//M0//QN1//SUB//DL0//EQ

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

//X

(i) Cyanohydrin: gem-Hydroxynitrile compounds in which cyanide and hydroxyl groups are attached with the same carbon atom are called cyanohydrins. These are prepared by addition of HCN to aldehydes or ketones in a basic medium.
(ii) Acetal: gem-Dialkoxy compounds in which the two alkoxy groups are present on the same carbon atom are called acetals. These are prepared by the reaction of aldehyde with two equivalents monohydric alcohol in presence of dry HCl gas.
When dihydric alcohol is used cyclic acetal is formed
(iii) Semicarbazone: Semicarbazone comes from aldehydes and ketones, which is formed by the condensation reaction between a ketone or aldehyde and semicarbazide.
(iv) Aldol: An aldol is a β-hydroxy aldehyde or ketone. It is produced in the presence of a base by the condensation reaction of two molecules of the same or a single molecule, each of two different aldehydes or ketones.
(v) Hemiacetal: gem-alkoxyalcohol compounds are called hemiacetals. These are prepared by the reaction of aldehyde with one equivalent monohydric alcohol in presence of dry HCl gas.
(vi) Oxime: Oximes are produced when aldehydes and ketones react with hydroxyl amine in acidic medium.
(vii) Ketal: Ketals are produced when a ketone is heated with dihydric alcohols like ethylene glycol in presence of dry HCl.
(viii) Imine: The chemical compounds which have double bond between carbon-nitrogen are called Imines. These are prepared by the reaction of aldehydes or ketones with ammonia derivatives.
= N –– Z = N –– Z + H 2 O
Z = alkyl, aryl, –NH 2, –OH, –NHC 6 H 5, – NHCONH 2, etc.
(ix) 2, 4-DNP-derivative: 2,4-Dinitrophenyl hydrazine are produced when aldehydes or ketones react with 2,4-dinitrophenyl hydrazine in weak acidic medium.
(2, 4-DNP derivatives are used for identification and characterisation of aldehydes and ketones.)
(x) Schiff’s base: Aldehydes and ketones in the presence of a residue of acid on treatment with primary aliphatic or aromatic amines yield a base of a Schiff.
e.g.,
CH 3 CH = NCH 2 CH 3 + H 2 O

//M0//QN2//SUB//DL0

Name the following compounds according to IUPAC system of nomenclature.

(i) CH 3 CH (CH 3) CH 2 CH 2 CHO
(ii) CH 3 CH 2 COCH(C 2 H 5)CH 2 CH 2 CI
(iii) CH 3 CH = CHCHO
(iv) CH 3 COCH 2 COCH 3
(v) CH 3 CH(CH 3)CH 2 C(CH 3)2 COCH 3
(vi) (CH 3)3 CCH 2 COOH
(vii) OHCC 6 H 4 CHO-p

//X

(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-enal
(iv) Pentan-2, 4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde

//M0//QN3//SUB//DL0//EQ

Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p, p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

//X

(i) 3-Methylbutanal:
(ii) p–Nitropropiophenone:
(iii) p-Methylbenzaldehyde:
(iv) 4-Methylpent-3-en-2-one:
(v) 4-Chloropentan-2-one:
(vi) 3-Bromo-4-phenylpentanoic acid:
(vii) p, p’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid:

//M0//QN4//SUB//DL0//EQ

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. (i) CH 3 CO(CH 2)4 CH 3 (ii) CH 3 CH 2 CHBrCH 2 CH(CH 3)CHO (iii) CH 3(CH 2)5 CHO (iv) Ph-CH=CH-CHO (v) (vi) PhCOPh

//X

No. | Structure | IUPAC name | Common name
(i) | CH 3 CO(CH 2)4 CH 3 | Heptan-2-one | –
(ii) | CH 3 CH 2 CHBrCH 2 CH(CH 3)CHO | 4-Bromo-2- methylhexanal | γ-Bromo a-methyl capro aldehyde
(iii) | CH 3(CH 2)5 CHO | Heptanal | –
(iv) | Ph-CH=CH-CHO | 3-Phenylprop- 2-enal | b-Phenyl acrolein
(v) |  | Cyclopentane- carbaldehyde | –
(vi) | PhCOPh | Diphenyl- methanone | Benzophenone

//M0//QN5//SUB//DL0//EQ

Draw structures of the following derivatives. (i) The 2, 4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehyde dimethyl acetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde

//X

(i)
(ii)
(iii)
(iv)
(v) (vi)

//M1//QN6//SUB//DL0//EQ

Predict the products formed when cyclohexane-carbaldehyde reacts with following reagents. (i) PhMgBr and then H3O+ (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric

acid

[Topic 8.4] [1 Mark Each]

//X

(i) Reaction of cyclohexanecarbaldehyde with PhMgBr and then H 3 O+:
(ii) Reaction of cyclohexane carbaldehyde with Tollen's reagent:
(iii) Reaction of cyclohexane carbaldehyde with semicarbazine and weak acid:
(iv) Reaction of cyclohexane carbaldehyde with excess ethanol and Acid:
(v) Reaction of cyclohexane carbaldehyde with zinc amalgam and dil. HCl

//M0//QN7//SUB//DL0//EQ

Which of the following compound undergoes to aldol condensation? Which compound give reaction with Cannizzaro process and which compound can not give reactions? 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

//X

2-methylpentanal, cyclohexanone, 1-phenyl-propanone and phenylacetaldehyde contains one or more α-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given below:
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
Methanal, benzaldehyde and 2,2-dimethylbutanal do not contain a-hydrogen and hence undergo Canizzaro reaction. The reactions and the structures of the expected products are given below:

(ix) 2,2-dimethylbutanal

Benzophenone (iv) is a ketone having no α-hydrogen while butan-1-ol (viii) is an alcohol. Both of these neither undergo aldol condensation nor cannizzaro reaction.

//M0//QN8//SUB//DL0//EQ

How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid

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(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid

//M0//QN9//SUB//DL0//EQ

Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which act as electrophile.

//X

(a) Propanal acts as both nucleophile as well as electrophile.
(b) Propanal as electrophile and butanal as nucleopile.
(c) Butanal as electrophile and propanal as nucleophile.
(d) Butanal acts as both nucleophile as well as an electrophile.

//M0//QN10//SUB//DL0//EQ

An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

//X

The compound having molecular formula C9H10O forms 2, 4-DNP derivative and reduces Tollen’s reagent.
Therefore, the given compound must be an aldehyde.
Again, the compound gives 1, 2 benzene dicarboxylic acids and undergoes Cannizzaro reaction followed by oxidation.
Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted.
Hence, the compound is found to be 2-ethylbenzaldehyde.
The given reactions can be explained by the following equations.

//M0//QN11//SUB//DL0//EQ

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.

//X

A is an organic compound with a molecular formula C 8 H 16 O 2. This gives a carboxylic acid (B) and alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound (A) must be an ester.
Further, oxidation of alcohol (C) with chromic acid gives acid (B). Thus, (B) and (C) must contain an equal number of carbon atoms.
A total of 8 carbon atoms are present in compound (A), each of (B) and (C) contain 4 carbon atoms.
Again, alcohol (C) gives but-1-ene on dehydration. Therefore, (C) is of straight-chain and hence, it is butan-1-ol.
On oxidation, Butan-1-ol gives butanoic acid. Hence, acid (B) is butanoic acid.
Hence, the ester with molecular formula C 8 H 16 O 2 is butyl butanoate.
All the given reactions can be explained by the following equations.

//M0//QN12//SUB//DL0

Arrange the following compounds in increasing order of their property as indicated :(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)(ii) CH 3 CH 2 CH(Br)COOH, CH 3 CH(Br) CH 2 COOH, (CH 3)2 CHCOOH, CH 3 CH 2 CH 2 COOH

(acid strength)(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

//X

(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN–. Therefore, the reactivity with HCN decreases, when the negative charge on the compound increases. The +I effect increases in the given compound. Steric hindrance also increases in the same. Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) The stability of the carboxyl ion increases by any group that helps to stabilize the negative charge, will increase the strength of the acid. Thus, groups having −I effect will increase the strength of the acids and groups having +I effect will decrease the strength of the acids. In the given compounds, Br− group has −I effect and −CH3 group has +I effect. Thus, acids containing Br− are stronger.
Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence,
CH 3 CH 2 CH 2 COOH is a stronger acid than (CH 3)2 CHCOOH.
Also, as the distance increases, +I effect grows weaker. Hence, CH 3 CH 2 CH(Br)COOH is a stronger acid than CH 3 CH(Br)CH 2 COOH.
Hence, the strengths of the given acids increase as:
(CH 3)2 CHCOOH < CH 3 CH 2 CH 2 COOH < CH 3 CH(Br)CH 2 COOH < CH 3 CH 2 CH(Br) COOH
(iii) The strength of acid is decreased by the electron donating group, while the strengths of acid increases by electron-withdrawing groups. As methoxy group is an electron-donating group, benzoic acid is a stronger acid than 4-methoxybenzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths of acid.
As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strength of the given acids increases as:
4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid

//M0//QN13//SUB//DL0//EQ

Give simple chemical tests to distinguish between the following pairs of compounds. (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal

//X

(i) Propanal and Propanone: Propanal and Propanone can be distinguish by iodoform test. At least one methyl group should be present in aldehydes and ketones linked to the carbonyl carbon atom to respond to iodoform test. They are oxidized by sodium hypoiodite (NaOI) to give iodoforms. Propanone being a methyl ketone responds to this test, but propanal does not.
CH 3 COCH 3 + 3NaOI → CHI 3↓ + CH 3 COONa + 2NaOH
Iodoform (Yellow ppts.)
(ii) Acetophenone and Benzophenone:
Acetophenone and Benzophenone can be distinguish by Iodoform test. C 6 H 5 COCH 3 + 3NaOI → CHI 3↓ + C 6 H 5 COONa + 2NaOH
Iodoform
CH 3 CH 2 CHO + NaOl → X (No reaction)
(Yellow ppts.)
C 6 H 5 COC 6 H 5 + 3NaOI → No yellow ppts.
(iii) Phenol and Benzoic acid: Phenol and benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test: Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet coloration.
6C 6 H 5 OH + FeCl 3 → [Fe(OC 6 H 5)]3– + 3H+ + 3Cl
Phenol Iron-phenol complex (violet coloration)
A buff coloured ppt of ferric benzoate is produced when benzoic acid reacts with neutral FeCl3.

3C 6 H 5 OH + FeCl 3 → (C 6 H 5 COO)3 Fe + 3HCl

Benzoic acid Ferric benzoate (Buff coloured ppt)
(iv) Benzoic acid and Ethylbenzoate: Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test.
Sodium bicarbonate test: Brisk effervescence is produced when acids react with NaHCO3 due to the evolution of CO2 gas.
Benzoic acid being an acid respond to this test, but ethylbenzoate does not.
C 6 H 5 COOH + NaHCO 3 → C 6 H 5 COONa + H 2 O + CO 2 ↑ + H 2 O
Benzoic acid Sodium benzoate
C 6 H 5 COOC 2 H 5 + NaHCO 3 → No effervescence due to evolution of CO 2 gas
(v) Pentan-2-one and Pentan-3-one: Pentan-2-one and pentan-3-one can be distinguished by iodoform test.
Iodoform test: Pentan-2-one responds to this test as it is a methyl ketone. But pentan-3-one not being a methyl ketone does not respond to this test.
(vi) Benzaldehyde and Acetophenone: Benzaldehyde and acetophenone can be distinguished by
Iodoform Test.
Iodoform Test: A yellow precipitate of iodoform is given by acetophenone (a methyl ketone) when it undergoes oxidation by sodium hypoiodite (NaOI). But benzaldehyde does not respond to this test.
(vii) Ethanal and Propanal: Ethanal and propanal can be distinguished by iodoform test.
Iodoform test: Carbonyl carbon atom having at least one methyl group in aldehydes and ketones responds to the iodoform test. Also, Ethanal having one methyl group linked to the carbonyl carbon atom responds to this test. But, there are no methyl group linked to the carbonyl carbon atom in propanal and thus, it does not respond to this state.
CH 3 CHO + 3NaOI →
HCOONa + CHI 3 + 2NaOH
CH 3 CH 2 CHO + NaOl → X
Ethanal Sodium Iodoform
methanoate (yellow ppt)

//M0//QN14//SUB//DL0//EQ

How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom: (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde.

//X

(i)
(ii)
(iii)
(iv)
(v)

//M0//QN15//SUB//DL0//EQ

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 l-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to a-Hydroxyphenylacetic acid (ix) Benzoic acid to m-Nitrobenzyl alcohol

//X

(i) Propanone to propene:
(ii) Benzoic acid to benzaldehyde:
(iii) Ethanol to 3-hydroxy butanal:
(iv) Benzene to m-nitroacetophenone:
(v) Benzaldehyde to benzophenone:
(vi) Bromobenzene to 1-phenylethanol:
(vii) Benzaldehyde to 3-phenylpropan-1-ol:
(viii) Benzaldehyde to a-hydroxyphenylacetic acid:
(ix) Benzoic acid to m-nitrobenzyl alcohol:

//M0//QN16//SUB//DL0//EQ

Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation

//X

(i) Acetylation: When an organic compound is introduced with an acetyl functional group, it is known as acetylation. Bases such as pyridine, dimethylaniline, etc. are present when this process is carried out. An acetyl group is substituted with an active hydrogen atom in this process. Acetylating agents such as acetyl chloride and acetic anhydride are commonly used in the process.
For example, acetylation of ethanol produces ethyl acetate.
CH 3 CH 2 OH + CH 3 COCl CH 3 COOCH 2 CH 3 + HCl
Ethanol Acetyl Chloride Ethyl acetate
(ii) Cannizzaro reaction: Aldehydes which do not have an a-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
(iii) Cross aldol condensation: When aldol condensation is carried out between two different aldehydes
and / or ketones, it is called cross aldol condensation. If both of them contain a-hydrogen atoms, it gives a mixture of four products.
(iv) Decarboxylation: The reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime is called decarboxylation. When aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed also results in decarboxylation. This electrolytic process is known as Kolbe’s electrolysis.
R–COONa R–H + Na 2 CO 3

//M0//QN17//SUB//DL0//EQ

Complete each synthesis by giving missing starting material, reagent or products. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)

//X

(i)
(ii)
(iii) C 6 H 5 CHO
C 6 H 5 CH = NNHCONH 2 + H 2 O
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)

//M0//QN18//SUB//DL0//EQ

Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in

good yield but 2,2,6-trimethyl cyclo hexanone 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.

//X

(i) Cyanohydrins are formed by cyclohexanones according to the following equation.
In this case, there will not be any steric hindrance, hence the nucleophile CN– can easily attack. However, in the case of 2, 2, 6-trimethylcyclohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN− cannot attack effectively.
(ii) Semicarbazide undergoes resonance involving only one of the two −NH 2 groups, which is attached directly to the carbonyl-carbon atom.
Therefore, the electron density on −NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other −NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.
RCOOH + R’OH RCOOR’ + H 2 O
Carboxylic acid Alcohol Ester Water
If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

//M0//QN19//SUB//DL0//EQ

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

//X

% of carbon = 69.77 %
% of hydrogen = 11.63 %
% of oxygen = [100 − (69.77 + 11.63)] %
= 18.6 %

Element | Atomic mass (gm⁄mol) | Percentage (%) | Atomic ratio = | Simple ratio | NO. of atoms
C | 12 | 69.77 | = 5.81 | = 5.0 | 5
H | 1 | 11.63 | = 11.63 | = 10.0 | 10
O | 16 | 18.6 | = 1.16 | = 1.0 | 1

Therefore, the empirical formula of the compound is C5H10O. Now, the empirical formula mass of the compound can be given as:
= 5 × 12 + 10 × 1 + 1 × 16
= 86 gm/mol
Molecular mass of the compound = 86 gm/mol
Therefore, the molecular formula of the compound is given by C5H10O.
Tollen’s reagent is not reduced by the given compound, hence it is not an aldehyde.
A positive iodoform test is given by the compound and also forms sodium hydrogen sulphate addition products. Since the compound is not an aldehyde, it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence, the given compound is pentan−2−one.

//M0//QN20//SUB//DL0

Give reason: carboxylic acids are more acidic than alcohols and phenols.OR Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

//X

Refer Que. no. 48 Page no. 181
Class 12 Chemistry (Part 2) 013