Q.Give the structures and IUPAC names of the products expected from the following reactions:

Concept: IUPAC Nomenclature of alcohols formed via reduction, hydration, and Grignard addition.

(a) Catalytic reduction of butanal

Butanal ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$) is reduced by $\text{H}_2$/Ni to a primary alcohol.

Step: The aldehyde group ($-\text{CHO}$) becomes $-\text{CH}_2\text{OH}$.

Product: Butan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$).

(b) Hydration of propene in the presence of dilute $\text{H}_2\text{SO}_4$

Propene ($\text{CH}_3\text{CH}=\text{CH}_2$) adds water following Markovnikov’s rule.

Step: $\text{H}^+$ adds to the less substituted carbon; $\text{OH}$ adds to the more substituted carbon.

Product: Propan-2-ol ($\text{CH}_3\text{CH}(\text{OH})\text{CH}_3$).

(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis …

Let’s understand the why behind each reaction before writing the names. IUPAC nomenclature is systematic: we name the longest carbon chain containing the principal functional group (here, the –OH group of alcohols), number it to give the –OH the lowest locant, and name substituents alphabetically.

(a) Catalytic reduction of butanal

Butanal is an aldehyde with four carbons: $\ce{CH3CH2CH2CHO}$. Catalytic reduction (using $\ce{H2}$ with a metal catalyst like Ni, Pt, or Pd) adds hydrogen across the carbonyl double bond ($\ce{C=O}$). The $\ce{C=O}$ becomes $\ce{C-OH}$, turning the aldehyde into a primary alcohol.

1. The carbonyl carbon in butanal is at the end of the chain (carbon 1). Reduction gives a –$\ce{CH2OH}$ group at that end.
2. The chain remains four carbons long, with the –OH on carbon 1.
3. The IUPAC name: the parent alkane is butane; replace the ‘e’ with ‘ol’ and indicate the position of –OH. Since it’s at the end, the locant is 1. So the name is butan-1-ol.

(b) Hydration of propene in the presence of dilute sulphuric acid

Propene is $\ce{CH3CH=CH2}$. Hydration adds $\ce{H2O}$ across the double bond, following Markovnikov’s rule: the hydrogen (from $\ce{H2O}$) attaches to the carbon with more hydrogens already, and the –OH attaches to the other carbon.

1. In propene, the double bond is between carbons 1 and 2. Carbon 1 has two hydrogens (terminal), carbon 2 has one hydrogen.
2. Markovnikov addition: $\ce{H}$ goes to carbon 1, $\ce{OH}$ goes to carbon 2.
3. The product is $\ce{CH3CH(OH)CH3}$, which is propan-2-ol (the –OH on the middle carbon of a three-carbon chain).
4. IUPAC name: parent chain is propane, –OH on carbon 2 → propan-2-ol.

(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis

Propanone (acetone) is $\ce{CH3COCH3}$, a ketone. Methylmagnesium bromide ($\ce{CH3MgBr}$) is a Grignard reagent — a strong nucleophile and base. The reaction proceeds in two steps:

1. Nucleophilic addition: The $\ce{CH3^-}$ (from the Grignard) attacks the electrophilic carbonyl carbon of propanone. The $\ce{C=O}$ bond breaks, and the oxygen picks up a $\ce{MgBr}$ group, forming an alkoxide intermediate: $\ce{(CH3)3C-O-MgBr}$.
2. Hydrolysis: Adding water (or dilute acid) protonates the alkoxide oxygen, giving the alcohol: $\ce{(CH3)3C-OH}$. …

Here is the clear solution method for each reaction, following the Concept-First Approach.

Method: Retrosynthetic Analysis & Functional Group Transformation

This method works backwards from the reagent to identify the functional group change, then applies IUPAC rules to name the product.

(a) Catalytic reduction of butanal

Step 1: Identify the functional group and the reaction type.

• Reactant: Butanal is an aldehyde ($-CHO$).
• Reagent: Catalytic reduction ($H_2$/Ni, Pd, or Pt) means hydrogenation.
• Concept: Aldehydes reduce to primary alcohols.

Step 2: Write the structural transformation.

• Butanal: $CH_3CH_2CH_2CHO$
• After reduction: $CH_3CH_2CH_2CH_2OH$

Step 3: Apply IUPAC nomenclature for the product.

• Parent chain: 4 carbons (but-).
• Suffix: -ol (alcohol).
• Position of -OH: Carbon 1.
• IUPAC Name: Butan-1-ol

Final Answer: Butan-1-ol ($CH_3CH_2CH_2CH_2OH$)

(b) Hydration of propene in the presence of dilute sulphuric acid

Step 1: Identify the functional group and the reaction type.

• Reactant: Propene is an alkene ($C=C$).
• Reagent: Dilute $H_2SO_4$ + $H_2O$ (hydration).
• Concept: This follows Markovnikov’s rule: the $H$ adds to the carbon with more $H$ atoms, and the $OH$ adds to the other carbon.

Step 2: Write the structural transformation.

• Propene: $CH_3-CH=CH_2$
• $H$ adds to $CH_2$ (more H's), $OH$ adds to $CH$.
• Product: $CH_3-CH(OH)-CH_3$

Step 3: Apply IUPAC nomenclature.

• Parent chain: 3 carbons (prop-).
• Suffix: -ol.
• Position of -OH: Carbon 2.
• IUPAC Name: Propan-2-ol

Final Answer: Propan-2-ol ($CH_3CH(OH)CH_3$)

(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis

Step 1: Identify the functional group and the reaction type.

• Reactant: Propanone is a ketone ($C=O$).
• Reagent: $CH_3MgBr$ (Grignard reagent) followed by $H_3O^+$ (hydrolysis). …

Here are the common mistakes students make on this exact IUPAC nomenclature and reaction mechanism question, along with how to avoid each.

General Mistake: Forgetting to Identify the Functional Group First

The Mistake: Students rush to name the product without first identifying the functional group created by the reaction. For example, they might see "catalytic reduction" and assume it's still an aldehyde.

How to Avoid: Before writing the name, always ask: "What functional group does this reaction produce?"

• Catalytic reduction of an aldehyde → Primary alcohol.
• Hydration of an alkene → Alcohol (follows Markovnikov's rule).
• Grignard reaction with a ketone → Tertiary alcohol.

(a) Catalytic reduction of butanal

Common Mistake 1: Writing the wrong parent chain.

• Error: Naming the product as "butan-1-ol" but drawing the structure of butan-2-ol (or vice versa).
• Why it happens: Students confuse the position of the $-$OH group. Butanal has the $-$CHO group at the end. Reduction adds H₂ across the C=O, giving $-$CH₂OH at the terminal carbon.
• How to avoid: Draw the structure of butanal first: $\ce{CH3CH2CH2CHO}$. The carbonyl carbon is C1. After reduction, C1 becomes $\ce{CH2OH}$. The product is butan-1-ol (n-butanol).

Common Mistake 2: Incorrect IUPAC numbering.

• Error: Writing "1-butanol" instead of "butan-1-ol" (or vice versa, depending on the exam board). In IUPAC, the locant (number) comes before the suffix.
• How to avoid: Use the format: parent chain + locant + suffix. For butan-1-ol, the locant "1" is placed before "-ol". (Some older systems use "1-butanol", but modern IUPAC prefers "butan-1-ol". Check your textbook's convention.)

Correct Answer:

• Structure: $\ce{CH3CH2CH2CH2OH}$
• IUPAC Name: Butan-1-ol (or 1-butanol)

(b) Hydration of propene in the presence of dilute sulphuric acid

Common Mistake 1: Ignoring Markovnikov's rule.

• Error: Writing the product as propan-1-ol ($\ce{CH3CH2CH2OH}$) instead of propan-2-ol.
• Why it happens: Students think the $-$OH adds to the less substituted carbon (anti-Markovnikov). But in acid-catalysed hydration, the hydrogen adds to the carbon with more hydrogens (Markovnikov), so the $-$OH goes to the more substituted carbon.
• How to avoid: For propene ($\ce{CH3CH=CH2}$), the double bond is between C1 and C2. The hydrogen (from H⁺) adds to C1 (which has 2 H's), and the $-$OH adds to C2 (which has 1 H). Product: propan-2-ol.

Common Mistake 2: Writing the wrong IUPAC name for the alcohol.

• Error: Writing "isopropyl alcohol" or "2-propanol" instead of "propan-2-ol".
• How to avoid: Use the systematic IUPAC name. The parent chain is propane. The $-$OH is on carbon 2. Hence, propan-2-ol.

Correct Answer:

• Structure: $\ce{CH3CH(OH)CH3}$
• IUPAC Name: Propan-2-ol (or 2-propanol)

(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis

Common Mistake 1: Forgetting that the Grignard reagent adds to the carbonyl carbon.

• Error: Writing the product as a ketone or an aldehyde.
• Why it happens: Students think the Grignard reagent replaces the oxygen or something else. Actually, the $\ce{CH3MgBr}$ adds the methyl group ($\ce{CH3-}$) to the carbonyl carbon, and after hydrolysis, you get an alcohol. …