Q.The following is not an appropriate reaction for the preparation of t-butyl ethyl ether. $\mathrm{C_2H_5ONa \; + \; CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-Cl \longrightarrow CH_3-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OC_2H_5}$

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 …

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. …

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). …

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 $X=\mathrm{CH_3CH_2CH_2CH_3}$. …

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: $X = (CH_3)_3C\text{-}O\text{-}CH_2CH_3$. …

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). …

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). …

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. …