Q.Arrange the following hydrogen halides in order of their decreasing reactivity with propene.

Concept and Intuition

Ozonolysis cleaves a C=C bond and replaces it with two C=O bonds — each alkene carbon becomes a carbonyl carbon, retaining its other two substituents. So the two products, put back together end-to-end at their carbonyl carbons (removing the O's and joining the carbons with a double bond), reconstruct the alkene. HBr addition to an alkene in presence of peroxide follows the anti-Markovnikov rule (via a free-radical mechanism): Br ends up on the carbon that already carries more hydrogens (less substituted, i.e. terminal) carbon. The Wurtz reaction couples two molecules of an alkyl halide with sodium metal, joining the two carbons that held the halogen and forming a new C–C bond, doubling the chain.

Step-by-Step Solution

1. Butanone is $CH_3–CO–CH_2–CH_3$; its carbonyl carbon carries a $CH_3$ and a $C_2H_5$ group. Methanal is $H_2C=O$; its carbonyl carbon carries two H's (i.e., it was a terminal $=CH_2$).
2. Joining these two carbonyl carbons back into a C=C bond gives $X$: $CH_2=C(CH_3)(C_2H_5)$, i.e. 2-methylbut-1-ene, $C_5H_{10}$ — matches the given formula.
3. HBr + peroxide → anti-Markovnikov addition: Br goes to the terminal $=CH_2$ carbon (more H's), H goes to the substituted carbon. So $Y = BrCH_2–CH(CH_3)–CH_2–CH_3$ (1-bromo-2-methylbutane).
4. Wurtz reaction: $2\,RBr + 2Na \rightarrow R–R + 2NaBr$. The two $CH_2Br$ carbons couple directly, doubling the chain: $Z = CH_3–CH_2–CH(CH_3)–CH_2–CH_2–CH(CH_3)–CH_2–CH_3$ (3,6-dimethyloctane), 10 carbons total.
5. Classify every carbon in $Z$ by how many other carbons it is bonded to: …

Concept and Intuition

This problem tests two classic reaction pathways for adding HBr to an unsymmetrical alkene, and then a Wurtz coupling step. The twist is that the number of primary carbons in the final products depends entirely on which regioisomer of the alkyl bromide is formed first.

• Markovnikov addition (ionic mechanism, no peroxides): The hydrogen adds to the less substituted carbon of the double bond, so the bromine ends up on the more substituted carbon. For propene, this gives 2‑bromopropane (isopropyl bromide).
• Anti‑Markovnikov addition (radical mechanism, with benzoyl peroxide): The bromine radical adds to the less substituted carbon, so the hydrogen ends up on the more substituted carbon. This gives 1‑bromopropane (n‑propyl bromide).

Then, each bromide undergoes the Wurtz reaction (2 R–Br + 2 Na → R–R + 2 NaBr) in dry ether. The coupling product’s carbon skeleton is simply two alkyl groups joined together. Counting primary carbons (carbons bonded to exactly one other carbon) in the resulting alkane gives the answer.

Step‑by‑Step Reasoning

1. Identify the major product from propene + HBr (no peroxide) Propene is $\ce{CH3-CH=CH2}$. In the absence of peroxides, HBr adds via an ionic mechanism: the proton attacks the less substituted end (the $\ce{CH2}$ group) to form a more stable secondary carbocation. Bromide then attacks the carbocation.

This is 2‑bromopropane (isopropyl bromide). It is the major product (A).

2. Identify the major product from propene + HBr with benzoyl peroxide Benzoyl peroxide initiates a radical chain. The bromine radical adds to the less substituted carbon of the double bond (the terminal $\ce{CH2}$), giving a more stable secondary radical. The radical then abstracts a hydrogen from HBr, yielding the anti‑Markovnikov product.

This is 1‑bromopropane (n‑propyl bromide). It is the major product (B).

3. Wurtz reaction of 2‑bromopropane (A) to give C Two molecules of 2‑bromopropane react with sodium in dry ether:

The product is 2,3‑dimethylbutane. Its structure:

Count primary carbons: The two end $\ce{CH3}$ groups and the two $\ce{CH3}$ branches are each bonded to exactly one other carbon. That gives 4 primary carbons.

4. Wurtz reaction of 1‑bromopropane (B) to give D …

Concept and Intuition

Reaction I: $PBr_3$ converts an alcohol $R-OH$ to $R-Br$ by first forming an alkyl dibromophosphite intermediate, which is then displaced by bromide in an $S_N2$-like step at the original carbinol carbon. For a primary alcohol like 1-propanol, there is no possibility of rearrangement — the product is simply 1-bromopropane, $CH_3CH_2CH_2Br$, straight-chained.

Reaction II: Ordinarily, HBr adds to an alkene following Markovnikov's rule (H to the carbon with more H's already, Br to the more substituted carbon), because the reaction proceeds through the more stable carbocation. But in the presence of a peroxide like benzoyl peroxide, $(C_6H_5COO)_2$, the mechanism switches to a free-radical chain mechanism: peroxide homolysis generates a radical that abstracts a Br from HBr, generating $Br^\bullet$, which adds to the alkene at the less hindered (terminal) carbon to give the more stable secondary radical intermediate — this is the peroxide effect / Kharasch effect, and it results in anti-Markovnikov addition. For propene ($CH_2=CH-CH_3$), Br ends up on the terminal $CH_2$ carbon, giving $CH_3CH_2CH_2Br$ — again straight-chain 1-bromopropane.

Step-by-Step Solution

1. Reaction I: 1-propanol ($CH_3CH_2CH_2OH$) + $PBr_3$ → $S_N2$-type substitution at the primary carbon, no rearrangement → X = $CH_3CH_2CH_2Br$ (straight chain). …

Concept and Intuition

Polar protic solvents stabilize carbocations by hydrogen bonding/solvation, which is exactly what an SN1 mechanism needs (rate-determining ionization to a carbocation). Only a sufficiently stable carbocation (tertiary $>$ secondary $\gg$ primary) makes SN1 the preferred pathway. So the question is really asking: which alkyl bromide $\mathrm{C_4H_9Br}$ is tertiary?

Step-by-Step Solution

1. The only tertiary $\mathrm{C_4H_9Br}$ isomer is tert-butyl bromide, $(\mathrm{CH_3})_3\mathrm{CBr}$ — this ionizes readily to the stable tertiary carbocation, favouring SN1 hydrolysis in polar protic solvents.
2. Option (A): isobutylene $\mathrm{CH_2=C(CH_3)_2}$ + HBr (no peroxide) proceeds via Markovnikov ionic addition — H adds to the terminal $\mathrm{CH_2}$, Br⁻ attacks the more stable tertiary carbocation intermediate, giving $(\mathrm{CH_3})_3\mathrm{CBr}$ directly.
3. Option (B): isobutylene + HBr/peroxide gives anti-Markovnikov addition (radical, Kharasch effect) — Br ends up on the primary carbon, giving isobutyl bromide (primary), not the target. …

Concept and Intuition

The peroxide effect only reverses regiochemistry for HBr addition (radical chain mechanism), giving the anti-Markovnikov product; direct acid-catalysed hydration of an alkene proceeds through the more stable carbocation (Markovnikov). Combining a primary alkyl halide with a sodium alkoxide is the classic Williamson ether synthesis (SN2), which works cleanly only when the halide is primary (avoids competing elimination).

Step-by-Step Solution

1. Propene + $\mathrm{HBr}$, $(\mathrm{C_6H_5CO})_2\mathrm{O_2}$ (peroxide): radical (anti-Markovnikov) addition places Br on the terminal (less substituted) carbon, giving X = 1-bromopropane, $\mathrm{CH_3CH_2CH_2Br}$ (primary bromide).
2. Propene + $\mathrm{H_2O/H^+}$: acid-catalysed Markovnikov hydration proceeds via the more stable secondary carbocation, giving Y = propan-2-ol, $(\mathrm{CH_3})_2\mathrm{CHOH}$ (isopropanol).
3. X and Y "react together with Na": sodium metal first reacts with the alcohol Y to generate sodium isopropoxide, $(\mathrm{CH_3})_2\mathrm{CHO^-Na^+}$ (+ $\mathrm{H_2}$). …

Concept and Intuition

Normally HBr adds Markovnikov (H to the carbon with more H's, Br to the more substituted carbon) via a carbocation mechanism. But in the presence of peroxides, a radical chain mechanism operates instead (the Kharasch/peroxide effect), and Br$^\bullet$ adds first to the terminal, less hindered carbon of the double bond — because this generates the more stable (more substituted) carbon radical at the other position. The final product therefore has Br on the terminal carbon: anti-Markovnikov addition.

Step-by-Step Solution

1. The alkene is $\mathrm{C_6H_5-CH_2-CH=CH_2}$ (allylbenzene); the double bond is between the middle and terminal carbons.
2. Under peroxide, $\mathrm{Br^\bullet}$ adds to the terminal $\mathrm{=CH_2}$ carbon (less hindered), producing the radical $\mathrm{C_6H_5-CH_2-\overset{\bullet}{C}H-CH_2Br}$ (a secondary radical, further stabilised by the adjacent benzylic-type chain).
3. This radical abstracts H from another HBr molecule, regenerating $\mathrm{Br^\bullet}$ and giving the final product $\mathrm{C_6H_5-CH_2-CH_2-CH_2-Br}$. …

Concept and Intuition

The clue "2,4-DNP positive but iodoform negative" pins down Z's functional class precisely: it must be a carbonyl compound (aldehyde or ketone), but NOT one bearing a $CH_3CO-$ group or a $CH_3CH(OH)-$ group. That single clue is enough to distinguish between the two possible regiochemical outcomes of adding HBr to styrene.

Step-by-Step Solution

1. Styrene is $C_6H_5-CH=CH_2$. Two very different additions of HBr are possible depending on the reagent:
   • Plain $HBr$ (ionic, Markovnikov): $H^+$ adds to the terminal $CH_2$, forming the highly stable benzylic carbocation on the ring-attached carbon; $Br^-$ attacks there, giving $C_6H_5CHBrCH_3$ (Markovnikov product, Br next to the ring).
   • $HBr$ with peroxide $(C_6H_5CO)_2O_2$ (radical, anti-Markovnikov / "peroxide effect"): $Br^\bullet$ adds first to the terminal $CH_2$ (leaving the more stable benzylic radical on the ring carbon), and after H-abstraction the net product is $C_6H_5CH_2CH_2Br$ — Br on the far (primary) carbon.
2. If X were plain $HBr$, Y = $C_6H_5CHBrCH_3$ → hydrolysis (SN1, benzylic) → $C_6H_5CH(OH)CH_3$ (1-phenylethanol) → oxidation → $C_6H_5COCH_3$ (acetophenone). But acetophenone is a methyl ketone and gives a positive iodoform test — this contradicts the given fact that Z is iodoform-negative. So X ≠ plain HBr. …

Concept and Intuition

Propene, $CH_3-CH=CH_2$, undergoes Markovnikov addition (dil. $H_2SO_4$/water, or direct HX) to put the new group on the more substituted (middle) carbon, giving 2-substituted products (isopropanol, 2-chloropropane). To land the substituent on the terminal carbon instead (1-chloropropane), you need the anti-Markovnikov route — hydroboration–oxidation — which places $-OH$ on the terminal carbon, then a separate step to swap that $-OH$ for $-Cl$.

Step-by-Step Solution

1. Option A: $(BH_3)_2$ then $H_2O_2/OH^-$ is hydroboration–oxidation. Boron adds to the less-substituted (terminal) carbon (anti-Markovnikov), so oxidation gives the terminal alcohol, 1-propanol, $CH_3CH_2CH_2OH$.
2. Treating 1-propanol with $HCl/ZnCl_2$ (Lucas-type reagent) replaces $-OH$ with $-Cl$ at the same carbon, giving 1-chloropropane, $CH_3CH_2CH_2Cl$ — exactly the target.
3. Option B proposes the same first step but then "NaCl" as the second reagent — $NaCl$ has no mechanism to convert an alcohol's $-OH$ into $-Cl$ (chloride ion alone isn't nucleophilic/electrophilic enough to displace an unactivated $-OH$); no reaction occurs. Rejected. …

Concept and Intuition

Normally HBr adds to an alkene by Markovnikov's rule (H to the carbon with more H's, Br to the more substituted carbon, via a carbocation intermediate). But in the presence of peroxides, the mechanism switches to a free-radical chain addition (the Kharasch/peroxide effect), which reverses the regiochemistry — Br adds to the terminal (less substituted) carbon, giving the primary bromide instead. The Wurtz-Fittig reaction then couples an aryl halide with an alkyl halide using sodium metal in dry ether, forming a new C–C bond directly (no rearrangement), preserving the alkyl chain as-is.

Step-by-Step Solution

1. Propene, $CH_2=CH-CH_3$, reacts with HBr in presence of peroxide $(C_6H_5CO)_2O_2$ — this triggers the anti-Markovnikov (free-radical) addition mechanism.
2. The bromine radical adds to the terminal $CH_2$ carbon (less hindered, forms the more stable secondary radical intermediate at the middle carbon), ultimately placing Br on the terminal carbon: $X = CH_3CH_2CH_2Br$ (1-bromopropane, a primary bromide).
3. $X$ is then treated with bromobenzene ($C_6H_5Br$) and sodium metal in dry ether — this is the Wurtz-Fittig reaction, which directly couples the alkyl group of $X$ to the phenyl ring of bromobenzene. …

Concept and Intuition

In the presence of peroxides, HBr adds to alkenes via a free-radical mechanism, placing Br on the less substituted carbon (anti-Markovnikov), which is the opposite regiochemistry to the normal ionic (Markovnikov) mechanism that places Br on the more substituted carbon. This distinction between the two mechanisms only matters — i.e., only gives a genuinely different product — when the two carbons of the C=C double bond are not equivalent (different substitution). If the alkene is symmetrical about the double bond, both mechanisms deposit Br at chemically equivalent positions, so the anti-Markovnikov product is identical to the normal (Markovnikov) product.

Step-by-Step Solution

1. Propene, $CH_3-CH=CH_2$: unsymmetrical (one carbon has a methyl + H, the other has 2 H's). Markovnikov gives 2-bromopropane; anti-Markovnikov (peroxide) gives 1-bromopropane — genuinely different products.
2. 1-Butene, $CH_3CH_2-CH=CH_2$: unsymmetrical (terminal alkene). Markovnikov gives 2-bromobutane; anti-Markovnikov gives 1-bromobutane — different products.
3. 2-Butene, $CH_3-CH=CH-CH_3$: symmetrical — both alkene carbons carry an identical methyl group. Whichever carbon Br adds to, the product is 2-bromobutane either way. So there is no observable anti-Markovnikov product distinct from the normal product — the peroxide effect makes no difference here. …

Concept and Intuition

In the presence of peroxides, HBr adds to an alkene by a free-radical mechanism (Kharasch effect), which reverses the usual Markovnikov regiochemistry: the bromine ends up on the less substituted (terminal) carbon because the more stable, more substituted radical forms first. A Grignard reagent reacting with $\mathrm{CO_2}$ and then aqueous acid is a standard one-carbon homologation to a carboxylic acid. $\mathrm{PCl_5}$ is a routine way to convert $-\mathrm{COOH}$ to $-\mathrm{COCl}$. Finally, Rosenmund reduction uses a poisoned Pd catalyst (Pd-$\mathrm{BaSO_4}$, often with a quinoline/sulfur poison) so that hydrogenation of the acid chloride stops cleanly at the aldehyde stage instead of continuing to the alcohol/alkane.

Step-by-Step Solution

1. Styrene $\mathrm{Ph{-}CH{=}CH_2}$ + HBr, peroxide $\Rightarrow$ anti-Markovnikov addition $\Rightarrow$ $X = \mathrm{Ph{-}CH_2{-}CH_2{-}Br}$ (2-bromoethylbenzene).
2. $X$ + Mg / dry ether $\Rightarrow$ Grignard reagent $\mathrm{Ph{-}CH_2{-}CH_2{-}MgBr}$.
3. Grignard + $\mathrm{CO_2}$, then $\mathrm{H_3O^+}$ hydrolysis $\Rightarrow$ $Y = \mathrm{Ph{-}CH_2{-}CH_2{-}COOH}$ (3-phenylpropanoic acid). …

Concept and Intuition

Electrophilic addition of $HX$ to an unsymmetrical alkene follows Markovnikov's rule: the proton adds first, generating whichever carbocation is more stable, and the halide then attacks that carbocation. When one end of a double bond is attached to an aromatic ring, protonating the other carbon creates a carbocation directly on the ring-attached (benzylic) carbon — and this cation is strongly stabilised by resonance delocalisation into the aromatic ring. That stabilisation vastly outweighs the modest stability of an ordinary secondary alkyl cation, so the benzylic pathway dominates.

Step-by-Step Solution

1. Number the alkene carbons: $C_6H_5$–$CH$(=C1)=$CH$(=C2)–$CH_3$(C3).
2. Two possible protonation modes: (i) $H^+$ adds to C1 (ring carbon) $\Rightarrow$ cation on C2, a plain secondary alkyl cation; (ii) $H^+$ adds to C2 $\Rightarrow$ cation on C1, a benzylic cation stabilised by the phenyl ring's $\pi$ system.
3. The benzylic cation (mode ii) is far more stable, so this pathway is strongly preferred. …