Question

Three isomeric compound M, N and P ($C_5H_{10}O$) give the following tests:

(i) M and P react with sodium bisulfite to form an adduct (ii) N consumes 1 mol of bromine and also gives trubidity with conc. HCl / anhydrous $ZnCl_2$ after prolong heating (iii) M reacts with excess of iodine in alkaline solution to give yellow crystalline compound with a characteristic smell. (iv) p-Rosaniline treated with sulphur dioxide developes pink colour on shaking with P.

The structures of M, N and P, respectively can be:

• A. M: Pentan-2-one, N: Pent-4-en-1-ol, P: Pentanal

Answer 1

Answer: (A)

M: Pentan-2-one, N: Pent-4-en-1-ol, P: Pentanal

Answer 2

The molecular formula of the three isomeric compounds M, N, and P is $C_5H_{10}O$. The degree of unsaturation is $(2 \times 5 + 2 - 10)/2 = 1$. This means each compound contains either one double bond (C=C or C=O) or one ring.

Let's analyze the given tests:

(i) M and P react with sodium bisulfite ($NaHSO_3$) to form an adduct. This reaction is characteristic of aldehydes and most methyl ketones. This indicates that M and P are either aldehydes or methyl ketones.

(ii) N consumes 1 mol of bromine and also gives turbidity with conc. HCl / anhydrous $ZnCl_2$ (Lucas test) after prolonged heating.

• Consuming 1 mol of bromine indicates the presence of one C=C double bond.
• Giving turbidity with Lucas reagent indicates the presence of an alcohol group. Prolonged heating suggests it is a primary alcohol (tertiary react immediately, secondary within minutes).

So, N is an unsaturated primary alcohol with one C=C double bond.

(iii) M reacts with excess of iodine in alkaline solution to give yellow crystalline compound with a characteristic smell. This is the Iodoform test. The Iodoform test is positive for compounds containing a methyl ketone group ($CH_3-CO-$) or a methyl carbinol group ($CH_3-CH(OH)-$). Since M also reacts with sodium bisulfite (indicating aldehyde or ketone), M must be a methyl ketone ($CH_3-CO-R'$). $M$ has 5 carbons, so $R'$ must be a $C_3H_7$ group. The possible methyl ketones with formula $C_5H_{10}O$ are Pentan-2-one ($CH_3COCH_2CH_2CH_3$) and 3-Methylbutan-2-one ($CH_3COCH(CH_3)_2$). Both are methyl ketones and would give a positive iodoform test and react with sodium bisulfite.

(iv) p-Rosaniline treated with sulphur dioxide (Schiff's reagent) develops pink colour on shaking with P. Schiff's test is a specific test for aldehydes. Since P also reacts with sodium bisulfite (indicating aldehyde or ketone), P must be an aldehyde. P has 5 carbons, so it is a $C_5H_{10}O$ aldehyde. Possible aldehydes are Pentanal ($CH_3CH_2CH_2CH_2CHO$), 2-Methylbutanal ($CH_3CH_2CH(CH_3)CHO$), 3-Methylbutanal ($(CH_3)_2CHCH_2CHO$), and 2,2-Dimethylpropanal ($(CH_3)_3CCHO$). All aldehydes react with sodium bisulfite and give a positive Schiff's test.

Based on the analysis:

• M is a methyl ketone ($C_5H_{10}O$).
• N is an unsaturated primary alcohol ($C_5H_{10}O$).
• P is an aldehyde ($C_5H_{10}O$).

Let's consider the options provided in the question. Assuming the options are presented as (M, N, P).

Let's check the option that corresponds to the correct answer (Option A in the original problem context, which is not explicitly shown here, but inferred from the solution format).

Assume Option A is:

M: Pentan-2-one ($CH_3COCH_2CH_2CH_3$) N: Pent-4-en-1-ol ($CH_2=CHCH_2CH_2CH_2OH$) P: Pentanal ($CH_3CH_2CH_2CH_2CHO$)

Check M: Pentan-2-one is a methyl ketone ($C_5H_{10}O$). It reacts with $NaHSO_3$ and gives a positive iodoform test. This fits the criteria for M.

Check N: Pent-4-en-1-ol ($CH_2=CHCH_2CH_2CH_2OH$) is an unsaturated primary alcohol ($C_5H_{10}O$). It has one C=C bond (reacts with 1 mol $Br_2$) and a primary alcohol group ($CH_2OH$). Primary alcohols give the Lucas test upon prolonged heating. This fits the criteria for N.

Check P: Pentanal ($CH_3CH_2CH_2CH_2CHO$) is an aldehyde ($C_5H_{10}O$). It reacts with $NaHSO_3$ and gives a positive Schiff's test. This fits the criteria for P.

Since this set of structures satisfies all the given conditions, it is the correct answer.