NEB Grade 12 Organic Chemistry Important Question Collection With Solution

Most Important Question from Organic Chemistry for NEB Grade 12 Board Exam 2079

Most Important Question from Organic Chemistry for NEB Grade 12 Board Exam 2079:

HALOALKANE:

a) Write down any three methods of Preparation of Haloalkanes.

Answer:

1. From alcohols: Haloalkanes can be prepared by treating alcohols with hydrogen halides (HX) such as HCl, HBr, or HI in the presence of a catalyst such as zinc chloride (ZnCl₂).

R-OH + HX → R-X + H₂O

2. From alkenes: Haloalkanes can also be prepared by the addition of hydrogen halides (HX) to alkenes.

RCH=CH₂ + HX → RCH₂CH₂X

3. From alkanes: Haloalkanes can be prepared by the free radical halogenation of alkanes in the presence of halogen (Cl₂, Br₂) and sunlight or heat.

RH + X₂ → RX + HX

b) Dehydrohalogenation Reaction

Answer: Dehydrohalogenation is a type of elimination reaction in which a hydrogen halide is removed from a molecule to form an alkene. The reaction is typically carried out in the presence of a strong base such as sodium hydroxide (NaOH) or potassium hydroxide (KOH).

R-X + NaOH → R-H + NaX R-H → R=CH₂ + HX

c) Name Reaction: Wurtz Reaction, Markovnikov's rule, and Anti Markovnikov's rule

Answer:

1. Wurtz Reaction: Wurtz reaction is a coupling reaction of two alkyl halides to form a higher alkane. The reaction is carried out in the presence of sodium (Na) or potassium (K) metal in dry ether. 2R-X + 2Na → R-R + 2NaX
2. Markovnikov's rule: Markovnikov's rule states that in the addition of a protic acid HX to an alkene, the hydrogen atom of HX will add to the carbon atom of the alkene that has the greatest number of hydrogen atoms, while the halogen atom will add to the other carbon atom of the alkene.

RCH=CH₂ + HX → RCH₂CH₂X

3. Anti-Markovnikov's rule: Anti-Markovnikov's rule is the opposite of Markovnikov's rule, stating that in the addition of a protic acid HX to an alkene, the hydrogen atom of HX will add to the carbon atom of the alkene that has the least number of hydrogen atoms, while the halogen atom will add to the other carbon atom of the alkene. The reaction is carried out in the presence of a peroxide (ROOR) such as benzoyl peroxide (C₆H₅COOC₆H₅).

RCH=CH₂ + HX → R-CH₂-CH₂X

d)      Preparation of chloroform, chemical properties of chloroform.

HALOARENE:

1. Write any two methods of preparation of chlorobenzene/ haloarenes.
2. Why is it difficult to undergoes nucleophilic substitution reaction Haloarene than haloalkane.
3. Formation of DDT
4. Name Reaction: Wurtz Fittig Reactio, Fitig Reaction, Friedel Craft’s alkylation, Friedel crafts acylation

ALCOHOL:

1. Victor Meyer's Method
2. Any Three Method for Preparation of Alcohol
3. Name Reaction: Oxo Process and Fermentation Process
4. Dehydration of Alcohol (Action with Conc.H₂SO₄ at different Temperature)
5. Dehydration (Action with Cupper)
6. Oxidation of all three-degree alcohol
7. Iodoform test, esterification test

PHENOL:

1. Any three methods of preparation of Phenol
2. Any three method of preparation of phenol
3. Laboratory test of phenol (Action with FeCl₃)
4. Name Reaction: Fries Rearrangement, Kolbe’s reaction, Riemer-Thiemann Reaction, Phenolphthalein and Bakelite formation Reaction
5. Action of Phenol with formaldehyde

ETHER:

1. Williamson’s Ether Synthesis
2. Isomerism of ether
3. Action with conc. H₂SO₄ at 140⁰C
4. Action with HI, HCl (Excess and Cold)
5. Preparation of Anisole

ALDEHYDE AND KETONE:

1. Preparation of aldehyde and ketone
2. Rosemond Reduction, Gem di-haloalkane from alcohol
3. Reduction of Carboxylic Acid (H₂/Li)
4. Chemical Properties: Action with sodium bisulphate, Action with HCN, Ammonia Derivatives, Tollen’s Test, Fehling Test, Aldol Condensation, Cannizzaro Reaction
5. Formation of Urotropine
6. What is formalin?
7. Action with phenol, Action with ammonia, action with Bakelite
8. Perkin’s Condensation, Benzoin Condensation (alc. KCN)

CARBOXYLIC ACID:

1. Preparation of Carboxylic Acid (Any Three Method)
2. Chemical Properties:
3. Anomalous Behavior of Formic Acid: Fehling Test, Benedict Test, Tollens’s Test, and action with acidified KMnO4.
4. Derivative of Carboxylic Acid: Preparation from Carboxylic Acid Only
5. Chemical Properties for Derivative of Carbocyclic Acid: Hydrolysis, action with ammonia, alkanamine, alcohol and reduction

Most Important: Claisen Condensation

NITRO COMPOUND:

1. Isomerism
2. Reduction of Nitroalkane: Catalytic Reduction, Acidic Reduction, Neutral Reduction
3. Test of Primary Secondary and Tertiary Nitroalkane

AROMATIC NITRO COMPOUND:

1. Preparation from Benzene
2. Reduction of Nitrobenzene: Catalytic Reduction, Acidic Reduction, Neutral Medium Reduction, Electrolytic Reduction

AMINO COMPOUNDS:

1. Preparation of amine: From Haloalkane
2. Hoffman’s Method for separation of Amino Compound of different degree
3. Action with HNO₂
4. Action with haloalkane

ANILINE:

1. Preparation of Nitrobenzene
2. Basic Nature of Aniline
3. Chemical Properties: Action with Acid Chloride, Halogenation, Nitration, Sulponation
4. Name Reaction: Diazotization, Coupling Reaction, Carbylamine Reaction

ORGANOMETALLIC COMPOUNDS:

1. Preparation of Grignard Reagent
2. Preparation of all three-degree alcohol by using suitable Grignard reagent
3. Action with acid halides, ester, and CO₂

What Happen When Type Question?

a)            Chloroform is heated with silver powder.

Ans: Ethyne is obtained when chloroform is heated with silver powder.

\[2CHC{l_3} + 6Ag \to CH \equiv CH + 6AgCl\]

b)            Chloroform is treated with acetone is presence of base.

Ans: Chloretone is obtained when chloroform is treated with acetone in presence of base.

\[CHC{l_3} + C{H_3} - \mathop C\limits^{\mathop {||}\limits^O }  - C{H_3}\mathop  \to \limits^{KOH} C{H_3} - \mathop C\limits_{\mathop |\limits_{CC{l_3}} }^{\mathop |\limits^{OH} }  - C{H_3}(Chloretone)\]

c)            Chloroform is heated with conc nitric acid?

Ans: \[CHC{l_3}{\rm{ }} + {\rm{ }}Conc.{\rm{ }}HN{O_3} \to CC{l_3} - N{O_2}(Chloropicrin) + {H_2}O\]

d)            Chloroform is boiled with aqueous KOH solution.

Ans: Potassium Format is obtained when chloroform is treated with aq. KOH.

\[CHC{l_3}{\rm{ }} + {\rm{ }}aq. KOH \to HCOOK + {\rm{ 3KCL +  2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

e)            Acetic anhydride is reduced with LiAIH₄?

Ans: Ethyl alcohol is formed when acetic anhydride is reduced in presence of LiAlH₄.

CH₃​CO−O−COCH₃​(Acetic anhydride) +LiAlH₄​→C₂​H₅​−OH (Ethyl Alcohol)

f)             Formaldehyde is treated with ammonia.

Ans: Urotropine (Hexamethyl tetramine) is obtained when formaldehyde is treated with ammonia.

6HCHO+4NH₃​⟶(CH₂​)₆​N₄​ +6H₂​O

g)            Nitrobenzene is reduced with LIAIH₄?

Ans: Azobenzene is obtained on reducing nitrobenzene in presence of LiAlH₄.

h)            Propanone is treated with HCN followed by hydrolysis?

Ans: Hydrogen cyanide is added to propanone to give acetone Cyanohydrin. On further hydrolysis of acetone cyanohydrin, we get 2-hydroxy 2-methyl propanoic acid.

$\begin{array}{l}C{H_3} - \mathop C\limits^{\mathop {||}\limits^O }  - C{H_3}{\rm{  +  HCN}} \to C{H_3} - \mathop C\limits_{\mathop |\limits_{CN} }^{\mathop |\limits^{OH} }  - C{H_3}(acetone{\rm{ }}Cyanohydrin)\\C{H_3} - \mathop C\limits_{\mathop |\limits_{CN} }^{\mathop |\limits^{OH} }  - C{H_3}\mathop  \to \limits^{{H_2}O/{H^ + }} C{H_3} - \mathop C\limits_{\mathop |\limits_{COOH} }^{\mathop |\limits^{OH} }  - C{H_3}(2 - hydroxy{\rm{ }}2 - methyl{\rm{ }}propanoic{\rm{ }}acid)\end{array}$

i)              Alkane amide is heated with Bromine in presence of aq. KOH

Ans: When Alkene amide is heated with broom in in presence of aqueous KOH it gives primary aminde having one carbon less than parent amide.

\[C{H_3} - \mathop C\limits^{\mathop {||}\limits^O }  - N{H_2}{\rm{  +  B}}{{\rm{r}}_{\rm{2}}}{\rm{ }}\mathop  \to \limits^{aq.{\rm{ }}KOH} 2KBr{\rm{  + }}C{H_3} - N{H_2} + {K_2}C{O_3} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

j)              Calcium acetate is heated.

Ans: When calcium acetate ((CH₃​COO)₂​Ca) is heated, acetone (CH₃​−CO−CH₃​) and calcium carbonate (CaCO₃​) are obtained.

\[{(CH3COO)_2}Ca\mathop { \to }\limits^\Delta  C{H_3} - CO - C{H_3} + CaC{O_3}\]

k)            Aniline is treated with aqueous bromine.

Ans: Bromine will get decolorized and a white precipitate of 2,4,6-tribromoaniline is formed

l)              Phenol is heated with Zn dust.

Ans: Phenol get reduced into benzene on action with zinc dust.

m)         Ethanol is treated with NaOH in presence of Iodine:

Ans: When ethanol is gently warmed with acquiescence and I₂ crystal it forms lemon eloquent of iodoform with hospital smell.

CH₃-CH₂-OH (Ethanol) + 6NaOH +4I₂  → CHI­­₃ (Iodoform)+5NaI+ 5H₂O +HCOONa

Give Reason type:

a) Phenol is more acidic than alcohol.

Answer: Phenol is more acidic than alcohol due to the presence of an electron withdrawing group (-OH) attached directly to the aromatic ring. This group stabilizes the conjugate base formed after losing a proton, making it easier to donate a proton.

b) Boiling point of alcohol is higher than Isomeric ether.

Answer: The boiling point of alcohol is higher than isomeric ether due to the presence of hydrogen bonding between alcohol molecules. Ether molecules do not have hydrogen bonding, which makes it easier for them to evaporate, leading to a lower boiling point.

c) Haloarene is less reactive than haloalkane towards nucleophilic substitution.

Answer: Haloarenes are less reactive than haloalkanes towards nucleophilic substitution because the aromatic ring in haloarenes stabilizes the intermediate formed during the reaction, making it difficult for the nucleophile to attack the electrophilic carbon atom.

d) Nitro group is a meta directing group and benzene ring deactivator.

Answer: Nitro group is a meta directing group because it withdraws electron density from the ring by inductive and resonance effects. This deactivates the ring towards electrophilic substitution reactions.

e) It is dangerous to distil an old sample of ether.

Answer: It is dangerous to distill an old sample of ether because it can form explosive peroxides upon exposure to air. These peroxides can accumulate in the distillation apparatus and can explode upon heating or shock.

f) Chloroform is stored in colored bottles.

Answer: Chloroform is stored in colored bottles because it is photosensitive and can decompose upon exposure to light. The decomposition can lead to the formation of toxic and corrosive gases.

g) Boiling point of o-nitro phenol is lower than p-nitro phenol.

Answer: The boiling point of o-nitrophenol is lower than p-nitrophenol due to intramolecular hydrogen bonding between the hydroxyl group and the nitro group in the ortho isomer. This reduces the intermolecular forces, leading to a lower boiling point.

h) Anisole gives phenol and iodomethane with HI but reverse is not feasible.

Answer: This is because HI cleaves the O–CH3 bond in anisole to form phenol and iodomethane. However, the reverse reaction is not feasible as the O–CH3 bond in iodomethane is stronger than the O–H bond in phenol, making it difficult to break.

i) Formic acid gives Tollens’s reagent test and Fehling's solution test though it is not aldehyde.

Answer: Formic acid is an organic acid and does not have an aldehyde functional group. However, it gives Tollens’s reagent test and Fehling's solution test due to its ability to reduce these reagents. This is because formic acid undergoes oxidative decarboxylation to produce carbon monoxide and water, which then react with Tollens’s reagent or Fehling's solution to give a positive test result.

j) Chloroacetic acid is stronger acid than acetic acid.

Answer: Chloroacetic acid is stronger than acetic acid because the chloro substituent is electron-withdrawing and destabilizes the carboxylate anion, making it easier for the acid to donate a proton. This results in a higher acidity for chloroacetic acid compared to acetic acid.

k) Ethyl iodide gives ethyl cyanide with alcoholic KCN but ethyl isocyanide with alcoholic AgCN.

Answer: This is because alcoholic KCN is a nucleophile that attacks the electrophilic carbon of the ethyl iodide to form an intermediate, which then undergoes hydrolysis to give ethyl cyanide. On the other hand, alcoholic AgCN is a weak nucleophile that forms a complex with the ethyl iodide, which then undergoes a rearrangement reaction to form ethyl isocyanide.

l) P-methoxy aniline is stronger base than aniline but p-nitro-aniline is a weaker base than aniline. Answer: P-methoxy aniline is a stronger base than aniline because the methoxy group is an electron-donating group that increases the electron density on the amino group, making it more basic. On the other hand, p-nitro-aniline is a weaker base than aniline because the nitro group is an electron-withdrawing group that decreases the electron density on the amino group, making it less basic.

m) Ketones are less reactive towards nucleophiles than aldehydes.

Answer: This is because ketones have two electron-withdrawing carbonyl groups that stabilize the carbonyl carbon, making it less susceptible to nucleophilic attack. In contrast, aldehydes have only one electron-withdrawing carbonyl group, making the carbonyl carbon more vulnerable to nucleophilic attack.

n) Amino group of aniline is protected before nitration.

Answer: The amino group of aniline is protected before nitration to prevent unwanted reactions with nitration reagents. This is because the amino group is a nucleophile that can react with the nitration reagents and form undesired products. Therefore, the amino group is protected by acylation or benzoylation to make it less reactive towards the nitration reagents.

o) P₂O₅ is not used as dehydrating agent for the preparation of anhydrous formic acid.

Answer: P₂O₅ is not used as a dehydrating agent for the preparation of anhydrous formic acid because it reacts vigorously with formic acid to form carbon monoxide and phosphoric acid. Instead, other dehydrating agents like sulfuric acid or phosphorus trichloride are used.

p) In haloarenes, halo group is ortho-para directing group and ring deactivator.

Answer: The halo group in haloarenes is an electron-withdrawing group that deactivates the benzene ring towards electrophilic substitution reactions. However, it is also an ortho-para directing group, meaning that it directs incoming electrophiles to the ortho and para positions relative to the halo group due to its ability to stabilize the intermediate through resonance.

q) OH, group in phenol is ortho-para directing and ring activating towards electrophilic substitution reaction.

Answer: The OH group in phenol is an electron-donating group that activates the benzene ring towards electrophilic substitution reactions. It is also an ortho-para directing group, meaning that it directs incoming electrophiles to the ortho and para positions relative to the OH group due to its ability to stabilize the intermediate through resonance. This makes phenol more reactive towards electrophilic substitution reactions compared to benzene.

Note: All these students are recommended to go through this name reaction before the examination.