Q.Show how will you synthesise:

1. 1-Phenylethanol from an alkene — The key idea is electrophilic addition of water to styrene (phenylethene). Treat styrene with dilute $H_2SO_4$ (acid-catalysed hydration). The proton adds to the less substituted carbon (Markovnikov addition), giving a benzylic carbocation, which is trapped by water. Deprotonation yields 1-phenylethanol.
2. Cyclohexylmethanol from an alkyl halide via $S_N2$ — Use cyclohexylmethyl bromide (or chloride) with a strong nucleophile like $OH^-$ (e.g., aqueous NaOH). The $S_N2$ attack occurs at the primary carbon, inverting configuration, to give cyclohexylmethanol. …

Let’s tackle each synthesis by thinking about the functional group transformation needed. An alcohol can be made from an alkene via electrophilic addition of water (hydration), or from an alkyl halide via nucleophilic substitution ($S_N1$ or $S_N2$). The trick is to match the carbon skeleton and the position of the –OH group.

(i) 1-phenylethanol from a suitable alkene

1-phenylethanol has the structure: $\ce{Ph-CH(OH)-CH3}$. The –OH is on the carbon next to the benzene ring. The obvious alkene precursor is styrene (phenylethene), $\ce{Ph-CH=CH2}$.

Why? Hydration of an alkene follows Markovnikov’s rule: the hydrogen adds to the less substituted carbon, and the –OH adds to the more substituted carbon. In styrene, the double bond is between a benzylic carbon and a terminal carbon. The benzylic carbon is more substituted (and also stabilises a carbocation well). So water adds to give the tertiary-like benzylic alcohol — exactly 1-phenylethanol.

The reaction: $\ce{Ph-CH=CH2 + H2O ->[H2SO4] Ph-CH(OH)-CH3}$.

(ii) cyclohexylmethanol using an alkyl halide by an $S_N2$ reaction

Cyclohexylmethanol is $\ce{C6H11-CH2OH}$. The –OH is on a primary carbon (the –CH2– group attached to the ring). For an $S_N2$ reaction, we need a good leaving group on a primary carbon, and a strong nucleophile.

The alkyl halide must be cyclohexylmethyl halide, e.g., $\ce{C6H11-CH2Br}$ (or chloride/iodide). The nucleophile is hydroxide ion ($\ce{OH-}$), which attacks the primary carbon from the back, displacing the halide.

$S_N2$ works beautifully here because the carbon is primary and unhindered (the cyclohexyl ring is bulky, but the –CH2– group is still accessible). The reaction: $\ce{C6H11-CH2Br + NaOH ->[H2O] C6H11-CH2OH + NaBr}$. …

Here is the clear, concept-first solution for each synthesis, with the method named and steps explained.

(i) 1-phenylethanol from a suitable alkene

Method: Acid-catalysed hydration (Electrophilic addition of water, following Markovnikov’s rule).

Concept: The alkene acts as a nucleophile. The electrophile is $\ce{H+}$ from dilute acid. The more stable carbocation intermediate (here, benzylic) determines the regiochemistry.

Steps:

1. Choose the alkene: Use styrene (phenylethene, $\ce{C6H5-CH=CH2}$).
2. Reaction conditions: Treat styrene with dilute $\ce{H2SO4}$ (or $\ce{H3PO4}$) in aqueous medium.
3. Mechanism:
   • Protonation of the double bond occurs at the terminal carbon (the less substituted end), forming a stabilised benzylic carbocation ($\ce{C6H5-CH+-CH3}$).
   • Water (a nucleophile) attacks the carbocation.
   • Deprotonation yields 1-phenylethanol ($\ce{C6H5-CH(OH)-CH3}$).

Key result: The $\ce{-OH}$ group attaches to the more substituted carbon (the benzylic carbon), following Markovnikov’s rule.

(ii) cyclohexylmethanol from an alkyl halide by an $S_N2$ reaction

Method: Nucleophilic substitution ($S_N2$) using a primary alkyl halide and a strong nucleophile.

Concept: $S_N2$ requires a good leaving group and a strong, sterically unhindered nucleophile. The carbon bearing the leaving group must be primary (or methyl) for a clean backside attack.

Steps:

1. Choose the alkyl halide: Use chloromethylcyclohexane (or bromomethylcyclohexane), $\ce{C6H11-CH2-X}$ (where X = Cl or Br). This is a primary halide — ideal for $S_N2$.
2. Choose the nucleophile: Use hydroxide ion ($\ce{OH-}$) from a strong base like aqueous $\ce{NaOH}$ or $\ce{KOH}$.
3. Reaction conditions: Heat under reflux in a polar aprotic solvent (e.g., acetone) or aqueous ethanol to favour $S_N2$.
4. Mechanism:
   • The $\ce{OH-}$ ion attacks the carbon bearing the halogen from the opposite side (backside attack).
   • The leaving group (halide) departs in a single concerted step.
   • Product: cyclohexylmethanol ($\ce{C6H11-CH2OH}$).

Key result: Inversion of configuration occurs at the reaction centre (though not stereochemically relevant here since the carbon is not chiral).

(iii) pentan-1-ol using a suitable alkyl halide

Method: Nucleophilic substitution ($S_N2$) using a primary alkyl halide and a strong nucleophile. …

Here are the common mistakes students make for each part of this question, along with the correct conceptual approach to avoid them.

(i) 1-phenylethanol from a suitable alkene

Common Mistake 1: Using the wrong alkene (e.g., ethenylbenzene/styrene).

• Why it’s wrong: Students often pick styrene ($C_6H_5CH=CH_2$). Hydration of styrene (via acid-catalyzed addition) follows Markovnikov’s rule. The $H^+$ adds to the less substituted carbon (the $CH_2$ end), placing the positive charge on the more stable benzylic carbon. This gives 1-phenylethanol as the major product. While this can work, the question asks for a suitable alkene. The most direct and unambiguous choice is ethenylbenzene (styrene).
• How to avoid: Always check the regiochemistry. For 1-phenylethanol, the OH must be on the benzylic carbon. The alkene must have the double bond directly attached to the benzene ring so that Markovnikov hydration places the OH there.

Common Mistake 2: Using hydroboration-oxidation.

• Why it’s wrong: Hydroboration-oxidation gives anti-Markovnikov addition. If you use styrene, you get 2-phenylethanol ($C_6H_5CH_2CH_2OH$), not 1-phenylethanol.
• How to avoid: Remember the rule: Markovnikov for acid-catalyzed hydration (or oxymercuration-demercuration); anti-Markovnikov for hydroboration. Match the reagent to the desired product.

Correct Synthesis:

• Alkene: Ethenylbenzene (styrene)
• Reagent: $H_2O/H_2SO_4$ (or $Hg(OAc)_2/H_2O$ followed by $NaBH_4$ for better yields without rearrangement).
• Reaction: Electrophilic addition of water across the double bond.

(ii) Cyclohexylmethanol from an alkyl halide by $S_N2$

Common Mistake 1: Choosing the wrong alkyl halide (e.g., 1-bromocyclohexane).

• Why it’s wrong: 1-bromocyclohexane is a secondary alkyl halide. $S_N2$ reactions are very slow on secondary substrates and practically impossible on tertiary ones due to steric hindrance. The reaction would be extremely inefficient.
• How to avoid: For a reliable $S_N2$, you need a primary alkyl halide. The carbon bearing the leaving group must be $CH_3$ or $RCH_2X$. Here, the target is cyclohexylmethanol ($C_6H_{11}CH_2OH$). The carbon with the OH is a primary carbon attached to the ring. So, the alkyl halide must be cyclohexylmethyl halide (e.g., $C_6H_{11}CH_2Br$).

Common Mistake 2: Using a strong base instead of a good nucleophile.

• Why it’s wrong: $S_N2$ requires a strong nucleophile, not necessarily a strong base. Using $OH^-$ (e.g., aqueous NaOH) is fine, but students sometimes use bulky bases like $t-BuO^-$ which are poor nucleophiles and favor elimination (E2).
• How to avoid: Use a small, strong nucleophile. For making an alcohol, the nucleophile is $OH^-$ (from NaOH or KOH in water/alcohol). The solvent should be polar aprotic (e.g., acetone) or a polar protic solvent (water/ethanol) — though aprotic speeds up $S_N2$.

Correct Synthesis:

• Alkyl halide: (Bromomethyl)cyclohexane ($C_6H_{11}CH_2Br$)
• Reagent: $NaOH$ (aq) or $KOH$ (aq)
• Reaction: $S_N2$ substitution.

(iii) Pentan-1-ol using a suitable alkyl halide

Common Mistake 1: Using 1-bromopentane with $OH^-$ — but forgetting about elimination.

• Why it’s wrong: 1-bromopentane is a primary alkyl halide, so $S_N2$ is fast. However, if you use a strong base like $OH^-$ at high temperature, elimination (E2) can compete, giving pent-1-ene as a byproduct. This reduces yield. …