Chapter 08: d and f - Block Elements - Target Publications · 2018. 12. 14. · 5 Chapter 10:...

1

Chapter 08: d and f - Block Elements

Q.1. How is the variability of oxidation states of the transition elements different from that of the

non-transition elements? Illustrate with examples. (NCERT) Ans: i. The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in

such a way that their oxidation states differ from each other by unity. eg. Fe2+ and Fe3+, Cu+ and Cu2+, etc. ii. In case of non-transition elements, the oxidation states differ by units of two. eg. Pb2+ and Pb4+, Sn2+ and Sn4+, etc. iii. In transition elements, the higher oxidation states are more stable for heavier elements in a group. eg. In group 6, Mo (VI) and W (VI) are more stable than Cr (VI). iv. In p-block elements, the lower oxidation states are more stable for heavier members due to inert pair

effect. eg. In group 14, Pb (II) is more stable than Pb (IV). Q.2. Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with: i. iodide ii. iron (II) solution iii. H2S. (NCERT) Ans: Potassium dichromate, K2Cr2O7 is a strong oxidizing agent and is used as a primary standard in volumetric

analysis involving oxidation of iodides, ferrous ions and S2 ions, etc. i. It oxidizes potassium iodide to iodine. 2 3

2 7 2 2Cr O 14H 6I 2Cr 3I 7H O ii. It oxidizes iron (II) salt to iron (III) salt. 2 2 3 3

2 7 2Cr O 14H 6Fe 2Cr 6Fe 7H O iii. It oxidizes H2S to S. 2 2 3

2 7 2Cr O 14H 3S 3S 2Cr 7H O Q.3. Decide which of the following atomic numbers are the atomic numbers of the inner transition

elements: 29, 59, 74, 95, 102, 104. (NCERT) Ans: Inner transition elements have atomic numbers from 58 to 71 and 90 to 103. Hence, the atomic numbers 59,

95 and 102 are the atomic numbers of the inner transition elements. Q.4. Write the electronic configurations of the elements with the atomic numbers 61, 91 and 101.

(NCERT) Ans: The electronic configurations are given in the following table.

Element Symbol Atomic Number Electronic Configuration Promethium Pm 61 [Xe] 4f5 5d0 6s2 Protactinium Pa 91 [Rn] 5f2 6d1 7s2 Mendelevium Md 101 [Rn] 5f13 6d0 7s2

Q.5. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this

statement by giving some examples from the oxidation state of these elements. (NCERT) Ans: Lanthanoids show limited number of oxidation state, viz., +2, +3 and +4 (out of which +3 is most

common). This is because of large energy gap between 4f and 5d-subshells. The dominant oxidation state of actinoids is also +3 but they show a number of other oxidation states also. eg. uranium (Z = 92) and plutonium (Z = 94), show +3, +4, +5 and +6, neptunium (Z = 93) shows +3, +4, +5, +6 and +7 etc. This is due to small energy difference between 5f, 6d and 7s-subshells of the actinoids.

d and f-Block Elements08

2

Std. XII Sci.: Perfect Chemistry - II

Q.6. Compare the chemistry of actinoids with that of the lanthanoids with special reference to : (NCERT)

i. electronic configuration ii. oxidation state iii. atomic size and ionic sizes iv. chemical reactivity Ans:

Characteristics Lanthanoids Actinoids i. Electronic

configuration It may be represented by [Xe]4f014 5d0 or 1 6s2

It may be represented by [Rn]5f014 6d0 or 1 7s2

ii. Oxidation state Show +3 oxidation state only, except in few cases where it is +2 or +4. They never show more than +4 oxidation state.

Show higher oxidation states such as +4, +5, +6, +7 also in addition to +3 oxidation state.

iii. Atomic and ionic sizes The ionic radii of M3+ ions in lanthanoids series show a regular decrease in size of ions with increase in atomic number. This decrease is known as lanthanoid contraction.

There is a greater and gradual decrease in the size of atoms or M3+ ions across the series. This greater decrease is known as actinoid contraction.

iv. Chemical reactivity These are less reactive metals and form oxides, sulphides, nitrides, hydroxides and halides etc. These also form H2 with acids. They show a lesser tendency for complex formation.

These are highly reactive metals especially in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temperature. They show a greater tendency for complex formation.

3

Chapter 09: Coordination Compounds

Q.1. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4

solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why? (NCERT)

Ans: i. 1 : 1 molar mixture of FeSO4 and (NH4)2SO4 forms a double salt, FeSO4.(NH4)2SO4.6H2O (Mohr’s salt) which exists only in solid state. In aqueous solution, it dissociates into ions Fe2+, NH4

+ and SO4

2. Hence, it gives the test of Fe2+ ion. ii. 1 : 4 molar mixture of CuSO4 and ammonia forms a coordination compound, [Cu(NH3)4]SO4. In

aqueous solution, it retains its identity and does not dissociate into ions. Hence, it does not give the test of Cu2+ ion.

Q.2. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution

of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution? (NCERT)

Ans: i. When excess of aqueous KCN is added to an aqueous solution of copper sulphate, the coordination entity formed is [Cu(CN)4]2.

[Cu(H2O)4]2+ + 4CN [Cu(CN)4]2 + 4H2O ii. [Cu(CN)4]2 is a coordination compound and hence, retains its identity in the aqueous solution. It does

not dissociate to give free Cu2+ ions. Hence, no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution.

Coordination Compounds 09

4


Q.1. Write the structures of the following compounds: i. 2-Chloro-3-methylpentane ii. 4-tert-Butyl-3-iodoheptane iii. 1,4-Dibromobut-2-ene iv. 1-Chloro-4-ethylcyclohexane (NCERT) v. 2-(2-Chlorophenyl)-1-iodooctane vi. 2-Bromobutane Ans: i. ii. iii. iv. 1 2

v. I CH2 CH (CH2)5CH3 vi. Q.2. Write the isomers of compound having formula C4H9Br. (NCERT) Ans: The isomers of compound having formula C4H9Br are as follows: i. H3C CH2 CH2 CH2 Br ii. 1-Bromobutane iii. iv. Q.3. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical

chlorination yields: i. a single monochloride ii. three isomeric monochlorides iii. four isomeric monochlorides (NCERT) Ans: i. In neopentane (C5H12), all the H-atoms are equivalent and hence, on photochemical chlorination

yields only one type of monohalogen derivative.

Cl

C2H5

Cl2

1

CH3 – CH CH – CH2 – CH3

Cl

CH3

2 3 4 51

1 H2C – CH = CH – CH2

Br Br

2 3 4

CH3 CH CH2 CH3 1 2 3 4

Br

2-Bromobutane Br

H3C CH2 CH CH3

H3C C Br

CH3

CH3

2-Bromo-2-methylpropane1-Bromo-2-methylpropaneCH3

H3C CH CH2Br

5 67H3C – CH2 – CH2 – C CH – CH2 – CH3

CH3

H3C – C – CH3

H

4 1 23

I

Halogen Derivatives of Alkanes and Arenes 10

5

Chapter 10: Halogen Derivatives of Alkanes and Arenes

ii. In n-Pentane(C5H12), replacement of a,b or c type of H-atom leads to formation of three different types of monohalogen derivatives which are isomers of each other.

iii. In isopentane (C5H12), replacement of a, b, c or d type of H-atoms leads to formation of four different

types of monohalogen derivatives which are isomers of each other. Q.4. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with

sodium ethoxide in ethanol and identify the major alkene: i. 1-Bromo-1-methylcyclohexane ii. 2-Chloro-2-methylbutane iii. 3-Bromo-2,2,3-trimethylpentane (NCERT) Ans: i. In 1-Bromo-1-methylcyclohexane, the -hydrogen atoms on either side of the Br atoms are

equivalent, therefore, only one type of product i.e., 1-methylcyclohexene is formed. 2 5 2 5C H ONa/C H OH

HBr 1-Bromo-1-methylcyclohexane ii. 2-Chloro-2-methylbutane has two different sets of equivalent -hydrogen atoms and hence, in

principle can give two alkanes (I and II). But according to Saytzeff’s rule, more highly substituted alkene (II), being more stable, is the major product.

iii. 3-Bromo-2,2,3-trimethylpentane has two different sets of -hydrogen atoms and hence, in principle,

can give two alkenes (I and II). But according to Saytzeff's rule, more highly substituted alkene (II), being more stable, is the major product.

CH3 C CH3

CH3

CH3

Neopentane (2,2-Dimethylpropane)

CH3 CH2 CH2 CH2 CH3

a b c b a

n-Pentane(Pentane)

CH3 CH CH2 CH3

a b c d

Isopentane (2Methylbutane)

CH3 a

CH2 = C CH2CH3 + CH3 C = CHCH3

CH3 CH3

2-Methylbut-1-ene (I) 2-Methylbut-2-ene (II) (Minor product) (Major product)

2 5 2 5C H ONa /C H OHHCl

2-Chloro-2-methylbutane

CH3 C

Cl

CH2CH3

CH3

Br H

CH3

H

H H

CH3

1-Methylcyclohexene

6


Q.5. Identify A, B, C and R in following: i. ii. (NCERT) Ans: i. + Mg Dryether ii. R Br + Mg Dry ether R Mg Br D O2 CH3CHCH3 Since, D gets attached to same C-atom on which MgX was present. Therefore, R CH3 CH CH3 Thus, structure of ‘C’ is

Q.6. Write the structure of the major organic product in each of the following reactions: i. CH3CH2CH2Cl + NaI Dry acetone

Heat ii (CH3)3CBr + KOH EthanolHeat

iii. CH3CH(Br)CH2CH3 + NaOH Water iv. CH3CH2Br + KCN aq.ethanol v. C6H5ONa + C2H5Cl vi. CH3CH2CH2OH + SOCl2 vii. CH3CH2CH = CH2 + HBr Peroxide viii. CH3CH = C(CH3)2 + HBr

(NCERT) Ans: i. CH3CH2CH2Cl + NaI Dryacetone, heat

Finkelstein reaction CH3CH2CH2I + NaCl

Br

Cyclohexyl bromide

Cyclohexyl magnesium

bromide (A)

+ Mg(OH)Br

Cyclohexane(B)

MgBr H O2

D Alkyl

bromideAlkyl

magnesiumbromide

Br + Mg Dryether A H O2 B

D

R Br + Mg Dryether C D O2 CH3CHCH3

C2H5ONa/C2H5OH HBr

2-Ethyl-3,3-dimethylbut-1-ene (I) (Minor product)

H3C

H3C 4

CH2 1

CH3 – C – C CH2 – CH33 2

C2H5ONa/C2H5OH HBr

3,4,4-Trimethylpent-2-ene (II)(Major product)

CH3 – C – C = CH – CH3

H3C

H3C 5 2

CH3 1 3 4

CH3 – C – C CH2 – CH3

3-Bromo-2,2,3-trimethylpentane

H3C

H3C 1 4

CH3 5

Br

3 2

MgBrIsopropyl magnesium bromide

CH3 CH CH3

1-Iodopropane1-Chloropropane

7

Chapter 10: Halogen Derivatives of Alkanes and Arenes

ii. (CH3)3CBr + KOH Ethanol, Dehydrohalogenation

+ KBr + H2O iii. + NaOH Water

Hydrolysis

iv. CH3CH2Br + KCN aq.ethanol + KBr v. C6H5ONa+ + C2H5Cl Williamson 's

synthesis + NaCl

vi. CH3CH2CH2OH + SOCl2 Reflux

Δ + HCl + SO2 vii. CH3CH2CH = CH2 + HBr Peroxide

Anti-Markownikoff's rule

viii. CH3CH = C CH3 + HBr Markownikoff's rule

Q.7. What happens when: i. n-butyl chloride is treated with alcoholic KOH? ii. bromobenzene is treated with Mg in the presence of dry ether? iii. ethyl chloride is treated with (aq) KOH? iv. methyl bromide is treated with sodium in the presence of dry ether? v. methyl chloride is treated with KCN? (NCERT) Ans: i. CH3CH2CH2CH2Cl + KOH(alc.) CH3CH2CH = CH2 + KCl + H2O ii. iii. iv. v.

But-1-enen-Butyl chloride

EthaneMethyl bromide

2CH3Br + 2Na Dry etherWurtzreaction

CH3CH3 + 2NaBr

Ethyl chloride EthanolCH3CH2Cl + KOH(aq) CH3CH2OH + KCl + H2O

C6H5Br + Mg Dry ether C6H5MgBrBromobenzene Phenyl magnesium

bromide

Methyl chloride

(alc.) Methyl cyanide

CH3Cl + KCN CH3CN + KCl

CH3CH2CN

C6H5OCH2CH3PhenetoleSodium phenoxide

CH3CH2CH2Cl 1-Chloropropane

1-Bromoethane

Propan-1-ol

Propane nitrile

But-1-ene CH3CH2CH2CH2Br

1-Bromobutane (Major product)

CH3CHCH2CH3

Br 2-Bromobutane

CH3 CH CH2 CH3

OHButan-2-ol

2-Bromo-2-methylpropane CH3 C = CH2

CH3

2-Methylpropene

CH32-Methylbut-2-ene

CH3CH2 C CH3

CH3

Br 2-Bromo-2-methylbutane

8


Q.8. Arrange the compounds of each set in order of reactivity towards 2NS displacement: i. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane. ii. 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane. iii. 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,

1-Bromo-3-methylbutane. (NCERT) Ans: i. Since, due to steric reasons, the order of reactivity in 2NS reactions follows the order : 1 > 2 >3,

therefore, order of reactivity of the given alkyl bromide is 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane. ii. Since, due to steric reasons, the order of reactivity of alkyl halides in 2NS reaction follows the order:

1 > 2 > 3, therefore, the order of reactivity of the given alkyl bromides is 1-Bromo-3-methylbutane (1) > 2-Bromo-3-methylbutane (2) > 2-Bromo-2-methylbutane (3). iii.

Since, in case of 1 alkyl halides, steric hindrance increases in the order: n-alkyl halides, alkyl halide

with a substituent at position other than the -position, one substituent at the -position, two substituents at the -position, therefore, the reactivity decreases in the same order. Thus, the reactivity of the given alkyl bromides decreases in the order:

1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane. Q.9. Predict the order of reactivity of the following compounds in 1NS and 2NS reactions. i. The four isomeric bromobutanes. ii. C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br (NCERT) Ans: i. CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr 1N(S )

CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr 2N(S )

CH3 C CH2 CH3

Br

CH3

2-Bromo-2-methylbutane (3)

CH3 CH CH2 CH2 CH3

Br 2-Bromopentane (2)

CH3 CH2 CH2 CH2 CH2 Br 1-Bromopentane (1)

CH3 CH CH2 CH2 Br

CH3


1 2 3 4

CH3 C CH2 CH3

Br

CH3


1 3 2 4

CH3 CH CH CH3

Br

CH3


1 3 2 4

CH3 C CH2 Br

CH3

CH3

1-Bromo-2,2-dimethylpropane

(1 with two -methyl groups)

CH3 CH2 CH2 CH2 Br 1-Bromobutane

(1 With no branching)

CH3 CH CH2 CH2 Br

CH3

1-Bromo-3-methylbutane

(1 with one methyl group at -position)

CH3 CH2 CH CH2 Br

CH3

1-Bromo-2-methylbutane

(1 with one -methyl group)

9

Chapter 10: Halogen Derivatives of Alkanes and Arenes Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more

stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH – group. Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in 1NS reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary bromide. Hence, the above order is followed in 1NS . The reactivity in 2NS reactions follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.

ii. C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br 1N(S )

C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br < C6H5CH(CH3)Br < C6H5CH2Br 2N(S ) Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more

stable than obtained from C6H5CH(CH3)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in 1NS reactions. A phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than C6H5CH(CH3)Br in 2NS reaction.

Q.10. Which compound in each of the following pairs will react faster in 2NS reaction with OH? i. CH3Br or CH3I ii. (CH3)CCl or CH3Cl (NCERT) Ans: i. Since I ion is a better leaving group than Br ion, therefore, CH3I reacts faster than CH3Br in 2NS

reaction with OH ion. ii. On steric grounds, 1 alkyl halides are more reactive than tert-alkyl halides in 2NS reactions.

Therefore, CH3Cl will react at a faster rate than (CH3)3CCl in a 2NS reaction with OH ion. Q.11.Write the structures of the following organic halogen compounds: i. p-Bromochlorobenzene ii. Perfluorobenzene (NCERT) Ans: i. ii. Q.12. Draw the structures of major monohalo products in each of the following reactions: i. ii.

(NCERT) Ans: i. Only alcoholic but not phenolic OH groups are replaced by Cl on heating with HCl. ii.

Cl

Brp-Bromochlorobenzene

+ HCl heat CH2OH

HO

CH2CH3

2Br ,heatorUV light

O2N

CH CH3

O2N1

4

Br

4-(1-Bromoethyl)nitrobenzene

CH2CH3 2Br ,heat or

UVlightO2N4-Ethylnitrobenzene

CH2OH

+ HCl heatHO4(Hydroxymethyl)phenol

CH2Cl

HO1

4

4-Chloromethylphenol

FF

F

F

F F

Perfluorobenzene

10


Q.1. Write structures of the compounds whose IUPAC names are as follows: i. Cyclohexylmethanol ii. 3-Cyclohexylpentan-3-ol iii. Cyclopent-3-en-1-ol iv. 3-(Chloromethyl)pentan-1-ol (NCERT) Ans: i. ii. iii. iv. Q.2. Name the following compounds according to IUPAC system: i. ii. iii. iv. v. (NCERT) Ans: i. ii. iii. iv.

CH3 OH

CH2OH

CH3 CH CH2 CH – CH – CH3

OH

H2C = CH – CH – CH2 CH2 CH3

CH3 CH2 CH CH CH CH3

CH2Cl CH3

CH2OH

Hex-1-en-3-olOH

H2C = CH – CH – CH2 CH2 CH3

CH3

CH3 – C = C – CH2OH

Br

3-Chloromethyl-2-isopropylpentan-1-ol

CH3 CH2 CH CH CH CH3

CH2Cl CH3

CH2OH

2,5-Dimethylhexane-1,3-diolCH3 OH

CH2OH

CH3 CH CH2 CH – CH – CH3

OH

Br

3-Bromocyclohexanol

OH

Br

3-(Chloromethyl)pentan-1-ol

HO CH2 CH2 CH CH2

CH2 CH3

Cl

CH2 OH

Cyclohexylmethanol

Cyclopent-3-en-1-ol

OH3-Cyclohexylpentan-3-ol

CH3 CH2 C CH2 CH3

OH

Alcohols, Phenols and Ethers11

11

Chapter 11: Alcohols, Phenols and Ethers

v. Q.3. Write the IUPAC names of the following compounds: i. ii. iii. iv. (NCERT) Ans: i. ii. iii. iv. Q.4. Draw the structures of all isomeric alcohols of molecular formula C5H12O, give their IUPAC names

and classify them as primary, secondary and tertiary alcohols. (NCERT) Ans: Eight isomers are possible with molecular formula C5H12O. These are: i. CH3 CH2 CH2 CH2 CH2 OH ii. CH3 CH2 CH2 CH CH3 Pentan-1-ol (1 alcohol) Pentan-2-ol (2 alcohol) iii. iv. v. vi. vii. viii.

2-Bromo-3-methylbut-2-en-1-olCH3

CH3 – C = C – CH2OH

Br

OH

OH

CH3 CH2 CH CH2 CH3 Pentan-3-ol (2 alcohol)

CH3 CH2 CH CH2 – OH

2-Methylbutan-1-ol (1 alcohol)

CH3

CH3 C CH2 – CH3 2-Methylbutan-2-ol (3 alcohol)

CH3

OH

CH3 CH CH2 – CH2 OH

3-Methylbutan-1-ol (1 alcohol)

CH3

CH3 CH CH CH3 3-Methylbutan-2-ol (2 alcohol)

CH3 OH

CH3 C CH2 – OH 2,2-Dimethylpropan-1-ol (1 alcohol)

CH3

CH3

2,2,4-Trimethylpentan-3-ol

OH

CH3 CH CH C CH3

CH3

CH3

CH3

5-Ethylheptane-2,4-diol

OH

H3C CH CH2 CH CH CH2 CH3

OH C2H5

Butane-2,3-diol

OH

CH3 CH CH CH3

OHPropane-1,2,3-triol

OH

HO CH2 CH CH2 OH

OH

CH3 CH CH C CH3

CH3

CH3

CH3 OH

H3C CH CH2 CH CH CH2 CH3

OH C2H5

OH

CH3 CH CH CH3

OH OH

HO CH2 CH CH2 OH

12


Q.5. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?

i. ii. (NCERT) Ans: i. In the preparation of isobutyl alcohol (CH3 CH – CH2OH) from methanal, CH2 OH is the part which comes from methanal and the remaining part comes from the Grignard reagent. ii. In the preparation of cyclohexylmethanol from methanal, CH2 OH is the part which comes from

methanal and the remaining part comes from Grignard reagent. Q.6. Give structures of the products you would expect when each of the following alcohol reacts with: i. HCl – ZnCl2 ii. HBr iii. SOCl2. a. Butan-1-ol. b. 2-Methylbutan-2-ol. (NCERT) Ans: i. HCl – ZnCl2: a. Butan-1-ol being a primary alcohol does not react with Lucas reagent (HCl – ZnCl2) at room

temperature. However, cloudiness appears only upon heating. CH3CH2CH2CH2OH conc HCl

anhydrous ZnCl , Heat2 CH3CH2CH2CH2Cl

Butan-1-ol 1-Chlorobutane b. But 2-methylbutan-2-ol (3) reacts at room temperature giving cloudiness immediately. CH3 – C – CH2 – CH3 conc HCl

Room temperature CH3 – C – CH2 – CH3

2-Methylbutan-2-ol 2-Chloro-2-methylbutane (Cloudiness) ii. HBr: Both alcohols react with HBr to give corresponding alkyl bromides. a. CH3CH2CH2CH2OH + HBr CH3CH2CH2CH2Br Butan-1-ol 1-Bromobutane

Cyclohexyl magnesium

bromide

H C H + Dryether C HOH

H + Mg(OH)Br

Formaldehyde (Methanal) Addition compound

Cyclohexylmethanol

O Mg Br CH2OH H O Mg Br

H

CH2OHCH3 CH CH2OH

CH3

CH3

Isopropyl magnesium bromide

H C H + CH3 CH Mg Br Dryether C HOH

HCH3 CH CH2 OH+ Mg(OH)Br

Formaldehyde(Methanal)

Isobutyl alcohol (2-Methylpropan-1-ol)

CH3

O

CH3

Addition compound CH3

H

CH CH3

O Mg Br

H

CH3|

| OH

CH3|

| Cl

13

Chapter 11: Alcohols, Phenols and Ethers

b. CH3 – C – CH2 – CH3 + HBr CH3 – C – CH2 – CH3 2-Methylbutan-2-ol 2-Bromo-2-methylbutane iii. SOCl2: Both alcohols react to give corresponding alkyl chlorides. a. CH3CH2CH2CH2OH + SOCl2 Pyridine

reflux CH3CH2CH2CH2Cl + SO2 + HCl Butan-1-ol 1-Chlorobutane b. CH3 – C – CH2 – CH3 + SOCl2 Pyridine

reflux CH3 – C – CH2 – CH3 + SO2 + HCl 2-Methylbutan-2-ol 2-Chloro-2-methylbutane Q.7. Write IUPAC names of the following compounds: i. ii. iii. iv. v. (NCERT) Ans: i. 2-Methylphenol ii. 4-Methylphenol iii. 2,5-Dimethylphenol iv. 2,6-Dimethylphenol v. 2,3-Diethylphenol Q.8. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.

(NCERT) Ans: Three isomers of monohydric phenols with molecular formula C7H8O are possible. They are as follows: Q.9. Write chemical reaction for the preparation of phenol from chlorobenzene. (NCERT) Ans:

CH3|

|OH

CH3|

| Br

OH C2H5

C2H5

OH

2-Methylphenol

CH3

OH

3-Methylphenol

CH3

OH

4-MethylphenolCH3

Cl

Chlorobenzene

+ 2NaOH 2

613K,300atmNaCl, H O 2 2

3

dil.HCl, NaClor H O CO ,

NaHCO

ONa

Sodium phenoxide

OH

Phenol

CH3 OH

CH3

OH

CH3

OH CH3

CH3 OH

CH3

CH3|

|OH

CH3 |

|Cl

14


Q.10. Give the equations of reactions for the preparation of phenol from cumene. (NCERT) Ans: Q.11. Suppose you are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of

phenol using these reagents. (NCERT) Ans: C6H6 2 4conc.H SO /

(Sulphonation) C6H5SO3H

2 3

fused NaOHNa SO C6H5ONa 2 4H SO

Hydrolysis C6H5OH Benzene Benzene sulphonic acid Sodium phenoxide Phenol Q.12. Give the IUPAC names of the following ethers: i. C2H5OCH2 CH CH3 ii. CH3 O CH2CH2Cl | CH3 iii. O2N C6H4 OCH3 (p) iv. CH3 CH2 CH2 OCH3 v. vi. vii. CH3 O CH2 CH CH3 viii. C6H5 O C7H15 (n ) | CH3 ix. CH3 CH2 O CH CH2 CH3 (NCERT) | CH3 Ans: i. 1-Ethoxy-2-methylpropane ii. 2-Chloro-1-methoxyethane iii. 4-Nitroanisole iv. 1-Methoxypropane v. 4-Ethoxy-1,1-dimethylcyclohexane vi. Ethoxybenzene vii. 1-Methoxy-2-methylpropane viii. 1-Phenoxyheptane ix. 2-Ethoxybutane Q.13. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and

3-methylpentan-2-ol. (NCERT) Ans:

+ O2

H3C C H

Cumene or (Isopropyl benzene or 2-Phenylpropane)

CH3

H3C C O O H

Cumene hydroperoxide

CH3

2 4dil.H SO + CH3 C CH3

Acetone

OH

Phenol

OCobalt naphthenate

423K(alkalinemedium)

(Air)

OC2H5

H3C CH3

OC2H5

CH3CH2OH HBr CH3CH2Br

BromoethaneEthanol

2CH3CH2 CH CH OH Na 2CH3CH2 CH CH O Na+ + H2

Sodium 3-methylpentan-2-oxide3-Methylpentan-2-ol

| | CH3 CH3

| | CH3 CH3

15

Chapter 11: Alcohols, Phenols and Ethers Q.14. How is 1-propoxypropane synthesized from propan-1-ol? Write mechanism of this reaction.

(NCERT) Ans: It can also be prepared by dehydration of propan-1-ol with conc. H2SO4 at 413K. Q.15. Write the names of reagents and equations for the preparation of the following ethers by

Williamson’s synthesis: i. 1-Propoxypropane ii. Ethoxybenzene iii. 2-Methoxy-2-methylpropane iv. 1-Methoxyethane. (NCERT) Ans: i. 1-Propoxypropane: CH3CH2CH2ONa+ + CH3CH2CH2 Br HeatCH3CH2CH2 O CH2CH2CH3 + NaBr Sodium propoxide 1-Bromopropane 1-Propoxypropane ii. Ethoxybenzene: + CH3CH2 Br Heat + NaBr Sodium phenoxide Bromoethane Ethoxybenzene iii. 2-Methoxy-2-methylpropane: CH3 CH3 | | CH3 C O Na+ + CH3 Br Heat CH3 C OCH3 + NaBr | Bromomethane | CH3 CH3 Sodium 2-methyl-2-propoxide 2-Methoxy-2-methylpropane iv. 1-Methoxyethane: CH3CH2ONa+ + CH3 Br Heat CH3CH2 O CH3 + NaBr Sodium ethoxide Bromomethane 1-Methoxyethane

O Na+ OCH2CH3

CH3CH2CH2ONa+ + CH3CH2CH2 Br Dryether CH3CH2CH2OCH2CH2CH3 + NaBr

1-PropoxypropaneSodium n-propoxide

CH3CH2 CH CH O Na+ + CH3CH2Br CH3CH2 CH CH OCH2CH3 + NaBr

Sodium 3-methylpentan-2-oxide 2-Ethoxy-3-methylpentane

| | CH3 CH3

| | CH3 CH3

Bromoethane

CH3CH2CH2OH P,Br2 CH3CH2CH2Br + H3PO3

Propan-1-ol 1-Bromopropane

CH3CH2CH2OH Na CH3CH2CH2ONa + 12

H2 Propan-1-ol Sodium

n-propoxide

1-PropoxypropaneCH3CH2CH2OH H

413K

CH3CH2CH2OCH2CH2CH3

Propan-1-ol

16


Q.16. Predict the product of the following reactions: i. CH3 CH2 CH2 O CH3 + HBr ii. + HBr iii. 2 4

3

conc.H SOconc.HNO

iv. (CH3)3C OC2H5 HI (NCERT) Ans: i. Both the alkyl groups attached to the oxygen atom are primary, therefore attack of Br – ion occurs on

the smaller methyl group leading to the formation of Propan-1-ol and Bromomethane. CH3 CH2 CH2 O CH3 + HBr 373K CH3 CH2 CH2 OH + CH3 Br n-Propyl methyl ether Propan-1-ol Bromomethane ii. Due to resonance, C6H5-O bond has some double bond character and hence, is stronger than OC2H5

bond. Therefore, the cleavage of the weaker OC2H5 bond occurs to yield Phenol and Bromoethane. + HBr 373K + CH3CH2 Br Ethoxybenzene Phenol Bromoethane iii. Due to the strong +R effect of the OC2H5 group, it is activating as well as o, p-directing . Therefore

nitration of ethoxybenzene will give a mixture of 2-Ethoxynitrobenzene. and 4-Ethoxynitrobenzene. 2 4

3

conc.H SOconc.HNO +

Ethoxybenzene 4-Ethoxynitrobenzene 2-Ethoxynitrobenzene (Major product) (Minor product)

iv. Since tert-Butyl carbocation is much more stable than ethyl carbocation, reaction follows 1NS mechanism leading to the formation of tert-butyl iodide and ethanol as shown below:

HI H+ + I Step I: Formation of carbocation: Step II: Formation of tert-Butyl iodide: CH3 CH3 | | I + CH3 C + Fast CH3 C I | | CH3 CH3 tert-Butyl iodide

OC2H5

OC2H5 OH

OC2H5

CH3 C O CH2CH3 H

CH3 C O CH2CH3 1NS

Slow CH3 C+ + CH3CH2OH

CH3

CH3

CH3

CH3 H CH3

CH3

+

tert-Butyl ethyl ether

Ethanol

tert-Butyl carbocation

OC2H5 OC2H5

NO2 1

2 OC2H5

O2N

4

1

tert-Butyl carbocation

17

Chapter 12: Aldehydes, Ketones and Carboxylic Acids

Q.1. Write the structures of products of the following reactions: i. ii. (C6H5CH2)2Cd + 2CH3COCl iii. H3C C C H

2Hg ,H SO2 4

iv. (NCERT) Ans: i. Benzene reacts with C2H5COCl, to yield propiophenone (Friedel Craft’s acylation). ii. Dibenzyl cadmium reacts with acetyl chloride to give Benzyl methyl ketone. iii. Propyne, in presence of hot dil. H2SO4 and HgSO4, forms acetone.

iv. Oxidation of methyl group by chromyl chloride gives a chromium complex. Acid hydrolysis of the

chromium complex gives corresponding aldehyde.

+ C2H5 Cl Anhyd AlCl3CS2

O

C

i. CrO Cl2 2ii .H O3

CH3

NO2

CrO Cl2 2CS2

H O3

CH3

4-Nitrotoluene 4-NitrobenzaldehydeNO2

CH(OCrOHCl2)2

NO2

CHO

NO2 Chromium complex

+ H5C2 Cl Anhyd AlCl3CS2

O

C

O

C C2H5 + HCl

Benzene Propiophenone Propanoylchloride

2CH3 C Cl + (C6H5CH2)2Cd 2CH3 C CH2C6H5 + CdCl2 O O

Acetyl chloride Dibenzyl cadmium Benzyl methyl ketone

H3C C CH + H2O 2 4 4dil.H SO HgSO333K

H3C C = CH2 Tautomerism CH3 C CH3

OHProp-1-en-2-ol

AcetonePropyne

O

Aldehydes, Ketones and Carboxylic Acids 12

18


Q.2. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:

i. Ethanal, Propanal, Propanone, Butanone ii. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. (Hint: Consider steric effect and electronic effect.) (NCERT) Ans: i. Ethanal, Propanal, Propanone, Butanone: a. Due to inductive (+I) effect aldehydes have more electrophilic carbonyl carbon than ketones. b. Hence aldehydes are more susceptible (to the attack of nucleophile) than ketones. c. Hence the reactivity of propanal and ethanal is higher than that of butanone and propanone. d. As the steric hindrance increases, reactivity decreases because the attack of the nucleophile to

the carbonyl carbon becomes more difficult. e. Hence the reactivity of propanal is lower than that of ethanal. Also the reactivity of butanone is

lesser than that of propanone. f. The increasing order of reactivity in nucleophilic addition reactions is, Butanone < Propanone < Propanal < Ethanal. ii. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone: a. The effect of methyl group at para position of aromatic aldehydes is not significant. Hence the

reactivity of benzaldehyde and p-tolualdehyde is comparable. b. The nitro group at para position is an electron withdrawing group and makes the carbonyl

carbon more electrophilic. Hence p-nitrobenzaldehyde is more reactive than benzaldehyde. c. Aldehydes have more electrophilic carbonyl carbon than ketones. Hence aldehydes are more

subceptible (to the attack of nucleophile) than ketones. d. hence benzaldehyde is more reactive than acetophenone. e. The increasing order of reactivity in nucleophilic addition reactions is, Acetophenone < p-Tolualdehyde , Benzaldehyde < p-Nitrobenzaldehyde. Q.3. Predict the products of the following reactions: i. ii. iii. iv. (NCERT) Ans: i. Cyclopentanone reacts with hydroxyl amine to form oxime. ii. Cyclohexanone reacts with 2,4-dinitro phenyl hydrazine to form 2,4-dinitro phenyl hydrazone.

+ HO NH2 H

O

+ H2N NH

O2N

NO2 O

+ HO NH2 H + H2O

O

Hydroxylamine

Cyclopentanone

N OH

Cyclopentanoneoxime

R CH = CH CHO + H2N C NH NH2 H

O

C O

CH3 + CH3CH2NH2 H

O2N O

Cyclohexanone

+ H2N NH NO2 + H2O

2,4-Dinitro phenyl hydrazone

N NH

O2N

NO2

2,4-Dinitro phenyl hydrazine

19


iii. , unsaturated aldehyde reacts with semicarbazide (H2NCONHNH2) to form semicarbazone. iv. Acetophenone reacts with ethyl amine to form an imine Q.4. Draw structures of the following derivatives: i. The 2,4-dinitrophenylhydrazone of benzaldehyde

ii. Cyclopropanone oxime iii. Acetaldehydedimethylacetal iv. The semicarbazone of cyclobutanone v. The ethylene ketal of hexan-3-one vi. The methyl hemiacetal of formaldehyde (NCERT) Ans: i. 2,4-dinitrophenylhydrazone of benzaldehyde ii. Cyclopropanone oxime iii. Acetaldehydedimethylacetal iv. The semicarbazone of cyclobutanone v. The ethylene ketal of hexan-3-one vi. The methyl hemiacetal of formaldehyde Q.5. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents: i. PhMgBr and then H3O+ ii. Tollen’s reagent iii. Semicarbazide and weak acid iv. Excess ethanol and acid v. Zinc amalgam and concentrated hydrochloric acid (NCERT) Ans: i. The reaction of cyclohexanecarbaldehyde with PhMgBr followed by acid hydrolysis gives an alcohol.

NO2

CH = N NH NO2

CH3 CH O CH3 O CH3

O H3C CH2 C CH2 CH2 CH3

O

CH2 H2C

N OH

N NH C NH2

O

Acetophenone

C O

CH3 + CH3CH2NH2 H

Ethylamine

C N CH2 CH3

CH3

Substituted imine

HO C O CH3

H

H

HR CH = CH CH = N C NH NH2 + H2O

O

Semicarbazide , Unsaturated aldehyde Semicarbazone

O

R CH = CH CHO + H2N C NH NH2

CHO

1. PhMgBr2. H O3

C Ph

OH

H Cyclohexane Carbaldehyde

1-Cyclohexyl-1-phenylmethanol

20


ii. Oxidation of cyclohexane carbaldehyde with Tollen’s reagent gives cyclohexane carboxylate ion. iii. The reaction of cyclohexane carbaldehyde with semicarbazide and weak acid gives a semicarbazone. iv. The reaction of cyclohexane carbaldehyde with excess ethanol and acid gives an acetal. v. Zinc amalgam and concentrated hydrochloric acid reduces CHO group to CH3 group

(Clemmensen’s reduction). Q.6. Give simple chemical tests to distinguish between the following pairs of compounds: i. Acetophenone and Benzophenone ii. Phenol and Benzoic acid iii. Benzoic acid and ethyl benzoate iv. Benzaldehyde and acetophenone v. Ethanal and propanal (NCERT) Ans: i. Acetophenone and Benzophenone: Acetophenone being a methyl ketone when treated with NaOI

(I2 / NaOH) gives yellow precipitate of iodoform whereas benzophenone does not. ii. Phenol and Benzoic acid: Phenol is a weak acid it does not react with weak base NaHCO3 whereas

benzoic acid is a strong acid. It reacts with NaHCO3 to form a sodium salt alongwith evolution of CO2.

Cyclohexane carbaldehyde

CHO

+ H2NNH C NH2 H

CH = NNH C NH2

O

Semicarbazide

O

Cyclohexane carbaldehyde semicarbazone

CHOZinc

amalgamconc.HCl

CH3


1-Methyl cyclohexane

CHO

Tollen's reagent

C O

O


Cyclohexane carboxylate ion

OC2H5Cyclohexane carbaldehyde

CHO

excessethanolacid

C

H

OC2H5

Cyclohexane carbaldehydediethyl acetal

PhCOCH3 + 3NaOI PhCOONa + CHI3 + 2NaOH Acetophenone Sodium

benzoateIodoform

(yellow precipitate)

PhCOPh + 3NaOI No yellow precipitate Benzophenone

+ NaHCO3(aq) No reaction

OH

Sodium bicarbonatePhenol

21


iii. Benzoic acid and ethyl benzoate: Benzoic acid is a carboxylic acid and reacts with NaHCO3 to form

a sodium salt alongwith evolution of CO2. Ethylbenzoate is an ester. It does not react with NaHCO3. iv. Benzaldehyde and acetophenone: Benzaldehyde being an aldehyde reduces Tollen’s reagent to

shining silver mirror whereas acetophenone being a ketone does not. v. Ethanal and propanal: Ethanal contains CH3CO group. Hence on treatment with NaOI (I2 / NaOH)

gives yellow precipitate of iodoform whereas propanal does not. Q.7. i. Write structural formulae and names of four possible aldol condensation products from

propanal and butanal. ii. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

(NCERT) Ans: i. ii.

+ NaHCO3(aq) + H2O + CO2

C OH

Sodium bicarbonateBenzoic acid

O

C ONa

O

Sodium benzoate

+ NaHCO3(aq) + H2O + CO2

C OH

Sodium bicarbonateBenzoic acid

O

C ONa

O

Sodium benzoate

+ NaHCO3(aq) No reactionCOCH2CH3

Sodium bicarbonateEthyl benzoate

O

PhCHO + 2[Ag(NH3)2]+ + 3OH PhCOO + 2Ag + 4NH3 + 2H2OBenzaldehyde Tollen’s

reagent Benzoate

ion Silver metal

PhCOCH3 Tollen 's reagent No silver mirrorAcetophenone

CH3CH2CHO + CH3CH2CH2CHO i. dil.NaOH

ii. H O2

CH3CH2CH2C = C CHO

CH3

H

Propanal (Nucleophile)

Butanal (Electrophile)

2-Methylhex-2-enal

CH3CH2CHO + CH3CH2CH2CHO i. dil NaOH

ii. H O2

CH3CH2C = C CHO

CH2CH3

H

Propanal (Electrophile)

Butanal (Nucleophile)

2-Ethylpent-2-enal

CH3CHO + 3NaOI HCOONa+ + CHI3 + 2NaOHEthanal Sodium

formateIodoform (yellow

precipitate)CH3CH2CHO + 3NaOI No yellow precipitate

Propanal

22

Std. XII Sci.: Perfect Chemistry - II iii.

iv.

Q.8. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction

and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction:

i. Methanal ii. 2-Methylpentanal iii. Benzaldehyde iv. Benzophenone v. Cyclohexanone vi. 1-Phenylpropanone vii. Phenylacetaldehyde viii. Butan-1-ol ix. 2,2-Dimethylbutanal (NCERT) Ans:

Compound Reaction Product i. HCHO

Methanal Cannizzaro H3C OH + HCOONa+

Methanol Sodium formate

ii. Aldol condensation

iii. Cannizzaro

iv. PhCOPh Benzophenone

Neither

v. Aldol condensation

Benzaldehyde

CHO

+

Benzyl alcohol

CH2OH

Sodium benzoate

COONa+

O Cyclohexanone

O

2-Methylpentanal CH3CH2CH2CHCHO

CH3

CH3CH2CH2CHO + CH3CH2CH2CHO i. dil NaOH

ii. H O2

CH3CH2CH2C = C CHO

CH2CH3

H

Butanal (Electrophile)

Butanal (Nucleophile)

2-Ethylhex-2-enal

CH3CH2CHO + CH3CH2CHO i. dil NaOH

ii. H O2

CH3CH2C = C CHO

CH3

H

Propanal (Electrophile)

Propanal (Nucleophile)

2-Methylpent-2-enal

2,4-Dimethyl-2-propylhept-3-enal

CH3CH2CH2 C

CH3

HC

CH2

C CH3

CH2

CH3

CHO

23


vi. Ph CH2COCH3

1-Phenyl propanone Aldol condensation

vii. Ph CH2CHO Phenyl acetaldehyde

Aldol condensation

viii. CH3CH2CH2CH2OH Butan-1-ol

Neither

ix. Cannizzaro reaction

Q.9. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative reduces Tollen’s

reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. (NCERT)

Ans: i. The organic compound (A) is 2-Ethylbenzaldehyde ii. Its molecular formula is C9H10O. iii. Since it is an aldehyde, it can form 2,4-DNP derivative and can reduce Tollens’ reagent. It can also

undergo Cannizzaro reaction, as it does not contain H atom. iv. On vigorous oxidation, both CHO group and the ethyl side chain are oxidized to COOH groups to

give 1,2-benzene dicarboxylic acid HOOC C6H4 COOH. Q.10. What is meant by the following terms? Give an example of the reaction in each case. i. Cyanohydrin ii. Acetal iii. Semicarbazone iv. Aldol v. Hemiacetal vi. Oxime vii. Ketal viii. Imine ix. 2,4-DNP-derivative x. Schiff’s base (NCERT) Ans: i. Cyanohydrin: It is a compound in which cyano and hydroxyl groups are present on same carbon

atom.

CH3CH2C CH2OH

CH3

CH3

2,2-Dimethylbutanol

2,2-Dimethylbutanoic acid

CH3CH2C COOH

CH3

CH3

+

CH2CH3

CHO

CH3CHO HCN CH3 C CN Ethanal

Acetaldehyde cyanohydrin

OH

H

4-Methyl-3,5-diphenylpent-3-en-2-one Ph C = C CH2 Ph

CH3CO

CH3

2,4-Diphenylbut-2-enal PhCH2C = C Ph

CHO

2,2-Dimethylbutanal

CH3 CH2 C CHO

CH3

CH3

24


ii. Acetal: It contains two alkoxy groups, one alkyl group and one H atom on the same carbon atom. iii. Semicarbazone: It is the condensation product of an aldehyde or ketone with semicarbazide. iv. Aldol: It is hydroxy aldehyde or ketone obtained by the condensation of two aldehyde or ketone

molecules in presence of a dilute alkali. v. Hemiacetal: It contains one hydroxyl and one alkoxy group on the same carbon atom. vi. Oxime: It is the condensation product of an aldehyde or ketone with hydroxyl amine. vii. Ketal: It contains two alkoxy groups and two alkyl groups on the same carbon atom. viii. Imine: It contains group. It is the condensation product of aldehydes and ketones with ammonia derivatives.

CH3COCH3 + H O2 C Acetone

Ethylene glycol

CH2 OH

CH2 OH

CH2 CH2

CH3 H3C

OO

Ketal

CH3COCH3 + NH3 CH3 C CH3

Imine

NH

Acetone Ammonia

CH3COCH3 + H2NNHCONH2 C + H2O

CH3H3C

NNHCNH2

O

Acetone semicarbazone

Acetone Semicarbazide

CH3CHO + H2NOH CH3 C = NOH

H

Ethanal Hydroxylamine

Oxime

C = N

CH3CHO + H O3

C Ethanal

Ethylene glycol

CH2 OH

CH2 OH

CH2 CH2

HH3C

OO

Acetal

CH3CHO + CH3CHO dil. NaOH CH3 CH CH2 CHO

OH

3-Hydroxy butanal (aldol)Ethanal Ethanal

CH3CHO + CH3OH Dry HCl gas CH3 C OCH3

OH

HAcetaldehyde Methanol

Hemiacetal

25


ix. 2,4-DNP derivative: Also known as 2,4-dinitro phenyl hydrazone, it is condensation product of an aldehyde or ketone with

2,4-dinitro phenyl hydrazine (2,4-DNP). x. Schiff’s base: It is azomethine obtained by the condensation of aldehydes and ketones with primary

amines. Q.11. Give the IUPAC names of the following compounds: i. PhCH2CH2COOH ii. (CH3)2C = CHCOOH iii. iv. (NCERT) Ans: i. PhCH2CH2COOH ii. (CH3)2C = CHCOOH 3-Phenylpropanoic acid 3-Methylbut-2-enoic acid iii. iv. Q.12. Show how each of the following compounds can be converted to benzoic acid. i. Ethyl benzene ii. Acetophenone iii. Bromobenzene iv. Phenylethene (Styrene) (NCERT) Ans: i. Oxidation of ethyl benzene with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid. ii. Oxidation of acetophenone with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid.

KMnO4KOH H O3

CH2CH3 COOK COOH

Benzoic acidEthyl benzene Potassium benzoate

KMnO4KOH H O3

COOK COOH

Benzoic acidAcetophenone

COCH3

Potassiumbenzoate

CH3COCH3 + H O3 Acetone

NH

N = C

NO2

NO2 2,4-DNP derivative

CH3

CH3NH

NH2

NO2

NO2 2,4-dinitrophenyl

hydrazine

NO2

COOH NO2

O2N

CH3

COOH

CH3

COOH

2-Methylcyclopentanecarboxylic acid

NO2

COOH NO2

O2N2,4,6-Trinitrobenzoic acid

CH3CHO + C2H5NH2 H

CH3CH = NCH2CH3 + H2O Ethanal Ethanamine Schiff’s base

26


iii. Bromobenzene reacts with magnesium to form Grignard reagent which attacks carbon dioxide to form an intermediate which on acid hydrolysis gives benzoic acid.

iv. Oxidation of phenyl ethene with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid. Q.13.An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to

give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved. (NCERT)

Ans: The organic compound A is butyl butanoate Q.14.Complete each synthesis by giving missing starting material, reagent or products. i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. (NCERT)

CH3CH2CH2CH2OH [O] OxidationChromic acid(H CrO )2 4

CH3CH2CH2COOH Butanol

(C) Butanoic acid

(B)

CH3CH2CH2CH2OH Dehydration CH3CH2CH = CH2 Butanol

(C) But-1-ene

KMnO4KOH, heat

CH2CH3 COOH

COOH

SOCl2heat

C

O

Ag(NH )3 2

CHO

O CHO

COOH

NaCN/HCl

C6H5CHO + CH3CH2CHO

dil.NaOH CH3COCH2COOC2H5 (i ) NaBH4(ii ) H

OH CrO3 CH2 CHO

(i) O3(ii ) Zn H O2 2 O

C6H5CHO H NCONHNH2 2

KMnO4KOH H O3

Benzoic acid Phenyl ethene

CH = CH2

COOK COOH

Potassiumbenzoate

Phenyl magnesium bromide

Mg O C O HOHdil.HCl

Benzoic acidBromo benzene

Br MgBrOMgBr

O COOH

CH3CH2CH2COOCH2CH2CH2CH3 dilH SO2 4hydrolysis CH3CH2CH2COOH + CH3CH2CH2CH2OH

Butyl butanoate[C8H16O2](A)

Butanoic acid (B)

Butanol(C)

27


Ans: i. ii. iii. iv. v. vi. vii. viii. ix. x. xi.

C H COCl6 5AlCl3

C

O

Benzene Benzophenone

Ag(NH )3 2OH

CHO

O

COO

O

CH3COCH2COOC2H5i. NaBH4ii. H CH3CCH2COOC2H5

H

OH

OH CrO3 O

Cyclohexanol Cyclohexanone

CH2 i. BH3ii. H O /OH2 2iii.PCC

CHO

i. O3ii. Zn H O2 2 O

Potassiumbenzoate

Ethyl benzene

KMnO4KOH, heat

CH2CH3 COOK

C6H5CHO + CH3CH2CHO

i.dil.NaOHii. C6H5CH = C CHO

CH3

COOH

COOH

SOCl2heat

COCl

COClPhthalic acid Phthaloyl chloride

C6H5CHO H NCONHNH2 2 C6H5CH = NNHC NH2

O

Benzaldehyde Benzaldehyde semicarbazone

CHO

COOH NaCN/HCl

C

COOH H

CN

Cyanohydrin

OH

1-Formylbenzoic acid

28


Q.15.Give plausible explanation for each of the following: i. Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. ii. There are two NH2 groups in semicarbazide. However, only one is involved in the formation of

semicarbazones. iii. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an

acid catalyst, the water or the ester should be removed as soon as it is formed. (NCERT) Ans: i. In cyclohexanone, the carbonyl carbon is not hindered. Hence, the nucleophile CN ion can easily

attack the carbonyl carbon. However in 2,2,6-trimethyl cyclohexanone, the carbonyl carbon is sterically hindered due to presence of methyl groups. Hence, the nucleophile CN ion cannot easily attack the carbonyl carbon.

ii. Out of two NH2 groups in semicarbazide, one is a part of amide functional group. In this, the lone pair on nitrogen atom is involved in the resonance with carbonyl group. Hence, this NH2 group cannot act as a nucleophile. Hence, it is not involved in the formation of semicarbazones.

iii. The formation of esters from a carboxylic acid and alcohol in the presence of an acid catalyst is a reversible reaction. If water or ester is not removed as soon as it is formed, the ester will hydrolyse to give starting materials carboxylic acid and alcohol.

Q.16.How will you bring about the following conversions in not more than two steps? i. Propanone to Propene ii. Benzoic acid to Benzaldehyde iii. Ethanol to 3-Hydroxybutanal iv. Benzene to m-Nitroacetophenone v. Benzaldehyde to Benzophenone vi. Bromobenzene to 1-Phenylethanol vii. Benzaldehyde to 3-Phenylpropan-1-ol viii. Benzaldehyde to -Hydroxyphenylacetic acid (NCERT) Ans: i. Propanone to Propene: ii. Benzoic acid to Benzaldehyde: iii. Ethanol to 3-Hydroxybutanal: iv. Benzene to m-Nitroacetophenone: v. Benzaldehyde to Benzophenone:

Benzoyl chloride

Benzoic acid

SOCl2SO , HCl2 H ,Pd/BaSO2 4

COCl

Benzaldehyde

CHOCOOH

Acetophenone Benzene

(CH CO) O3 2Anhyd.AlCl3

Conc.HNO3Conc.H SO2 4

COCH3 COCH3

NO2

m-Nitroacetophenone

CH3CH2OH Cu,573K CH3CHO dil.NaOHAldol

condensation CH3 CH CH2CHO

Ethanol

OH

3-Hydroxybutanal Ethanal

CH3COCH3 NaBH ,CH OH4 3reductionCH3 CH CH3 Conc.H SO2 4

443K CH3CH = CH2 Propanone

OH

Propan-2-ol Propene

C6H5CHO K Cr O2 2 7H C6H5COOH CaCO3 (C6H5COO)2Ca Distil C6H5COC6H5

Benzaldehyde Benzoic acid Benzophenone Calcium benzoate

29


vi. Bromobenzene to 1-Phenylethanol: vii. Benzaldehyde to 3-Phenylpropan-1-ol: viii. Benzaldehyde to -Hydroxyphenylacetic acid:

Benzaldehyde

+ CH3CHO dil NaOH,crossaldolcondensation

H , Ni2Catalytic

hydrogenation

3-Phenyl prop-2-enal

CH = CHCHO CHO

Acetaldehyde 3-Phenylpropan-1-ol

CH2CH2CH2OH

Benzaldehyde

NaCN,HClpH 9 10 H ,H O2

Hydrolysis

-Hydroxy phenyl acetic acid

CHO CH CN

OH

CHCOOH

OH

Benzaldehyde cyanohydrin

Phenyl magnesium bromide

Bromobenzene

Mgdryether i. CH CHO3

ii. H O3

MgBr

1-Phenyl ethanol

CH CH3Br

OH

www.targetpublications.org/std-12th-perfect-chemistry-2-notes-science-mh-board.html

Date post:	05-Aug-2021
Category:	Documents
Upload:	others
View:	2 times
Download:	0 times

Chapter 08: d and f - Block Elements - Target Publications · 2018. 12. 14. · 5 Chapter 10:...

Documents