Q: Q2. Draw the structures of the following compounds.

The structures of the given compounds are as follows:\n\n(i) α-Methoxypropionaldehyde:\n - Propionaldehyde is CH₃CH₂CHO.\n - α-position is the carbon adjacent to the aldehyde group.\n - Structure: CH₃-CH(OCH₃)-CHO\n\n(ii) 3-Hydroxybutanal:\n - Butanal is a 4-carbon aldehyde.\n - Hydroxyl group at position 3.\n - Structure: CH₃-CH(OH)-CH₂-CHO\n\n(iii) 2-Hydroxycyclopentanecarbaldehyde:\n - Cyclopentanecarbaldehyde means a -CHO group directly attached to a cyclopentane ring.\n - Hydroxyl group at position 2 of the cyclopentane ring (relative to the -CHO group).\n - Structure: A cyclopentane ring with a -CHO group attached to one carbon, and an -OH group attached to the adjacent carbon.\n\n(iv) 4-Oxopentanal:\n - Pentanal is a 5-carbon aldehyde.\n - 'Oxo' indicates a ketone group (C=O) at position 4.\n - Structure: CH₃-CO-CH₂-CH₂-CHO\n\n(v) Di-sec-butyl ketone:\n - Ketone (C=O) with two sec-butyl groups attached.\n - A sec-butyl group is -CH(CH₃)CH₂CH₃.\n - Structure: (CH₃CH₂CH(CH₃))-CO-(CH(CH₃)CH₂CH₃)\n\n(vi) 4-Fluoracetophenone:\n - Acetophenone is C₆H₅-CO-CH₃.\n - Fluoro group at position 4 (para) on the phenyl ring.\n - Structure: F-C₆H₄-CO-CH₃ (where F is para to -CO-CH₃)\n\n(vii) 3-Cyclohexylpentan-3-ol:\n - Pentan-3-ol is a 5-carbon chain with an -OH group at position 3.\n - Cyclohexyl group attached to position 3.\n - Structure: CH₃CH₂-C(OH)(C₆H₁₁)-CH₂CH₃ (where C₆H₁₁ is a cyclohexyl group)\n\n(viii) Cyclopent-3-en-1-one:\n - Cyclopentanone is a 5-membered cyclic ketone.\n - Double bond at position 3 (relative to the ketone carbon as 1).\n - Structure: A 5-membered ring with a ketone at C1 and a double bond between C3 and C4.\n\n(ix) p-Bromobenzaldehyde:\n - Benzaldehyde is C₆H₅-CHO.\n - Bromo group at para position on the phenyl ring.\n - Structure: Br-C₆H₄-CHO (where Br is para to -CHO)

Q: Q5. Arrange the following compounds in increasing order of their boiling points. Ethanal, Propanal, Propan-1-ol, Butan-1-ol.

The boiling point of a compound depends primarily on the strength of intermolecular forces of attraction and molecular mass. Generally, higher molecular mass leads to higher boiling point due to increased van der Waals forces.\n\nLet's analyze the given compounds:\n1. **Ethanal (CH₃CHO)**: Molecular mass = 44 g/mol. Contains a polar carbonyl group, leading to dipole-dipole interactions and weak van der Waals forces.\n2. **Propanal (CH₃CH₂CHO)**: Molecular mass = 58 g/mol. Also contains a polar carbonyl group, leading to dipole-dipole interactions and stronger van der Waals forces than ethanal due to higher molecular mass.\n3. **Propan-1-ol (CH₃CH₂CH₂OH)**: Molecular mass = 60 g/mol. Contains a hydroxyl (-OH) group, enabling strong intermolecular hydrogen bonding, in addition to dipole-dipole and van der Waals forces.\n4. **Butan-1-ol (CH₃CH₂CH₂CH₂OH)**: Molecular mass = 74 g/mol. Contains a hydroxyl (-OH) group, enabling strong intermolecular hydrogen bonding. It has a higher molecular mass than propan-1-ol, leading to stronger van der Waals forces and thus a higher boiling point.\n\n**Comparison of Intermolecular Forces:**\n* Aldehydes (Ethanal, Propanal) exhibit dipole-dipole interactions and van der Waals forces.\n* Alcohols (Propan-1-ol, Butan-1-ol) exhibit strong hydrogen bonding, which is significantly stronger than dipole-dipole interactions.\n\nTherefore, alcohols will have higher boiling points than aldehydes of comparable molecular masses.\n\n**Order based on molecular mass and intermolecular forces:**\n* Ethanal (MM=44, dipole-dipole) < Propanal (MM=58, dipole-dipole)\n* Propan-1-ol (MM=60, H-bonding) < Butan-1-ol (MM=74, H-bonding)\n\nComparing aldehydes and alcohols:\n* Propanal (MM=58) vs Propan-1-ol (MM=60): Propan-1-ol has hydrogen bonding, so its boiling point will be much higher than propanal despite similar molecular mass.\n\n**Increasing order of boiling points:**\nEthanal < Propanal < Propan-1-ol < Butan-1-ol

Q: Q6. Give the structures of the following compounds.\n(i) α-Methoxypropionaldehyde\n(ii) 3-Hydroxybutanal\n(iii) 2-Hydroxycyclopentane carbaldehyde\n(iv) 4-Oxopentanal\n(v) Di-sec-butyl

The structures of the given compounds are as follows:\n\n(i) **α-Methoxypropionaldehyde**\n * Parent chain: Propionaldehyde (3-carbon aldehyde).\n * α-position: The carbon atom adjacent to the carbonyl carbon (C-2).\n * Methoxy group (-OCH₃) at α-position.\n * Structure: CH₃-CH(OCH₃)-CHO\n\n(ii) **3-Hydroxybutanal**\n * Parent chain: Butanal (4-carbon aldehyde).\n * Hydroxy group (-OH) at C-3.\n * Structure: CH₃-CH(OH)-CH₂-CHO\n\n(iii) **2-Hydroxycyclopentane carbaldehyde**\n * Parent group: Cyclopentane carbaldehyde (CHO group directly attached to a cyclopentane ring).\n * Hydroxy group (-OH) at C-2 of the cyclopentane ring (relative to the CHO group as C-1).\n * Structure: A cyclopentane ring with a -CHO group and an -OH group on the adjacent carbon.\n\n(iv) **4-Oxopentanal**\n * Parent chain: Pentanal (5-carbon aldehyde).\n * Oxo group (=O, ketone) at C-4.\n * Structure: CH₃-CO-CH₂-CH₂-CHO\n\n(v) **Di-sec-butyl ketone**\n * Ketone functional group (-CO-).\n * Two sec-butyl groups attached to the carbonyl carbon.\n * sec-butyl group: CH₃-CH₂-CH(CH₃)-\n * Structure: (CH₃-CH₂-CH(CH₃))-CO-(CH(CH₃)-CH₂-CH₃)\n\n(vi) **4-Fluoroacetophenone**\n * Parent compound: Acetophenone (C₆H₅-CO-CH₃, a benzene ring with -CO-CH₃ group).\n * Fluoro group (-F) at para (4th) position on the benzene ring relative to the -CO-CH₃ group.\n * Structure: F-C₆H₄-CO-CH₃ (where F is para to COCH₃)

Q: Q11. How will you convert the following compounds into benzoic acid?\n(i) Ethylbenzene\n(ii) Acetophenone\n(iii) Bromobenzene\n(iv) Phenyl ethene (Styrene)

The conversion of the given compounds into benzoic acid can be achieved through various synthetic routes involving oxidation, Grignard reaction, and haloform reaction.\n\n**(i) Ethylbenzene to Benzoic acid**\nEthylbenzene can be converted to benzoic acid by vigorous oxidation of the alkyl side chain using strong oxidizing agents like alkaline potassium permanganate, followed by acidification.\n\n**(ii) Acetophenone to Benzoic acid**\nAcetophenone, being a methyl ketone, can be converted to benzoic acid via the haloform reaction (specifically iodoform reaction), which yields sodium benzoate, followed by acidification.\n\n**(iii) Bromobenzene to Benzoic acid**\nBromobenzene can be converted to benzoic acid using a Grignard reagent. Bromobenzene is first converted to phenylmagnesium bromide, which then reacts with carbon dioxide, followed by hydrolysis.\n\n**(iv) Phenyl ethene (Styrene) to Benzoic acid**\nPhenyl ethene (Styrene) can be converted to benzoic acid by vigorous oxidative cleavage of the carbon-carbon double bond and subsequent oxidation of the resulting fragments using a strong oxidizing agent like alkaline potassium permanganate, followed by acidification.

Q: Q15. An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzened

The organic compound with molecular formula C9H10O is 2-ethylbenzaldehyde (o-ethylbenzaldehyde).

Question 1

Q1. Name the following compounds according to IUPAC system of nomenclature:

Accepted Answer

The IUPAC names for the given compounds are as follows:

(i) CH₃CH(CH₃)CH₂CH₂CHO: 4-Methylpentanal
   - Longest carbon chain containing the aldehyde group is 5 carbons (pentanal).
   - Methyl group is at position 4.

(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₃: 3-Ethylpentan-2-one
   - Longest carbon chain containing the ketone group is 5 carbons (pentan).
   - Ketone group is at position 2 (pentan-2-one).
   - Ethyl group is at position 3.

(iii) CH₃CH(OH)CH₂CH₂COOH: 4-Hydroxypentanoic acid
   - Longest carbon chain containing the carboxylic acid group is 5 carbons (pentanoic acid).
   - Hydroxyl group is at position 4.

(iv) O₂NCH₂CH₂CH₂COOH: 4-Nitrobutanoic acid
   - Longest carbon chain containing the carboxylic acid group is 4 carbons (butanoic acid).
   - Nitro group is at position 4.

(v) CH₃CH(CH₃)CH₂C(CH₃)₂COOH: 3,3,4-Trimethylpentanoic acid
   - Longest carbon chain containing the carboxylic acid group is 5 carbons (pentanoic acid).
   - Methyl groups are at positions 3, 3, and 4.

(vi) (CH₃)₂CHCOCH(Cl)CH₃: 3-Chloro-2,4-dimethylpentan-2-one
   - Longest carbon chain containing the ketone group is 5 carbons (pentan).
   - Ketone group is at position 2 (pentan-2-one).
   - Chloro group is at position 3.
   - Methyl groups are at positions 2 and 4.

(vii) CH₃CH₂C(Br)(CH₃)CH₂COOH: 3-Bromo-3-methylpentanoic acid
   - Longest carbon chain containing the carboxylic acid group is 5 carbons (pentanoic acid).
   - Bromo group is at position 3.
   - Methyl group is at position 3.

(viii) (CH₃)₂C=CHCHO: 3-Methylbut-2-enal
   - Longest carbon chain containing the aldehyde group and double bond is 4 carbons (butenal).
   - Double bond is at position 2 (but-2-enal).
   - Methyl group is at position 3.

(ix) C₆H₅CH₂CH₂COOH: 3-Phenylpropanoic acid
   - Longest carbon chain containing the carboxylic acid group is 3 carbons (propanoic acid).
   - Phenyl group is at position 3.

(x) CH₃CH=CHCOOH: But-2-enoic acid
   - Longest carbon chain containing the carboxylic acid group and double bond is 4 carbons (butenoic acid).
   - Double bond is at position 2 (but-2-enoic acid).

(xi) OHC-C₆H₄-CHO (p-): Benzene-1,4-dicarbaldehyde (or p-Benzenedicarbaldehyde)
   - Benzene ring with two aldehyde groups at para positions.

(xii) CH₃CH₂CH(Br)CH₂CH(CH₃)CHO: 4-Bromo-2-methylhexanal
   - Longest carbon chain containing the aldehyde group is 6 carbons (hexanal).
   - Bromo group is at position 4.
   - Methyl group is at position 2.

Question 2

Q2. Draw the structures of the following compounds.

Accepted Answer

The structures of the given compounds are as follows:

(i) α-Methoxypropionaldehyde:
    - Propionaldehyde is CH₃CH₂CHO.
    - α-position is the carbon adjacent to the aldehyde group.
    - Structure: CH₃-CH(OCH₃)-CHO

(ii) 3-Hydroxybutanal:
    - Butanal is a 4-carbon aldehyde.
    - Hydroxyl group at position 3.
    - Structure: CH₃-CH(OH)-CH₂-CHO

(iii) 2-Hydroxycyclopentanecarbaldehyde:
    - Cyclopentanecarbaldehyde means a -CHO group directly attached to a cyclopentane ring.
    - Hydroxyl group at position 2 of the cyclopentane ring (relative to the -CHO group).
    - Structure: A cyclopentane ring with a -CHO group attached to one carbon, and an -OH group attached to the adjacent carbon.

(iv) 4-Oxopentanal:
    - Pentanal is a 5-carbon aldehyde.
    - 'Oxo' indicates a ketone group (C=O) at position 4.
    - Structure: CH₃-CO-CH₂-CH₂-CHO

(v) Di-sec-butyl ketone:
    - Ketone (C=O) with two sec-butyl groups attached.
    - A sec-butyl group is -CH(CH₃)CH₂CH₃.
    - Structure: (CH₃CH₂CH(CH₃))-CO-(CH(CH₃)CH₂CH₃)

(vi) 4-Fluoracetophenone:
    - Acetophenone is C₆H₅-CO-CH₃.
    - Fluoro group at position 4 (para) on the phenyl ring.
    - Structure: F-C₆H₄-CO-CH₃ (where F is para to -CO-CH₃)

(vii) 3-Cyclohexylpentan-3-ol:
    - Pentan-3-ol is a 5-carbon chain with an -OH group at position 3.
    - Cyclohexyl group attached to position 3.
    - Structure: CH₃CH₂-C(OH)(C₆H₁₁)-CH₂CH₃ (where C₆H₁₁ is a cyclohexyl group)

(viii) Cyclopent-3-en-1-one:
    - Cyclopentanone is a 5-membered cyclic ketone.
    - Double bond at position 3 (relative to the ketone carbon as 1).
    - Structure: A 5-membered ring with a ketone at C1 and a double bond between C3 and C4.

(ix) p-Bromobenzaldehyde:
    - Benzaldehyde is C₆H₅-CHO.
    - Bromo group at para position on the phenyl ring.
    - Structure: Br-C₆H₄-CHO (where Br is para to -CHO)

Question 3

Q3. Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Accepted Answer

To arrange the compounds in increasing order of their boiling points, we need to consider the types and strengths of intermolecular forces present in each compound.

**Compounds and their Intermolecular Forces:**
1.  **CH₃CH₂CH₃ (Propane):** This is an alkane. It is a non-polar molecule and exhibits only weak London Dispersion Forces (LDFs). These are the weakest intermolecular forces.
2.  **CH₃OCH₃ (Dimethylether):** This is an ether. It is a polar molecule due to the bent structure around the oxygen atom, leading to a net dipole moment. It exhibits dipole-dipole interactions in addition to London Dispersion Forces. Dipole-dipole interactions are stronger than LDFs.
3.  **CH₃CHO (Ethanal/Acetaldehyde):** This is an aldehyde. It is a polar molecule due to the highly polar carbonyl (C=O) group. It exhibits strong dipole-dipole interactions (stronger than those in ethers due to the greater polarity of the C=O bond) and London Dispersion Forces. However, it cannot form hydrogen bonds with itself.
4.  **CH₃CH₂OH (Ethanol):** This is an alcohol. It is a polar molecule and contains a hydroxyl (-OH) group. This allows it to form strong intermolecular hydrogen bonds, in addition to dipole-dipole interactions and London Dispersion Forces. Hydrogen bonding is the strongest type of intermolecular force among these compounds.

**Order of Intermolecular Force Strength:**
London Dispersion Forces < Dipole-dipole interactions < Hydrogen bonding

**Boiling Point Order:**
Stronger intermolecular forces require more energy to overcome, resulting in higher boiling points.

Therefore, the increasing order of boiling points is:

**CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH**

**Explanation:**
*   **Propane (CH₃CH₂CH₃):** Lowest boiling point due to only weak London Dispersion Forces.
*   **Dimethylether (CH₃OCH₃):** Higher boiling point than propane due to additional dipole-dipole interactions, but no hydrogen bonding.
*   **Ethanal (CH₃CHO):** Higher boiling point than dimethylether due to stronger dipole-dipole interactions arising from the more polar carbonyl group, but still no hydrogen bonding.
*   **Ethanol (CH₃CH₂OH):** Highest boiling point due to the presence of strong intermolecular hydrogen bonding, which requires significantly more energy to break.

Question 4

Q4. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
(i) Ethanal, Propanal, Propanone, Butanone
(ii) Benzaldehyde, p-Tolual

Accepted Answer

Reactivity of aldehydes and ketones towards nucleophilic addition reactions is primarily governed by two factors: 1. **Steric Hindrance:** As the number and size of alkyl groups around the carbonyl carbon increase, steric hindrance increases, making it more difficult for the nucleophile to attack the carbonyl carbon. This decreases reactivity. 2. **Electronic Effects:** The carbonyl carbon (C=O) is electrophilic due to the electron-withdrawing oxygen. Any factor that increases the partial positive charge on the carbonyl carbon will increase its electrophilicity and thus its reactivity towards nucleophilic attack. Electron-donating groups (like alkyl groups) decrease the electrophilicity by pushing electron density towards the carbonyl carbon, thereby decreasing reactivity. Electron-withdrawing groups increase electrophilicity and reactivity. Generally, aldehydes are more reactive than ketones due to less steric hindrance and stronger positive charge on the carbonyl carbon (one alkyl group vs. two alkyl groups). --- **(i) Ethanal, Propanal, Propanone, Butanone** Let's analyze each compound: * **Ethanal (CH₃CHO):** Has one methyl group and one hydrogen atom attached to the carbonyl carbon. Least steric hindrance among the given compounds and only one electron-donating alkyl group. * **Propanal (CH₃CH₂CHO):** Has one ethyl group and one hydrogen atom attached to the carbonyl carbon. More steric hindrance than ethanal due to the larger ethyl group, and a stronger electron-donating effect from the ethyl group compared to methyl. * **Propanone (CH₃COCH₃):** Has two methyl groups attached to the carbonyl carbon. More steric hindrance than aldehydes and two electron-donating methyl groups, which reduce the electrophilicity of the carbonyl carbon more effectively than one alkyl group. * **Butanone (CH₃COCH₂CH₃):** Has one methyl group and one ethyl group attached to the carbonyl carbon. Most steric hindrance among the given compounds (due to the ethyl group) and two electron-donating alkyl groups (methyl and ethyl), making it the least electrophilic. **Order of Reactivity (Decreasing):** Aldehydes > Ketones Within aldehydes: Smaller alkyl group > Larger alkyl group Within ketones: Smaller alkyl groups > Larger alkyl groups **Increasing order of reactivity in nucleophilic addition reactions:** **Butanone < Propanone < Propanal < Ethanal** **Explanation:** * **Butanone** is a ketone with one methyl and one ethyl group, leading to maximum steric hindrance and electron donation, thus least reactive. * **Propanone** is a ketone with two methyl groups, less steric hindrance and electron donation than butanone, making it more reactive than butanone. * **Propanal** is an aldehyde with one ethyl group, less steric hindrance and electron donation than ketones, making it more reactive than propanone. * **Ethanal** is an aldehyde with one methyl group, least steric hindrance and electron donation, making it the most reactive. --- **(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.** Let's analyze each compound, focusing on steric and electronic effects on the carbonyl carbon. * **Benzaldehyde (C₆H₅CHO):** Has a phenyl group and a hydrogen atom. The phenyl group is weakly electron-withdrawing by induction but can donate electrons by resonance, making the overall effect complex but generally less electron-donating than alkyl groups. It has less steric hindrance than ketones. * **p-Tolualdehyde (CH₃-C₆H₄-CHO):** Has a methyl group at the para position of the phenyl ring. The methyl group is electron-donating (+I effect and hyperconjugation). This electron-donating effect is transmitted through the benzene ring to the carbonyl carbon, decreasing its electrophilicity and thus reactivity compared to benzaldehyde. * **p-Nitrobenzaldehyde (O₂N-C₆H₄-CHO):** Has a nitro group at the para position of the phenyl ring. The nitro group is strongly electron-withdrawing (-M and -I effect). This strong electron-withdrawing effect increases the partial positive charge on the carbonyl carbon, making it more electrophilic and thus most reactive. * **Acetophenone (C₆H₅COCH₃):** This is a ketone, with a phenyl group and a methyl group attached to the carbonyl carbon. Ketones are generally less reactive than aldehydes due to greater steric hindrance and stronger electron-donating effect from the additional alkyl group (methyl in this case) compared to a hydrogen atom in aldehydes. **Order of Reactivity (Decreasing):** * Aldehydes are generally more reactive than ketones. * For substituted benzaldehydes, electron-withdrawing groups increase reactivity, while electron-donating groups decrease reactivity. **Increasing order of reactivity in nucleophilic addition reactions:** **Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde** **Explanation:** * **Acetophenone:** Being a ketone, it is the least reactive due to greater steric hindrance and electron-donating effect of the methyl group compared to aldehydes. * **p-Tolualdehyde:** The methyl group at the para position is electron-donating, which decreases the electrophilicity of the carbonyl carbon, making it less reactive than benzaldehyde. * **Benzaldehyde:** More reactive than p-tolualdehyde as it lacks the electron-donating methyl group. * **p-Nitrobenzaldehyde:** The nitro group at the para position is strongly electron-withdrawing, which significantly increases the electrophilicity of the carbonyl carbon, making it the most reactive among these compounds.

Question 5

Q5. Arrange the following compounds in increasing order of their boiling points. Ethanal, Propanal, Propan-1-ol, Butan-1-ol.

Accepted Answer

The boiling point of a compound depends primarily on the strength of intermolecular forces of attraction and molecular mass. Generally, higher molecular mass leads to higher boiling point due to increased van der Waals forces.

Let's analyze the given compounds:
1.  **Ethanal (CH₃CHO)**: Molecular mass = 44 g/mol. Contains a polar carbonyl group, leading to dipole-dipole interactions and weak van der Waals forces.
2.  **Propanal (CH₃CH₂CHO)**: Molecular mass = 58 g/mol. Also contains a polar carbonyl group, leading to dipole-dipole interactions and stronger van der Waals forces than ethanal due to higher molecular mass.
3.  **Propan-1-ol (CH₃CH₂CH₂OH)**: Molecular mass = 60 g/mol. Contains a hydroxyl (-OH) group, enabling strong intermolecular hydrogen bonding, in addition to dipole-dipole and van der Waals forces.
4.  **Butan-1-ol (CH₃CH₂CH₂CH₂OH)**: Molecular mass = 74 g/mol. Contains a hydroxyl (-OH) group, enabling strong intermolecular hydrogen bonding. It has a higher molecular mass than propan-1-ol, leading to stronger van der Waals forces and thus a higher boiling point.

**Comparison of Intermolecular Forces:**
*   Aldehydes (Ethanal, Propanal) exhibit dipole-dipole interactions and van der Waals forces.
*   Alcohols (Propan-1-ol, Butan-1-ol) exhibit strong hydrogen bonding, which is significantly stronger than dipole-dipole interactions.

Therefore, alcohols will have higher boiling points than aldehydes of comparable molecular masses.

**Order based on molecular mass and intermolecular forces:**
*   Ethanal (MM=44, dipole-dipole) < Propanal (MM=58, dipole-dipole)
*   Propan-1-ol (MM=60, H-bonding) < Butan-1-ol (MM=74, H-bonding)

Comparing aldehydes and alcohols:
*   Propanal (MM=58) vs Propan-1-ol (MM=60): Propan-1-ol has hydrogen bonding, so its boiling point will be much higher than propanal despite similar molecular mass.

**Increasing order of boiling points:**
Ethanal < Propanal < Propan-1-ol < Butan-1-ol

Question 6

Q6. Give the structures of the following compounds.
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl

Accepted Answer

The structures of the given compounds are as follows:

(i) **α-Methoxypropionaldehyde**
   *   Parent chain: Propionaldehyde (3-carbon aldehyde).
   *   α-position: The carbon atom adjacent to the carbonyl carbon (C-2).
   *   Methoxy group (-OCH₃) at α-position.
   *   Structure: CH₃-CH(OCH₃)-CHO

(ii) **3-Hydroxybutanal**
   *   Parent chain: Butanal (4-carbon aldehyde).
   *   Hydroxy group (-OH) at C-3.
   *   Structure: CH₃-CH(OH)-CH₂-CHO

(iii) **2-Hydroxycyclopentane carbaldehyde**
   *   Parent group: Cyclopentane carbaldehyde (CHO group directly attached to a cyclopentane ring).
   *   Hydroxy group (-OH) at C-2 of the cyclopentane ring (relative to the CHO group as C-1).
   *   Structure: A cyclopentane ring with a -CHO group and an -OH group on the adjacent carbon.

(iv) **4-Oxopentanal**
   *   Parent chain: Pentanal (5-carbon aldehyde).
   *   Oxo group (=O, ketone) at C-4.
   *   Structure: CH₃-CO-CH₂-CH₂-CHO

(v) **Di-sec-butyl ketone**
   *   Ketone functional group (-CO-).
   *   Two sec-butyl groups attached to the carbonyl carbon.
   *   sec-butyl group: CH₃-CH₂-CH(CH₃)-
   *   Structure: (CH₃-CH₂-CH(CH₃))-CO-(CH(CH₃)-CH₂-CH₃)

(vi) **4-Fluoroacetophenone**
   *   Parent compound: Acetophenone (C₆H₅-CO-CH₃, a benzene ring with -CO-CH₃ group).
   *   Fluoro group (-F) at para (4th) position on the benzene ring relative to the -CO-CH₃ group.
   *   Structure: F-C₆H₄-CO-CH₃ (where F is para to COCH₃)

Question 7

Q7. Write the structures of products of the following reactions:
(i) CH₃–CH₂–CH₂–COCl + H₂/Pd–BaSO₄
(ii) C₆H₅–CH₃ + CrO₃/acetic anhydride, then H₃O⁺
(iii) CH₃–CN + (i) DIBAL-H (ii)

Accepted Answer

Here are the structures of the main products for each reaction:

(i) **CH₃–CH₂–CH₂–COCl + H₂/Pd–BaSO₄**
    *   **Reaction Type:** Rosenmund reduction.
    *   **Description:** Acid chlorides are catalytically hydrogenated over palladium poisoned by barium sulfate (BaSO₄) to yield aldehydes. BaSO₄ acts as a poison to prevent further reduction of the aldehyde to alcohol.
    *   **Product:** Butanal (CH₃–CH₂–CH₂–CHO)

(ii) **C₆H₅–CH₃ + CrO₃/acetic anhydride, then H₃O⁺**
    *   **Reaction Type:** Etard reaction.
    *   **Description:** Toluene (methylbenzene) is oxidized by chromyl chloride (CrO₂Cl₂) or chromium trioxide (CrO₃) in acetic anhydride to form a chromium complex, which upon hydrolysis yields benzaldehyde. The use of acetic anhydride prevents further oxidation to benzoic acid.
    *   **Product:** Benzaldehyde (C₆H₅–CHO)

(iii) **CH₃–CN + (i) DIBAL-H (ii) H₂O**
    *   **Reaction Type:** Reduction of nitrile to aldehyde using DIBAL-H (Diisobutylaluminium hydride).
    *   **Description:** DIBAL-H is a mild reducing agent that reduces nitriles to imines, which upon hydrolysis give aldehydes. The reaction is carried out at low temperatures (-78°C) to prevent over-reduction to primary amines.
    *   **Product:** Ethanal (CH₃–CHO)

(iv) **CH₂=CH–CH₂CN + (i) DIBAL-H (ii) H₂O**
    *   **Reaction Type:** Reduction of nitrile to aldehyde using DIBAL-H.
    *   **Description:** Similar to (iii), DIBAL-H selectively reduces the nitrile group to an aldehyde without affecting the carbon-carbon double bond.
    *   **Product:** But-3-enal (CH₂=CH–CH₂–CHO)

(v) **C₆H₅–CO–CH₃ + C₆H₅–CHO + NaOH**
    *   **Reaction Type:** Crossed Aldol condensation.
    *   **Description:** Acetophenone (C₆H₅–CO–CH₃) has α-hydrogens, while benzaldehyde (C₆H₅–CHO) does not. In the presence of a base (NaOH), acetophenone forms an enolate which attacks the carbonyl carbon of benzaldehyde. This is followed by dehydration to form an α,β-unsaturated ketone.
    *   **Product:** 1,3-Diphenylprop-2-en-1-one (Chalcone) (C₆H₅–CH=CH–CO–C₆H₅)

(vi) **CH₃–CH₂–COOH + Cl₂/Red P, H₂O**
    *   **Reaction Type:** Hell-Volhard-Zelinsky (HVZ) reaction.
    *   **Description:** Carboxylic acids having an α-hydrogen are halogenated at the α-position when treated with chlorine or bromine in the presence of a small amount of red phosphorus. Water is used in the workup.
    *   **Product:** 2-Chloropropanoic acid (CH₃–CH(Cl)–COOH)

Question 8

Q8. Show how will you synthesise the following compounds from appropriate alkenes.
(i) 1-Phenylethanol
(ii) Cyclohexylmethanol
(iii) Pentan-1-ol
(iv) Pentan-2-ol

Accepted Answer

The synthesis of alcohols from alkenes typically involves hydration reactions. Depending on the desired regioselectivity (Markovnikov vs. anti-Markovnikov), different reagents are used:\n\n*   **Markovnikov's Rule:** Hydrogen adds to the carbon with more hydrogens, and the hydroxyl group adds to the carbon with fewer hydrogens. Achieved by acid-catalyzed hydration (H₂O/H⁺) or oxymercuration-demercuration (Hg(OAc)₂/H₂O, then NaBH₄).\n*   **Anti-Markovnikov's Rule:** Hydrogen adds to the carbon with fewer hydrogens, and the hydroxyl group adds to the carbon with more hydrogens. Achieved by hydroboration-oxidation ((i) BH₃.THF (ii) H₂O₂/OH⁻).\n\nLet's synthesize each compound:\n\n### (i) 1-Phenylethanol\n\n1-Phenylethanol is a secondary alcohol where the -OH group is attached to the carbon bearing the phenyl group and one methyl group. This corresponds to a Markovnikov addition product.\n\n*   **Target Structure:** C₆H₅-CH(OH)-CH₃\n*   **Appropriate Alkene:** Styrene (Phenylethene, C₆H₅-CH=CH₂)\n\n**Synthesis Pathway:**\nStyrene undergoes acid-catalyzed hydration (addition of water across the double bond) following Markovnikov's rule. The -OH group adds to the more substituted carbon (the one attached to the phenyl group), and -H adds to the less substituted carbon.\n\n**Reaction:**\nC₆H₅-CH=CH₂ + H₂O $\xrightarrow{H^+}$ C₆H₅-CH(OH)-CH₃\n\n### (ii) Cyclohexylmethanol\n\nCyclohexylmethanol is a primary alcohol where a -CH₂OH group is attached to a cyclohexane ring. This requires an anti-Markovnikov addition if starting from an alkene where the -CH₂OH group is formed from a terminal carbon.\n\n*   **Target Structure:** Cyclohexane-CH₂OH\n*   **Appropriate Alkene:** Methylenecyclohexane (Cyclohexylidenemethane)\n\n**Synthesis Pathway:**\nMethylenecyclohexane undergoes hydroboration-oxidation. This reaction proceeds via anti-Markovnikov addition, placing the -OH group on the less substituted carbon of the double bond, which in this case is the exocyclic methylene carbon.\n\n**Reaction:**\nCyclohexane=CH₂ + (i) BH₃.THF (ii) H₂O₂/OH⁻ $\longrightarrow$ Cyclohexane-CH₂OH\n\n### (iii) Pentan-1-ol\n\nPentan-1-ol is a primary alcohol. To obtain a primary alcohol from an alkene, anti-Markovnikov addition of water is required.\n\n*   **Target Structure:** CH₃-CH₂-CH₂-CH₂-CH₂-OH\n*   **Appropriate Alkene:** Pent-1-ene (CH₃-CH₂-CH₂-CH=CH₂)\n\n**Synthesis Pathway:**\nPent-1-ene undergoes hydroboration-oxidation. The -OH group adds to the terminal carbon (C-1), which is the less substituted carbon of the double bond, following anti-Markovnikov's rule.\n\n**Reaction:**\nCH₃-CH₂-CH₂-CH=CH₂ + (i) BH₃.THF (ii) H₂O₂/OH⁻ $\longrightarrow$ CH₃-CH₂-CH₂-CH₂-CH₂-OH\n\n### (iv) Pentan-2-ol\n\nPentan-2-ol is a secondary alcohol. To obtain a secondary alcohol from a terminal alkene, Markovnikov addition of water is required. If starting from an internal alkene, Markovnikov addition can also lead to a secondary alcohol.\n\n*   **Target Structure:** CH₃-CH(OH)-CH₂-CH₂-CH₃\n*   **Appropriate Alkene:** Pent-1-ene (CH₃-CH₂-CH₂-CH=CH₂) or Pent-2-ene (CH₃-CH₂-CH=CH-CH₃)\n\n**Synthesis Pathway (using Pent-1-ene):**\nPent-1-ene undergoes acid-catalyzed hydration. The -OH group adds to the more substituted carbon (C-2), following Markovnikov's rule.\n\n**Reaction (using Pent-1-ene):**\nCH₃-CH₂-CH₂-CH=CH₂ + H₂O $\xrightarrow{H^+}$ CH₃-CH(OH)-CH₂-CH₂-CH₃\n\n**Alternative Synthesis Pathway (using Pent-2-ene):**\nPent-2-ene also undergoes acid-catalyzed hydration. The -OH group can add to either C-2 or C-3, but due to symmetry and stability of carbocation intermediates, pentan-2-ol is a major product, along with pentan-3-ol.\n\n**Reaction (using Pent-2-ene):**\nCH₃-CH₂-CH=CH-CH₃ + H₂O $\xrightarrow{H^+}$ CH₃-CH(OH)-CH₂-CH₂-CH₃ (and CH₃-CH₂-CH(OH)-CH₂-CH₃)

Question 9

Q11. How will you convert the following compounds into benzoic acid?
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenyl ethene (Styrene)

Accepted Answer

The conversion of the given compounds into benzoic acid can be achieved through various synthetic routes involving oxidation, Grignard reaction, and haloform reaction.

**(i) Ethylbenzene to Benzoic acid**
Ethylbenzene can be converted to benzoic acid by vigorous oxidation of the alkyl side chain using strong oxidizing agents like alkaline potassium permanganate, followed by acidification.

**(ii) Acetophenone to Benzoic acid**
Acetophenone, being a methyl ketone, can be converted to benzoic acid via the haloform reaction (specifically iodoform reaction), which yields sodium benzoate, followed by acidification.

**(iii) Bromobenzene to Benzoic acid**
Bromobenzene can be converted to benzoic acid using a Grignard reagent. Bromobenzene is first converted to phenylmagnesium bromide, which then reacts with carbon dioxide, followed by hydrolysis.

**(iv) Phenyl ethene (Styrene) to Benzoic acid**
Phenyl ethene (Styrene) can be converted to benzoic acid by vigorous oxidative cleavage of the carbon-carbon double bond and subsequent oxidation of the resulting fragments using a strong oxidizing agent like alkaline potassium permanganate, followed by acidification.

Question 10

Q12. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid

Accepted Answer

Simple chemical tests can be used to distinguish between the given pairs of compounds based on their characteristic functional groups and reactivity.

**(i) Propanal and Propanone**
Propanal is an aldehyde, while propanone is a ketone. Aldehydes are easily oxidized, while ketones are not. This difference in reactivity forms the basis of several distinguishing tests.
*   **Tollen's Test:** Propanal (aldehyde) will give a positive Tollen's test, forming a silver mirror on the inner walls of the test tube due to the reduction of Ag+ ions to metallic silver. Propanone (ketone) will not react.
    *   Propanal: CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH- → CH3CH2COO- + 2Ag(s) + 4NH3 + 2H2O
    *   Propanone: No reaction
*   **Fehling's Test:** Propanal will give a positive Fehling's test, producing a red precipitate of cuprous oxide (Cu2O). Propanone will not react.
    *   Propanal: CH3CH2CHO + 2Cu2+ (from Fehling's solution) + 5OH- → CH3CH2COO- + Cu2O(s) + 3H2O
    *   Propanone: No reaction

**(ii) Acetophenone and Benzophenone**
Acetophenone is a methyl ketone (contains a -COCH3 group), whereas benzophenone does not. The presence of the methyl ketone group allows acetophenone to undergo the iodoform test.
*   **Iodoform Test:** Acetophenone will give a positive iodoform test, producing a yellow precipitate of iodoform (CHI3) when treated with iodine and sodium hydroxide. Benzophenone, lacking the -COCH3 group, will not give this test.
    *   Acetophenone: C6H5COCH3 + 3I2 + 4NaOH → C6H5COONa + CHI3(s) + 3NaI + 3H2O
    *   Benzophenone: No reaction

**(iii) Phenol and Benzoic acid**
Phenol is a weak acid, while benzoic acid is a stronger carboxylic acid. They also differ in their reactivity with ferric chloride.
*   **Sodium Bicarbonate Test:** Benzoic acid, being a stronger acid than carbonic acid, reacts with sodium bicarbonate (NaHCO3) to produce carbon dioxide gas, observed as brisk effervescence. Phenol is too weak an acid to react with NaHCO3.
    *   Benzoic acid: C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2(g)
    *   Phenol: No reaction
*   **Ferric Chloride Test:** Phenol reacts with neutral ferric chloride solution to give a characteristic violet, blue, or green coloration due to the formation of a colored complex. Benzoic acid does not give this test.
    *   Phenol: 6C6H5OH + FeCl3 → [Fe(OC6H5)6]3- + 3H+ + 3HCl (Colored complex)
    *   Benzoic acid: No characteristic color change

**(iv) Benzoic acid and Ethyl benzoate**
Benzoic acid is a carboxylic acid, while ethyl benzoate is an ester. Carboxylic acids are acidic and react with bases, while esters are neutral.
*   **Sodium Bicarbonate Test:** Benzoic acid will react with sodium bicarbonate to produce carbon dioxide gas (brisk effervescence). Ethyl benzoate, being an ester, is neutral and will not react.
    *   Benzoic acid: C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2(g)
    *   Ethyl benzoate: No reaction
*   **Litmus Test:** Benzoic acid will turn blue litmus paper red. Ethyl benzoate will have no effect on litmus paper.

**(v) Pentan-2-one and Pentan-3-one**
Pentan-2-one is a methyl ketone (contains a -COCH3 group), whereas pentan-3-one does not.
*   **Iodoform Test:** Pentan-2-one will give a positive iodoform test, producing a yellow precipitate of iodoform (CHI3) when treated with iodine and sodium hydroxide. Pentan-3-one will not give this test.
    *   Pentan-2-one: CH3CH2CH2COCH3 + 3I2 + 4NaOH → CH3CH2CH2COONa + CHI3(s) + 3NaI + 3H2O
    *   Pentan-3-one: No reaction

**(vi) Benzaldehyde and Acetophenone**
Benzaldehyde is an aldehyde, while acetophenone is a methyl ketone. They can be distinguished by tests for aldehydes and the iodoform test.
*   **Tollen's Test:** Benzaldehyde (aldehyde) will give a positive Tollen's test (silver mirror). Acetophenone (ketone) will not react.
    *   Benzaldehyde: C6H5CHO + 2[Ag(NH3)2]+ + 3OH- → C6H5COO- + 2Ag(s) + 4NH3 + 2H2O
    *   Acetophenone: No reaction
*   **Iodoform Test:** Acetophenone (methyl ketone) will give a positive iodoform test (yellow precipitate of CHI3). Benzaldehyde does not have a methyl ketone group and will not react.
    *   Acetophenone: C6H5COCH3 + 3I2 + 4NaOH → C6H5COONa + CHI3(s) + 3NaI + 3H2O
    *   Benzaldehyde: No reaction

**(vii) Ethanal and Propanal**
Both ethanal and propanal are aldehydes, so they will both give positive Tollen's and Fehling's tests. However, ethanal is a methyl aldehyde (contains a -CHO group directly attached to a methyl group), which allows it to undergo the iodoform test.
*   **Iodoform Test:** Ethanal will give a positive iodoform test, producing a yellow precipitate of iodoform (CHI3) when treated with iodine and sodium hydroxide. Propanal, lacking the -COCH3 or -CH(OH)CH3 group, will not give this test.
    *   Ethanal: CH3CHO + 3I2 + 4NaOH → HCOONa + CHI3(s) + 3NaI + 3H2O
    *   Propanal: No reaction

Question 11

Q12.10. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid

Accepted Answer

Here are simple chemical tests to distinguish between the given pairs of compounds:

**(i) Propanal and Propanone**

*   **Principle:** Aldehydes (like propanal) are easily oxidized, while ketones (like propanone) are generally resistant to oxidation under mild conditions.
*   **Test:** Tollens' Test or Fehling's Test
    *   **Tollens' Test:** Add Tollens' reagent (ammoniacal silver nitrate solution) to both compounds and warm gently.
        *   **Propanal:** Gives a positive test, forming a silver mirror on the inner walls of the test tube or a black precipitate of silver. (RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag(s) + 4NH₃ + 2H₂O)
        *   **Propanone:** Gives a negative test; no silver mirror or precipitate is formed.
    *   **Fehling's Test:** Add Fehling's solution (a mixture of Fehling's A and B) to both compounds and heat.
        *   **Propanal:** Gives a positive test, forming a reddish-brown precipitate of cuprous oxide (Cu₂O). (RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O(s) + 3H₂O)
        *   **Propanone:** Gives a negative test; no reddish-brown precipitate is formed.

**(ii) Acetophenone and Benzophenone**

*   **Principle:** Acetophenone is a methyl ketone (contains the CH₃CO- group) and can undergo the iodoform reaction, while benzophenone does not.
*   **Test:** Iodoform Test
    *   **Procedure:** Add iodine solution (I₂) and sodium hydroxide solution (NaOH) to both compounds and warm gently.
        *   **Acetophenone:** Gives a positive test, forming a yellow precipitate of iodoform (CHI₃) with a characteristic antiseptic smell. (C₆H₅COCH₃ + 3I₂ + 4NaOH → C₆H₅COONa + CHI₃(s) + 3NaI + 3H₂O)
        *   **Benzophenone:** Gives a negative test; no yellow precipitate is formed.

**(iii) Phenol and Benzoic acid**

*   **Principle:** Carboxylic acids are stronger acids than phenols and react with sodium bicarbonate to produce carbon dioxide gas. Phenols do not.
*   **Test:** Sodium Bicarbonate Test
    *   **Procedure:** Add a saturated solution of sodium bicarbonate (NaHCO₃) to both compounds.
        *   **Benzoic acid:** Gives a positive test, producing brisk effervescence due to the evolution of carbon dioxide gas. (C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑)
        *   **Phenol:** Gives a negative test; no effervescence is observed.
*   **Alternative Test:** Ferric Chloride Test
    *   **Procedure:** Add a few drops of neutral ferric chloride (FeCl₃) solution to an aqueous solution of each compound.
        *   **Phenol:** Gives a positive test, producing a characteristic violet, blue, green, or purple coloration due to the formation of a complex. (6C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3H⁺ + 3HCl)
        *   **Benzoic acid:** Gives a negative test (or a buff-colored precipitate of ferric benzoate if concentrated, which is distinct from the phenol color).

**(iv) Benzoic acid and Ethyl benzoate**

*   **Principle:** Benzoic acid is a carboxylic acid and reacts with sodium bicarbonate, while ethyl benzoate (an ester) does not.
*   **Test:** Sodium Bicarbonate Test
    *   **Procedure:** Add a saturated solution of sodium bicarbonate (NaHCO₃) to both compounds.
        *   **Benzoic acid:** Gives a positive test, producing brisk effervescence due to the evolution of carbon dioxide gas. (C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑)
        *   **Ethyl benzoate:** Gives a negative test; no effervescence is observed.

**(v) Pentan-2-one and Pentan-3-one**

*   **Principle:** Pentan-2-one is a methyl ketone (contains the CH₃CO- group) and can undergo the iodoform reaction, while pentan-3-one does not.
*   **Test:** Iodoform Test
    *   **Procedure:** Add iodine solution (I₂) and sodium hydroxide solution (NaOH) to both compounds and warm gently.
        *   **Pentan-2-one:** Gives a positive test, forming a yellow precipitate of iodoform (CHI₃) with a characteristic antiseptic smell. (CH₃CH₂CH₂COCH₃ + 3I₂ + 4NaOH → CH₃CH₂CH₂COONa + CHI₃(s) + 3NaI + 3H₂O)
        *   **Pentan-3-one:** Gives a negative test; no yellow precipitate is formed.

**(vi) Benzaldehyde and Acetophenone**

*   **Principle:** Benzaldehyde is an aldehyde and gives a positive Tollens' test, while acetophenone is a ketone and gives a positive iodoform test.
*   **Test 1: Tollens' Test**
    *   **Procedure:** Add Tollens' reagent to both compounds and warm gently.
        *   **Benzaldehyde:** Gives a positive test, forming a silver mirror or black precipitate. (C₆H₅CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → C₆H₅COO⁻ + 2Ag(s) + 4NH₃ + 2H₂O)
        *   **Acetophenone:** Gives a negative test.
*   **Test 2: Iodoform Test**
    *   **Procedure:** Add iodine solution (I₂) and sodium hydroxide solution (NaOH) to both compounds and warm gently.
        *   **Acetophenone:** Gives a positive test, forming a yellow precipitate of iodoform (CHI₃). (C₆H₅COCH₃ + 3I₂ + 4NaOH → C₆H₅COONa + CHI₃(s) + 3NaI + 3H₂O)
        *   **Benzaldehyde:** Gives a negative test.

**(vii) Ethanal and Propanal**

*   **Principle:** Ethanal (acetaldehyde) is the only aldehyde that contains the CH₃CO- group (or CH₃CH(OH)- group after reduction) and thus gives a positive iodoform test. Propanal does not.
*   **Test:** Iodoform Test
    *   **Procedure:** Add iodine solution (I₂) and sodium hydroxide solution (NaOH) to both compounds and warm gently.
        *   **Ethanal:** Gives a positive test, forming a yellow precipitate of iodoform (CHI₃) with a characteristic antiseptic smell. (CH₃CHO + 3I₂ + 4NaOH → HCOONa + CHI₃(s) + 3NaI + 3H₂O)
        *   **Propanal:** Gives a negative test; no yellow precipitate is formed.

Question 12

Q12.9. Give the plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two -NH2 gro

Accepted Answer

Here are the plausible explanations for each of the given statements:

**(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.**

Cyanohydrin formation is a nucleophilic addition reaction where the cyanide ion (CN⁻) attacks the electrophilic carbonyl carbon. The rate and yield of nucleophilic addition reactions are significantly affected by steric hindrance around the carbonyl carbon.

*   **Cyclohexanone:** The carbonyl carbon in cyclohexanone is relatively less sterically hindered. The two α-hydrogens and the ring structure allow for easier approach of the nucleophile (CN⁻) to the carbonyl carbon. Thus, it forms cyanohydrin in good yield.
*   **2,2,6-trimethylcyclohexanone:** This compound has three bulky methyl groups (two at position 2 and one at position 6) in close proximity to the carbonyl carbon. These methyl groups create significant steric hindrance, making it difficult for the nucleophile (CN⁻) to approach and attack the carbonyl carbon effectively. This steric bulk repels the incoming nucleophile, thereby hindering the nucleophilic addition reaction and resulting in a very poor yield of cyanohydrin.

**(ii) There are two -NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.**

Semicarbazide (NH₂-CO-NH-NH₂) has two primary amine (-NH₂) groups and one secondary amine (-NH-) group. The two primary amine groups are:
1.  One directly attached to the carbonyl group (-CO-NH₂).
2.  One attached to the nitrogen which is further attached to the carbonyl group (-NH-NH₂).

Only the primary amino group directly attached to the nitrogen that is *not* adjacent to the carbonyl group (i.e., the terminal -NH₂ of the -NH-NH₂ part) is involved in semicarbazone formation. This is because of the following reasons:

*   **Resonance effect:** The -NH₂ group directly attached to the carbonyl group (-CO-NH₂) has its lone pair of electrons delocalized through resonance with the carbonyl group. This resonance makes the nitrogen atom less nucleophilic (less available for attack on the carbonyl carbon of the aldehyde/ketone) and less basic (lone pair is shared). The resonance structure can be represented as: NH₂-C(=O)-NH-NH₂ ↔ NH₂=C(-O⁻)-NH-NH₂. This delocalization reduces the electron density on this nitrogen, making it a poor nucleophile.
*   **Steric hindrance:** While less significant than the resonance effect, the proximity to the bulky carbonyl group can also contribute slightly to reduced reactivity.

In contrast, the lone pair of electrons on the terminal nitrogen of the -NH-NH₂ group is not involved in resonance with the carbonyl group. Therefore, its lone pair is more available for donation, making it more nucleophilic and basic. This is the nitrogen that attacks the carbonyl carbon of the aldehyde or ketone to form the semicarbazone.

**(iii) Acetaldehyde is more reactive than propanal in nucleophilic addition reactions.**

The reactivity of aldehydes and ketones towards nucleophilic addition reactions is governed by two main factors:

1.  **Steric hindrance:** Smaller alkyl groups around the carbonyl carbon lead to less steric hindrance, allowing easier approach of the nucleophile.
2.  **Electronic effects (Inductive effect):** Electron-donating alkyl groups increase the electron density on the carbonyl carbon, making it less electrophilic and thus less susceptible to nucleophilic attack. Conversely, electron-withdrawing groups increase electrophilicity.

*   **Acetaldehyde (CH₃CHO):** Has one methyl group and one hydrogen atom attached to the carbonyl carbon. The methyl group exerts a +I (positive inductive) effect, pushing electron density towards the carbonyl carbon, making it slightly less electrophilic. However, the hydrogen atom is very small and offers minimal steric hindrance.
*   **Propanal (CH₃CH₂CHO):** Has one ethyl group and one hydrogen atom attached to the carbonyl carbon. The ethyl group (CH₃CH₂-) is larger than a methyl group and also exerts a stronger +I effect compared to a methyl group. The stronger +I effect increases the electron density on the carbonyl carbon more than in acetaldehyde, making it less electrophilic. Additionally, the ethyl group offers more steric hindrance than a methyl group.

Both the increased steric hindrance and the stronger electron-donating inductive effect of the ethyl group in propanal make its carbonyl carbon less accessible and less electrophilic compared to acetaldehyde. Hence, acetaldehyde is more reactive than propanal in nucleophilic addition reactions.

**(iv) p-nitrobenzaldehyde is more reactive towards nucleophilic addition reactions than benzaldehyde.**

Similar to the previous point, the reactivity of carbonyl compounds towards nucleophilic addition reactions depends on the electrophilicity of the carbonyl carbon. Electron-withdrawing groups increase electrophilicity, while electron-donating groups decrease it.

*   **Benzaldehyde (C₆H₅CHO):** The phenyl group is weakly electron-withdrawing by induction but can also donate electrons by resonance to the carbonyl group, making the carbonyl carbon slightly less electrophilic compared to aliphatic aldehydes.
*   **p-nitrobenzaldehyde (O₂N-C₆H₄-CHO):** The nitro group (-NO₂) is a strong electron-withdrawing group, both through its -I (inductive) effect and especially its -M (mesomeric/resonance) effect. When present at the para position, the -NO₂ group effectively withdraws electron density from the benzene ring, and consequently, from the carbonyl carbon through the conjugated system. This withdrawal of electron density makes the carbonyl carbon significantly more positive (more electrophilic) and thus more susceptible to attack by nucleophiles.

Therefore, due to the strong electron-withdrawing effect of the nitro group at the para position, the carbonyl carbon in p-nitrobenzaldehyde is more electrophilic and hence more reactive towards nucleophilic addition reactions than benzaldehyde.

**(v) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.**

Friedel-Crafts reactions (alkylation and acylation) are electrophilic aromatic substitution reactions that require the benzene ring to be activated or at least not strongly deactivated. The reaction proceeds via the attack of an electrophile (e.g., R⁺ or RCO⁺) on the electron-rich benzene ring.

*   **Aromatic carboxylic acids (Ar-COOH):** The carboxyl group (-COOH) is a strong electron-withdrawing group. It withdraws electron density from the benzene ring primarily through its -M (mesomeric/resonance) effect and also through its -I (inductive) effect. This strong electron withdrawal significantly deactivates the benzene ring, making it highly electron-deficient.
*   **Deactivation:** A deactivated benzene ring is much less susceptible to attack by electrophiles. The electrophiles required for Friedel-Crafts reactions (e.g., carbocations or acylium ions) are not strong enough to react with such a deactivated ring.
*   **Lewis Acid Interaction:** Furthermore, the carboxyl group has a lone pair of electrons on its oxygen atoms. These lone pairs can coordinate with the Lewis acid catalyst (e.g., AlCl₃) used in Friedel-Crafts reactions. This coordination forms a complex (e.g., Ar-COOH·AlCl₃), which further deactivates the ring and also consumes the catalyst, preventing it from generating the necessary electrophile. This complexation effectively poisons the catalyst.

Due to both the strong deactivating effect of the carboxyl group on the benzene ring and the complexation of the carboxyl group with the Lewis acid catalyst, aromatic carboxylic acids do not undergo Friedel-Crafts reactions.

Question 13

Q13. Give the chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv

Accepted Answer

Here are the chemical tests to distinguish between the given pairs of compounds:

**(i) Propanal and Propanone**
*   **Test:** Tollen's Test (Silver Mirror Test)
*   **Procedure:** Add Tollen's reagent (ammoniacal silver nitrate solution) to both compounds and warm gently.
*   **Observation:**
    *   **Propanal (an aldehyde):** A bright silver mirror is formed on the inner side of the test tube. This is because aldehydes are oxidized to carboxylate ions, and Ag⁺ ions are reduced to metallic silver (Ag).
        `CH3CH2CHO + 2[Ag(NH3)2]⁺ + 3OH⁻ → CH3CH2COO⁻ + 2Ag↓ + 4NH3 + 2H2O`
    *   **Propanone (a ketone):** No silver mirror is formed. Ketones do not react with Tollen's reagent.

**(ii) Acetophenone and Benzophenone**
*   **Test:** Iodoform Test
*   **Procedure:** Warm each compound with an aqueous solution of iodine and sodium hydroxide (or NaOI).
*   **Observation:**
    *   **Acetophenone (a methyl ketone, C6H5COCH3):** A yellow precipitate of iodoform (CHI3) with a characteristic antiseptic smell is formed.
        `C6H5COCH3 + 3NaOI → C6H5COONa + CHI3↓ + 2NaOH`
    *   **Benzophenone (not a methyl ketone, C6H5COC6H5):** No yellow precipitate is formed.

**(iii) Phenol and Benzoic acid**
*   **Test:** Sodium Bicarbonate Test
*   **Procedure:** Add a small amount of sodium bicarbonate (NaHCO3) solution to both compounds.
*   **Observation:**
    *   **Benzoic acid (a carboxylic acid):** Brisk effervescence of carbon dioxide (CO2) gas is observed. Carboxylic acids are strong enough to react with NaHCO3.
        `C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2↑`
    *   **Phenol:** No effervescence is observed. Phenols are less acidic than carbonic acid and do not react with NaHCO3.

**(iv) Benzoic acid and Ethyl benzoate**
*   **Test:** Sodium Bicarbonate Test
*   **Procedure:** Add a small amount of sodium bicarbonate (NaHCO3) solution to both compounds.
*   **Observation:**
    *   **Benzoic acid:** Brisk effervescence of CO2 gas is observed.
        `C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2↑`
    *   **Ethyl benzoate (an ester):** No reaction occurs, and no effervescence is seen.

**(v) Pentan-2-one and Pentan-3-one**
*   **Test:** Iodoform Test
*   **Procedure:** Warm each compound with an aqueous solution of iodine and sodium hydroxide.
*   **Observation:**
    *   **Pentan-2-one (a methyl ketone, CH3COCH2CH2CH3):** A yellow precipitate of iodoform (CHI3) is formed.
    *   **Pentan-3-one (not a methyl ketone, CH3CH2COCH2CH3):** No yellow precipitate is formed.

**(vi) Benzaldehyde and Acetophenone**
*   **Test 1: Tollen's Test**
    *   **Benzaldehyde (an aldehyde):** Forms a silver mirror.
    *   **Acetophenone (a ketone):** No reaction.
*   **Test 2: Iodoform Test**
    *   **Acetophenone (a methyl ketone):** Forms a yellow precipitate of iodoform.
    *   **Benzaldehyde (not a methyl ketone):** No reaction.

**(vii) Ethanal and Propanal**
*   **Test:** Iodoform Test
*   **Procedure:** Warm each compound with an aqueous solution of iodine and sodium hydroxide.
*   **Observation:**
    *   **Ethanal (CH3CHO):** Gives a positive iodoform test because it contains the CH3-C=O group. A yellow precipitate of iodoform is formed.
        `CH3CHO + 3NaOI → HCOONa + CHI3↓ + 2NaOH`
    *   **Propanal (CH3CH2CHO):** Does not contain the required group and thus gives a negative iodoform test. No yellow precipitate is formed.

Question 14

Q14. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii

Accepted Answer

The preparation of the given compounds from benzene can be carried out as follows:

**(i) Methyl benzoate**
This synthesis involves three steps: Friedel-Crafts alkylation, oxidation, and esterification.
*   **Step 1: Friedel-Crafts Alkylation:** Benzene is treated with methyl chloride (CH3Cl) in the presence of anhydrous AlCl3 to form Toluene.
*   **Step 2: Oxidation:** Toluene is oxidized using a strong oxidizing agent like alkaline KMnO4 or CrO3-H2SO4 to form Benzoic acid.
*   **Step 3: Esterification:** Benzoic acid is treated with methanol (CH3OH) in the presence of a small amount of concentrated H2SO4 to form Methyl benzoate.

**(ii) m-Nitrobenzoic acid**
This synthesis involves introducing the carboxyl group first, as it is a meta-directing group.
*   **Step 1: Preparation of Benzoic acid:** Benzene is first converted to Toluene (Friedel-Crafts alkylation) and then oxidized to Benzoic acid as shown in part (i).
*   **Step 2: Nitration:** Benzoic acid is treated with a nitrating mixture (conc. HNO3 + conc. H2SO4) to introduce a nitro group at the meta position, yielding m-Nitrobenzoic acid.

**(iii) p-Nitrobenzoic acid**
For the para-isomer, the nitro group must be introduced first while the alkyl group is present, as the methyl group is ortho, para-directing.
*   **Step 1: Preparation of Toluene:** Benzene is converted to Toluene via Friedel-Crafts alkylation with CH3Cl/anhyd. AlCl3.
*   **Step 2: Nitration:** Toluene is nitrated using conc. HNO3 + conc. H2SO4. This produces a mixture of ortho and para-Nitrotoluene. The p-isomer is the major product and can be separated by fractional crystallization.
*   **Step 3: Oxidation:** The p-Nitrotoluene is then oxidized with a strong oxidizing agent like alkaline KMnO4 or chromic acid to yield p-Nitrobenzoic acid.

**(iv) Phenylacetic acid**
This synthesis involves increasing the carbon chain length by one carbon atom.
*   **Step 1: Preparation of Benzyl chloride:** Benzene is first converted to Toluene. Toluene then undergoes free-radical chlorination with Cl2 in the presence of UV light to form Benzyl chloride.
*   **Step 2: Nucleophilic Substitution:** Benzyl chloride is treated with aqueous ethanolic potassium cyanide (KCN) to form Phenylacetonitrile (Benzyl cyanide).
*   **Step 3: Hydrolysis:** Phenylacetonitrile is subjected to acid hydrolysis (H⁺/H2O) to convert the cyanide group (-CN) into a carboxylic acid group (-COOH), yielding Phenylacetic acid.

**(v) p-Nitrobenzaldehyde**
This synthesis follows a similar path to p-Nitrobenzoic acid, but the final oxidation step is milder to stop at the aldehyde stage.
*   **Step 1: Preparation of Toluene:** Benzene to Toluene (Friedel-Crafts alkylation).
*   **Step 2: Nitration:** Toluene to p-Nitrotoluene (Nitration, followed by separation of the para isomer).
*   **Step 3: Etard Reaction:** p-Nitrotoluene is treated with chromyl chloride (CrO2Cl2) in a non-polar solvent like CS2, followed by hydrolysis. This specific reaction oxidizes the methyl group to an aldehyde group, yielding p-Nitrobenzaldehyde.

Question 15

Q15. An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzened

Accepted Answer

The organic compound with molecular formula C9H10O is 2-ethylbenzaldehyde (o-ethylbenzaldehyde).

Question 16

Q16. Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation

Accepted Answer

### (i) Acetylation
**Definition:** Acetylation is the process of introducing an acetyl group (CH₃CO-) into a molecule. This is typically achieved by reacting a compound containing an active hydrogen atom (e.g., in -OH, -NH₂, -SH groups) with an acetylating agent such as acetic anhydride [(CH₃CO)₂O] or acetyl chloride (CH₃COCl) in the presence of a base (like pyridine) or an acid catalyst.

**Mechanism/Reaction:** The acetyl group replaces the active hydrogen atom. For example, alcohols and amines undergo acetylation.

**Example:** Acetylation of ethanol with acetic anhydride:
CH₃CH₂-OH + (CH₃CO)₂O → CH₃CH₂-O-CO-CH₃ + CH₃COOH
(Ethanol) + (Acetic Anhydride) → (Ethyl Acetate) + (Acetic Acid)

Acetylation of aniline with acetic anhydride:
C₆H₅-NH₂ + (CH₃CO)₂O → C₆H₅-NH-CO-CH₃ + CH₃COOH
(Aniline) + (Acetic Anhydride) → (Acetanilide) + (Acetic Acid)

**Uses:** Acetylation is used to protect -OH or -NH₂ groups, to determine the number of hydroxyl groups in a compound (e.g., in carbohydrates), and in the synthesis of pharmaceuticals (e.g., aspirin).

### (ii) Cannizzaro reaction
**Definition:** The Cannizzaro reaction is a disproportionation reaction undergone by aldehydes which do not possess an α-hydrogen atom (e.g., formaldehyde, benzaldehyde, pivaldehyde). In this reaction, two molecules of the aldehyde react in the presence of a strong base (like concentrated NaOH or KOH) to produce one molecule of a primary alcohol and one molecule of a carboxylic acid salt.

**Mechanism/Reaction:**
1.  **Nucleophilic attack:** The hydroxide ion attacks the carbonyl carbon of one aldehyde molecule. 2.  **Hydride transfer:** A hydride ion is transferred from the tetrahedral intermediate to the carbonyl carbon of a second aldehyde molecule. This is the rate-determining step. 3.  **Proton transfer:** The alkoxide ion formed abstracts a proton from the carboxylic acid, leading to the formation of the alcohol and the carboxylate ion.

**General Reaction:**
2 R-CHO + Conc. NaOH → R-CH₂OH + R-COONa

**Example:** Cannizzaro reaction of benzaldehyde:
2 C₆H₅-CHO + Conc. NaOH → C₆H₅-CH₂OH + C₆H₅-COONa
(Benzaldehyde) → (Benzyl Alcohol) + (Sodium Benzoate)

**Cross-Cannizzaro reaction:** If two different aldehydes (both lacking α-hydrogens) are used, it's called a cross-Cannizzaro reaction. Typically, formaldehyde is oxidized to formate, and the other aldehyde is reduced to alcohol.

### (iii) Cross aldol condensation
**Definition:** Cross aldol condensation (or mixed aldol condensation) is an aldol condensation reaction that occurs between two different carbonyl compounds (aldehydes or ketones), at least one of which must contain α-hydrogen atoms. If both carbonyl compounds contain α-hydrogens, it leads to a mixture of four products, making it less synthetically useful unless one reactant is much more reactive or lacks α-hydrogens.

**Conditions:** The reaction is carried out in the presence of a dilute base (like NaOH, KOH, Ba(OH)₂) or an acid.

**Example:** Cross aldol condensation between benzaldehyde (no α-H) and acetophenone (has α-H):

C₆H₅-CHO + CH₃-CO-C₆H₅  
(Benzaldehyde) + (Acetophenone)

In the presence of dilute NaOH, the α-hydrogen of acetophenone is abstracted to form an enolate, which then attacks the carbonyl carbon of benzaldehyde. Subsequent dehydration yields the α,β-unsaturated ketone.

C₆H₅-CHO + CH₃-CO-C₆H₅  --[Dil. NaOH, Heat]--> C₆H₅-CH=CH-CO-C₆H₅ + H₂O
(Benzaldehyde) + (Acetophenone) → (Chalcone / 1,3-Diphenylprop-2-en-1-one) + (Water)

**Key Strategy:** To obtain a single major product in cross aldol condensation, one of the reactants should ideally lack α-hydrogens (e.g., formaldehyde, benzaldehyde) and be more reactive towards nucleophilic attack, while the other reactant has α-hydrogens and is added slowly to the mixture.

### (iv) Decarboxylation
**Definition:** Decarboxylation is a chemical reaction that removes a carboxyl group (-COOH) from a carboxylic acid, releasing carbon dioxide (CO₂) and forming a compound with one less carbon atom. It is essentially the reverse of carboxylation.

**Conditions:** Carboxylic acids generally undergo decarboxylation upon heating, especially if they have an electron-withdrawing group at the α-position or are β-keto acids or malonic acids. Simple monocarboxylic acids require strong heating with soda lime (a mixture of NaOH and CaO).

**Mechanism/Reaction (Soda Lime Decarboxylation):**
R-COOH + NaOH --[CaO, Heat]--> R-H + Na₂CO₃

**Example:** Decarboxylation of sodium acetate with soda lime:
CH₃-COONa + NaOH --[CaO, Heat]--> CH₄ + Na₂CO₃
(Sodium Acetate) + (Sodium Hydroxide) → (Methane) + (Sodium Carbonate)

**Example (β-keto acid):** β-keto acids readily decarboxylate upon heating, often without a catalyst, due to the formation of a stable six-membered cyclic transition state involving the enol form.

CH₃-CO-CH₂-COOH --[Heat]--> CH₃-CO-CH₃ + CO₂
(Acetoacetic acid) → (Acetone) + (Carbon Dioxide)

**Uses:** Decarboxylation is an important reaction in organic synthesis for shortening carbon chains and in biological processes (e.g., in the Krebs cycle).

Chapter 12: Aldehydes, Ketones and Carboxylic Acids