether phenol can be dehydrated on heating with catalytic amounts of conc. H2SO4.

    Solution
    Resonance structures of phenol:

    Due to resonance, carbon-oxygen bond develops a slight double bond character. Therefore, it is not easy to replace —OH in phenols by Cl.
    No, phenols cannot be dehydrated on heating with small amount of conc. H2SO4. Instead, phenols with conc. H2SO4 undergoes sulphonation.

    Question 296
    CBSEENCH12008308

    Account for the following:
    Alcohols with three or less carbons are water-soluble while alcohols with five or more carbons are insoluble.

    Solution
    The solubility of alcohols in water is due to the presence of intermolecular hydrogen bonding. As the molecular mass increases, the solubility in water decreases. Greater carbon content makes the alcohols less water solubility.
    Question 297
    CBSEENCH12008309

    Account for the following:
    tert-butanol is more volatile than n-butanol.

    Solution
    tert-butanol and n-butanol have the same molecular mass but t-butanol is more branched. As branching increases, Van der Waals forces increases due to decrease in surface area, t-butanol has a lower boiling point than n-butanol. Hence t-butanol is more volatile than n-butanol.
    Question 298
    CBSEENCH12008310

    To cleave the C—O bond in alcohols small amount of acid is added. Explain.

    Solution
    When we put a few drops of acid, the oxygen of alcohol is converted to oxonium ion which weakens C—O bond. Due to the development of positive charge on oxygen, water molecule is eliminated after the cleavage of C—O bond and carbocation is formed.

    Question 302
    CBSEENCH12008314
    Question 303
    CBSEENCH12008315
    Question 304
    CBSEENCH12008316
    Question 305
    CBSEENCH12008317

    What is meant by Absolute alcohol?

    Solution

    Absolute alcohol: This is ethyl alcohol of about 100 percent purity. It is made from rectified spirit by keeping it for a few days with quick lime.
    Later on, it is distilled to get absolute alcohol (about 99% pure).
    Alternatively, the rectified spirit is mixed with benzene and distilled when a ternary mixture of water, benzene and alcohol distils over. When all the water distils out then a binary mixture of benzene and alcohol distils at a slightly higher temperature. After all the excess of benzene distils over, then absolute alcohol distils over.

    Question 306
    CBSEENCH12008318

    What is meant by Methylated spirit?

    Solution

    Methylated spirit: Since ethyl alcohol can be used for drinking purposes, it is heavily taxed in the form of excise duty. On the other hand, ethyl alcohol used for industrial purposes is duty free, industrial alcohol can be used for drinking purposes. In order to make rectified spirit unfit for drinking purposes. it is denatured by mixing some poisonous substances like methyl alcohol.
    Such alcohol (denatured) is known as methylated spirit. It is extensively used in paints, varnishes, medicines and as a solvent.

     
    Question 307
    CBSEENCH12008319

    What is meant by Denatured alcohol?

    Solution
    Denatured alcohol: Ethanol is made unfit for drinking puposes by the addition of the methanol, pyridine etc. These materials are added to make sure that it will not be used for beverage purposes. The process is called denaturing and the alcohol thus obtained is called denatured alcohol. Denatured alcohol can be manufactured and sold duty free for industrial use.
    Question 308
    CBSEENCH12008320

    Explain the following:
    Cyclohexanol is more soluble in water than 1-hexanol.

    Solution
    The alkyl group in cyclohexanol is more compact than the alkyl group in hexanol-1. The OH group of cyclohexanol is more exposed and is more available for hydrogen bonding with water. Hence, cyclohexanol is more soluble in water than hexanol-1
    Question 309
    CBSEENCH12008321

    Explain the following:
    Propane-1, 3-diol is more soluble in water than propan-1-ol.

    Solution
    In propane-1,3-diol, there are more OH groups present which increases the number of H-bonds between water and the diol. Hence, propane-1, 3-diol is more soluble in water than propan-1-ol.

    Question 310
    CBSEENCH12008322

    How can we distinguish between 1°, 2° and 3° alcohols by the reaction with Cu at 573 K?

    Solution
    1° alcohol on reaction with Cu/573 K gives an aldehyde with same number of carbon atoms
    RCH subscript 2 OH space rightwards arrow from 573 space straight K to Cu of space RCHO space plus straight H subscript 2 left parenthesis dehydrogenation right parenthesis
    2° alcohol on reaction with Cu/573 K gives a ketone with same number of carbon atom.

    3° alcohol undergo dehydration with Cu/573 K to form an alkene

    Question 311
    CBSEENCH12008323

    What happens when phenol reacts with bromine in solvent of low polarity like CS2or CHCl3 at low temperature? Give a mechanism for the reaction.

    Solution
    Phenol reacts with Br2/CS2 to give monobromo derivatives.

    The mechanism of the reaction is as follows.

    Question 312
    CBSEENCH12008324

    When 3-methylbutan-2-ol is heated with HBr, the following reaction takes place:





    Give a mechanism for this reaction.

    Solution

    The given reaction involve the following mechanism,
    Step 1: Protonation to 3-methylbutan-2-ol

    Step 2: Formation of secondary carbocation by the elimination of a water molecule.

    Step 3 : Rearrangement by the hydride-ion shift

    Step 4: Nucleophilic attack

    Question 317
    CBSEENCH12008329

    Explain giving reasons:
    (i) Boiling point of an alcohol is higher than the corresponding alkane.

    (ii) Lower alcohols are soluble in water, while the higher alcohols are not.

    (iii) Boiling point of mono hydric alcohols increases with increase in carbon atoms.

    Solution
    (i) The boiling points of alcohols are higher than those of the corresponding alkanes, due to the intermolecular association of a large number of alcohol molecules through hydrogen bonding. Thus

    Since a large amount of energy is needed to break the extensive hydrogen bonds, which hold a large number of alcohol molelcules together, so the boiling points of alcohols are significantly higher than the corresponding alkanes, in which no such hydrogen bonding can take place.

    (ii) The solubility of an alcohol in water depends upon its capacity to form hydrogen bonds with water molecules. However, the non-polar hydrocarbon part of the alcohol molecule is unable to form hydrogen bonds with other. Consequently, the lower alcohols are soluble in water, due to the formation of hydrogen bonds between highly polarised hydroxyl group (-OH), present in alcohols as well as water. However, in higher alcohols the hydrocarbon part becomes larger in size and, therefore, the solubility effect due to hydroxyl groups (-OH), of alcohol is out weighed by the counter large ‘repulsive effect’ of non-polar hydrocarbon part of the alcohol molelcule. Hence, higher alcohols become in soluble in water. The large inductive effect (+I effect) of the hydrocarbon part (or alkyl group) in higher alcohols also reduces the polar character of hydroxyl group (–OH) in higher alcohols, which in turn reduces their capacity to form hydrogen bonding with water.

    (iii) With the increase in molar mass of alcohol, the intermolecular Van der Waals forces of attraction increases, thereby the boiling point also increases. Thus b.p. of C4H5OH > C3H7OH > C2H5OH > CH3OH


    Question 318
    CBSEENCH12008330

    Which reaction of phenol with CHCl3 in NaOH followed by acidification gives salicyaldehyde? Suggest a mechanism for this reaction.

    Solution
    In Reimer-Tiemann reaction, phenol reacts with CHCl3 in NaOH followed by acidification gives salicyaldehyde.

    The mechanism for the reaction is


    Question 320
    CBSEENCH12008332

    How will you convert phenol into Biphenyl?

    Solution

    Conversion of biphenyl from phenol

    Question 321
    CBSEENCH12008333

    How will you convert phenol into Acetophenone?

    Solution

    Conversion of acetophenone from phenol:

    Question 322
    CBSEENCH12008334

    Draw resonating structures for phenol showing the correct position of positive and negative charges and then account for the following:

    (i) The ring of phenol is more easily oxidised than that of benzene.

    (ii) Phenol is nitrated more easily than benzene.

    Solution
    The resonating structures for phenol are:


    (i) The OH group is strongly activating group and increases the electron density on benzene and making the ring of phenol very electron rich. As a result, it can readily donate electrons to an oxidising agents.

    (ii) Nitration of phenol is an electrophilic substitution reaction. OH group increases the electron density on benzene ring making it more electron rich. As a result, NO2+ ion can easily attack on the ring of phenol.


    Question 323
    CBSEENCH12008335

    How is ethanol obtained commercially by fermentation process?

    Solution
    Ethanol is obtained commercially by the fermentations of sugar present in molasses, sugarcane, grapes or starch.
    The suger is first converted into a mixture of glucose and fructose in the presence of an enzyme, invertase.
    stack bold C subscript bold 12 bold H subscript bold 22 bold O subscript bold 11 with bold C bold a bold n bold e bold space bold s bold u bold g bold a bold r below bold space bold plus bold space bold H subscript bold 2 bold O bold space bold rightwards arrow with bold I bold n bold v bold e bold r bold t bold a bold s bold e on top bold space stack bold C subscript bold 6 bold H subscript bold 12 bold O subscript bold 6 with bold G bold l bold u bold c bold o bold s bold e below bold space bold plus bold space stack bold C subscript bold 6 bold H subscript bold 12 bold O subscript bold 6 with bold F bold r bold u bold c bold t bold o bold s bold e below

    Glucose and fructose undergo fermentation in anaerobic conditions, in the presence of enzyme, zymase to form ethanol.
    bold C subscript bold 6 bold H subscript bold 12 bold O subscript bold 6 bold space bold rightwards arrow with bold Zymase on top bold space stack bold 2 bold C subscript bold 2 bold H subscript bold 5 bold OH with bold ethanol below bold plus bold 2 bold CO subscript bold 2
    Question 324
    CBSEENCH12008336

    Solubility of alcohol in water is due to  ________.

    Solution
    hydrogen bonding  
    Question 327
    CBSEENCH12008339

    o-Hydroxy benzoic acid is called __________.

    Solution

    salicyclic acid

    Question 328
    CBSEENCH12008340
    Question 329
    CBSEENCH12008341

    100% pure ethanol is called _________.

    Solution
    absolute alcohol
    Question 331
    CBSEENCH12008343

    IUPAC name of cumene is _________.

    Solution
    2-phenyl propane
    Question 332
    CBSEENCH12008344
    Question 333
    CBSEENCH12008345

    CH3CH2OH is a ___________ alcohol.

    Solution

    primary

    Question 335
    CBSEENCH12008347
    Question 337
    CBSEENCH12008349
    Question 342
    CBSEENCH12008354
    Question 348
    CBSEENCH12008360

    The compound which is least soluble in water is



    • CH subscript 3 CH subscript 2 OH


    • CH2 = CH—OCH2CH3
    • CH3—O—CH3

    Solution

    C.

    CH2 = CH—OCH2CH3

     

    Question 349
    CBSEENCH12008361







    Solution

    C.



    Question 350
    CBSEENCH12008362
    Question 354
    CBSEENCH12008366

    The reaction of   
      

    with HBr gives

    Solution

    B.

    Question 355
    CBSEENCH12008367

    In the reaction:


    Solution

    C.

    Question 356
    CBSEENCH12008368

    Write IUPAC name of:

    Solution

    Propane 1,2,3-triol

    Question 357
    CBSEENCH12008369

    Write IUPAC name of:

    Solution

    2,3 dibromo-1 chloro pentanol

    Question 358
    CBSEENCH12008370

    Write IUPAC name of:



    Solution

    2-Methylpropan-1-ol

    Question 359
    CBSEENCH12008371

    In absence of kerosene oil can we store sodium metal in diethyl ether or ethyl alcohol or anhydrous hexanol?

    Solution

    Sodium metal can be store in hydrocarbon and ether but it cannot be store in alcohol because it easily react with it.  

    Question 360
    CBSEENCH12008372

    How is CO converted into methanol?

    Solution

    Conversion of CO to methanol.
    CO + CO2 + 7H2 --> 2 CH3OH + 2H2 + H2O

    Question 361
    CBSEENCH12008373

    Which of the following will have a higher boiling point and why? CH3NH2 or CH3OH.

    Solution

     

    Molecular mass of methanol is 32g/mol and molecular mass of methylamine is 31g/mol. Since methanol have higher molecular mass thus it have higher boiling point. Also, CH3OH has very strong H-bonding between its molecules. This is due to the greater electronegativity of oxygen atom. Such strong association of molecules leads to rise in the boiling point. CH3NH2 has comparitively lesser H-bonding due to lesser electronegativity of N atom and hence lesser boiling atom. 

    Question 362
    CBSEENCH12008374

    Write one example of each:
    (i) a nucleophile,  (ii) an electrophile.

    Solution

    All molecules or ions with a free pair of electrons or at least one pi bond can act as nucleophiles. Because nucleophiles donate electrons, they are by definition Lewis bases. For example Cynaide ion CN- or Hydroxyl ion OH- are nucleophilies.

    Electrophiles are positively charged or neutral species having vacant orbitals that are attracted to an electron rich centre. It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile. Because electrophiles accept electrons, they are Lewis acids. For example mercury cation Hg+.

    Question 363
    CBSEENCH12008375

    How would you differentiate between phenol and carboxylic acid by simple chemical test?

    Solution

    Carboxylic acid react with aqueous sodium bicarbonate and carboxylate salt and carbon dioxide. Phenol does not give this test.

    Question 364
    CBSEENCH12008376

    Which Grignard reagent should react with acetone to get 2-phenyl propan-2-ol?

    Solution

    Acetone react with Phenylmagnesium bromide to form 2-phenyl propan-2-ol. 

    Question 365
    CBSEENCH12008377

    Which compound is formed when benzaldehyde is reduced with LiAlH4 ?

    Solution

    When benzaldehyde react with LiAlH4, it form benzyl alcohol.

    Question 366
    CBSEENCH12008378

    How would you convert tertiary butyl chloride to tertiary butyl alcohol?

    Solution

    Conversion of tertiary butyl alcohol from tertiary butyl chloride.

    Question 367
    CBSEENCH12008379

    How many σ-bonds are present in 3-methyl phenol?

    Solution


    6 carbon atoms of the ring form a sigma bond with the neighbouring carbon atoms of the ring. Therefore 6 sigma bonds. Another carbon atom forms a sigma bond with the carbon of -CH3 group which in turn forms a sigma bond with each of the 3 hydrogen atoms. 4 more sigma bonds.
    4 carbon atoms of the ring form a sigma bond with hydrogen atoms. 4 more sigma bonds.
    -OH form 2 sigma bond.
    Total =6+2+4+4 =16 sigma bonds. 

    Question 368
    CBSEENCH12008380

    Define fermentation.

    Solution

    Fermentation is define as the chemical breakdown of a substance by bacteria, yeasts, or other microorganisms, typically involving effervescence and the giving off of heat.

     

    Question 369
    CBSEENCH12008381

    What happens when ethylamine is treated with nitrous acid?

    Solution

    When ethyl amine react with nitrous acid, it form ethyl diazonium salt which is being unstable, liberate nitrogen gas and ethanol.
    bold CH subscript bold 3 bold CH subscript bold 2 bold NH subscript bold 2 bold space bold rightwards arrow with bold NaNO subscript bold 2 bold space bold plus bold space bold HCl bold space bold space on top bold left square bracket bold CH subscript bold 3 bold CH subscript bold 2 bold N to the power of bold plus bold Cl to the power of bold minus bold right square bracket bold space

bold left square bracket bold CH subscript bold 3 bold CH subscript bold 2 bold N to the power of bold plus bold Cl to the power of bold minus bold right square bracket bold space bold space bold rightwards arrow with bold space bold space bold space bold space bold space bold H subscript bold 2 bold O bold space bold space bold space bold space bold space on top bold CH subscript bold 3 bold CH subscript bold 2 bold OH bold space bold plus bold N subscript bold 2 bold space bold plus bold HCl

    Question 370
    CBSEENCH12008382

    Give one use of phenyl ether.

    Solution

    The main application of diphenyl ether is as a eutectic mixture with biphenyl, used as a heat transfer medium.

    Question 371
    CBSEENCH12008383

    Discuss the Williamson's synthesis for the preparation of ethers.

    Solution

    Williamson's important method laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.
    R-x +R-O-Na ---> R-O-R' +NaX

    Question 372
    CBSEENCH12008384
    Question 373
    CBSEENCH12008385

    What happens when primary, secondary and tertiary alcohols are treated with copper powder at 573 K?

    Solution

    When the vapours of a primary or a secondary alcohol are passed over heated copper at 573K, dehydrogenation takes place and an aldehyde or a ketone is formed. In case of tertiary alcohols, it goes dehydration.

    Question 374
    CBSEENCH12008386

    How is phenol obtained commercially from benzene?

    Solution

    Benzene react with chlorine in the presenc of ferric chloride which form chlorobezene, which on further reaction with sodium hydroxide gives sodium phenoxide. Then sodium phenoxide acidified and form phenol.

    Question 379
    CBSEENCH12008391

    Write the IUPAC names of the following:
    (CH3)2CHCH2CH2OH

    Solution

    3-methyl butan-1-ol

    Question 380
    CBSEENCH12008392

    Write the IUPAC names of the following:

    Solution

    2- phenylethanol

    Question 381
    CBSEENCH12008393

    Write the IUPAC names of the following:



    Solution

    4-chloro 2,3 dimethyl hexanol

    Question 382
    CBSEENCH12008394

    Why is that phenol is acidic and hexanol is neutral towards a solution of NaOH?

    Solution

    Phenols are more reactive than alcohols. This is due to the fact that that the phenoxide ion produced by the loss of proton from phenol is highly stabilized by resonance while there is no such resonance stabilization of the cation formed by loss of protons from hexanol or any other alcohol.

    Question 383
    CBSEENCH12008395
    Question 385
    CBSEENCH12008397

    Phenol is a very weak acid. What substitutions in the molecule can make it stronger acid and a weaker acid and why?

    Solution

    Phenol is very weak acid. Phenol can lose a hydrogen ion and forms phenoxide ion. The negative charge on the oxygen atom is delocalised around the ring and thus stblilised. Hence lesser the acidity.

    The resonance structures of phenoxide ions explain the delocalization of negative charge. In case of substituted phenols, acidity of phenols increases in the presence of electron withdrawing group. This is due to the stability of the phenoxide ion generated. The acidity of phenols further increases if these groups are attached at ortho and para positions. This is due to the fact that the negative charge in phenoxide ion is mainly delocalized at ortho and para positions of the attached benzene ring. On the other hand, the acidity of phenols decreases in presence of electron donating groups as they prohibit the formation of phenoxide ion.

    Question 386
    CBSEENCH12008398

    Explain giving reasons, why:
    Phenol has a smaller dipole moment than methanol?

    Solution

    Phenol has a smaller dipole moment (1.54 D) than methanol. The smaller dipole moment of phenol is due to the electron withdrawing effect of phenyl group (whereas alkyl group in alcohols has electron releasing effect).
    In phenol, the lone pairs of electrons of the oxygen atom are delocalised across the aromatic ring (via electron transfer known as resonance) this reduces the polarity of the C-O bond. Methanol, which has no aromatic ring and the lone pair of electrons from the oxygen are not delocalised meaning that the C-O bond, in this case, has a higher polarity.

    Question 388
    CBSEENCH12008400

    Write chemical equations for what happens when:

    (i) 2-methyl-2-proponal vapours are passed over heated copper.

    (ii) A Triglyceride is treated with sodium hydroxide solution.

    Solution

    i) When 2-methyl-2-proponal is treated with copper then isobutlene is formed.

    ii)  When Triglyceride is treated with sodium hydroxide, it forms glycerol.

    Question 389
    CBSEENCH12008401

    Give reasons for the following:
     Phenol does not undergo protonation readily.

    Solution

    The lone pair on oxygen O-H in phenol is being shared with benzene ring through resonance. thus, a lone pair is not fully present on oxygen and hence phenols do not undergo protonation reactions.

    Question 390
    CBSEENCH12008402

    Which alcohol is known as wood alcohol? How will you prepare it commercially? Give its one use?

    Solution

    Methanol CH3OH is known as wood alcohol. methanol produced by catalytic hydrogenation on carbon monoxide at high pressure and temperature and in the presence of ZnO-Cr2O3 catalyst.
    CO subscript 2 space plus 2 straight H subscript 2 space rightwards arrow from stack 200 minus 300 space atm space with 573 minus 673 straight K below to ZnO minus Cr subscript 2 straight O subscript 3 of space CH subscript 3 OH
    Methanol is colourless liquid and highly poisonous in nature.
    Use:
    Methanol is used as a solvent in paints.

    Question 391
    CBSEENCH12008403

    Discuss the Relative reactivity of alcohols towards Lucas reagent.

    Solution

    Primary alcohols on reaction with Lucas reagents (Conc. HCl/anhy.ZnCl2) does not forms white cloudiness or turbidity at room temperature.

    RCH2OH + Lucas Regent -->No Reaction

    Secondary alcohol : Secondary alcohol on reaction with locals reagent (Conc. HCl/anhy.ZnCl2) forms white cloudiness or turbidity appears with in about 5 minutes.

    R2CHOH+ Lucas reagent --> R2CHCl


    Tertiary alcohol: Tertiary alcohol on reaction with Lucas Reagent (Conc. HCl/anhy.ZnCl2) forms white cloudiness or turbidity immediately.

    R3C-OH +lucas --> R3CCl

    Question 392
    CBSEENCH12010056

    How do you convert the following?

    Phenol to anisole

    Solution

    Phenol to Anisole

    Question 393
    CBSEENCH12010057

    How do you convert the following?

    Propan-2-ol to 2-methylpropan-2-ol

    Solution
    Propan-2-ol to 2-methylpropan-2-ol
    Question 394
    CBSEENCH12010058

    How do you convert the following?

    Aniline to Phenol

    Solution

    Aniline to Phenol

    Question 395
    CBSEENCH12010059

    Write the mechanism of the following reaction: 2CH3-CH2-OH -->  CH3-CH2-O-CH2-CH3?

    Solution

    2CH3-CH2-OH --->  CH3-CH2-O-CH2-CH3


    The given reaction follows SN2 mechanism as shown below:

    Question 397
    CBSEENCH12010089

    Write the IUPAC name of the compound.

    Solution

    The IUPAC name of the given organic compound is 3-Hydroxybutan-1-oic acid. 

    Question 398
    CBSEENCH12010090

    Which of the following isomers is more volatile: o-nitrophenol or p-nitrophenol?

    Solution

    In o-nitro phenol, -OH is linked to –NO2 by means of intramolecular H-bonding so, it is highly volatile.

     Explanation:

    Stronger intermolecular forces would make the substance less volatile. So we can see intermolecular H-bonding in P-nitro phenol, so P-nitro phenol is less volatile, where as in o-nitro phenol intra H-bonding takes places that’s why it is highly volatile.

    Question 399
    CBSEENCH12010156

    Explain the mechanism of the following reaction: 

    Solution

    The mechanism of the reaction is given below:

    Question 400
    CBSEENCH12010157

    How will you convert 

    Propene to Propan-2-ol?

    Solution

    To convert from Propene to Propan-2-ol, the addition of H2SO4 takes place in accordance with Markovnikov ‘s  rule i.e.

    Question 401
    CBSEENCH12010158

    How will you convert 

    Phenol to 2, 4, 6- trinitrophenol?

    Solution

    When concentrated nitric acid is added to phenol in the presence of sulphuric acid it gives 2, 4, 6-trinitrophenol.

    Question 402
    CBSEENCH12010179

    Although phenoxide ion has more number of resonating structures than Carboxylate ion, Carboxylic acid is a stronger acid than phenol. Give two reasons.

    Solution

    On losing a proton, carboxylic acids forms carboxylate ion and phenol forms phenoxide ion as follows:

    RCOO-Carboxylate ion

     

    Phenoxide ion

     

    Now, the negative charge is delocalized in both molecules as follows:

    The conjugate base of the carboxylic acid has two resonance structures in which negative charge in delocalized over two oxygen atoms (since O is more electronegative than C) which stabilises the carboxylate ion.

    On the other hand, in phenoxide ion the charge is delocalized over entire molecule on the less electronegative atom (Carbon), thus resonance of phenoxide is not important in comparison to resonance in carboxylate ion.

    Further, in carboxylate ion, the negative charge is effectively delocalized over two oxygen atoms whereas it is less effectively delocalized over one oxygen atom and a less electronegative carbon atom.

    Thus, Phenol is less acidic than carboxylic acids. In other words, carboxylic acids are stronger acids than phenol

    Question 403
    CBSEENCH12010181
    Question 404
    CBSEENCH12010182

    Give simple chemical tests to distinguish between the following pairs of compounds:

    (i) Ethanal and Propanal

    (ii) Benzoic acid and Phenol

    Solution

    (i) Distinguish test between ethanal and propanal:

     Iodoform Test: Ethanal gives iodoform test.

     CH3CHO + 4NaOH + 3I2 ---> CHI3 (Yellow ppt.) + HCOONa + 3NaI + 3H2O
    Propanal does not give this test.
    CH3CH2CHO + 4NaOH + 3I2 ---> No Reaction.

     

    (ii) Distinguish test between Benzoic acid and Phenol:

     NaHCO3Test: When Benzoic acid reacts with NaHCO3, the brisk effervescence of CO2 gas evolved.

     

     
    Phenol does not give this test.

    C6H5OH + NaHCO3 -->  No Reaction

    Question 408
    CBSEENCH12010211

    Write chemical equation for the following conversions:

    (i) Nitrobenzene to benzoic acid.

    (ii) Benzyl chloride to 2-phenylethanamine.

    (iii) Aniline to benzyl alcohol.

    Solution

    (i) Nitrobenzene to benzoic acid

    ii) Benzyl chloride to 2-phenylethanamine

    (iii) Aniline to benzyl alcohol

    Question 409
    CBSEENCH12010237

    How would you obtain 

    (i) Picric acid (2, 4, 6-trinitrophenol) from phenol, 

    (ii) 2-Methylpropene from 2-methylpropanol? 

    Solution

    (i) Phenol on reaction with concentrated HNO3 results in the formation of 2,4,6-Trinitrophenol or Picric acid. 

    ii) 2-methyl propene can be obtained from 2-methyl propanol by the reaction of the later with alc. KOH.

    Question 410
    CBSEENCH12010272

    How do you convert the following:

    (i) C6H5CONH2 to C6H5NH2

    (ii) Aniline to phenol

    (iii) Ethanenitrile to ethanamine

    Solution

    Ti) o convert Benzamide to aniline:

    Benzamide on heating with a mixture of Br₂ in presence of NaOH or KOH (i.e. NaOBr or KOBr) is given aniline.

     

    ii) Aniline first converts to diazonium salt using NaNO2/HCl at 0-5 degree C, Then diazonium ion reacts with the water at 100 degrees C to produced phenol

     

    (iii)

     

     

    Question 412
    CBSEENCH12010292

    Ethanal is soluble in water. Why?

    Solution

    Ethanal CH3CHO is soluble in water because they form a hydrogen bond with water.

    Question 413
    CBSEENCH12010301

    Explain the mechanism of the following reaction: 

     

     

    Solution

    Mechanism of the following reaction is given below:

     

     

    step 1: Formation of protonated alcohol

     

     

     

    Step 2: Formation of carbocation

     

     

     

    Step 3: Formation of ethene by removal of proton

     

    Question 414
    CBSEENCH12010338

    Explain the following behaviours:

    (i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

    (ii) Ortho-nitro phenol is more acidic than ortho-methoxyphenol.

    Solution

    (i) As alcohol is a polar solvent .it form hydrogen bonding with water molecules while another hydrocarbon of comparable molecular masses does not form hydrogen bonding due to being non- polar i.e.

    Alcohols undergo intermolecular H-bonding while hydrocarbons do not.

    Hence, they are more soluble in water than the hydrocarbons of comparable molecular masses.

     

    (ii)

    The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. As a result, it is easier to lose a proton.

    Also, the o-nitro phenoxide ion formed after the loss of protons is stabilised by resonance.

    Hence, ortho-nitrophenol is a stronger acid.

    On the other hand, a methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily.

    For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

    Question 415
    CBSEENCH12010339

    Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol.

    Solution

    Some reactive alkenes undergo direct hydration in the presence of mineral acids which act as catalysts. The addition of water to the double bond takes place in accordance with Markonikoff’s rule.

    The mechanism of reaction involves the following three steps:

    (i) Protonation of alkene to form carbocation by electrophilic attack of H3O+.

     

    (ii) Nucleophilic attack of water of carbocation.

     

    (iii) Deprotonation to from an alcohol.

     

    Question 416
    CBSEENCH12010367
    Question 418
    CBSEENCH12010389

    Write the IUPAC name of the given compound:

    Solution

    2- Phenylethanol

    Question 419
    CBSEENCH12010417

    How would you convert ethanol to ethene?

    Solution

    When dehydration of ethanol is carried out at a higher temperature (443 K), ethene is formed. The following chemical reaction takes place.

    Question 420
    CBSEENCH12010457
    Question 421
    CBSEENCH12010458

    Write the mechanism of the following reaction:

    CH3CH2OH ---> HBrCH3CH2Br + H2O

    Solution

    Mechanism of the given reaction is the following:
    Wired Faculty CHEN12116621

    Question 423
    CBSEENCH12010536
    Question 424
    CBSEENCH12010537

    Give simple chemical tests to distinguish between the following pairs of
    compounds :
    Ethanol and Phenol

    Solution

    Ethanol gives Iodoform test but phenol does not.
    stack CH subscript 3 CH subscript 2 OH space with Ethanol below plus space 4 NaOI space rightwards arrow with space on top stack CHI subscript 3 space with Iodoform space left parenthesis yellow space ppt. right parenthesis below plus HCOONa space plus space NaI space plus straight H subscript 2 straight O space plus 2 NaOH

    Question 425
    CBSEENCH12010538

    Give simple chemical tests to distinguish between the following pairs of
    compounds :
    Propanol and 2-methylpropan-2-ol

    Solution

    Lucas’ reagent Test
    Propanol – Primary alcohol
    2-methylpropan-2-ol – Tertiary alcohol

    Question 431
    CBSEENCH12010544

    Write the IUPAC name of the following:

     

    Solution

    IUPAC name the given compound is, 

    3,3 - dimethylpentan-2-ol

    Question 432
    CBSEENCH12010551
    Question 434
    CBSEENCH12010675

    Arrange the following compounds in order of decreasing acidity

    • II>IV>I>III

    • I>II>III>IV

    • III>I>II>IV

    • IV>III>I>II

    Solution

    C.

    III>I>II>IV

    Electron withdrawing group increases the acidic strength of the compounds by destabilising and stabilising the phenoxide ion formed respectively.
    The presence of an electron withdrawing group increases the acidic strength by stabilising the phenoxide ion. On other hands, the presence of electron releasing group destabilises the phenoxide ion and decrease the acidic strength.
    NO2>Cl>CH3>OCH3

    Question 435
    CBSEENCH12010676

    A solution of (–) –1 – chloro –1 – phenylethane is toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of

    • carbanion

    • Carbene

    • Carbocation

    • free radical

    Solution

    C.

    Carbocation


    Because of the planar structure of carbonation, both retention and inversion can take place.
    Question 436
    CBSEENCH12010707

    Iodoform can be prepared from all except

    • ethyl methyl ketone

    • isopropyl alcohol

    • 3-methyl-2-butanone

    • Isobutyl alcohol

    Solution

    D.

    Isobutyl alcohol

    Iodoform is given by
    1) methyl ketones R-CO-CH3
    2) alcohols of the type R-CH(OH)CH3
    where R can be hydrogen also

    Question 439
    CBSEENCH12010737

    From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is

    • 1-Butanol

    • 2-Butanol

    • 2-Methylpropan-2-ol

    • 2-Methylpropanol

    Solution

    C.

    2-Methylpropan-2-ol

    Formation of tertiary butyl carbocation is the most rapid compared to other lesser stable ones with Lucas reagent.

    Question 440
    CBSEENCH12010742

    The main product of the following reaction is

    straight C subscript 6 straight H subscript 5 CH subscript 2 CH left parenthesis OH right parenthesis CH left parenthesis CH subscript 3 right parenthesis subscript 2 space rightwards arrow with Conc. space straight H subscript 2 SO subscript 4 on top space ?

    Solution

    A.

    Question 445
    CBSEENCH12010821

    The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly

    • benzoyl chloride

    • benzyl chloride

    • o-and p-chlorotoluene

    • m-chlorotoluene

    Solution

    C.

    o-and p-chlorotoluene

    Due to o- and p- directing nature of CH3 group. 

    Question 446
    CBSEENCH12010831

    In the following sequence of reactions, 

    CH subscript 3 CH subscript 2 OH space rightwards arrow with space space space straight P space plus space straight I subscript 2 on top straight A space rightwards arrow from ether to Mg of straight B space rightwards arrow with HCHO on top space straight C space rightwards arrow with straight H subscript 2 straight O on top space straight D comma space then compound ‘D’ is

    • butanal

    • n-butyl alcohol

    • n-propyl alcohol

    • propanal

    Solution

    C.

    n-propyl alcohol

    CH subscript 3 CH subscript 2 OH space rightwards arrow with straight P space plus space straight I subscript 2 on top space stack CH subscript 3 CH subscript 2 straight I with left parenthesis straight A right parenthesis below space rightwards arrow from dry space ether to Mg of space stack CH subscript 3 CH subscript 2 MgI with left parenthesis straight B right parenthesis below
rightwards arrow with HCHO on top space stack CH subscript 3 CH subscript 2 CH subscript 2 OMgI with left parenthesis straight C right parenthesis below space rightwards arrow with straight H subscript 2 straight O on top stack space CH subscript 3 CH subscript 2 CH subscript 2 OH with left parenthesis straight D right parenthesis below space plus space Mg left parenthesis OH right parenthesis straight I
    therefore the compound D is n-propyl alcohol
    Question 449
    CBSEENCH12010866
    Question 453
    CBSEENCH12010921

    The reaction



    is fastest when X is

    • Cl

    • NH2

    • OC2H5

    • OCOR

    Solution

    A.

    Cl

    Conjugated acid of Cl- is a stronger acid i.e. HCl.

    Question 455
    CBSEENCH12010975

    On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is

    • CH3COOC2H5 + NaCl

    • CH3Cl + C2H5COONa

    • CH3COCl + C2H5OH + NaOH

    • CH3COONa + C2H5OH

    Solution

    A.

    CH3COOC2H5 + NaCl

    Aqueous NaCl is neutral hence there is no reaction between ethyl acetate and aqueous NaCl

    Question 458
    CBSEENCH12010978

    Which of the following undergoes reaction with 50% sodium hydroxide solution to give the corresponding alcohol and acid?

    • Phenol

    • Benzoic acid

    • Butanal

    • Benzaldehyde

    Solution

    D.

    Benzaldehyde

     A + NaOH  → alcohol  + acid
    Thus, it is Cannizzaro reaction
    A is thus aldehyde without H at alpha - carbon
    (as C6H5CHO.HCHO)
    2C6H5CHO + NaOH  → C6H5CH2OH + C6H5COONa

    Question 459
    CBSEENCH12010979

    Among the following compound which can be dehydrated very easily is

    Solution

    C.

    Dehydration of alcohol is in order 1°  < 2°  < 3°. 

    Question 460
    CBSEENCH12010988

    Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2 to form product B. A and B are respectively:

    Solution

    B.

    This step involves acid-base reaction at first followed by SNAE reactions:

    Step 1:

     

    Step 2:

    This step is called bromination.

    Mechanism of above reaction is given as,

    Question 463
    CBSEENCH12011031

    The reaction


    Can be classified as

    • Alcohol formation reaction

    • Dehydration reaction

    • Williamson alcohol synthesis reaction

    • Williamson ether synthesis reaction

    Solution

    D.

    Williamson ether synthesis reaction

    The formation of ether from alcohol in te presence of base followed by alkylation is known as Williamson ether synthesis reaction.

    Question 464
    CBSEENCH12011082

    The reaction,


    is called,

    • Williamson synthesis

    • Williamson continuous etherification process

    • Etard reaction

    • Gattermam -Koch reaction

    Solution

    A.

    Williamson synthesis

    The reaction of alkyl halides with sodium alkoxide or sodium phenoxide to form ethers is called Williamson synthesis. Here, in this reaction alkyl halide should be primary and alkoxide e.g. should be bulkier as shown below,
    R-X     +         R'-ONa       --->         R-O-R' + NaX
    alkyl halide      sodium alkoxide          ether

    Question 465
    CBSEENCH12011108

    Which of the following will not be soluble in sodium hydrogen carbonate ?

    • 2,4,6 trinitrophenol

    • Benzoic acid

    • o-nitrophenol

    • Benzenesulphonic acid

    Solution

    C.

    o-nitrophenol


    While 2,4,6-trinitrophenol, benzoic acid and benzene sulphonic acid are soluble in NaHCO3.This reaction is possible in the forward direction if acid is more acidic than H2CO3.o-nitrophenol is less acidic than H2CO3.Hence, it is not soluble in sodium hydrogen carbonate.
    Question 467
    CBSEENCH12011136

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

    Solution

    C.

    The ether, which gives more stable carbocation, gives CH3OH as one of the product with hot concentrated HI. the order of stability of carbocation is
    30 > 20 > 10
    Thus,   gives CH3 OH as one of the reaction. The reaction proceeds as

    Question 468
    CBSEENCH12011169

    In the following reaction,

    Solution

    A.

    Question 469
    CBSEENCH12011183

    Which of the following compounds can be used as antifreeze in automobile radiators?

    • Methyl alcohol

    • Glycol

    • Nitrophenol

    • Ethyl alcohol

    Solution

    B.

    Glycol

    Glycol is widely used as an antifreeze in automobile radiators.

    Question 471
    CBSEENCH12011253

    Which of the following is not the product of dehydration of 

    Solution

    B.

    When intermediate carbocation is stable, no rearrangement takes place in carbonation,

    Question 472
    CBSEENCH12011264

    Given are cyclohexanol (I), acetic acid (II), 2,4,6-trinitropenol (III) and phenol (IV). In these, the order of decreasing acidic character will be

    • III > II> IV> I

    • II > III> I > IV

    • II > III> IV> I 

    • III > IV> II >  I

    Solution

    A.

    III > II> IV> I

    Higher the tendency to give a proton, higher is the acidic character, and tendency to lose a proton depends upon the stability of intermediate, ie, carbanion formed.
    2,4,6 - trinitrophenol after the loss of a proton gives 2,4,6 -trinitrophenoxide ion which is stabilised by resonance -I effect and -M effect thus a most acidic among the given compounds. Phenol after losing a proton form phenoxide ion which is also stabilised by resonance, -M and -I effects but is less stabilised as compared to 2,4,6 trinitrophenoxide ions. Thus, it is less acidic as compared to 2,4,6 trinitrophenol. (CH3COOH) after losing a proton give acetate CH3OO- ion which is stabilised by only resonance. However, it is more resonance stabilised as compared to a phenoxide ion, thus more acidic as compared to phenol. 2,4,6 tri-nitrophenol, however, is more acidic than acetic acid due to the presence of three electron withdrawing  - NO2 groups. cyclohexanol gives an anion that is least stable among the given, thus, it is least acidic.
    Hence, the correct order of acidic strength is 2,4,6 -trinitrophenol > acetic acid > phenol > cyclohexanol
    III> II> IV> I

    Question 473
    CBSEENCH12011266

    Which one of the following compounds has the most acidic nature? 

    Solution

    B.

    The presence of electron withdrawing substituent increases the acidity while electron releasing substituent, when present, decrease the acidity.
    Phenyl is an electron withdrawing substituent while -CH3 is an electron releasing substituent, Moreover, phenoxide ion is more resonance stabilised as compared to benzoyl oxide ion, thus, release proton more easily. That's why is a strong acid among the given.

    Question 474
    CBSEENCH12011270

    Which one is most reactive towards electrophilic reagent?

    Solution

    A.

    Electron withdrawing substituent deactivates the benzene nucleus towards electrophilic substitution while electron releasing substituent activates the ring towards electrophilic substitution.
    Among the given -OH has the higher electron donating tendency and thus, activates the ring more towards electrophilic substitution. Hence

    is more reactive towards electrophilic reagents.

    Question 475
    CBSEENCH12011288

    Among the following four compounds

    A) Phenol 
    B) Methyl Phenol
    C) meta - nitrophenol
    D) para-nitrophenol

    the acidity order is

    • D > C > A > B

    • C > D > A> B

    • A > D > C > B

    • B > A > C> D 

    Solution

    A.

    D > C > A > B

    An electron withdrawing group (-I showing group like -NO2, -CN) stabilises the phenoxide ion, thus when present, increases the acidity of phenol. On the other hand, the electron releasing groups (+I showing group like -CH3, -C2H5), when present, decrease the acidity of phenol by destabilising phenoxide ion. Hence, the correct order of acidity of given compound is
    p-nitrophenol > m-nitrophenol > phenol > methyl phenol

    Question 476
    CBSEENCH12011346

    which one of the following is most reactive towards electrophilic attack? 


    Solution

    C.

    Electron releasing groups stabilise the intermediate cation whereas electron withdrawing groups destabilise it.
    The order of electron withdrawing group is:
    NO2 > Cl >OH> C6H5
    Hence is more reactive towards electrophilic substitution because -OH group least destabilise the carbocation. Further -OH group is strongly activating group while -NO2 and -Cl are deactivating group.

    Question 477
    CBSEENCH12011371

    The reaction: 


    Which of the following compounds will be formed ?

    Solution

    D.

    When conc. HI or HBr reacts with either, the corresponding alcohol and iodide is formed when there is a case of mixed ethers, the halogen atom attaches to the smaller alkyl group, due to steric effect.

    Question 478
    CBSEENCH12011385

    The order of decreasing reactivity towards an electrophilic reagent, for the following:

    i) Benzene
    ii) Toluene
    iii) Chlorobenzene
    iv) Phenol

    would be:

    • (i) > (ii) > (iii) > (iv)

    • (ii) > (iv) > (i) > (iii)

    • (iv) > (iii) > (ii) >(i)

    • (iv) > (ii) > (i) > (iii)

    Solution

    D.

    (iv) > (ii) > (i) > (iii)

    Benzene having any activating group i.e, OH, R etc., undergoes electrophilic substitution very easily as compared to benzene itself. Thus toluene (C6H5CH3), phenol (C6H5OH) undergo electrophilic substitution very readily than benzene.Chlorine with +E and +M effect deactivating the ring due to strong -I effect. So, it is difficult to carry out the substitution  in chlorobenzene than in benzene, so correct order is 
    Phenol > Toluene > Benzene > Chlorobenzene

    Question 479
    CBSEENCH12011400

    The major organic product in the reaction,

    CH3 -O- CH(CH3)2 + HI → Product is:

    • CH3OH + (CH3)2 + CHI

    • ICH2OCH(CH3)2

    • CH3O CI(CH3)2

    • CH3I + (CH3)2CHOH

    Solution

    D.

    CH3I + (CH3)2CHOH

    CH subscript 3 minus straight O minus CH left parenthesis CH subscript 3 right parenthesis subscript 2 space plus space HI space rightwards arrow with 373 space straight K on top space straight H subscript 3 Cl space plus space left parenthesis CH subscript 3 right parenthesis CHOH
    In the case of unsymmetrical ether, the alkyl halide is always formed from a smaller alkyl group. This happens so because I- ion being larger in size approaches smaller alkyl group to avoid steric hindrance.
    Question 480
    CBSEENCH12011417

    Which one is the most acidic compound?

    Solution

    D.

    –NO2 group has very strong –I & –R effects

    Question 482
    CBSEENCH12011432

    Identify A and predict the type of reaction

    Solution

    A.

    Question 485
    CBSEENCH12011457

    In the reaction,

    The electrophile involved is

    • Dichloromethyl cation (C+HCl2)

    • Formyl cation (C+HO)

    • Dichlorocarbene (:CCl2)

    • Dichloromethyl anion (C-HCl2)

    Solution

    C.

    Dichlorocarbene (:CCl2)

    Reimer-Tiemann reaction. The electrophile formed is :CCl2 (Dichlorocarbene) according to the following reaction

    CHCl3 + OH- CCl3  + H2OCCl3      :CCl2 Electrophile+ Cl-

    Question 486
    CBSEENCH12011475

    The product (P) is

    Solution

    A.

    This is an example of Williamson's synthesis.

    Question 487
    CBSEENCH12011507

    Aqueous 10% NaHCO3 solution is used as a reagent for identifying 'A'. Which of the following compounds yield 'A' on hydrolysis?

    • CH3COOC2H5

    • C2H5-COO-C2H5

    • CH3CHO

    • CH3CH2OH

    Solution

    A.

    CH3COOC2H5

    NaHCO3 is basic in nature so it is used in the identification of acid only. In the given compound only ester (CH3COOC2H5) on hydrolysis gives acid as

    CH3COOC2H5 + H2O  CH3COOHAcetic acid + C2H5OHEthyl alcohol

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