Alcohols: Preparation & Properties

Hint: Primary alcohols react with HCl via $SN_2$ mechanism, which is slow and inefficient.

Reactions (I) and (IV) can yield alkyl halide due to the presence of anhydrous $ZnCl_2$ as it acts a catalyst. Reaction (III) is also possible because halogenation will take place by $SN_1$ mechanism and alcohol is tertiary resulting in stable carbocation.

Primary alcohols react with HCl via $SN_2$ mechanism, which is slow and inefficient due to:

• Poor leaving group (-OH is a bad leaving group).

• HCl alone cannot easily protonate the OH group enough to make it a good leaving group.

• No carbocation is formed (since primary carbocations are unstable).

🔹 Why ZnCl₂ is needed (Lucas reagent):

• ZnCl₂ is a Lewis acid that coordinates with the -OH group.

• Converts the -OH into a better leaving group by forming a complex like:
  
  R-OH···ZnCl₂ → R-O⁺-ZnCl₂⁻

• This allows Cl⁻ from HCl to displace the complex and form the alkyl halide.

Hint: Ether reacts with HI and forms phenol and methyl iodide

Explanation:
When phenylmethyl ether reacts with HI then phenol and methyl iodide are formed as a product.
Step-1:
In presence of acid ether can be dissociated. Start with protonating the oxygen of the ether (by the acid of course).
The lone pair of electrons on the oxygen atom of the ether reacts with a proton (H⁺) from HI.
Step-2:
I⁻ can only ever attack the carbon of the methyl group. The other carbon which is part of the benzene ring is not
at all approachable because the delocalized π electron cloud blocks and prevents nucleophiles from attacking the
sp2 carbon.

The reaction is as follows:

Mechanism: 

Hint: In Williamson synthesis reaction, alkoxide reacts with alkyl chloride and form ether as a product.
 

Explanation:
The formation of ether by reaction of alcohol in the presence of base followed by alkylation is known as Williamson ether synthesis. 

Hint: If the compound contains ${\mathrm{CH}}_3-\stackrel{}{\mathrm{C}}\stackrel{\mathrm{OH}}{\mathrm{I}}-$ unit then it will give an iodoform test.

In iodoform reaction, I₂, and NaOH, or NaOI reacts with alcohol which contains ${\mathrm{CH}}_3-\stackrel{}{\mathrm{C}}\stackrel{\mathrm{OH}}{\mathrm{I}}-$ unit. When alcohol reacts with I₂, and NaOH, or NaOI then sodium carboxylate and yellow ppt of CHI₃ is obtained. The reaction is as follows:

Hint: The reaction is Hydroboration-oxidation

The hydroboration oxidation reaction is an organic chemical reaction that is employed for the conversion of alkenes into

alcohol. This is done via a two-step process which includes a hydroboration step and an oxidation step.

Hydroboration oxidation reaction mechanism can be considered as an anti-Markovnikov

a reaction. In this reaction, the hydroxyl group attaches to the carbon which is less substituted. The reaction is as follows:

$C{H}_3-C{H}_2-CH=C{H}_2\underset{2.}{\overset{}{\to }}$ $H_{2}O,$ $H_2{O}_2,OH1.$ $B{H}_3.THFC{H}_{3}C{H}_{2}C{H}_{2}C{H}_2$ $OH$

                                                              (Hydroboration-oxidation)

When Grignard reagent reacts with ketone then tertiary alcohol is obtained as a product. Grignard reagent acts as a nucleophile and carbonyl carbon of ketone acts as an electrophile. The mechanism of the reaction is as follows:

The product contains four carbon atoms chain as a parent chain. The structure with numbering is as follows:

Hence the name of the compound is 2-Methylbutan-2-ol.

Hint: Ketone reaction with the Grignard reagent and form tertiary alcohol.

Explanation:

Ketones when reacting with Grignard reagents produces tertiary alcohols. The reaction is an example of a nucleophilic addition reaction.

The reaction is as follows:

The product is  tert-Butyl alcohol.