Q.Predict the major product of acid catalysed dehydration of

Concept & Intuition

Ozonolysis is a reaction that cleaves a carbon–carbon double bond and replaces each of the two doubly bonded carbons with a carbonyl group (C=O). If the carbon originally had two alkyl groups attached (i.e., it was a disubstituted carbon in the double bond), it becomes a ketone; if it had one or zero alkyl groups, it becomes an aldehyde or formaldehyde.

Here, the product is acetone, which is a ketone with the structure (CH₃)₂C=O. That means the original alkene must have had a carbon in the double bond that was attached to two methyl groups — i.e., a (CH₃)₂C= fragment. The other half of the double bond can be anything, but the overall formula is C₅H₁₀, so we need to check which of the given alkenes has that exact structural feature.

Step-by-step reasoning

1. Identify the structural requirement for acetone formation Acetone = (CH₃)₂C=O. This comes from a carbon in the double bond that had two methyl groups attached. So the alkene must contain the fragment

(the other carbon can have H or alkyl groups).

2. Examine each option

   • (A) 2-methyl-2-butene Structure: CH₃–C(CH₃)=CH–CH₃ The double bond is between C2 and C3. C2 has two methyl groups (one from the main chain, one as a branch) → exactly the (CH₃)₂C= fragment. Ozonolysis:

 Acetone + acetaldehyde. ✓

• (B) 2-methyl-1-butene

  Structure: CH₂=C(CH₃)–CH₂–CH₃

  The double bond is terminal (C1=C2). C1 has two H’s → would give formaldehyde (H₂C=O), not acetone. C2 has one methyl and one ethyl → would give a ketone, but not acetone (it would be butanone). ✗

• (C) 3-methyl-1-butene

  Structure: CH₂=CH–CH(CH₃)₂

  Terminal double bond again: C1 gives formaldehyde, C2 gives isobutyraldehyde. No acetone. ✗

• (D) Cyclopentane

  This is a saturated alkane (no double bond), so ozonolysis does not occur. ✗

3. Confirm the molecular formula 2-methyl-2-butene is C₅H₁₀ (5 carbons, 10 hydrogens) — matches perfectly.

Concept & Intuition

The reaction is acid‑catalyzed hydration of an alkyne using Hg²⁺ as a catalyst (the classic “oxymercuration” of alkynes). For an internal alkyne like 2‑butyne (CH₃–C≡C–CH₃), the two triple‑bond carbons are equally substituted (both secondary). Markovnikov’s rule says the –OH adds to the more substituted carbon, but here both are equally substituted, so the enol formed is symmetrical. That enol immediately tautomerizes to the more stable keto form — a ketone, not an aldehyde. The key is that terminal alkynes give aldehydes; internal alkynes give ketones.

Step‑by‑Step Reasoning

1. Identify the substrate

   2‑butyne is CH₃–C≡C–CH₃. It is an internal alkyne (the triple bond is not at the end of the chain).

2. Reaction conditions

   Dilute H₂SO₄ with HgSO₄ provides Hg²⁺ ions, which coordinate to the triple bond, making it more electrophilic. Water then attacks the activated π‑bond.

3. Regiochemistry (Markovnikov addition)

   For an unsymmetrical alkyne, the –OH ends up on the more substituted carbon. Here both carbons are equally substituted (each has one methyl group), so the water can add to either carbon with equal probability. The immediate product is an enol:

(the exact placement of the double bond is the same either way because the molecule is symmetric).

4. Tautomerization The enol is unstable and rapidly undergoes keto‑enol tautomerization. The hydrogen on the carbon adjacent to the –OH shifts to the oxygen, and the double bond moves to become a carbonyl:

This is butanone (also called methyl ethyl ketone).

5. Check the options
   • (A) Butanal is an aldehyde — would come from a terminal alkyne.
   • (B) 1‑butanol is a primary alcohol — not formed here.
   • (C) Butanone is a ketone — matches our product.
   • (D) 2‑butanol is a secondary alcohol — would require full reduction, not hydration.

Concept and Intuition

Pentene is an alkene with one carbon‑carbon double bond. The characteristic reaction of alkenes is electrophilic addition — here, bromine (Br₂) adds across the double bond. Each molecule of pentene consumes exactly one molecule of Br₂. So the problem reduces to a simple stoichiometric calculation: find how many moles of pentene are in 5 g, then the same number of moles of Br₂ is needed, and convert that to grams.

A common mistake is to forget that Br₂ is diatomic (Br₂, not Br), so its molar mass is 160 g/mol, not 80 g/mol. We’ll watch for that.

Step‑by‑Step Solution

1. Determine the molecular formula of pentene The name “pentene” tells us it has 5 carbon atoms and one double bond. For an alkene with one double bond, the general formula is CₙH₂ₙ. So for n = 5:

2. Calculate the molar mass of pentene Atomic masses: C = 12 u, H = 1 u.

3. Find the number of moles in 5 g of pentene

4. Relate moles of pentene to moles of bromine The addition reaction is:

The mole ratio is 1:1. Therefore:

5. Calculate the mass of Br₂ required Bromine is diatomic: Br₂. Its molar mass is:

So the mass of Br₂ is:

6. Match with the given options The value 11.4 g corresponds to option (C).

Concept & Intuition

Ozonolysis of an alkene adds ozone across the C=C bond, forming an ozonide. Reductive workup (e.g., with Zn/H₂O or dimethyl sulfide) cleaves the ozonide into two carbonyl compounds. The key rule:

• A vinylic carbon with two hydrogens (terminal =CH₂) becomes formaldehyde (HCHO).
• A vinylic carbon with one hydrogen (=CHR) becomes an aldehyde (RCHO).
• A vinylic carbon with no hydrogens (=CR₂) becomes a ketone (RCOR).

Thus, to get only ketones, every vinylic carbon in the starting alkene must be of the =CR₂ type — no =CH₂ or =CHR groups allowed.

Step-by-step analysis

1. Option (A): 1,2-butadiene

   Structure: CH₂=C=CH–CH₃. This is an allene (cumulated diene), not a simple alkene. Reductive ozonolysis of an allene cleaves both double bonds, giving a mixture: one terminal =CH₂ yields formaldehyde, the central carbon becomes CO₂ or a ketone depending on substitution, and the other end gives an aldehyde. Definitely not only ketones.

   Result: mixture includes aldehydes.

2. Option (B): 1,4-cyclohexadiene

   Structure: a six-membered ring with two double bonds at positions 1 and 4. Each double bond is of the type –CH=CH– (one H on each vinylic carbon). Reductive ozonolysis cleaves both double bonds, producing dialdehydes (specifically, a linear dialdehyde with four carbons between the two aldehyde groups). No ketones.

   Result: only aldehydes.

3. Option (C): But-2-ene

   Structure: CH₃–CH=CH–CH₃. Each vinylic carbon has one H and one alkyl group (=CHR). Reductive ozonolysis gives two molecules of acetaldehyde (CH₃CHO), an aldehyde.

   Result: only aldehydes.

4. Option (D): 2,3-dimethylbut-2-ene

   Structure: (CH₃)₂C=C(CH₃)₂. Each vinylic carbon has two methyl groups and no hydrogen (=CR₂). Reductive ozonolysis cleaves the double bond to give two molecules of acetone (CH₃COCH₃), a ketone.

   Result: only ketones.

The key concept here is electrophilic addition to alkenes. Alkenes are electron‑rich (due to the π bond) and act as nucleophiles. When HBr approaches, the H⁺ (an electrophile) attacks first, forming a carbocation. The stability of that carbocation determines which carbon the Br⁻ ends up on. Under normal conditions (no peroxides, no light), the reaction follows Markovnikov’s rule: the hydrogen adds to the less‑substituted carbon, and the bromine adds to the more‑substituted carbon. Here, that gives 2‑bromopropane.

Let’s walk through it step by step:

1. Identify the substrate and reagent

   Propene is $\ce{CH3-CH=CH2}$, an unsymmetrical alkene. HBr is a polar molecule with a partial positive charge on H and partial negative on Br. No special conditions (peroxide, light, catalyst) are shown — so we assume ionic, not radical, mechanism.

2. First step: electrophilic attack

   The π electrons of the double bond attack the H⁺ of HBr. This forms a carbocation and a bromide ion. Two possible carbocations could form:

   • Primary carbocation (if H⁺ adds to the middle carbon): $\ce{CH3-CH2-CH2+}$ (less stable)
   • Secondary carbocation (if H⁺ adds to the terminal carbon): $\ce{CH3-CH+-CH3}$ (more stable, because it’s stabilised by two alkyl groups)
   Tip

   The secondary carbocation is about 40 kJ/mol more stable than the primary one. This energy difference is the driving force for Markovnikov addition.

3. Second step: nucleophilic attack

   The bromide ion (Br⁻) quickly attacks the positively charged carbon of the carbocation. In the secondary carbocation, that carbon is the middle carbon (C‑2). So Br⁻ attaches there, giving $\ce{CH3-CHBr-CH3}$ — exactly the product shown.

4. Classify the reaction type

   • Electrophilic addition: The alkene adds a molecule across the double bond, with the first step being attack by an electrophile (H⁺). This matches perfectly.
   • Free radical substitution: Requires radicals (e.g., from peroxides or UV light) and typically replaces a hydrogen, not adds across a double bond.
   • Nucleophilic addition: Alkenes are not electrophilic enough for direct nucleophilic attack; this would require a carbonyl or similar.
   • Free radical addition: Would occur if peroxides were present (anti‑Markovnikov product), but here the product is Markovnikov, and no radical initiator is shown.
   Watch out

   A classic pitfall: if you see HBr + alkene and no conditions, many students jump to “free radical addition” because they remember the peroxide effect. But the peroxide effect requires explicit presence of peroxides or light. Without them, the ionic Markovnikov mechanism dominates.

5. Confirm the product matches

   The figure shows $\ce{CH3-CHBr-CH3}$ (Br on the middle carbon). That is exactly the Markovnikov product. So the reaction is electrophilic addition.

Let’s examine each statement carefully, using the principles of organic reaction mechanisms and structural theory.

1. Statement (A): “The major product formed when 2-methylpropene reacts with dilute H₂SO₄ is tert-butyl alcohol.”

   • 2-Methylpropene is (CH₃)₂C=CH₂. In dilute H₂SO₄, the alkene undergoes electrophilic addition of water (hydration). The mechanism: the alkene protonates to form the more stable carbocation. Protonation can occur at either carbon of the double bond.
     • Protonation at the terminal CH₂ gives a tertiary carbocation: (CH₃)₃C⁺ (very stable).
     • Protonation at the central carbon gives a primary carbocation: (CH₃)₂CH–CH₂⁺ (much less stable).
   • The reaction proceeds exclusively via the tertiary carbocation, which then reacts with water to give tert-butyl alcohol, (CH₃)₃COH.
   • This is a textbook example of Markovnikov addition. So statement (A) is correct.

2. Statement (B): “The electrophilic addition to an unsymmetrical alkene always occurs through the formation of a more stable carbanion intermediate.”

   • This is false on two counts. First, the intermediate in electrophilic addition is a carbocation, not a carbanion. Second, the regioselectivity is governed by the stability of the carbocation (Markovnikov’s rule), not a carbanion.
   • A carbanion would be involved in nucleophilic addition, not electrophilic addition. So (B) is incorrect.

3. Statement (C): “Between the two alkenes — (i) (CH₃)₂C=CH–CH₃ and (ii) C₆H₅–CH=CH–CH₂–CH₃ — compound (i) will show geometrical isomerism.”

   • Geometrical isomerism (cis/trans or E/Z) requires that each carbon of the double bond has two different substituents.
     • For (i): (CH₃)₂C=CH–CH₃. The left carbon has two methyl groups (identical), so it cannot show geometrical isomerism.
     • For (ii): C₆H₅–CH=CH–CH₂–CH₃. The left carbon has H and C₆H₅ (different), the right carbon has H and CH₂CH₃ (different). So (ii) can show geometrical isomerism.
   • Thus statement (C) is incorrect.

4. Statement (D): “Greater the number of alkyl groups attached to the double bonded carbon atoms, the less stable is the alkene.”

   • In reality, alkyl groups are electron-donating and stabilize the double bond via hyperconjugation and inductive effects. More alkyl substituents mean greater stability.
   • For example, tetramethylethylene (four alkyl groups) is more stable than ethylene (zero alkyl groups). So the statement is the opposite of the truth. Hence (D) is incorrect.

Two clues pin down the alcohol. The ethane released with the Grignard reagent gives its molar mass; the positive Tollens (silver-mirror) test on the oxidation product shows that product is an aldehyde, so the alcohol must be primary.

1. Molar mass from the gas volume. A monohydric alcohol reacts with ethyl magnesium iodide as:

One mole of alcohol releases one mole of ethane. At STP, 1 mole occupies $22400\ \text{cm}^3$, so:

Thus moles of alcohol $= 0.1$, and:

2. Identify the carbon count. For $\text{C}_n\text{H}_{2n+2}\text{O}$:

So the alcohol is $\text{C}_5\text{H}_{12}\text{O}$ — a pentanol isomer.

3. Use the Tollens test to fix the structure.

   PCC oxidises primary alcohols to aldehydes and secondary alcohols to ketones; only an aldehyde gives a silver mirror. So the alcohol is primary. Of the options, butan-2-ol (A) and pentan-2-ol (C) are secondary (they give ketones), and 2,2-dimethylethan-1-ol (D) is only a $\text{C}_4$ name with molar mass 74 — it fails the mass check. The primary $\text{C}_5$ alcohol is 2,2-dimethylpropan-1-ol (B).

4. Confirm the molar mass of the answer.

   $\text{C}_5\text{H}_{12}\text{O}$: $5(12) + 12(1) + 16 = 88\ \text{g/mol}$ — matches exactly.

Concept and intuition:

The molecular formula C₃H₄Br₄ suggests a saturated four‑bromine derivative of propane. When treated with Zn dust and heat, vicinal dibromides lose Br₂ to form alkenes; here, with four bromines, two successive eliminations occur, yielding a triple bond. The product [B] is therefore an alkyne. Then, Hg²⁺/H⁺ hydration of terminal alkynes gives methyl ketones (Markovnikov addition), and methyl ketones give the iodoform test. So [C] must be a methyl ketone with three carbons — propanone (acetone). Let’s verify step by step.

1. Identify the starting compound C₃H₄Br₄ With 3 carbons and 4 bromines, the only possible saturated structure is 1,1,2,2‑tetrabromopropane:

(Other arrangements like 1,1,1,2‑tetrabromopropane would have a different H count; here H₄ fits exactly.)

2. Reaction with Zn dust (dehalogenation) Zn dust removes vicinal bromine pairs. The first elimination gives an alkene:

The second elimination (still with excess Zn) removes the remaining two bromines:

So [B] is propyne (methylacetylene).

3. Hydration of [B] with Hg²⁺/H⁺ Hg²⁺/H⁺ adds water across the triple bond following Markovnikov’s rule: the OH ends up on the more substituted carbon. For propyne:

The enol tautomerizes immediately to the more stable keto form:

Thus [C] is propanone (acetone).

4. Iodoform test confirmation Acetone (CH₃COCH₃) has a methyl group attached to the carbonyl carbon, so it gives a positive iodoform test (yellow precipitate of CHI₃). This matches the problem statement.

Step 1 — A: sodium phenoxide.

Phenol is weakly acidic ($\mathrm{p}K_a \approx 10$) because the phenoxide left behind is resonance-stabilised. NaOH therefore deprotonates it:

Why this step is needed: the phenoxide ion is a far stronger activator of the ring than phenol itself (the full negative charge is delocalised onto the ortho and para carbons), making the ring nucleophilic enough to attack a weak electrophile like $\mathrm{CO_2}$ in the next step.

Step 2 — B: the Kolbe–Schmitt reaction → salicylic acid.

Sodium phenoxide is heated with $\mathrm{CO_2}$ under pressure (~400 K, 4–7 atm). The electron-rich ortho carbon attacks $\mathrm{CO_2}$ (electrophilic substitution on the ring); acidification with $\mathrm{H^+}$ then liberates the free acid:

(The ortho product dominates because the incoming carboxylate is held near the phenoxide oxygen — chelation with $\mathrm{Na^+}$.)

Step 3 — Y: acetylation with acetic anhydride → aspirin.

Salicylic acid still has a free phenolic $-\mathrm{OH}$. Acetic anhydride, $\mathrm{(CH_3CO)_2O}$, acetylates that $-\mathrm{OH}$ to an ester ($-\mathrm{OCOCH_3}$), leaving the $-\mathrm{COOH}$ untouched:

This is exactly the industrial preparation of aspirin, the analgesic/antipyretic (and antiplatelet) drug.

Step 4 — Eliminate the distractors.

• (A) Salicylaldehyde — that is the Reimer–Tiemann product (phenol + $\mathrm{CHCl_3}$/NaOH), not this route; and it would be intermediate $B$-like, never the final acetylated product.
• (C) Cumene — made by Friedel–Crafts alkylation of benzene with propene; it is a precursor of phenol, not a product of it.
• (D) Picric acid — 2,4,6-trinitrophenol, from nitration of phenol with conc. $\mathrm{HNO_3}$; no nitrating agent appears here.

ANSWER: B

The Kolbe–Schmitt reaction (commonly called Kolbe’s reaction) is a classic method to introduce a carboxyl group (–COOH) directly onto the aromatic ring of phenol. The key insight is that phenol itself is not reactive enough toward CO₂; you need the more nucleophilic phenoxide ion (from sodium phenate) to attack the electrophilic carbon of CO₂. The reaction proceeds under high pressure and temperature, and after acidification, yields ortho-hydroxybenzoic acid (salicylic acid).

Let’s walk through the reasoning step by step.

1. What the reaction actually does Kolbe’s reaction converts phenol into salicylic acid. The overall transformation is:

The carboxyl group attaches ortho to the –OH group. This is an electrophilic aromatic substitution where CO₂ acts as the electrophile.

2. Why sodium phenate is necessary

   Phenol (C₆H₅OH) is a weak acid. Treating it with NaOH gives sodium phenoxide (C₆H₅O⁻Na⁺). The phenoxide ion is far more electron-rich than phenol itself — the negative charge on oxygen is delocalised into the ring, making the ortho and para positions strongly nucleophilic. CO₂ is a weak electrophile, so only the activated phenoxide can attack it.

   Watch out

   A common mistake is to think phenol reacts directly with CO₂. It does not — the reaction requires the phenoxide ion. So the reactants are sodium phenate and CO₂, not phenol and CO₂.

3. The role of CO₂

   Carbon dioxide is the source of the carboxyl group. Under pressure (about 4–7 atm) and heat (125–150°C), CO₂ inserts into the ortho position of the phenoxide ring. The intermediate is a salicylate salt, which upon acidification yields salicylic acid.

4. Eliminating the other options

   • (A) Sodium phenate and CCl₄: CCl₄ is not involved in Kolbe’s reaction. It is used in the Reimer–Tiemann reaction (with CHCl₃) to introduce a formyl group.
   • (B) Phenol and CHCl₃: This is the Reimer–Tiemann reaction, which gives salicylaldehyde, not salicylic acid.
   • (D) Phenol and CCl₄: No standard reaction; CCl₄ does not carboxylate phenol. Only option (C) matches the actual reagents: sodium phenate and CO₂.

Step 1 — Convert the water mass to moles.

Step 2 — Compare with the $\text{XeF}_6$.

The whole question turns on this ratio — the stoichiometry of the water decides which product forms.

Step 3 — The hydrolysis ladder of $\text{XeF}_6$.

$\text{XeF}_6$ hydrolyses in stages, each water molecule replacing two F atoms by one O:

Step 4 — Pick the stage set by our 1 : 1 ratio.

With exactly one mole of water per mole of $\text{XeF}_6$, hydrolysis stops at the first stage:

So 0.1 mol of $\text{XeOF}_4$ (and 0.2 mol HF) is obtained.

Step 5 — Reject the others.

• (A) $\text{XeO}_3$ — would require 0.3 mol (5.4 g) of water; we only have 1.8 g. This is the trap for anyone who ignores the stoichiometry.
• (C) $\text{XeO}_2\text{F}_2$ — needs 0.2 mol (3.6 g) of water. Not available.
• (D) $\text{Xe}+\text{XeO}_3$ — that is the disproportionation of $\text{XeF}_6$ in basic solution / hydrolysis of $\text{XeF}_4$, not this reaction.

ANSWER: B

The key here is recognising what happens when an alcohol vapour is passed over heated copper at $573~\mathrm{K}$. This is a classic dehydration reaction — the copper acts as a catalyst, and the high temperature drives the elimination of water. For tertiary alcohols, the reaction follows Zaitsev’s rule, but because the alcohol is tertiary, the most substituted alkene is also the only possible one due to the carbon skeleton.

Let’s walk through it step by step.

1. Identify the alcohol. Tertiary butyl alcohol is $(CH_3)_3COH$ — a tertiary carbon (attached to three methyl groups) bonded to an –OH group.

2. Recall the reaction conditions. Passing alcohol vapours over heated copper at $573~\mathrm{K}$ is a standard method for dehydration. The copper catalyses the elimination of water, forming an alkene. This is similar to using concentrated $H_2SO_4$ at high temperature, but with a solid catalyst.

3. Determine the elimination product. Dehydration removes the –OH and a hydrogen from an adjacent carbon. In $(CH_3)_3COH$, the carbon bearing the –OH has no hydrogen atoms (it’s tertiary and fully substituted with methyl groups). So the hydrogen must come from one of the three methyl groups.

   Removing a hydrogen from any methyl group gives the same alkene: the double bond forms between the central carbon and that methyl carbon. The product is $(CH_3)_2C=CH_2$, which is 2-methylpropene (also called isobutylene).

4. Check the options.

   • (A) But-2-ene: $CH_3CH=CHCH_3$ — requires a four-carbon straight chain, not possible here.
   • (B) 2-Butanone: a ketone, not an alkene — no oxidation occurs under these conditions.
   • (C) 2-Methylpropene: $(CH_3)_2C=CH_2$ — matches our product.
   • (D) Butanal: an aldehyde, again not formed by simple dehydration.