Q.Complete the following acid-base reactions and name the products:

Concept and Intuition

Three very different strategies for making primary amines are being tested, and the key discipline is to count carbons through each transformation, since some routes preserve chain length and one (Hofmann bromamide degradation) deliberately removes a carbon.

• $\mathrm{AgNO_2}$ vs $\mathrm{NaNO_2}$: nitrite is an ambident nucleophile (can attack through O or N). With the softer, more covalent silver salt $\mathrm{AgNO_2}$, alkylation occurs preferentially through nitrogen, giving the nitroalkane as the major product (with $\mathrm{NaNO_2}$, the ionic character favours O-alkylation, giving the alkyl nitrite instead). Reducing a nitroalkane ($\mathrm{Fe/HCl}$) gives the primary amine with the same carbon skeleton.
• The cyanide route ($\mathrm{NaCN}$, then reduction) is a classic chain-extension: the nucleophilic $\mathrm{CN^-}$ displaces the halide, adding one carbon to the chain as a nitrile, and reducing the nitrile ($\mathrm{H_2/Ni}$ or $\mathrm{LiAlH_4}$) gives a primary amine that is one carbon longer than the starting halide.
• The Hofmann bromamide degradation ($\mathrm{Br_2/OH^-}$ on an amide) is the opposite: it removes the carbonyl carbon (as it leaves via an isocyanate/carbamate intermediate that loses $\mathrm{CO_2}$), giving a primary amine with one carbon fewer than the starting amide.

Step-by-Step Solution

1. I: $\mathrm{CH_3CH_2CH_2Cl}$ (3 carbons) + $\mathrm{AgNO_2}$ → $\mathrm{CH_3CH_2CH_2NO_2}$ (1-nitropropane, N-alkylation favoured by the covalent Ag–N bond). + $\mathrm{Fe/HCl}$ (reduction) → $\mathrm{CH_3CH_2CH_2NH_2}$ = n-propylamine. Matches.
2. II: $\mathrm{CH_3CH_2Cl}$ (2 carbons) + $\mathrm{NaCN}$ → $\mathrm{CH_3CH_2CN}$ (propanenitrile, 3 carbons after chain extension). + $\mathrm{H_2/Ni}$ → $\mathrm{CH_3CH_2CH_2NH_2}$ = n-propylamine. Matches.
3. III: $\mathrm{CH_3CH_2CONH_2}$ (propanamide, 3 carbons) + $\mathrm{Br_2/OH^-}$ (Hofmann degradation) → loses the carbonyl carbon as $\mathrm{CO_2}$, giving $\mathrm{CH_3CH_2NH_2}$ = ethylamine (2 carbons), not n-propylamine. Does not match.
4. So only I and II correctly give n-propylamine.

Common Mistakes

• Assuming Hofmann degradation preserves carbon count — it always removes one carbon from the amide.
• Missing the Ag$^+$ vs Na$^+$ nitrite distinction and assuming route I gives the nitrite ester (which on reduction would not cleanly give the amine in one obvious step).

ANSWER: B

Concept and Intuition

Several reagents convert $R-OH \to R-Cl$, but "preferred for pure product" is a purity argument, not a yield argument. The reagent of choice is the one whose by-products leave the flask on their own (as gases), so no work-up/purification is needed to remove them from the alkyl chloride.

Step-by-Step Solution

1. $R-OH + HCl/ZnCl_2 \to R-Cl + H_2O$ (Lucas reagent) — works well only for tertiary/some secondary alcohols and leaves $ZnCl_2$/water behind.
2. $R-OH + PCl_5 \to R-Cl + POCl_3 + HCl$ — the by-product $POCl_3$ is a liquid that must be removed by distillation, contaminating the crude product.
3. $R-OH + PCl_3 \to R-Cl + H_3PO_3$ — leaves phosphorous acid behind, again needing separation.
4. $3R-OH + SOCl_2 \to R-Cl + SO_2\uparrow + HCl\uparrow$ — both by-products are gases that simply bubble out, so the alkyl chloride is obtained directly in pure form with essentially no work-up.
5. Hence $SOCl_2$ is the textbook "preferred reagent for pure alkyl chloride."

Common Mistakes

• Picking $PCl_5$ because it is the most commonly taught halogenating agent — but the question specifically asks about purity, which is $SOCl_2$'s distinguishing feature.
• Forgetting that this reaction is normally run with a trace of pyridine to mop up the HCl and drive the reaction, without changing the by-product argument.

ANSWER: C

Concept and Intuition

Both reactions replace the diazonium group ($-N_2^+$) with a halogen via a radical/copper-mediated mechanism, and both give the same type of product (aryl halide + $N_2$). The distinguishing feature examiners test is the reagent form: Sandmeyer's reaction historically uses a cuprous halide salt (e.g. $Cu_2Cl_2$) generated in situ or added directly, while the Gattermann reaction is the simplified variant that uses copper powder (metallic Cu) together with the corresponding hydrohalic acid (HCl/HBr) — no cuprous salt needed.

Step-by-Step Solution

1. The reaction described is $ArN_2^+Cl^-$ treated with Cu (metal powder) and HCl.
2. Because the copper source is elemental copper powder (not a cuprous salt like $Cu_2Cl_2$), this specific reagent combination is named the Gattermann reaction, distinguishing it from the Sandmeyer reaction (which specifically uses cuprous chloride/cuprous bromide).
3. Both reactions replace $-N_2^+$ with $-Cl$ to give chlorobenzene, but the naming convention in exams hinges on the copper reagent identity.
4. Etard and Finkelstein reactions are unrelated: Etard converts toluene to benzaldehyde via $CrO_2Cl_2$; Finkelstein converts alkyl chlorides/bromides to alkyl iodides using NaI in acetone.

Common Mistakes

• Calling every diazonium-to-aryl-halide reaction 'Sandmeyer' regardless of whether Cu metal or a cuprous salt is used — the specific reagent (Cu powder here) is the Gattermann variant.
• Confusing this with Finkelstein (which is for alkyl, not aryl/diazonium, halide exchange).

ANSWER: D

Concept and Intuition

An isocyanide, $R-N\equiv C$, is hydrolysed under acidic aqueous conditions by nucleophilic attack of water at the terminal carbon, ultimately cleaving the C–N bond to release the alkyl group AS AN AMINE (with its nitrogen intact, unchanged connectivity to R) while the isocyanide carbon (originally bonded only to N and nothing else besides the lone pair/triple bond) ends up as formic acid, HCOOH. This is a key distinguishing test reaction of isocyanides vs their isomeric nitriles (which hydrolyse differently, to carboxylic acids + ammonia/amine on the OTHER side).

Step-by-Step Solution

1. Isobutyl isonitrile: $(CH_3)_2CHCH_2-N\equiv C$ — the isobutyl group ($(CH_3)_2CHCH_2-$) is attached to the isocyanide nitrogen.
2. Acid-catalysed hydrolysis proceeds: $R-NC + 2H_2O \xrightarrow{H^+} R-NH_2 + HCOOH$.
3. Applying this to isobutyl isonitrile: the alkyl group R = $(CH_3)_2CHCH_2-$ ends up as the primary amine $(CH_3)_2CHCH_2NH_2$ (isobutylamine), retaining the exact same carbon skeleton and nitrogen attachment.
4. The isocyanide carbon (the one triple-bonded to N in R–N≡C) is hydrolysed to formic acid, HCOOH.
5. So the hydrolysis products are $(CH_3)_2CHCH_2NH_2$ + $HCOOH$ — option (A).

Common Mistakes

• Confusing isonitrile hydrolysis (gives R-NH₂ + HCOOH) with nitrile hydrolysis (R-CN gives R-COOH + NH₃) — these are structurally different functional groups (isocyanide vs nitrile) with different hydrolysis outcomes.
• Changing the alkyl group's carbon skeleton in the amine product — it must remain exactly the same as in the starting isonitrile (isobutyl throughout).

ANSWER: A

Concept and Intuition

Alkyl halides react differently depending on the solvent/base used: aqueous KOH (with water as solvent) favours $S_N2$ nucleophilic substitution because $\mathrm{OH^-}$ acts predominantly as a nucleophile in the polar protic medium, while alcoholic KOH (KOH dissolved in ethanol, with much less free water) favours $E2$ elimination, as the base character of $\mathrm{OH^-}$/ethoxide dominates, abstracting a beta-hydrogen to form an alkene.

Step-by-Step Solution

1. Substrate: 1-chlorobutane, $\mathrm{CH_3CH_2CH_2CH_2Cl}$ -- a primary alkyl halide, Cl on C1.
2. Aqueous KOH: favours $S_N2$ substitution (primary halides are ideal $S_N2$ substrates -- unhindered backside attack). $\mathrm{OH^-}$ displaces $\mathrm{Cl^-}$ directly: product P = $\mathrm{CH_3CH_2CH_2CH_2OH}$ (1-butanol).
3. Alcoholic KOH: favours E2 elimination. The base removes a beta-hydrogen (a hydrogen on the carbon adjacent to the one bearing Cl).
4. In 1-chlorobutane, C1 bears Cl; the only carbon adjacent to C1 is C2 -- so the only beta-hydrogens available are those on C2. There is no other beta-carbon (C1 is a terminal/primary carbon), so there is only one possible elimination direction.
5. Removing an H from C2 and Cl from C1 forms the double bond between C1-C2: $\mathrm{CH_2{=}CHCH_2CH_3}$, i.e. but-1-ene -- this is the only alkene that can form; but-2-ene is not possible from this substrate.
6. So P = 1-butanol, Q = but-1-ene, matching option (C).

Common Mistakes

• Assuming Zaitsev's rule (more substituted alkene favoured) applies here to give but-2-ene -- Zaitsev's rule only matters when there is a choice of beta-hydrogens from different beta-carbons; here there is only one beta-carbon (C2), so only one alkene is geometrically possible.

ANSWER: C

Concept and Intuition

Potassium cyanide, being an ionic (predominantly carbon-nucleophilic) source of $CN^-$, substitutes alkyl halides via their carbon end to give alkyl cyanides (nitriles), $R$–$C\equiv N$, rather than isocyanides. Nitriles are the functional equivalent of a "masked carboxylic acid" — acidic (or basic) hydrolysis converts the $C\equiv N$ group all the way to $-COOH$ (via an amide intermediate).

Step-by-Step Solution

1. $CH_3Cl + KCN \rightarrow CH_3CN + KCl$ — this is $(A)$, methyl cyanide (acetonitrile).
2. Acidic hydrolysis of a nitrile: $CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$ (via the amide $CH_3CONH_2$ intermediate).
3. So $(B) = CH_3COOH$ (acetic acid).

Common Mistakes

• Using $AgCN$ instead of $KCN$ mentally — $AgCN$ (more covalent, nitrogen-nucleophilic) would give the isocyanide $CH_3NC$ instead, a completely different pathway.
• Stopping hydrolysis at the amide stage instead of carrying it through to the carboxylic acid under acidic/aqueous conditions.

ANSWER: C