Q.The following reaction is classified as: CH3CH2I + KOH(aq) → CH3CH2OH + KI.

The water-gas shift reaction is a crucial industrial process, particularly in the production of hydrogen. To understand it, let's first define "water gas" and then see how the "shift" reaction comes into play.

Water gas is a mixture of carbon monoxide ($\mathrm{CO}$) and hydrogen ($\mathrm{H_2}$), typically produced by passing steam over hot coke (carbon) at high temperatures:

$\mathrm{C(s) + H_2O(g) \xrightarrow{1270\,K} CO(g) + H_2(g)}$

While water gas itself is a valuable fuel, the goal in many industrial applications is to produce pure hydrogen. The water-gas shift reaction is employed to convert the carbon monoxide present in water gas into additional hydrogen. This is achieved by reacting the carbon monoxide with more steam.

Here's how we can identify the correct reaction:

1. Analyze Option (A):

   $\mathrm{C(s) + H_2O(g) \xrightarrow{1270\,K} CO(g) + H_2(g)}$

   This reaction represents the production of water gas itself, often called coal gasification or the Bosch process (when followed by the shift reaction). It is not the water-gas shift reaction, but rather the reaction that produces the water gas that will then undergo the shift.

2. Analyze Option (B):

   $\mathrm{CH_4(g) + H_2O(g) \xrightarrow{1270\,K,\,Ni} CO(g) + 3H_2(g)}$

   This reaction is known as steam reforming of methane. It's a primary industrial method for producing hydrogen from natural gas. While it produces $\mathrm{CO}$ and $\mathrm{H_2}$ (similar to water gas), it starts with methane, not carbon, and is a different process from the water-gas shift.

3. Analyze Option (C):

   $\mathrm{CO(g) + H_2O(g) \xrightarrow{673\,K,\,catalyst} CO_2(g) + H_2(g)}$ …

Concept & Intuition

Isopropyl alcohol is propan-2-ol, $\ce{(CH3)2CHOH}$. To get it, the carbon skeleton must have a three-carbon chain with the —OH group on the middle carbon. Each reaction here builds an alcohol from a carbonyl or alkene; we just check the product’s structure. The trick is that Grignard reactions add the alkyl group to the carbonyl carbon, so the alcohol’s —OH ends up on the carbon that was the carbonyl carbon. For isopropyl alcohol, that carbonyl carbon must become the secondary carbon of a three-carbon chain.

Step-by-step reasoning

1. Option (A): $\ce{CH2=CH-CH3} \xrightarrow{\ce{H2O}/\ce{H+}}$

   Acid-catalyzed hydration of propene follows Markovnikov’s rule: the H adds to the less substituted carbon, OH to the more substituted carbon.

   $\ce{CH2=CH-CH3 + H2O ->[H+] CH3-CH(OH)-CH3}$

   Product: propan-2-ol (isopropyl alcohol). So (A) gives it.

2. Option (B): $\ce{CH3CHO} \xrightarrow{\text{(i) } \ce{CH3MgBr} \atop \text{(ii) } \ce{H2O}}$

   Acetaldehyde ($\ce{CH3CHO}$) reacts with methylmagnesium bromide. The Grignard reagent adds $\ce{CH3-}$ to the carbonyl carbon.

   $\ce{CH3-C(=O)-H + CH3MgBr ->[H2O] CH3-C(OH)(CH3)-H}$

   That’s $\ce{(CH3)2CHOH}$ — again isopropyl alcohol. So (B) gives it.

3. Option (C): $(\ce{CH3})_2\ce{CO} \xrightarrow{\ce{NaBH4}}$

   Sodium borohydride reduces acetone (a ketone) to a secondary alcohol.

   $(\ce{CH3})_2\ce{C=O} \xrightarrow{\ce{NaBH4}} (\ce{CH3})_2\ce{CHOH}$

   Product: isopropyl alcohol. So (C) gives it.

4. Option (D): $\ce{CH2O} \xrightarrow{\text{(i) } \ce{C2H5MgBr} \atop \text{(ii) } \ce{H2O}}$ …

Concept & Intuition

Hydrogen gas is produced when a reducing agent (like a metal or carbon) reacts with water (steam) or when water itself is decomposed. But if oxygen is present (as in air), carbon preferentially burns to oxides, not hydrogen. The key is to check the products: if the reaction involves water or steam and a sufficiently reactive element, $H_2$ is likely; if it involves free oxygen, $H_2$ is not formed.

1. Reaction I: Tin with steam Tin is below iron in the reactivity series but still reacts with steam at high temperature. The reaction is:

Tin displaces hydrogen from water, producing hydrogen gas. So I gives $H_2$.

2. Reaction II: Steam over hot coke This is the classic water-gas reaction. Hot carbon (coke) reduces steam:

The mixture of carbon monoxide and hydrogen is called water gas. Clearly, II gives $H_2$.

3. Reaction III: Passage of air over hot coke …

The concept first: two ideas, used in reverse

Idea 1 — What $\mathrm{S_N1}$ tells us about $Y$.

The $\mathrm{S_N1}$ mechanism has two steps:

The rate-determining step is the ionisation, so the reaction is fast only when the carbocation $\mathrm{R^+}$ is stable. Carbocation stability, set by the electron-releasing inductive effect and hyperconjugation of alkyl groups, runs

Hence tertiary halides hydrolyse by $\mathrm{S_N1}$, primary halides by $\mathrm{S_N2}$, and secondary sit in between. If we are told $Y$ hydrolyses by $\mathrm{S_N1}$, then $Y$ is a $3^\circ$ bromide.

Idea 2 — What Markovnikov's rule tells us about $X \rightarrow Y$.

HBr (no peroxide) adds by the ionic route: $\mathrm{H^+}$ attaches to the doubly-bonded carbon that produces the more stable carbocation, and $\mathrm{Br^-}$ then bonds to that cationic carbon. So the Br ends up on the more substituted carbon.

Step-by-step: test each alkene

1. (A) Pent-2-ene, $\mathrm{CH_3CH{=}CHCH_2CH_3}$. Both alkene carbons are $\mathrm{CH}$; protonation either way gives a secondary carbocation, so $Y$ is a $2^\circ$ bromide. Not the best $\mathrm{S_N1}$ substrate. ✗
2. (B) Pent-1-ene, $\mathrm{CH_3CH_2CH_2CH{=}CH_2}$. Markovnikov gives $\mathrm{CH_3CH_2CH_2CHBrCH_3}$ via a secondary cation. ✗
3. (C) 3-Methylbut-1-ene, $(\mathrm{CH_3})_2\mathrm{CH{-}CH{=}CH_2}$. Direct Markovnikov protonation gives the secondary cation $(\mathrm{CH_3})_2\mathrm{CH{-}\overset{+}{C}H{-}CH_3}$ and hence a $2^\circ$ bromide. ✗ (It can rearrange by a hydride shift, but the question asks for the clean, direct answer.) …

Concept and intuition:

The sequence starts with a tertiary alkyl bromide. Alcoholic KOH under heat favors E2 elimination over substitution because the base is strong and the solvent is polar but not strongly nucleophilic. The only possible elimination gives 2‑methylpropene (isobutylene). Then HBr adds to this alkene. Since the alkene is symmetrical, Markovnikov addition places the bromine on the more substituted carbon, regenerating the original tertiary bromide. The key insight: the product of the second step is the same as the starting material, so the overall sequence is a “round trip.”

Step‑by‑step reasoning:

1. First step – E2 elimination The substrate is $(\mathrm{CH}_3)_3\mathrm{CBr}$, a tertiary alkyl halide. Alcoholic KOH is a strong base (ethoxide or hydroxide in ethanol) and the reaction is heated. Under these conditions, elimination (E2) dominates over substitution (SN2 is impossible on a tertiary carbon; SN1 would be slow without a good leaving group in a protic solvent, but here the strong base drives E2). The only β‑hydrogens are on the three methyl groups, all equivalent. Elimination removes a β‑hydrogen and the bromine, forming a double bond between C1 and C2:

So X is 2‑methylpropene (isobutylene).

2. Second step – electrophilic addition of HBr HBr adds to the alkene via a two‑step mechanism: protonation forms the more stable carbocation, then bromide attacks. The alkene is symmetrical: both carbons of the double bond are equally substituted (one is $\mathrm{CH}_2$, the other is $\mathrm{C}(\mathrm{CH}_3)_2$). Markovnikov’s rule says the hydrogen adds to the less substituted carbon, giving the more stable tertiary carbocation:

Then bromide ion attacks the carbocation: …

Dehydrohalogenation is an elimination reaction where a hydrogen halide (HX) is removed from an alkyl halide to form an alkene. This typically proceeds via an E2 mechanism with strong bases or E1 with weak bases. The rate depends on two critical factors:

Carbocation stability (or transition-state stability): The more substituted the carbon bearing the halogen, the more stable the developing positive character in the transition state. Tertiary > secondary > primary.

Leaving-group ability: Better leaving groups accelerate elimination. Bromide is a better leaving group than chloride because Br⁻ is more stable (larger, more polarizable) than Cl⁻.

Let me decode the structures and rank them:

1. Identify each compound:

   • A: $\ce{CH3-CH2-CH2-CH2-Br}$ — primary alkyl bromide
   • B: Branched with Br on a secondary carbon — secondary alkyl bromide
   • C: $\ce{CH3-CH2-CH2-CH2-Cl}$ — primary alkyl chloride
   • D: Branched with Cl on a tertiary carbon — tertiary alkyl chloride

2. Apply the carbocation stability principle:

   The carbon bearing the halogen determines reactivity. For elimination:

   • D (tertiary) forms the most stable transition state
   • B (secondary) is intermediate
   • A and C (both primary) are least reactive by this criterion

3. Apply the leaving-group effect:

   Between compounds with the same substitution:

   • Bromides (A, B) react faster than chlorides (C, D)

   But we must compare across both factors: …

The S$_\text{N}$2 reaction, or Substitution Nucleophilic Bimolecular reaction, is a fundamental type of organic reaction where a nucleophile replaces a leaving group on an electrophilic carbon atom. Understanding its mechanism, kinetics, and stereochemistry is key to predicting its outcome and reactivity. The "2" in S$_\text{N}$2 signifies that the rate-determining step involves two species: the substrate (alkyl halide) and the nucleophile.

Let's break down the characteristics of an S$_\text{N}$2 reaction to evaluate each statement.

1. Mechanism and Stereochemistry (Statement I):

   The S$_\text{N}$2 reaction is a concerted, one-step process. This means the bond formation between the nucleophile and the carbon, and the bond breaking between the carbon and the leaving group, occur simultaneously. The nucleophile attacks the carbon atom from the backside, directly opposite to the leaving group. This backside attack causes the configuration of the carbon atom to invert, much like an umbrella turning inside out in a strong wind. This phenomenon is known as Walden inversion.

   Important

   If the carbon atom undergoing substitution is chiral, an S$_\text{N}$2 reaction will always result in a product with the opposite configuration to the reactant. This is referred to as complete inversion of configuration.

   Therefore, statement I, "It proceeds with complete inversion of configuration," is correct.

2. Kinetics (Statement II):

   The rate of an S$_\text{N}$2 reaction depends on the concentration of both the alkyl halide (substrate) and the nucleophile. This is because both species are involved in the single, rate-determining step.

   The rate law for an S$_\text{N}$2 reaction is given by:

   $\text{Rate} = k[\text{R-X}][\text{Nu}^-]$

   Here, $k$ is the rate constant, $[\text{R-X}]$ is the concentration of the alkyl halide, and $[\text{Nu}^-]$ is the concentration of the nucleophile. Since the rate depends on the concentration of two species, the reaction is second order overall (first order with respect to the alkyl halide and first order with respect to the nucleophile).

   Therefore, statement II, "It is a first order reaction," is incorrect.

3. Intermediates (Statement III):

   As established, the S$_\text{N}$2 reaction is a concerted process. It does not involve the formation of any stable, isolable intermediate. Instead, it passes through a single, high-energy transition state. In this transition state, the carbon atom is simultaneously partially bonded to both the incoming nucleophile and the departing leaving group. The carbon atom is pentacoordinate (has five partial bonds) in this fleeting state.

   Watch out

   A common misconception is to confuse a transition state with an intermediate. An intermediate is a species that exists for a measurable period and can sometimes be isolated, whereas a transition state is a fleeting, high-energy arrangement of atoms at the peak of an energy barrier, which cannot be isolated. …

The key to this question lies in understanding why the usual Markovnikov rule is reversed. Normally, addition of HX to an alkene follows Markovnikov’s rule (the hydrogen attaches to the less substituted carbon). But with HBr and peroxides, the product is the opposite. The reason is a complete change in the reaction mechanism — from ionic to free radical.

Concept and intuition:

In the absence of peroxides, HBr adds via an electrophilic mechanism: the alkene acts as a nucleophile, attacking a proton (H⁺), forming a carbocation. The more stable carbocation (tertiary > secondary > primary) determines the product — that’s Markovnikov.

However, peroxides (like ROOR) decompose into oxygen-centered radicals. These radicals initiate a chain reaction where the bromine atom (Br•) becomes the attacking species, not H⁺. The bromine radical adds to the alkene in a way that forms the more stable carbon radical, which then abstracts a hydrogen from HBr. This leads to the anti-Markovnikov product.

Let’s walk through the mechanism step by step.

1. Initiation — Peroxide decomposition Peroxides (e.g., di-tert-butyl peroxide) undergo homolytic cleavage when heated or exposed to light:

Each alkoxy radical (RO•) is highly reactive and seeks to abstract a hydrogen atom.

2. Chain propagation — Generation of bromine radical The alkoxy radical abstracts a hydrogen from HBr:

Now we have a bromine atom (Br•), a neutral radical. This is the key reactive species.

3. Addition of bromine radical to propene The bromine radical adds to the double bond of propene. Which carbon does it attack?
   • If Br• adds to the terminal carbon (C1), the resulting radical is a secondary carbon radical (more stable).
   • If Br• adds to the internal carbon (C2), the resulting radical is a primary carbon radical (less stable). Radical stability follows the same order as carbocations: tertiary > secondary > primary. So the bromine radical adds to the less substituted carbon (the terminal one) to give the more stable radical:

This is the opposite regiochemistry from the ionic mechanism.

4. Hydrogen abstraction from HBr …

Concept & Intuition

$\mathrm{S_N1}$ reactions proceed via a two-step mechanism: first, the leaving group departs, forming a carbocation; then the nucleophile attacks. The rate-determining step is carbocation formation, so anything that stabilizes the carbocation speeds up the reaction. Conversely, a substrate that cannot form a reasonably stable carbocation will be very slow. Here, we compare four chlorides: we need to judge the stability of the carbocation that would result from losing $\mathrm{Cl^-}$.

1. Identify the carbocation each substrate would form

   • (A) $\mathrm{C_6H_5CH_2Cl}$ → benzylic carbocation $\mathrm{C_6H_5CH_2^+}$
   • (B) $\mathrm{C_6H_5Cl}$ → phenyl carbocation $\mathrm{C_6H_5^+}$
   • (C) $\mathrm{H_2C=CH-CH_2Cl}$ → allyl carbocation $\mathrm{H_2C=CH-CH_2^+}$
   • (D) $(\mathrm{C_6H_5})_2\mathrm{CHCl}$ → diphenylmethyl carbocation $(\mathrm{C_6H_5})_2\mathrm{CH^+}$

2. Rank carbocation stability

   • Benzylic and allylic carbocations are stabilized by resonance: the positive charge is delocalized into the π-system.
   • Diphenylmethyl has two phenyl rings, so even more resonance stabilization — it is the most stable here.
   • Phenyl carbocation is anti-aromatic (if you consider the empty p orbital in the ring, it would be a 4π-electron system in the cation form) and extremely unstable; it is not stabilized by resonance because the empty orbital is orthogonal to the π-system. In fact, it is so unstable that $\mathrm{S_N1}$ is essentially impossible for chlorobenzene.

3. Apply to $\mathrm{S_N1}$ reactivity

   • More stable carbocation → faster $\mathrm{S_N1}$ reaction.
   • Least stable carbocation → slowest (least reactive).
   • The phenyl carbocation is by far the least stable; therefore $\mathrm{C_6H_5Cl}$ is least reactive. …

The question tests a classic contrast in organic chemistry: the stereochemistry of alkyne reduction. An alkyne has a triple bond — two π bonds — and adding one mole of H₂ can stop at the alkene stage if we use a poisoned catalyst (Lindlar’s) or a dissolving metal reduction (Na/NH₃). The difference lies entirely in how the two hydrogen atoms add across the triple bond.

Lindlar’s catalyst (Pd/CaCO₃ poisoned with lead or quinoline) adsorbs H₂ on its surface and delivers both hydrogens to the same face of the alkyne — this is syn addition. The result is a cis-alkene. In contrast, sodium in liquid ammonia generates solvated electrons that add to the alkyne, forming a radical anion; the subsequent protonation steps force the two hydrogens to end up on opposite faces — anti addition — giving a trans-alkene.

Now let’s apply this to the specific molecule: pent-2-yne (CH₃–C≡C–CH₂CH₃). The triple bond is between carbons 2 and 3.

1. Identify the starting alkyne.

   The structure is $\ce{CH3-C#C-CH2CH3}$, which is pent-2-yne. It is an internal, unsymmetrical alkyne. The triple bond is not at the end, so both products will be disubstituted alkenes.

2. Reaction with Lindlar’s catalyst (P).

   $\ce{H2}$ in the presence of Lindlar’s catalyst adds both hydrogen atoms to the same side of the triple bond.

   The product is cis-pent-2-ene:

with the two hydrogens on the same side (cis geometry). So P is a cis-alkene.

3. Reaction with Na/liq. NH₃ (Q). …

The problem asks us to identify the final product Z in a multi-step organic reaction sequence. To do this, we need to determine the product of each individual step, understanding the type of reaction and the role of the reagents involved. This sequence demonstrates common transformations in organic chemistry: nucleophilic substitution, elimination, and electrophilic addition (hydration).

Concept and Intuition

1. Alcohol to Alkyl Halide (Substitution): Alcohols can be converted to alkyl halides by replacing the hydroxyl group ($-OH$) with a halogen atom. Reagents like PBr$_3$, PCl$_5$, SOCl$_2$, or HX are commonly used for this nucleophilic substitution. PBr$_3$ specifically replaces $-OH$ with $-Br$.
2. Alkyl Halide to Alkene (Elimination): Alkyl halides can undergo elimination reactions to form alkenes. This typically happens when a strong base is used, especially in an alcoholic solvent (e.g., alcoholic KOH). The base abstracts a proton from a carbon adjacent to the carbon bearing the halogen, and the halogen leaves as a halide ion, forming a double bond. This is known as dehydrohalogenation.
3. Alkene to Alcohol (Hydration): Alkenes can be converted back to alcohols by adding water across the double bond. This process, called hydration, can be achieved through various methods. Acid-catalyzed hydration, using dilute acid (like H$_2$SO$_4$) and heat, is one such method. The reaction proceeds via the formation of a carbocation intermediate, followed by nucleophilic attack of water and deprotonation.

Let's apply these concepts to the given reaction sequence.

Step-by-Step Solution

1. First Reaction: CH$_3$CH$_2$OH $\xrightarrow{\text{PBr}_3}$ X
   • The starting material is ethanol, CH$_3$CH$_2$OH, which is a primary alcohol.
   • PBr$_3$ (phosphorus tribromide) is a reagent used to convert alcohols into alkyl bromides. The hydroxyl group ($-OH$) is replaced by a bromine atom ($-Br$). This is a nucleophilic substitution reaction.
   • The reaction proceeds as follows:

*   Therefore, **X is bromoethane (CH$_3$CH$_2$Br)**.

2. Second Reaction: X $\xrightarrow{\text{alc. KOH}}$ Y

* The reactant is X, which we identified as bromoethane (CH$_3$CH$_2$Br).

* The reagent is alcoholic KOH (potassium hydroxide in an alcohol solvent). Alcoholic KOH is a strong base and promotes elimination reactions, specifically dehydrohalogenation, where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form an alkene.

* In this case, a hydrogen atom from the methyl group and the bromine atom from the ethyl group are removed.

* The reaction proceeds as follows:

*   Therefore, **Y is ethene (CH$_2$=CH$_2$)**. …