Q.Give structures of the products you would expect when each of the following alcohol reacts with

(A) Correct. Anisole undergoes Friedel–Crafts acylation with $CH_3COCl$/anhyd. $AlCl_3$; $-OCH_3$ is o/p-directing, giving mainly the para product 4-methoxyacetophenone. ✓

(B) Correct. Isopropyl methyl ether $(CH_3)_2CH\text{-}O\text{-}CH_3$ with conc. HI: $I^-$ attacks the less hindered carbon (the methyl) by SN2, giving $CH_3I$ + isopropyl alcohol. ✓

(C) Correct. tert-Butyl methyl ether: the tert-butyl group forms a stable 3° carbocation, so cleavage is SN1 — $I^-$ goes to the tert-butyl group giving 2-iodo-2-methylpropane (tert-butyl iodide) + methanol. ✓

(D) Incorrect. In the Williamson ether synthesis the alkyl halide must undergo SN2; an aryl halide such as bromobenzene does NOT react with sodium methoxide. Anisole is instead prepared from sodium phenoxide + methyl iodide (phenoxide is the nucleophile, methyl the SN2 substrate). So this statement is wrong.

The key to this problem is recognising that the reaction conditions (385 K, Cl₂ gas) favour free-radical substitution at the benzylic position, not electrophilic aromatic substitution. Chlorine gas at high temperature or in the presence of light abstracts a hydrogen from the methyl group of toluene, producing benzyl chloride. Then, sodium ethoxide (a strong nucleophile and a strong base) displaces the chlorine via an SN2 reaction, giving an ether. The product is therefore a benzyl ethyl ether — a benzene ring with a –CH₂–O–C₂H₅ substituent.

1. Identify the first reaction (chlorination) Toluene (C₆H₅–CH₃) reacts with Cl₂ at 385 K. This temperature is high enough to promote homolytic cleavage of Cl₂ into chlorine radicals. The benzylic C–H bond is weak (due to resonance stabilisation of the benzylic radical), so the radical chain reaction selectively replaces one hydrogen on the methyl group:

The product X is benzyl chloride.

2. Identify the second reaction (nucleophilic substitution) Sodium ethoxide (Na⁺ –OCH₂CH₃) is a strong nucleophile. Benzyl chloride has a primary carbon (the –CH₂Cl) that is also benzylic, making it highly reactive toward SN2 displacement. The ethoxide ion attacks the carbon, pushing out chloride:

The product Y is benzyl ethyl ether.

3. Match the structure to the options

   • Option (A): a benzene ring with a –CH₂–O–C₂H₅ side chain — exactly benzyl ethyl ether.
   • Option (B): a benzene ring with a methyl group and an ethoxy group in para positions — that would come from electrophilic substitution, not from this sequence.
   • Option (C): meta substitution pattern — also not formed here.
   • Option (D): ortho substitution pattern — again, not formed.

   Only option (A) shows the correct –CH₂–O–C₂H₅ group attached directly to the ring.

Concept & Intuition

This problem tests two classic organic reactions in sequence:

1. Diazonium salt hydrolysis – a way to replace an amino group (via diazotization) with a hydroxyl group.
2. Williamson ether synthesis – the reaction of an alkoxide (or phenoxide) with a primary alkyl halide to form an ether.

The key is to track the functional group transformations step by step, paying attention to the fact that allyl bromide is a primary halide with a double bond that remains untouched during the S_N2 reaction.

Step-by-step reasoning

1. Starting material: Benzene diazonium chloride Benzene diazonium chloride ($\mathrm{C}_6\mathrm{H}_5\mathrm{N}_2^+ \mathrm{Cl}^-$) is formed by diazotizing aniline. When warmed with water, it undergoes hydrolysis (a type of substitution where $\mathrm{N}_2$ is replaced by $\mathrm{OH}$).

The product is phenol.

2. Formation of the sodium salt of phenol Phenol is weakly acidic. Treating it with a base (like NaOH) gives sodium phenoxide:

This phenoxide ion is a strong nucleophile.

3. Reaction with allyl bromide Allyl bromide ($\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\mathrm{Br}$) is a primary alkyl halide. The phenoxide ion attacks the electrophilic carbon (the one bonded to Br) in an S_N2 reaction.

The product is phenyl allyl ether (allyl phenyl ether). The double bond remains intact because S_N2 does not affect π-bonds.

4. Matching with options

   • (A) $\mathrm{C}_6\mathrm{H}_5-\mathrm{CH}-(\mathrm{CH}_3)_2$ — isopropylbenzene, no oxygen.
   • (B) $\mathrm{C}_6\mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$ — exactly our product.
   • (C) $\mathrm{C}_6\mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ — saturated ether, would require hydrogenation.
   • (D) $\mathrm{C}_6\mathrm{H}_5-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ — propylbenzene, no oxygen.

   Clearly, (B) is correct.

Concept & Intuition

Williamson’s synthesis is the classic way to make ethers: an alkoxide ion (strong nucleophile/base) attacks an alkyl halide in an Sₙ2 reaction. The catch: Sₙ2 works best on primary (or methyl) halides; secondary works poorly; tertiary halides almost always undergo E2 elimination instead, because the bulky base prefers to pull a proton rather than attack the crowded carbon. So if the ether you want would have to come from a tertiary alkyl halide, Williamson’s synthesis fails. The trick is to check which alkyl group in the ether would come from the halide — and whether that halide is primary, secondary, or tertiary.

Step‑by‑step reasoning

1. Identify the two halves of each ether

   Williamson’s synthesis couples an alkoxide (R–O⁻) with an alkyl halide (R′–X). The ether is R–O–R′. We can choose which half is the alkoxide and which is the halide — but the halide must be one that can undergo Sₙ2.

2. Examine option (A): C₆H₅OCH₂CH₃

   • Phenoxide (C₆H₅O⁻) + ethyl halide (CH₃CH₂–X) works perfectly: ethyl halide is primary, so Sₙ2 is fast and no elimination.
   • Alternatively, ethoxide + phenyl halide would fail (aryl halides don’t do Sₙ2), but we can always choose the better route. So (A) is easily prepared.

3. Examine option (B): (C₂H₅)₂O

   • Ethoxide (CH₃CH₂O⁻) + ethyl halide (CH₃CH₂–X) — both primary. Classic, clean Sₙ2. (B) is easily prepared.

4. Examine option (C): (CH₃)₃COC(CH₃)₃

   • This is di‑tert‑butyl ether. To make it, we would need tert‑butoxide [(CH₃)₃CO⁻] + tert‑butyl halide [(CH₃)₃C–X].
   • The halide is tertiary: Sₙ2 is impossible (steric hindrance), and E2 elimination dominates, giving isobutene.
   • Could we swap roles? No — the other combination is the same (both groups are tertiary). So no Williamson route exists. (C) cannot be prepared.

5. Examine option (D): (CH₃)₃COCH₂CH₃

   • Here we have a tert‑butyl group and an ethyl group. We choose the primary ethyl halide as the electrophile: tert‑butoxide + ethyl halide.
   • Ethyl halide is primary → Sₙ2 works. The tert‑butoxide is bulky but still a good nucleophile; it attacks the primary carbon cleanly. (D) is easily prepared.

The problem asks you to trace the transformation of sodium salicylate through a sequence of reactions. The trick lies in understanding what each reagent does and, crucially, where the allyl group ends up — on oxygen or on the ring. Many students instinctively assume allyl bromide will attack the ring (Friedel–Crafts style), but here the conditions favour a simple ether formation.

1. Start with sodium salicylate [A]

   This is the sodium salt of salicylic acid: a benzene ring with an –OH group and a –COONa group in ortho positions. The –COONa is a carboxylate salt.

2. First reaction: (i) NaOH/CaO/Heat, then (ii) HCl(aq)

   The mixture of NaOH and CaO (soda lime) with heat is the classic decarboxylation condition. The carboxylate group (–COONa) is lost as CO₂, and the ring gains a hydrogen in its place.

   After heating, the product is phenol (C₆H₅OH). The subsequent treatment with aqueous HCl simply neutralises any remaining base and ensures the product is the free phenol.

   So [B] is phenol.

3. Second reaction: (i) NaOH(aq)

   Phenol is weakly acidic (pKa ≈ 10). Aqueous NaOH deprotonates it to form sodium phenoxide (C₆H₅O⁻ Na⁺).

   This step is important because the phenoxide ion is a much better nucleophile than neutral phenol.

   So the intermediate after this step is sodium phenoxide — but the scheme labels it again as [B] (the printed label says (B) on the second arrow). That means the intermediate before the final step is sodium phenoxide.

4. Third reaction: Allyl bromide

   Allyl bromide (CH₂=CH–CH₂Br) is an alkyl halide. Sodium phenoxide is a strong nucleophile. The reaction that follows is a Williamson ether synthesis: the phenoxide oxygen attacks the electrophilic carbon of allyl bromide (the CH₂Br carbon), displacing bromide.

   This gives phenyl allyl ether: C₆H₅–O–CH₂–CH=CH₂.

   The allyl group is attached to oxygen, not directly to the ring. No sodium remains in the final organic product because NaBr is formed as a byproduct.

5. Check the options

   • (A) shows –ONa and an allyl group directly on the ring (C-allylation). That would require a different mechanism (e.g., Claisen rearrangement, but that needs heat and is not indicated here).
   • (B) shows –ONa and a saturated propyl group on the ring — no double bond, so not allylation.
   • (C) shows –ONa and a saturated alkyl chain at the para position — again, no allyl group.
   • (D) shows an ether: the ring has –O–CH₂–CH=CH₂, with no sodium. This matches exactly.

[!TLDR]

The ether can be prepared by Williamson synthesis (Assertion correct), but the mechanism stated in the Reason (SN1 of a primary alkoxide on a tertiary halide) is wrong, so the answer is (C).

Concept

Williamson ether synthesis (CBSE/NCERT Class 12 Alcohols, Phenols and Ethers) is an $\mathrm{S_N2}$ reaction of an alkoxide on an alkyl halide. It works best when the halide is primary; with tertiary halides the alkoxide acts as a base and elimination (E2) dominates.

Solution

To make n-propyl tert-butyl ether, the correct pairing is tert-butoxide ($\mathrm{(CH_3)_3CO^-}$) + n-propyl bromide (primary halide), an $\mathrm{S_N2}$ reaction that gives a good yield. So the Assertion is correct.

The Reason claims the reaction proceeds by an $\mathrm{S_N1}$ attack of the primary alkoxide on the tert-alkyl halide. That combination is exactly the wrong choice: a bulky tertiary halide with a strong alkoxide base undergoes E2 elimination to give isobutylene, not the ether – and alkoxides do not drive an $\mathrm{S_N1}$ nucleophilic attack. Hence the Reason is incorrect.

Assertion correct, Reason incorrect $\Rightarrow$ option (C).

[!ANSWER]

(C) Assertion is correct but Reason is incorrect

Step 1 — The definition

An organometallic compound is one that contains at least one direct bond between a carbon atom of an organic group and a metal atom ($\mathrm{C-M}$). A compound merely containing both a metal and carbon is not enough — the bond itself must be C–M.

Step 2 — Examine each option

(A) $\mathrm{CH_3COONa}$ (sodium acetate): the sodium ion is associated with the carboxylate oxygen — $\mathrm{CH_3COO^-\ Na^+}$. The linkage is $\mathrm{Na-O}$. Not organometallic.

(B) $\mathrm{CH_3CH_2MgBr}$ (ethylmagnesium bromide): ✓

This is a Grignard reagent. The ethyl group is bonded directly to magnesium:

The $\mathrm{C-Mg}$ bond is strongly polarised $\mathrm{C}^{\delta-}\!-\!\mathrm{Mg}^{\delta+}$ (carbon is far more electronegative than Mg), which is precisely why the carbon behaves as a carbanion/nucleophile and attacks carbonyl groups. This is the organometallic compound.

(C) $\mathrm{(CH_3COO)_2Ca}$ (calcium acetate): again a carboxylate salt — the $\mathrm{Ca^{2+}}$ is held by the acetate oxygens, $\mathrm{Ca-O}$. Not organometallic.

(D) $\mathrm{CH_3ONa}$ (sodium methoxide): an alkoxide, $\mathrm{CH_3O^-\ Na^+}$ — the sodium sits on oxygen, $\mathrm{Na-O}$. Not organometallic.

Step 3 — The pattern

The three distractors are all metal–oxygen compounds. Only the Grignard reagent puts the metal directly onto carbon.

ANSWER: B

Step 1 — The concept: what makes a pair "functional isomers".

Two compounds are functional isomers if they:

1. have the SAME molecular formula, and
2. possess DIFFERENT functional groups.

Both conditions must hold. If the formulas match but the functional group is the same, the relationship is something else — metamerism (different alkyl groups either side of the same functional group) or chain isomerism, not functional isomerism. That distinction is precisely what this question tests.

Step 2 — Option (A): $CH_3COOH$ and $HCOOCH_3$.

• $CH_3COOH$ (ethanoic acid): $C_2H_4O_2$ — functional group = carboxylic acid ($-COOH$)
• $HCOOCH_3$ (methyl methanoate): $C_2H_4O_2$ — functional group = ester ($-COO-$)

Same formula ✓, different functional groups ✓ → IS a functional isomer pair (the classic acid–ester pair).

Step 3 — Option (B): $C_2H_5OC_2H_5$ and $C_3H_7OCH_3$. ← the answer

• $C_2H_5OC_2H_5$ (diethyl ether): $C_4H_{10}O$ — functional group = ether ($-O-$)
• $C_3H_7OCH_3$ (methyl propyl ether): $C_4H_{10}O$ — functional group = ether ($-O-$)

Same formula ✓ — but both are ETHERS, i.e. the same functional group ✗.

They differ only in how the 4 carbons are distributed on either side of the oxygen (2+2 versus 3+1). That is the textbook definition of METAMERISM, not functional isomerism.

→ NOT a pair of functional isomers. This is what the question asks for.

Step 4 — Option (C): $CH_3CH_2OH$ and $CH_3OCH_3$.

• $CH_3CH_2OH$ (ethanol): $C_2H_6O$ — functional group = alcohol ($-OH$)
• $CH_3OCH_3$ (dimethyl ether): $C_2H_6O$ — functional group = ether ($-O-$)

Same formula ✓, different functional groups ✓ → IS a functional isomer pair (the textbook alcohol–ether example).

Step 5 — Option (D): $CH_3CH_2NO_2$ and $H_2NCH_2COOH$.

• $CH_3CH_2NO_2$ (nitroethane): C₂, H₅, N, O₂ → $C_2H_5NO_2$ — functional group = nitro ($-NO_2$)
• $H_2NCH_2COOH$ (glycine): $H_2N-CH_2-COOH$ → C₂, H₅, N, O₂ → $C_2H_5NO_2$ — functional groups = amino + carboxylic acid

Same formula ✓, different functional groups ✓ → IS a functional isomer pair.

Step 6 — Collate.

The question asks which is NOT a pair of functional isomers → (B).

ANSWER: B

Step 1 — Identify the Grignard reagent from the figure.

The skeletal group drawn before MgBr is an isobutyl group, $(\mathrm{CH_3})_2\mathrm{CH{-}CH_2{-}}$. So the reagent is

Step 2 — What Grignards do with active hydrogen.

In $R{-}MgBr$ the carbon is strongly carbanionic ($R^{\delta-}$–$Mg^{\delta+}$), so $R^-$ is an extremely strong base. Any compound with an active (acidic) hydrogen — $\mathrm{H_2O}$, ROH, $\mathrm{RNH_2}$, RCOOH, $\mathrm{RC{\equiv}CH}$ — instantly protonates it:

Methylamine, $\mathrm{CH_3NH_2}$, has N–H bonds (active hydrogen). So the Grignard is destroyed, not added to.

Step 3 — Write the reaction.

The organic compound X = 2-methylpropane (isobutane), molecular formula $\mathrm{C_4H_{10}}$.

(Note: the alkyl group simply picks up the proton — the carbon skeleton is unchanged, so X keeps all four carbons.)

Step 4 — Count the isomers of X's molecular formula, $\mathrm{C_4H_{10}}$.

All structural (chain) isomers of $\mathrm{C_4H_{10}}$:

1. $\mathrm{CH_3CH_2CH_2CH_3}$ — n-butane (straight chain)
2. $(\mathrm{CH_3})_3\mathrm{CH}$ — isobutane / 2-methylpropane (branched)

No other arrangement of 4 carbons is possible (a 4-carbon skeleton offers only one branching site), and $\mathrm{C_4H_{10}}$ is saturated, so there are no positional/geometrical isomers.

Step 5 — The trap.

Students often try to make the Grignard add to the amine and count isomers of a bigger amine product. Grignards never add to N–H amines — the acid–base reaction is far faster.

ANSWER: A

Step 1 — Why diazonium ions are unstable in general.

$\mathrm{N_2}$ is an outstanding leaving group (an extremely stable neutral molecule), so any $\mathrm{R\!-\!N_2^+}$ is under constant pressure to expel $\mathrm{N_2}$ and leave a carbocation $\mathrm{R^+}$. Anything that stabilises the diazonium ion itself (rather than the departing carbocation) increases its stability.

Step 2 — The aryl case (B).

In benzenediazonium ion the $-\mathrm{N_2^+}$ is attached directly to the $sp^2$ ring carbon. The ring's $\pi$ system conjugates with the diazonium group, spreading the positive charge over the ring by resonance:

This resonance stabilisation is why benzenediazonium chloride can actually be isolated and stored at 0–5 °C, and why it is the workhorse of diazo-coupling and Sandmeyer reactions.

Step 3 — Why the others fail.

• (A) $\mathrm{CH_3N_2^+}$ and (C) $\mathrm{CH_3CH_2N_2^+}$ — alkyl groups offer no resonance; they decompose immediately, losing $\mathrm{N_2}$ to give very unstable primary carbocations.
• (D) $\mathrm{C_6H_5CH_2N_2^+}$ — benzyl. The ring is separated from $-\mathrm{N_2^+}$ by an $sp^3$ $\mathrm{CH_2}$, so it cannot delocalise the diazonium charge. Worse, the ring does stabilise the benzyl carbocation that forms on losing $\mathrm{N_2}$ — which makes it decompose even faster.

ANSWER: B