Q.Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.

The key idea is that alkenes, being electron-rich due to the $\pi$ bond, act as nucleophiles and attack an electrophile. This initiates a two-step electrophilic addition mechanism.

Step 1 (Slow): The $\pi$ electrons attack the electrophile (e.g., $\ce{H+}$ from $\ce{H2SO4}$ or $\ce{BH3}$). This forms a more stable carbocation intermediate (Markovnikov addition) or a cyclic intermediate (as in oxymercuration). …

The key to understanding this reaction is to see the alkene not as a passive double bond, but as a region of high electron density — a nucleophile waiting to happen. The $\pi$ electrons are loosely held and easily polarised, making them an attractive target for any electron-deficient species (an electrophile). When an acid like sulphuric acid ($\ce{H2SO4}$) is used, the electrophile is the proton ($\ce{H+}$). The entire process is a classic electrophilic addition, but with water as the final nucleophile, giving an alcohol instead of a haloalkane.

Let’s walk through the mechanism step by step.

1. Protonation of the alkene (the slow, rate-determining step) The $\pi$ bond of the alkene attacks a proton from the acid ($\ce{H2SO4}$ or $\ce{H3O+}$). This forms a $\sigma$ bond between one carbon and the hydrogen, while the other carbon is left with a positive charge — a carbocation intermediate. Why does this happen? The $\pi$ electrons are basic; they seek a positive centre. The proton is the simplest electrophile. For a generic alkene like propene ($\ce{CH3CH=CH2}$), the proton adds to the less substituted carbon (the one with more hydrogens) because that gives the more stable carbocation (tertiary > secondary > primary). This is Markovnikov’s rule in action.

If the proton added to the other carbon, we’d get a primary carbocation ($\ce{CH3CH2CH2+}$), which is much less stable and forms much more slowly.

2. Nucleophilic attack by water The carbocation is highly reactive and electron-deficient. Water, with its lone pairs on oxygen, acts as a nucleophile and attacks the positively charged carbon. This forms a protonated alcohol (an oxonium ion).

Notice that the oxygen now bears a positive charge because it donated a lone pair.

3. Deprotonation to give the alcohol The oxonium ion is acidic (the $\ce{O-H}$ bond is weakened by the positive charge). A nearby water molecule (or the conjugate base of the acid, like $\ce{HSO4-}$) abstracts a proton, regenerating the acid catalyst and yielding the neutral alcohol.

The $\ce{H3O+}$ can then go on to protonate another alkene molecule, making the process catalytic in acid. …

Method: Electrophilic Addition Mechanism (for hydration of alkenes to alcohols)

This is the standard acid-catalysed hydration mechanism — the most common exam method for converting alkenes to alcohols via electrophilic addition.

Step 1: Protonation of the alkene (slow, rate-determining step)

The alkene acts as a nucleophile. The $\pi$-electron cloud attacks the electrophilic proton ($\ce{H+}$) from the acid catalyst (usually dilute $\ce{H2SO4}$).

• The $\ce{H+}$ adds to the less substituted carbon of the double bond (Markovnikov’s rule).
• This forms a carbocation intermediate — the more stable carbocation is formed (tertiary > secondary > primary).

Why? The $\pi$ bond is electron-rich and easily polarised. The proton is a strong electrophile.

Step 2: Nucleophilic attack by water (fast)

Water (present in excess) acts as a nucleophile. The lone pair on oxygen attacks the positively charged carbocation.

• This forms a protonated alcohol (oxonium ion).

Why? The carbocation is electron-deficient; water donates a pair of electrons.

Step 3: Deprotonation (fast)

A water molecule (or the conjugate base $\ce{HSO4-}$) removes a proton from the oxonium ion.

• This regenerates the acid catalyst and yields the neutral alcohol.

Final Result …

Here is a breakdown of the common mistakes students make regarding the electrophilic addition mechanism for preparing alcohols from alkenes (specifically, acid-catalyzed hydration), along with how to avoid each.

The Core Concept (The "Why")

The reaction is acid-catalyzed hydration. The goal is to add a molecule of water ($H_2O$) across the double bond of an alkene to form an alcohol. The "trick" is that water is a poor electrophile, so we need a strong acid (like $H_2SO_4$ or $H_3PO_4$) to protonate the alkene first, making it a good electrophile.

Mistake #1: Forgetting the Catalyst is Regenerated

The Error: Students often write the mechanism as if the acid ($H^+$) is consumed. They show $H_2SO_4$ as a reactant that disappears, or they forget to show the final step where the catalyst is given back.

How to Avoid:

• Remember the role: The acid is a catalyst. It speeds up the reaction without being used up.
• Trace the $H^+$: The mechanism starts with an $H^+$ attacking the alkene. It must end with an $H^+$ being released.
• The final step is key: After the water molecule attacks the carbocation, you get a protonated alcohol ($ROH_2^+$). The last step is always a deprotonation (loss of $H^+$) to give the neutral alcohol and regenerate the acid catalyst.

Mistake #2: Ignoring Carbocation Stability (Markovnikov's Rule)

The Error: Students randomly place the $H^+$ on either carbon of the double bond, leading to the wrong major product. They forget that the reaction follows Markovnikov's rule.

How to Avoid:

• Always check the carbocation: The $H^+$ will add to the carbon that gives the more stable carbocation.
• Stability order: Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$) > Methyl.
• The rule: "The rich get richer." The hydrogen adds to the carbon that already has more hydrogens (the less substituted carbon), leaving the positive charge on the more substituted (more stable) carbon.
• Example: For propene ($CH_3-CH=CH_2$), the $H^+$ adds to the $CH_2$ group (giving a $2^\circ$ carbocation), not to the $CH$ group (which would give a less stable $1^\circ$ carbocation).

Mistake #3: Writing the Wrong Arrow-Pushing for the First Step

The Error: Students draw the arrow from the $H^+$ (from the acid) directly to the alkene carbon, or they show the alkene's pi bond attacking the $H^+$ incorrectly.

How to Avoid:

• The pi bond is the nucleophile: The electron-rich double bond (the $\pi$ electrons) attacks the electron-deficient proton ($H^+$).
• Correct arrow: Draw a curved arrow from the center of the double bond (or from one of the $\pi$ electrons) towards the $H^+$.
• Show the leaving group: The $H^+$ comes from the acid ($H_2SO_4$). The arrow should show the $\pi$ bond attacking the $H$ of the $H-O-SO_3H$ bond, and simultaneously, the $O-H$ bond breaks, giving the $HSO_4^-$ ion.

Mistake #4: Forgetting the Rearrangement Possibility

The Error: Students assume the carbocation formed is always the final one. They don't check if a hydride shift or alkyl shift can occur to form a more stable carbocation.

How to Avoid:

• Always check for rearrangement: After the first step, look at the carbocation. Can a neighboring carbon (with a $H$ or $CH_3$ group) shift its bond to make a more stable carbocation?
• The rule: If a $1^\circ$ carbocation can become a $2^\circ$, or a $2^\circ$ can become a $3^\circ$ via a 1,2-shift, it will happen. …