Q.Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?

Concept and Intuition

• (A) $(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3COCH_3$: this is Williamson ether synthesis. The rule of thumb is that the alkyl halide should be unhindered (methyl or primary) for a clean $S_N2$; the bulk of the alkoxide doesn't matter much since it's the nucleophile, not the electrophile. Methyl bromide is a perfect $S_N2$ substrate, so this reaction proceeds cleanly to give methyl tert-butyl ether. Correct.
• (B) $(C_2H_5)_2O \xrightarrow[\Delta]{excess\ HI} 2C_2H_5I + H_2O$: with excess hot HI, a symmetrical ether is cleaved completely, both alkyl-oxygen bonds broken, to give two equivalents of alkyl iodide. Correct textbook fact.
• (C) $(CH_3)_3COC_2H_5 \xrightarrow[\Delta]{HI} (CH_3)_3CI + C_2H_5OH$: this is a mixed ether with one tertiary and one primary alkyl group. Cleavage proceeds via $S_N1$ at the carbon that gives the more stable carbocation (tertiary), so the tert-butyl group leaves as the iodide and the ethyl-oxygen fragment is released as ethanol. This is the standard textbook outcome. Correct.
• (D) Aryl halides like bromobenzene have their C–X bond strengthened by resonance with the ring (partial double-bond character) and the carbon is $sp^2$, blocking backside $S_N2$ attack. Nucleophilic substitution on an unactivated aryl halide by a simple alkoxide under ordinary conditions simply does not happen — it requires either very forcing conditions (high temperature and pressure, as in phenol manufacture from chlorobenzene) or a strong base capable of a benzyne (elimination-addition) pathway (e.g. $NaNH_2$). Plain $CH_3ONa$ at ordinary conditions will not convert bromobenzene to anisole. This statement is therefore NOT correct.

Step-by-Step Solution

1. Check (A): unhindered methyl halide + alkoxide → clean Williamson synthesis. Correct.
2. Check (B): excess hot HI on a simple dialkyl ether → complete cleavage to 2 alkyl iodides. Correct.
3. Check (C): mixed tertiary/primary ether + HI → $S_N1$ cleavage gives tert-halide + alcohol. Correct.
4. Check (D): aryl halide + alkoxide under ordinary conditions → no reaction as drawn; aryl C–X bonds resist nucleophilic substitution. NOT correct.
5. So the false statement is (D).

Common Mistakes

• Treating aryl halides like alkyl halides for nucleophilic substitution purposes.
• Doubting (C) because it "looks incomplete" (ethanol, not ethyl iodide) — this is exactly the expected $S_N1$ outcome for a hindered mixed ether with limited HI/only one alkyl group forming a stable cation.

ANSWER: D

Concept and Intuition

Tertiary C-H bonds are the weakest and most easily oxidised C-H bonds in an alkane (the resulting radical/cation is most stabilised), so a strong oxidant like $KMnO_4$ selectively converts isobutane's one tertiary hydrogen into a tertiary alcohol rather than attacking the primary methyl hydrogens. Converting that alcohol to its sodium alkoxide and reacting with a primary alkyl halide (Williamson ether synthesis, $S_N2$ at the primary carbon of $CH_3Br$) builds the ether cleanly, since $S_N2$ works well on primary halides.

Step-by-Step Solution

1. Isobutane $(CH_3)_3CH$ + $KMnO_4$ → oxidation at the sole tertiary C-H → X = tert-butyl alcohol, $(CH_3)_3C-OH$ (2-methylpropan-2-ol).
2. X + Na → sodium tert-butoxide, $(CH_3)_3C-O^-Na^+$ (Na displaces the O-H proton).
3. Sodium tert-butoxide + $CH_3Br$ → Williamson ether synthesis (the alkoxide's oxygen performs $S_N2$ on the primary carbon of methyl bromide) → Y = $(CH_3)_3C-O-CH_3$, methyl tert-butyl ether.
4. Naming Y by IUPAC rules: the parent chain is propane with two substituents at C2 — methyl and methoxy. Substituent prefixes are cited in alphabetical order (methoxy before methyl, since "metho" precedes "methy"), giving 2-methoxy-2-methylpropane.

Common Mistakes

• Writing "2-methyl-2-methoxypropane" (option C) — same substituents, but the wrong (non-alphabetical) citation order, which is not the correct IUPAC name.
• Trying to oxidise a primary C-H of isobutane instead of recognising that the tertiary C-H is oxidised preferentially.

ANSWER: D

Concept and Intuition

This is a multi-step synthesis chaining three classic reactions: diazotisation/hydrolysis (amine → phenol via the diazonium salt), Williamson ether synthesis (phenoxide + alkyl halide → aryl alkyl ether), and electrophilic aromatic bromination directed by a strongly activating, ortho/para-directing methoxy group. Since anisole's methoxy substituent is already present on the ring, the incoming $\mathrm{Br^+}$ (from $\mathrm{Br_2}$ in acetic acid) attacks preferentially para (and some ortho), and para is reported as the major product due to less steric crowding.

Step-by-Step Solution

1. $\mathrm{C_6H_5NH_2} \xrightarrow{\mathrm{NaNO_2/HCl},\ 273K} \mathrm{C_6H_5N_2^+Cl^-}$ (benzenediazonium chloride) — standard diazotisation of a primary aromatic amine at 0–5°C.
2. $\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{\mathrm{H_2O},\ \Delta}$ phenol ($\mathrm{C_6H_5OH}$) + $\mathrm{N_2}$ + HCl — hydrolysis of the diazonium salt on warming. So X = phenol.
3. Phenol + NaOH → sodium phenoxide ($\mathrm{C_6H_5O^-Na^+}$); phenoxide + $\mathrm{CH_3Br}$ → anisole ($\mathrm{C_6H_5OCH_3}$) via $\mathrm{S_N2}$ (Williamson ether synthesis).
4. Anisole + $\mathrm{Br_2/CH_3COOH}$: the $-\mathrm{OCH_3}$ group strongly activates the ring and directs ortho/para. In a moderately mild brominating medium like acetic acid, the bulkier, less hindered para position is the major site of substitution.
5. Hence Y = 4-bromoanisole (Br and $-\mathrm{OCH_3}$ at para positions).

Common Mistakes

• Forgetting the hydrolysis step and thinking X is still the diazonium salt.
• Choosing the ortho product as major — while some ortho product forms, para is the reported major product for anisole bromination under these conditions due to sterics.
• Confusing the ether-forming step with simple methylation of the ring (option D) — the $-\mathrm{OCH_3}$ must remain as an ether, not be replaced by $-\mathrm{CH_3}$.

ANSWER: C

Concept and Intuition

A Grignard reagent $R\text{-}MgX$ behaves like a carbanion $R^-$. Any proton source acidic enough (even water, $pK_a\approx15.7$, is far more acidic than the alkane conjugate acid) instantly protonates it: $R\text{-}MgX + H_2O \to R\text{-}H + Mg(OH)X$. This is the classic way to convert a halide into the corresponding hydrocarbon — it is not how you make an alcohol (that needs the Grignard to attack a carbonyl carbon first).

Separately, when a nucleophile/base attacks an alkyl halide, the outcome ($S_N2$ vs $E2$) depends on both the base's bulk and the substrate's branching. Ethoxide ($\mathrm{NaOEt}$) is a comparatively small, strong base. Against a primary substrate with an easily accessible backside carbon (n-butyl bromide), steric hindrance to backside attack is minimal, so $S_N2$ substitution is the major pathway, not elimination.

Step-by-Step Solution

1. $\mathrm{CH_3CH_2CH_2CH_2Br} + Mg \xrightarrow{dry\ ether} \mathrm{CH_3CH_2CH_2CH_2MgBr}$ (Grignard reagent formed).
2. $\mathrm{CH_3CH_2CH_2CH_2MgBr} + H_2O \to \mathrm{CH_3CH_2CH_2CH_3} + Mg(OH)Br$ — protonolysis gives butane, so $X=\mathrm{CH_3CH_2CH_2CH_3}$.
3. $\mathrm{CH_3CH_2CH_2CH_2Br} + \mathrm{NaOEt} \to$ ethoxide's oxygen performs backside attack on the primary carbon (unhindered) $\Rightarrow S_N2$ substitution dominates.
4. Product: $\mathrm{CH_3CH_2CH_2CH_2\text{-}OC_2H_5}$ (di-alkyl ether), so $Y=\mathrm{CH_3CH_2CH_2CH_2OC_2H_5}$.

Common Mistakes

• Assuming Grignard + $H_2O$ gives an alcohol — it only does that if the Grignard first adds to a carbonyl compound; with plain water it just gives the alkane.
• Assuming any alkoxide + primary halide must give elimination — bulky bases (like $t$-BuO$^-$) favour elimination on branched substrates, but ethoxide on an unhindered primary carbon still substitutes cleanly.

ANSWER: A

Concept and Intuition

The classic trap in these paired reactions is to only look at how bulky the base/nucleophile is. What actually controls the outcome is whether the nucleophile can reach the back lobe of the C–LG bond:

• If the carbon bearing the leaving group is primary and unhindered (ethyl bromide), $S_N2$ remains fast and dominant even with a bulky base like $t$-butoxide, because the bulk of the base doesn't block approach to a completely open primary carbon nearly as much as branching at the substrate would.
• If the carbon bearing the leaving group is tertiary (tert-butyl bromide), backside attack is essentially impossible regardless of which base is used — the only viable pathway is $E2$ (proton abstraction from a $\beta$-carbon), so even a small, strong, ionic base like ethoxide gives elimination as the major product.

Step-by-Step Solution

1. Reaction 1: $(CH_3)_3CONa$ (base/nucleophile) + $\mathrm{CH_3CH_2Br}$ (primary substrate). The electrophilic carbon (in $CH_3CH_2Br$) is unhindered, so $t$-butoxide's oxygen performs $S_N2$ substitution: $X = (CH_3)_3C\text{-}O\text{-}CH_2CH_3$.
2. Reaction 2: $(CH_3)_3CBr$ (tertiary substrate) + $\mathrm{CH_3CH_2ONa}$ (base/nucleophile). Backside attack at the tertiary carbon is blocked by the three methyl groups, so $S_N2$ cannot occur; ethoxide instead removes a $\beta$-hydrogen from one of the methyls in an $E2$ process.
3. Elimination product: $Y = (CH_3)_2C=CH_2$ (isobutylene / 2-methylpropene).

Common Mistakes

• Assuming that because $t$-butoxide is "the bulky one," it must always favour elimination — its bulk matters far less when the substrate carbon itself is wide open (primary, no $\beta$-branching issue for backside attack).
• Forgetting that a tertiary substrate physically cannot undergo $S_N2$, so any base (bulky or not) reacting with it is forced toward $E2$ (or $S_N1/E1$, but with a strong base present, $E2$ dominates).

ANSWER: D

Concept and Intuition

Williamson synthesis is an $\mathrm{S_N2}$ reaction between an alkoxide and an alkyl halide. Two structural traps show up here: (1) tertiary alkyl halides are poor $\mathrm{S_N2}$ substrates — with a strong nucleophile/base like an alkoxide they instead undergo E2 elimination; and (2) aryl halides (like bromobenzene) cannot undergo $\mathrm{S_N2}$ at all because the aryl carbon is sp2, in-plane, and shielded, and the C–X bond is strengthened by ring resonance.

Step-by-Step Solution

1. For X, $\mathrm{CH_3CH_2C(CH_3)_2{-}O{-}CH_2CH_2CH_3}$: the tertiary-pentyl part must come in as the alkoxide (sodium tert-alkoxide), while the primary propyl part comes in as the halide (propyl bromide) — the primary halide undergoes clean $\mathrm{S_N2}$ with the bulky alkoxide, avoiding elimination.
2. For Y, phenyl propyl ether ($\mathrm{C_6H_5{-}O{-}CH_2CH_2CH_3}$): since aryl halides (bromobenzene) cannot undergo nucleophilic substitution, the correct route is phenoxide (from phenol) attacking the primary alkyl halide (propyl bromide).
3. Matching option (B): tertiary alkoxide + primary bromide (for X); phenol (phenoxide) + primary bromide (for Y) — consistent with both mechanistic requirements.
4. Options using a tertiary halide, or using bromobenzene as the electrophile, would fail (elimination or no reaction respectively).

Common Mistakes

• Pairing a tertiary alkyl halide with a primary alkoxide, which favours E2 elimination over substitution due to steric hindrance and base strength.
• Trying to react bromobenzene (an aryl halide) with an alkoxide, which simply does not proceed under normal Williamson conditions.

ANSWER: B

Concept and Intuition

The identity of a base/nucleophile controls whether an alkyl halide undergoes substitution or elimination. Sodium tert-butoxide, $(CH_3)_3CO^-Na^+$, is strongly basic (as a good alkoxide) but its oxygen is buried under three bulky methyl groups, making it a very poor nucleophile for SN2 (it cannot easily approach a carbon from the back side). This steric bulk is exactly why tert-butoxide is the textbook reagent of choice to force elimination (E2) rather than substitution even on primary alkyl halides, which would normally favour SN2 with a small nucleophile like hydroxide or ethoxide. For a primary bromide such as $-CH_2CH_2Br$, tert-butoxide therefore removes a $\beta$-hydrogen (E2) to give the terminal alkene $-CH=CH_2$, releasing Br$^-$.

Meanwhile, this same molecule also carries a tertiary, benzylic bromide, $-C(CH_3)_2Br$, directly on the ring. In problems of this kind, the answer choices make clear that the reaction is selective for the primary position over the tertiary one under the given conditions/timescale — the tertiary center is left as the unreacted starting bromide in the major product.

Step-by-Step Solution

1. Identify the base: Me$_3$CONa = sodium tert-butoxide — strong base, very weak/hindered nucleophile.
2. Recognize its standard textbook behaviour: promotes E2 elimination, especially valuable for driving primary alkyl halides toward alkenes instead of the substitution product they would normally favour with a small nucleophile.
3. Apply E2 to the primary bromide arm: $-CH_2-CH_2-Br \xrightarrow{-HBr} -CH=CH_2$ (a terminal vinyl group, the only possible elimination product from a 2-carbon primary chain).
4. Leave the tertiary benzylic C-Br bond as-is in the major product P (matching how the option set frames the outcome — one site reacts, the other survives).
5. Assemble product P: aromatic ring retaining $-C(CH_3)_2Br$ at one position and now bearing $-CH=CH_2$ at the meta position — option (C).

Common Mistakes

• Assuming the more substituted (tertiary) bromide must always react first — reactivity toward a given reagent depends on both electronic and steric/mechanistic compatibility, not substitution level alone.
• Picking a substitution (ether/alcohol) product for a bulky alkoxide base — tert-butoxide's whole teaching point is that it avoids acting as a good nucleophile, favouring E2 instead.

ANSWER: C

Concept and Intuition

Williamson ether synthesis works best when the alkyl halide is primary/unhindered (favours clean $S_N2$) — methyl halides are ideal because they have no $\beta$-hydrogens at all, so competing elimination is impossible regardless of how bulky the incoming alkoxide nucleophile is. The alkoxide's own carbon skeleton is untouched by the reaction (only its oxygen attacks); the methyl group from the alkyl halide simply attaches to that oxygen.

Step-by-Step Solution

1. Identify the nucleophile: the sodium alkoxide of the branched alcohol, $R_3C$–$O^-\,Na^+$ (with $R_3C$ its full carbon skeleton, retained unchanged throughout the reaction).
2. Identify the electrophile: $CH_3Cl$ — small, unhindered, no possibility of elimination (no $\beta$-H).
3. $S_N2$ attack: the alkoxide oxygen's lone pair displaces $Cl^-$ from the methyl carbon.
4. Product: the alkoxide's carbon skeleton remains attached to oxygen, which now also bears the new methyl group — an unsymmetrical ether, $R_3C$–$O$–$CH_3$.
5. Among the choices, this connectivity (skeleton retained – O – new $CH_3$) matches option (A).

Common Mistakes

• Expecting steric hindrance from the bulky alkoxide to force an elimination or a different mechanism — with methyl halide as the electrophile, there is no $\beta$-H for elimination to compete, so clean substitution dominates regardless of alkoxide bulk.

ANSWER: A