Test: Reaction of Alcohols - NEET MCQ

• The acidity of alcohols is influenced by the ability of the alkoxide ion formed after deprotonation to stabilize the negative charge.
• Electron-withdrawing groups (EWGs) increase acidity by stabilizing the alkoxide ion, while electron-donating groups (EDGs) decrease acidity.
• In the given alcohols, the presence of EWGs like halogens increases the acidity.
• Option B has a more acidic hydrogen due to the presence of an EWG, such as a halogen or an electronegative substituent, making it more acidic than others.

When alkenes undergo acid-catalyzed hydration (addition of water in the presence of an acid), the major product is formed following Markovnikov's rule. This rule states that the hydroxyl group (-OH) will add to the carbon with the most alkyl substituents (the more stable carbocation intermediate).

For the alkene (CH₃)₂C = CH₂:

• The double bond is between a tertiary carbon and a methylene (-CH₂).
• On acid hydration, the tertiary carbocation (CH₃)₃C⁺ is formed as an intermediate, which is highly stable.
• The hydroxyl group adds to the tertiary carbon, yielding tertiary butyl alcohol (CH₃)₃COH.

Greater the number of hydroxy groups , greater its a bility to form H-bonds with water, hence greater solubility.

Both (I) and (II) are 2° alcohols, less acidicthan (I) and (IV) where there is electron withdrawing chloride group is present. Between (I) and (II), (I) is less acidic as ring is considered here equivalent to two ethyl groups attached to α-carbon. (Ill) is more acidic than (IV) because in (III) Cl is at α-carbon while it is at β-carbon in (IV).

Dihydric alcohols are always more soluble in water than monohydric alcohol. Between (I) and (II), (I) is more soluble as it forms intermolecular H-bonds with water while (II) forms intramolecular H-bonds which decreases its ability to form intermolecular H-bonds with water.

• NaHCO₃ (Sodium bicarbonate): Phenol is acidic enough to react with NaHCO₃ to produce carbon dioxide, whereas ethanol is not acidic enough.
• Na (Sodium): Both ethanol and phenol react with sodium to release hydrogen gas.
• CH₃MgBr (Grignard reagent): Both react to form respective magnesium alkoxides/aryloxides.
• NaOH (Sodium hydroxide): Phenol is acidic and forms sodium phenoxide with NaOH, while ethanol does not react.

Thus, NaOH can distinguish ethanol from phenol.

In the below reaction , no bond with chiral carbon is disturbed, hence, configuration is retained in product.

 Phenolic—OH is neutralised here

Methanol is a liquid chemical with the formula CH3​OH. It is colourless, volatile, flammable and poisonous.

Methanol is made from the destructive distillation of wood and is chiefly synthesized from carbon monoxide and hydrogen.

It is more acidic compound bc2​ this molecule loss H⁺ ions very easily. as compared to compound [A]

It reacts very difficult to lucas reagent process via SN1 mechanism which depends on carbocation stability.

It is diacid base loss H⁺ easily.

No such reaction occur with CH₃OH.
Acidic KMnO₄ forms CO₂ with methanol but CH₃COOH with ethanol.

All these groups increases the stability of gauche conformer due to stable intramolecular H-bonding.

(c) PCI₅ brings about S_N2 reaction, chloro alkane with inverted configuration would be produced.
(d) 3-methyl-2-butanol is a secondary alcohol, will be oxidised to ketone with acidic dichromate solution.

The key point is formation of Y from ortho methoxy phenol by Reimer-Tiemann reaction.

The key point is formation of Y from ortho methoxy phenol by Reimer-Tiemann reaction.

The key point is formation of Y from ortho methoxy phenol by Reimer-Tiemann reaction.

(i) IHCI/ZnCI₂ (Lucas reagent) gives white turbidity.
(ii) NaOH does not give any visible change.

(iv) MnO₂ does not oxidises 2-propanol.
(v) NaNH2 gives effervescence of NH₃,
(vi) C₂H₅ MgBrgives C₂H₆(g)
(vii) NaH gives H₂(g)
(viii), (ix) and (x) all react with 2-propanol but no visible change would be observed.
Hence, only (i), (iii), (v), (vi) and (vii) gives visible change.

Three stereomers of final product are formed, a meso and a pair of enantiomers.

• PCC (Pyridinium Chlorochromate) is a mild oxidizing agent used to oxidize alcohols.
• Primary alcohols are oxidized to aldehydes, while secondary alcohols are oxidized to ketones.
• 2-Propanol (isopropanol) is a secondary alcohol, which means it can be oxidized to a ketone.
• Methanol and ethanol are primary alcohols and oxidized to aldehydes, not ketones.
• Benzyl alcohol is also a primary alcohol, thus forming aldehyde.
• Therefore, 2-Propanol (option C) is oxidized to a ketone by PCC.