Q.Write chemical equations for the following conversions:

The concept first

Before choosing a reagent, classify the two compounds:

• $\mathrm{C_6H_5N(CH_3)_2}$ ($N,N$-dimethylaniline): the nitrogen carries three carbon groups → tertiary amine, no N–H.
• $\mathrm{(C_2H_5)_2NH}$ (diethylamine): the nitrogen carries two carbon groups and one H → secondary amine, one N–H.

So you need a test that distinguishes $2^\circ$ from $3^\circ$ — i.e. a test that keys off the presence of an N–H bond, not off basicity.

Hinsberg's test does exactly that. Benzenesulphonyl chloride acylates the nitrogen, but only if there is an N–H available:

• $1^\circ$ amine → $\mathrm{C_6H_5SO_2NHR}$. The N–H left over is made acidic by the two electron-withdrawing $\mathrm{S{=}O}$ groups → the sulphonamide dissolves in KOH/NaOH.
• $2^\circ$ amine → $\mathrm{C_6H_5SO_2NR_2}$. No N–H remains, so the sulphonamide is a solid that is insoluble in alkali.
• $3^\circ$ amine → no reaction (no N–H to substitute); the amine remains as an oily layer, dissolving only on adding acid.

Step-by-step

1. Test option (D), $\mathrm{CHCl_3}$ + alc. KOH (carbylamine). It gives the foul-smelling isocyanide only with primary amines. Both compounds here are non-primary → both give a negative test → cannot distinguish. ✗
2. Test option (B), anhy. $\mathrm{ZnCl_2}$ + conc. HCl (Lucas reagent). This is a test for alcohols (distinguishing $1^\circ$/$2^\circ$/$3^\circ$ ROH by turbidity). Amines are not alcohols. ✗
3. Test option (C), $\mathrm{CrO_3/H_2SO_4}$. An oxidising mixture; it does not give two clearly different, characteristic observations for these two amines. ✗
4. Test option (A), $\mathrm{C_6H_5SO_2Cl}$ (Hinsberg).
   • With $\mathrm{(C_2H_5)_2NH}$ ($2^\circ$):

 a **solid sulphonamide, insoluble in aqueous KOH** — a clear positive observation.

• With $\mathrm{C_6H_5N(CH_3)_2}$ ($3^\circ$): no reaction — the amine is recovered unchanged. Two visibly different outcomes → the reagent distinguishes them. ✓

ANSWER: A

The concept: why alkali-solubility identifies the class of amine

When an amine reacts with a sulphonyl chloride (Hinsberg's reagent, here p-toluenesulphonyl chloride):

• 1° amine $\mathrm{RNH_2}$ gives $\mathrm{RNH{-}SO_2Ar}$. The N–H left on nitrogen sits between an electron-withdrawing $\mathrm{SO_2}$ group and the ring, so it is acidic — the sulphonamide dissolves in NaOH forming a salt.
• 2° amine $\mathrm{R_2NH}$ gives $\mathrm{R_2N{-}SO_2Ar}$. Nitrogen now has no hydrogen at all, nothing to ionise, so the product is insoluble in alkali.
• 3° amine does not react at all.

So "Z is insoluble in alkali" is a direct statement that Y is a secondary amine.

Step-by-step

1. Use the Hinsberg clue. Z insoluble in alkali $\Rightarrow$ Y has no N–H after sulphonylation $\Rightarrow$ Y is 2°.
2. Screen the options. $\mathrm{C_6H_5CH_2NH_2}$ (benzylamine) is a primary amine — its tosylamide $\mathrm{C_6H_5CH_2NH{-}SO_2C_6H_4CH_3}$ still has an acidic N–H and would dissolve in KOH. So options (A) and (C) are out. Y must be $\mathrm{C_6H_5NHCH_3}$.
3. Now identify the reduction. A secondary amine bearing an $\mathrm{N{-}CH_3}$ group is the hallmark product of reducing an isocyanide (carbylamine):

Contrast with a nitrile, $\mathrm{R{-}C\equiv N + 4[H] \rightarrow R{-}CH_2{-}NH_2}$, which always gives a primary amine — that is the benzylamine route, already ruled out.

4. Pick the reagent. $\mathrm{NaBH_4}$ is too mild for this reduction (it does not reduce nitriles/isocyanides), so option (B) fails. Catalytic hydrogenation, $\mathrm{H_2\,|\,Ni}$, is the standard reagent.

5. Confirm the sequence:

ANSWER: D

Concept & Intuition

This is a classic organic chemistry matching problem about carbonyl derivatives. Each reagent (X) reacts with a carbonyl compound (aldehyde or ketone) to give a specific nitrogen-containing derivative. The key is to recall the functional group of the product formed:

• Oximes come from hydroxylamine ($\mathrm{NH_2OH}$).
• Hydrazones come from hydrazine ($\mathrm{NH_2NH_2}$).
• Semicarbazones come from semicarbazide ($\mathrm{NH_2NHCONH_2}$).
• Schiff bases (imines) come from primary amines ($\mathrm{RNH_2}$), especially aromatic ones like aniline.

Let’s check each option step by step.

1. Option (A): $\mathrm{NH_2OH}$ → Hydrazone

   Hydroxylamine ($\mathrm{NH_2OH}$) reacts with a carbonyl to form an oxime (with a $\mathrm{C=NOH}$ group), not a hydrazone. Hydrazones come from hydrazine. So this is incorrect.

2. Option (B): $\mathrm{NH_2NH_2}$ → Semicarbazone

   Hydrazine ($\mathrm{NH_2NH_2}$) gives a hydrazone ($\mathrm{C=NNH_2}$). Semicarbazones require semicarbazide ($\mathrm{NH_2NHCONH_2}$). So this is incorrect.

3. Option (C): $\mathrm{C_6H_5NH_2}$ → Schiff base

   Aniline ($\mathrm{C_6H_5NH_2}$) is a primary aromatic amine. It reacts with aldehydes/ketones to form an imine (also called a Schiff base), with the general structure $\mathrm{C_6H_5N=CR_2}$. This is correct.

4. Option (D): $\mathrm{RNH_2}$ → Oxime

   A generic primary amine ($\mathrm{RNH_2}$) forms an imine (Schiff base), not an oxime. Oximes require hydroxylamine. So this is incorrect.

The concept first: why aliphatic diazonium salts die instantly

Diazotisation ($\ce{NaNO2 + HCl}$) makes the same $\ce{-N2+}$ group from any primary amine. The difference is stability. In an aryl diazonium salt the $\ce{-N2+}$ is conjugated with the ring, so at 273–278 K it survives long enough to be used in coupling/Sandmeyer reactions. In an alkyl diazonium ion there is no such delocalisation, and $\ce{N2}$ is an outstanding leaving group — so the ion falls apart the moment it forms, giving a carbocation that the solvent (water) captures. That is why primary aliphatic amines simply effervesce and give alcohols with nitrous acid.

Step-by-step

Step 1 — Identify X. Acetamide with bromine in aqueous alkali is the textbook Hofmann bromamide degradation:

So $\ce{X = CH3NH2}$ (methanamine). Notice the carbon count drops from 2 to 1 — the carbonyl carbon leaves as carbonate.

Step 2 — Diazotise X.

Step 3 — Let it decompose in water.

So $\ce{Y = CH3OH}$.

Step 4 — Rule out the others. (A) $\ce{CH3NH2}$ is X, not Y. (B) $\ce{CH3CONHBr}$ is only the intermediate N-bromoamide inside step 1. (D) would require ring/α-bromination, which $\ce{Br2/NaOH}$ on an amide does not do.

ANSWER: C

The concept first: why alkali-solubility fingerprints the amine class

p-Toluenesulphonyl chloride (a Hinsberg-type reagent) reacts with the amine's lone pair, replacing an $\ce{N-H}$ hydrogen with the bulky $\ce{-SO2Ar}$ group:

• Primary amine $\ce{R-NH2}$ → $\ce{R-NH-SO2Ar}$. One $\ce{N-H}$ remains. That hydrogen is acidic, because the resulting anion is stabilised by the strongly electron-withdrawing sulphonyl group. Hence the product dissolves in KOH/NaOH.
• Secondary amine $\ce{R2NH}$ → $\ce{R2N-SO2Ar}$. No $\ce{N-H}$ is left. There is no acidic proton to remove, so the product is insoluble in alkali. ✓
• Tertiary amine $\ce{R3N}$ → no reaction (there is no $\ce{N-H}$ to substitute in the first place), so no product Y at all.

The stem says a product Y forms and is insoluble in alkali. Both facts together force X to be secondary.

Step-by-step

Step 1 — Classify X. Product formed ⇒ not tertiary. Product alkali-insoluble ⇒ not primary. Therefore X is a $2^\circ$ amine.

Step 2 — Benzoylation (Schotten–Baumann). Benzoyl chloride $\ce{C6H5COCl}$ acylates the amine nitrogen:

The amide nitrogen ends up carrying both original alkyl groups plus the benzoyl group — i.e. it is a tertiary (N,N-disubstituted) amide with no $\ce{N-H}$.

Step 3 — Test each option against "benzoyl on a $2^\circ$ nitrogen".

• (A) $\ce{CH3CH2CH(NH2)COC6H5}$ — the benzoyl is on carbon, and a free $\ce{-NH2}$ survives. Wrong on both counts.
• (B) $\ce{CH3CH2CH2N(CH3)COCH2C6H5}$ — the acyl group here is $\ce{-COCH2C6H5}$ (phenylacetyl), not benzoyl $\ce{-COC6H5}$.
• (C) $\ce{CH3CH2NHCOC6H5}$ — this amide still has an $\ce{N-H}$, so it must have come from a primary amine ($\ce{CH3CH2NH2}$). But a primary amine would have given an alkali-soluble Y. Contradiction.
• (D) $\ce{CH3CH2N(CH3)COC6H5}$ — nitrogen carries ethyl + methyl + benzoyl: exactly what you get by benzoylating the secondary amine $\ce{CH3CH2-NH-CH3}$. ✓

Step 4 — Back-substitute and check. With $\ce{X = CH3CH2NHCH3}$, the sulphonamide would be $\ce{CH3CH2N(CH3)SO2C6H4CH3}$ — no $\ce{N-H}$, insoluble in alkali. ✓ Everything is consistent.

ANSWER: D

Concept & Intuition

Most alkali metal nitrates (like NaNO₃, KNO₃) decompose on strong heating to give the nitrite and oxygen:

$\mathrm{2\,MNO_3 \rightarrow 2\,MNO_2 + O_2}$.

But lithium is an exception. Because Li⁺ is very small, it has a high charge density and strongly polarises the nitrate ion. This destabilises the nitrate, causing it to break down further — lithium nitrite (LiNO₂) itself is unstable at high temperature and decomposes to lithium oxide, nitrogen dioxide, and oxygen. So the final products are not simply nitrite + O₂, but oxide + NO₂ + O₂.

Step-by-step reasoning

1. General trend for alkali metal nitrates

   For Na, K, Rb, Cs:

   $\mathrm{2\,MNO_3 \xrightarrow{\Delta} 2\,MNO_2 + O_2}$

   The nitrite is stable at the decomposition temperature.

2. Lithium’s anomaly

   Li⁺ is much smaller than other alkali ions. Its high polarising power weakens the N–O bonds in the nitrate ion, so decomposition occurs at a lower temperature and proceeds further.

3. First step – formation of nitrite

   Initially, LiNO₃ does form LiNO₂ and O₂:

   $\mathrm{2\,LiNO_3 \rightarrow 2\,LiNO_2 + O_2}$

   But LiNO₂ is not stable at the temperature needed for decomposition.

4. Second step – further decomposition of LiNO₂

   Lithium nitrite decomposes to lithium oxide, nitrogen dioxide, and oxygen:

   $\mathrm{2\,LiNO_2 \rightarrow Li_2O + NO_2 + NO}$

   However, NO reacts immediately with O₂ to give NO₂:

   $\mathrm{2\,NO + O_2 \rightarrow 2\,NO_2}$

5. Overall net reaction

   Combining the steps:

   $\mathrm{4\,LiNO_3 \rightarrow 2\,Li_2O + 4\,NO_2 + O_2}$

   Dividing by 2 gives the simplest form:

   $\mathrm{2\,LiNO_3 \rightarrow Li_2O + 2\,NO_2 + \frac{1}{2}O_2}$

   So the products are $\mathrm{Li_2O}$, $\mathrm{NO_2}$, and $\mathrm{O_2}$.