(a) Write a suitable chemical equation to complete each of the following transformations :
(i) Butan-1-ol to butanoic acid
(ii)4-Methylacetophenone to benzene-1,4-dicarboxylic acid
(b) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.
OR
(a) Give chemical tests to distinguish between:
(i) Propanol and propanone
(ii)Benzaldehyde and acetophenone
(b) Arrange the following compounds in an increasing order of their property as indicated :
(i) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) Benzoic acid, 3, 4-dinitrobenzoic acid, 4-Methoxy- benzoic acid (acid strength)

(iii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH (acid strength)

(a)

(i) To convert butanol to butanoic acid can be done by simple oxidation process using oxidising agents like acidified potassium permanganate or potassium dichromate solution.

(ii) In 4-methylacetophenone, there is one methyl group and one keto-methyl group. We need to convert both of these groups to carboxylic acid. This is done by acidic hydrolysis. So, first we come to an intermediate stage so that both the entities to be hydrolysed are similar and then we can hydrolyse them to carboxylic acids.

(b)

• Since the compound is responding to 2,4-DNP test then it surely must either be an aldehyde or a ketone.

• Also only aldehydes and not ketones respond to the silver mirror test by producing a shiny white precipitate of silver, the compound is an aldehyde or it has an aldehydic group.

From the information given,

• We get to know that the compound has a benzene ring. Also the compound responds to Cannizaro’s reaction, so the compound lacks an α-Hydrogen atom.

• Also it is given that on oxidation, we get 1,2-benzenedicarboxylic acid. So, the –CHO group is directly attached to benzene ring and the remaining carbon atoms (that is C₂H₅) is ortho substituted.

So the compound looks like:

o-ethyl benzaldehyde.

OR

(a)

(i) Iodoform Test : This test is given by compounds having a keto-methyl group. Propanone has a keto-methyl group in it so it responds to this test whereas propanol does not have one. Propanone on reacting with hot NaOH and iodine gives a yellow precipitate of iodoform while propanol does not.

(ii) Iodoform Test : This test is given by compounds having a keto-methyl group. Acetophenone has a keto-methyl group in it so it responds to this test whereas benzaldehyde does not have one. Acetophenone on reacting with hot NaOH and iodine gives a yellow precipitate of iodoform while propanol does not.

(b)

(i) When HCN reacts with a compound, the attacking species is a nucleophile CN^- . Therefore, as the negative charge on the carbon atom containing the -C=O group increases, its reactivity with HCN decreases. So compounds where +I showing groups are present more, their reactivity would be lesser. In the given compounds, the +I effect increases as shown below.

It can be observed that steric hindrance also increases as the number of +I showing groups imcrease. Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as: Methyl tert-butyl ketone < Acetone < Acetaldehyde.

(ii)

• After losing a proton, carboxylic acids gain a negative charge. So, any group that will help stabilise the negative charge will increase the stability of the carboxylate (COO^-) ion and as a result, will increase the strength of the acid.

• We know that electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids.

• As methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid than benzoic acid. But nitro group is an electron-withdrawing group so it will increase the strengths of acids.

• More the number of electron-withdrawing groups, more will be the acidic strength. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a stronger acid than benzoic acid and methoxybenzoic acid.

Hence, the strengths of the given acids increase as: 4-Methoxybenzoic acid < Benzoic acid < 3,4-Dinitrobenzoic acid

(iii) After losing a proton, carboxylic acids gain a negative charge. So, any group that will help stabilise the negative charge will increase the stability of the carboxylate (COO^-) ion and as a result, will increase the strength of the acid.

So, groups having +I effect will decrease the strength of the acids and groups having - I effect will increase the strength of the acids. We know - CH₃ group has +I effect and -Br group has - I effect. Thus, acids containing -Br are stronger. Also, the -I effect grows weaker as distance from the carboxylate ion increases. Hence, CH₃CH(Br)CH₂COOH is a weaker acid than CH₃CH₂CH(Br)COOH.

Hence, the strengths of the given acids increase as: (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH.