AP Inter 2nd Year Chemistry Study Material Chapter 12(b) Aldehydes, Ketones, and Carboxylic Acids

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids

Very Short Answer Questions

Question 1.
Arrange the following compounds in increasing order of their property indicated.

1. Acetaldehyde, Acetone, Methyl t. butyl ketone reactivity towards HCN.
2. Floroacetic acid, monochloroacetic acid, Acetic acid and Dichloroacetic acid (acid strength)

Answer:

1. Due to the presence of groups around the carbonyl group the reactivity of a compound depends on the steric hindrance.
   Greater the steric hindrance, less will be the reactivity of the compound. Reactivity towards HCN is in the following order.
   Methyl tertiary butyl ketone < Acetone < Acetaldehyde
2. Acid strength of given compounds is Dichloro acetic acid > fluoro acetic acid > chloro acetic acid > acetic acid.

Question 2.
Write the reaction showing a-halogenation of carboxylic acid and give its name.
Answer:
Carboxylic acids having a-hydrogens are halogenated at the a-position on treatment with chlorine or bromine in presence of small amount of red phosphorous to give a-halo carboxylic acids.
This reaction is named as Hell-volhard – Zelinsky (HvZ) reactions

Question 3.
Although phenoxide ion has more number of resonating structures thancarboxylate ion carboxylic acid is a stronger acid than phenol. Why ?
Answer:

• Phenoxide ion has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom.
• The negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas in phenoxide ion the negative charge less effectively delocalised over one oxygen atom and less electronegative carbon atoms.
  

Question 4.
How do you distinguish acetophenone and benzophenone?
Answer:
On idoform test Acetophenone gives positive, where as benzophenone (C₆H₅COC₆H₅) does not

Question 5.
Explain the position of electrophilic substitution In benzolc acid.
Answer:
Benzoic acid undergo electrophilic substitution reactions in which carboxyl group acts as a deactivating and meta dirécting group.

Question 6.
Write equations showing the conversion of
i) Acetic acid to Acetyl chloride
ii) Benzoic acid to Benzamide
Answer:
i) acetic acid reacts with PCl₃ (or) PCl₅ (or) SOCl₆ to form acetyl chloride

ii) Benzoic acid reacts with ammonia to form benzamide

Question 7.
An organic acid with molecular formula C₈H₈O₂ on decarboxylation forms Toluene. Identify the organic acid.
Answer:
The organic acid is phenyl acetic acid

Question 8.
List the reagents needed to reduce carboxylic acid to alcohol.
Answer:
The Reagents required to reduce carboxylic acid to alcohol are

1. LiAlH₄/Ether (or) B₂H₆
2. H₄O⁺

Question 9.
Write the mechanism of esterification.
Answer:
Mechanism of esterification of carboxylic acids : The esterification of carboxylic acids with alcohols is a kind of nucleophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Proton transfer in the tetrahedral intermediate converts the hydroxyl group into – ⁺OH₂ group, which, being a better leaving group, is eliminated as neutral water molecule. The protonated ester so formed finally loses a proton to give the ester.

Question 10.
Compare the acidic strength of acetic acid, Chloroacetic acid, benzoic acid and Phenol. [Mar. 14]
Answer:
Benzoic acid (C₆H₅COOH) > Chloro acetic acid (ClCH₂COOH) > Acetic acid (CH₃COOH) > Phenol (C₆H₅OH)

Short Answer Questions

Question 1.
Write the equations of any aldehyde with Fehlings reagent.
Answer:
Fehling’s reagent is mixture of two solutions Fehling’s A and Fehlings B.
Fehling’s A – aq. CuSO₄ solution
Fehling’s B – Sodium potassium tartarate (Rochelle salt)
Acetaldehyde reacts with Fehlings .eagent and gives a redbrown ppt.
Reaction:
CH₃ – CHO + 2CU⁺² + 5OH^– → RCOO^– + CU₂O + 3H₃O (Red – brown ppt)

Question 2.
What is Tollens reagent? Explain Its reaction with Aldehydes.
Answer:
Tollens Reagent : Freshly prepared ammonicai silver nitrate solution is called Tollens reagent.

On warming an aldehyde with Tollens reagent a bright silver mirror is produced due to formation of silver metal.
R – CHO + 2 [Ag(NH₃)₂]⁺ + 3OH^– → RCOO^– + 2Ag + 2H₂O + 4NH₃

Question 3.
Write the oxidation products of : Acetaldehyc, Acetone and Acetophenone.
Answer:
a) Acetaldehyde under goes oxidation to foim acetic acid.

b) Acetone undergoes oxidation to form acetic acid

c) Acetophenone undergoes oxidation to form benzoicacid and chloroform

Question 4.
Explain why Aldehydes and ketones undergoes nucleophilic addition while alkenes undergoes electrophilic addition though both are unsaturated compounds.
Answer:
The carbon – oxygen double bond in carbonyl compounds is polarised due to higher electronegativity of oxygen relative to carbon. Hence the carbonyl carbon is an electrophilic and carboxyl oxygen is a nucleophilic centre. So aldehydes, ketones undergoes nucleophilic addition reaction.

Alkenes contain which is a source of electron density so electrophiles add no to C = C to give addition products. Hence alkenes undergo electrophilic addition reaction.

Question 5.
Write the IUPAC names of the following:
i) CH₃CH₂CH(Br) CH₃COOH
ii) Ph. CH₃COCH₃COOH
iii) CH₃.CH (CH₃) CH₂COOC₂H₅
Answer:

Question 6.
Arrange the following In the increasing order of their acidic strength:
Benzolc acid, 4 – Methoxybenzoic acid, 4 – Nitrobenzoic acid and 4 – Methylbenzoic acid.
Answer:
Electron donating group (-OCH₃) decreases the acidic strength where as electron withdrawing group (NO₂) increases the same.
Increases order of acidic strength is:
4-Methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3, 4-dinitro benzoic acid.

Question 7.
DescrIbe the following: .
i) Cross aldol condensation
ii) Decarboxylation
Answer:
i) Cross Aldol Condensation : When aldol condensation ¡s earned out between two different aldehydes and (or) ketones, it is called cross aldol condensation.

If both the reactants contain α-hydrogen atoms, it gives a mixture of four products.

Ketones can also be used as one component in the cross aldol reactions

ii) Decarboxylatlon : Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with sodalizne (a mixutre of NaOH & CaO in ratio 3 : 1)
.-, This reaction is called decarboxylation

Question 8.
ExplaIn the role of electron withdrawing and electron releasing groups on the acidity of carboxylic acids.
Answer:

• Electron with drawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through decocalisation of the negative charge by inductive effect.
• Electron donating groups decreases the äcidity of carboxylic acids by destabilising the conjugate base.
  
  Eg : Cl^– is a electron with drawing group acidic strength order in case of chloro acetic acids
  CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH

Question 9.
Draw the structures of the following derivatives:
i) Acetaldehyde dimethyl acetal
ii) The ethylene ketal of hexan-3-one
iii) The methyl hemiacetal of formaldehyde.
Answer:
The structures of following are
i) Acetaldehyde dimethyl acetal

ii) Ethylene Ketal of hexan-3-one

iii) Methyl hemiacetal of formaldehyde

Question 10.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It doesnot reduce Tollens’ reagent but forms sodium hydrogensulphite adduct and gives +ve iodoform test. On vigorous oxidation forms ethanoic and propanoic acids. Write the possible structure of the compound.
Answer:
Step : 1 To determine the molecular formula of the compound.

Emperical formula of the given compound = C₅H₁₀O
Molecular formula = n × (emperical formula)
Where n = \(\frac{\text { Molecular mass of the compound }}{\text { Emperical formula mass of compound }}\)
Given, molecular mass = 86
Emperical formula mass of C₅H₁₀O = (12 × 5) + (10 × 1) + 16 = 60 + 10 + 16 = 86
n = \(\frac{86}{86}\) = 1
Molecular formula = 1 × (C₅H₁₀O)
∴ Molecular formula = C₅H₁₀O

Long Answer Questions

Question 1.
Explain the following terms. Give an example of the reaction in each case. [A.P. Mar. 18]
i) Cyanohydrin
ii) Acetal
iii) Semicarbazone
iv) Aldol
v) Hemiacetal
vi) Oxime
Answer:
i) Cyanohydrin
Aldehydes and ketones react withk hydrogen cyanide (HCN) forms addition products called (or) known as cyanohydrins.

ii) Acetal
In the presence of dry HCl gas, an aldehyde reacts with two equivalents of a monohydric alchol forms gem-dialkoxy compounds are known as acetals.
—> In acetal two alkoxy groups are present on the terminal C-atom.

iii) Semicarbazone
Aldehydes/ketones react with semicarbazide forms certain compounds called as senilcarbazones.
For example:

iv) Aldol
When an aldehyde ((or) ketone) having at least one a-hydrogen atom undergo a reaction in the presence of dilute alkali as catalyst to form aldol (or) β- hydroxy aldehydes ((or) ketals in case of ketones), the reaction is called aldol condensation.
For example:

v) Hemiacetat : In the presence of dry HCl gas an aldehyde reacts with one molecule of a monohydric alcohol forms gem-alkoxy alcohols. These are known as hemiacetals.
For example:

vi) Oxime: In weak acidic medium, an aldehyde ketone reacts with hydroxylamine forms products which are known as oxims.

Question 2.
Name the following compounds according to IUPAC system of nomenclature:
i) CH₃CH(CH₃)CH₂CH₂CHO
ii) CH₃CH₂COCH (C₂H₅)CH₂CH₂Cl
iii) CH₃CH = CHCHO
iv) CH₃COCH₂COCH₃
Answer:
IUPAC names of following

Question 3.
Draw the structures of the following compounds.
i) 3-Methylbutanal
ii) p-Nitropropiophenone
iii) p-Metylbenzaldehyde
iv) 3-Bromo-4-phenylpentanoic acid
Answer:
i) 3 – Methyl butanal

ii) p-Nitropropiophenone

Question 4.
Write the IUPAC names of the following ketones and Aldehydes. Wherever possible, give also common names.
i) CH₃CO(CH₂)₄ CH₃
ii) CH₃CH₂CHBrCH₂CH (CH₃)CHO
iii) CH₃(CH₂)₅CHO
iv) PhCH = CHCHO
v)
vi) PhCOPh
Answer:
i) CH₃CO(CH₂)₄ CH₃

IUPAC Name: Heptan-2-one
Common Name: Methnyl n-pentyl ketone

ii) CH₃CH₂CHBrCH₂CH (CH₃)CHO

IUPAC Name : 4 – bromo – 2 – methyl hexanal
Common Name : γ – bromo – α – methyl caproaldehyde

iii) CH₃(CH₂)₅CHO

IUPAC Name: Heptanal
Common Name i n – heptyi aldehyde

iv) Ph – CH =CH – CHO

IUPAC Name : 3 – Phenyl Prop-2-en-1-al
Common Name: β – phenyl acrolein

v)
IUPAC Name: Cyclopentane Carbaldehyde

vi) PhCOPh

IUPAC Name : Diphenyl methanone
Common Name : Benzophenone

Question 5.
Draw the structures of the following derivatives.
i) The 2, 4 – dinitrophenylhydrazone of benzaldehyde
ii) Cyclopropanone oxime
iii) Acetaldehyde hemiacetal
iv) The Semicarbazone of cyclobutanone
Answer:
i) The 2, 4 – dinitro phenyl hydrazone of benzaldehyde :

ii) Cyclopropanone oxime

iii) Acetaldehyde hemiacetal

iv) The Semicarbazone of cyclobutanone

Question 6.
Predict the products formed when Cyclohexanecarbaldehyde reacts with following reagents.
i) PhMgBr and then H₃O⁺
ii) Tollens reagent
iii) Semicarbazide and weak acid
iv) Zinc amalgam and dilute HCl
Answer:
The products are formed when cyclohexane carbaldehyde reacts with following.
i) Ph MgBr and the H₃O⁺

ii) Tollens reagent

iii) Semicarbazide and weak acid

iv) Zinc amalgam and dilute HCl

Question 7.
Which of the following compounds would Undergo aldol condensation ? Write the structures of the products expected.
i) 2-Methylpentanal
ii) 1-Phenylpropanone
iii) Phenyl acetaldehyde
iv) 2,2 – Dimethylbutanal
Answer:
Compounds having one (or) more a-H atoms undergo aedol condensation.
So from above only first three compounds having α-H atoms. Therefore they undergo aldol condensation. They are namely

1. 2-methyl Pentanal
2. 1 – Phenyl propanone
3. Phenyl acetaldehyde

i) 2 – Methyl Pentanal


ii) 1 – Phenyl propanone

iii) Phenyl acetaldehyde

Question 8.
An organic compound A(C₉H₁₀O) forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizaro reaction. On vigorous oxidation it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
Answer:

• The compouhd having molecular formula C₉H₁₀O forms a 2, 4 – DNP derivative and reduces Tollen’s reagent. So it is an aldehyde.
• It undergoes cannizaro reaction, so the aldehyde group should be directly attached to the benzene ring.
• On vigorous oxidation it gives 1,2- benzene dicarboxylic acid, so it should be an ortho substituted benzaldehyde. For molecular formula C₉H₁₀O, the possibility is only O-ethyl benzaldehyde.
• The equations for all reactions are given below.
  
  

Question 9.
How do you distinguish the following pairs of compounds ?
i) Propanal and propanone
ii) Acetophenone and benzophenone
iii) Phenol and benzoic acid
iv) Pentan-2-one and Pentan-3-one
Answer:
i) Propanal and Propanone
On idoform test propanone responds, but absence of CH₃CO – group in propanal (CH₃CH₂CHO) it does not respond.

ii) Acetophenone and benzophenone
Acetophenone gives positive idoform test whereas benzophenone (C₆H₅COC₆H₅) does not

iii) Phenol and benzoic acid
Benzoic acid reacts with sodium bicarbonate to produce effervescences of carbon dioxide where as phenol (C₆H₅OH) does not

iv) Pentan-2-one and Pentan – 3 – One
On idoform test Pentan – 2 – One-responds whereas Pentan – 3 – one (C₂H₅ COC₂H₅) does not

Question 10.
How are the following conversions carried in not more than two steps ?
i) Ethanol to 3-hydroxybutanal
ii) Bromobenzene to 1-Phenylethanol
iii) Benzaldehyde to ± Hydroxyphenylacetic acid
iv) Benzaldehyde to benzophenone
Answer:
i) Ethanol to 3-hydroxybutanal

iv) Benzaldehyde to benzophenone

Question 11.
Describe the following. [A.P. & T.S. Mar. 19, 16] [A.P. Mar. 18]
i) Acetylation
ii) Cannizaro reaction
iii) Cross aldol condensation
iv) Decarboxylation
Answer:
i) Acetylation : When active hydrogen atom of alcohol, phenol (or) an amine is replaced by acetyl (CH₃CO^–) group to form corresponding ester (or) amide, the reaction is known as acetylation.

Reagents used are acid chloride (or) acid anhydride in presence of a base like pyridine (or) dimethylaniline.
For example :

ii) Cannizaro reaction: On treating with concentrated alkali, aldehydes which do not have any α – hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction.
This reaction is called cannizaro reaction.

As a result, one molecule of aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
For example:

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 the reactants contain a – hydrogen atoms, it gives a mixture of four products.

Ketones can also be used as one component in the cross aldol reactions

iv) Decarboxylation
Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with sodalime (a mixutre of NaOH & CaO in ratio 3: 1). This reaction is called decarboxylation.

Question 12.
Complete each synthesis by giving the missing starting material, reagent or product.

Answer:

Question 13.
Explain how methyl ketones are distinguished from other ketones. Write the equations showing it.
Answer:
Oxidation of methyl ketones by haloform reactIon : Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidisedby sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform. This oxidation does not affect a carbon-carbon double bond, If present in the molecule. lodoform reaction with sodium hypoiodite is also used for detection of CH₃CO group or CH₃CH(OH) group which produces CH₃CO group on oxidation.

Question 14.
Write the equations showing the conversion of the following along with reagents.
i) 1-phenyipropane to Benzoic acid
ii) Benzamide to Benzoic acid
iii) Ethyl butanoate to Butanoic acid.
Answer:
i)

Question 15.
Write the products and reagents needed for the below given conversions
i) 3-Nitrobromobenzene to 3-Nitrobenzoic acid
ii) 4-Methyl’acetophenone to Benzene- 1-4-dicarboxylic acid
Answer:
i) 3-Nitrobromobenzene to 3-Nitrobenzoic acid:

ii) 4-Methyl’acetophenone to Benzene- 1-4-dicarboxylic acid

Textual Examples

Question 1.
Give names of the reagents ot bring about the following transformations:

1. Hexan – 1 – ol to hexanal
2. Cyclohexanol to cyclohexanone
3. p – Fluorotoluene to p – fluorobenzaldehyde
4. Ethanenitrile to ethanal
5. Allyl alcohol to propanal
6. But-2-ene to ethanal

Solution:

1. C₅H₅NH⁺ CrO₃Cl^– (PCC)
2. K₂Cr₂O₇ in acidic medium
3. CrO₃ in the presence of acetic anhydride/1. CrO₂Cl₂₅ 2. HOH
4. (Diisobutyl) aluminiinn hydride (DIBAL-H)
5. PCC
6. O₃/H₂O-Zn dust

Question 2.
Arrange the following compounds in the increasing order of their boiling points:
CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₅C₂-O-C₂H₅, CH₃CH₂CH₂CH₂CH₃
Solution:
The molecular masses of these compounds are in the range of 72 to 74. Since only butan-1- ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-l-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak vander Waals forces. Hence increasing order of boiling points of the given compounds is as follows :
CH₃CH₂CH₂CH₂CH₃ < H₅C₂-O-C₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH

Question 3.
Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Explain your answer.
Solution:
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic them carbon . atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Question 4.
An organic compound (A) with molecular formula C₈H₈O forms an orange-red precipitate with 2,4 – DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens or Fehlings reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.
Solution:
(A) forms 2, 4 – DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens or Fehling’s reagent, (A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyers reagent. This indicates the presence of unsaturation due to an aromatic, ring.

Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Question 5.
Write chemical reactions to affect the following transformations :
i) Butan-1-ol to butanoic acid
ii) Benzyl alcohol to phenylethanoic acid
iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid
iv) 4 – Methylacetophenone to benzene – 1, 4 – dicarboxylic acid
v) Cyclohexene to hexane-1, 6 – dioic acid
vi) Butanal to butanoic acid.
Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:


Answer:
Primary alcohols (1), (ii), (iii)
Secondary alcohols (iv) and (v)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols In the above examples.
Answer:
Allylic alcohols (ii) and (vi)

Question 3.
Name of the following compounds according to IUPAC system.

Solution:
i) 3-Chloromethyl – 2 isopropylpentan – 1 – ol
ii) 2, 5 – Dimethylhexane – 1, 3 – diol
iii) 3 – Bromocyclohexanol
iv) Hex- 1 -en-3-ol
v) 2-Bromo~3-methylbut-2-en-1-ol

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

Solution:

Question 5.
Write the structures of the products of the following reactions :

Solution:

Question 6.
Predict the major product of acid catalysed dehydration of
i) 1 – methylcyclohexanol and
ii) butan – 1 – ol
Solution:
i) 1 – Methylcyclohexene
ii) A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethyoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Solution:

Question 8.
Which of the following Is an appropriate set of reactants for the preparation of 1 – methoxy-4-nltrobenzene?

Solution:
ii)

Question 9.
Predict the products of the following reactions:

Solution:

Question 10.
Write the structures of the following compounds.
i) α-Methoxypropionaldehyde
ii) 3- Hydroxybutanal
iii) 2-Hydroxycyclopentane carbaldehyde
iv) 4-Oxopentanal
v) Di-Sec. butyl ketone
vi) 4 – Fluoroacetophenone
Solution:

Question 11.
Write the structures of products of the following reactions;

Solution:

Question 12.
Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₂OCH₃, CH₃CH₂CH₃
Solution:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Ethanal, Porpanal, Propanone, Butanone.
2. Benzaldehyde, p – Toualdehyde, p – Nitrobenzaldehyde, Acetophenone.

Solution:

1. Butanone < Propanone < Propanal < Ethanal
2. Acetophenone < p – Touladehyde, Benzaldehyde < p – Nitgrobenzaldehyde.
   Hint: Consider steric effect and electronic effect.

Question 14.
Predict the products of the following reactions :


Solution:

Question 15.
Give the IUPAC names of the following compounds:
i) Ph CH₂CH₂COOH
ii) (CH₃)₂C = CHCOOH

Solution:
i) 3-Phenylpropanoic acid
ii) 3 – Mehtylbut-2-enoic acid
iii) 2-Metylcyclopentanecarboxylicacid
iv) 2, 4, 6 – Trinitrobenzoic acid

Question 16.
Show how each of the following compounds can be converted to benzoic acid.
i) Ethylbenzene
ii) Acetophenone
iii) Bromobenzene
iv) Phenylethene (Styrene)
Solution:

Question 17.
Which acid of each pair shown here would you expect to be stronger?
i) CH₃CO₂H or CH₂FCO₂H
ii) CH₂FCO₂H or CH₂ClCO₂H
iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H

Solution:
i) CH₃COOH
ii) CH₂FCOOH
iii) CH₃CHFCH₂COOH
iv)