Chemistry Important Long Question for NEB Board Exam

 

1. a) Arrange the compound in the complete reaction sequence with the suitable reagent:
Aniline, Benzene diazonium Chloride , Benzonitrile, Benzamide, Benzoic Acid.
Solution:

b) Write the name of aldehyde which gives tollens test and shows aldol condensation.
Solution:

c) How is 2- Hydroxy propanoic acid obtained from Ethanal.
Solution:

2. a) Write the structure A, B and C in the following:
Ans:

Compound A is Aniline

Compound B is benzene diazonium chloride salt.

Compound C is iodobenzene.
b) What happened when C is heated with sodium metal in presence of dry ether?
Ans: When Iodobenzene is heated with sodium metal in presence of dry ether it gives biphenyl as shown in the reaction given below:

c) What product would you get when compound A and compound B are heated.

Ans: When compound A and compound B are heated, we will get P amino azobenzene.

3. a) An organic compound a which characteristic order, on treatment with NaOH Forms two compounds B & C which on oxidation with CrO₃ gives back compound A. Compound C is the sodium salt of acid. Compound C when heated with soda lime yields an aromatic hydrocarbon D. Deduce the structure of A, B, C and D. Write down the chemical equation for all the reaction taking place.

Solution:

b) Why NH₂ group of aniline is protected before nitration?

Solution:

To prevent unwanted side reactions and to ensure selective nitration at the ortho and/or para positions. Aniline contains a reactive amino group (-NH2) that can react with nitric acid during the nitration reaction, leading to the formation of undesirable by-products. For example, the amino group can be nitrated, leading to the formation of di- and tri-nitroanilines instead of mono-nitroanilines. These by-products are difficult to separate from the desired product and can reduce the overall yield of the reaction.

c) Write a product which is obtained by the reduction of acetic anhydride.

Solution:

CH₃​CO−O−COCH₃​ (acetic anhydride)+LiAlH₄​→C₂​H₅​−OH (Ethanol)

4. Give Reason: 

a) Electrophilic substitution in benzoic acid takes place at meta position.

Ans: Benzoic acid has a deactivating carboxylic group that decreases electron density at ortho and para positions. This makes the meta position more favorable for electrophilic substitution as it has higher electron density. Therefore, electrophiles attack at the meta position rather than ortho and para positions.

b) Carboxylic acids do not give characteristic reaction of carbonyl group.

Ans: The carbonyl carbon in carboxylic acids exhibits reduced electrophilicity due to resonance, which is why carboxylic acids do not show characteristic reactions of the carbonyl group.

(c) Carboxylic acids are higher boiling liquids than aldehydes, ketones, and alcohols of comparable molecular masses.

Ans: Carboxylic acids have strong intermolecular hydrogen bonding, leading to extensive molecular association. As a result, they have higher boiling points than aldehydes, ketones, and alcohols of similar molecular weight.

5. a) An organic compound a with molecular formula C₈H₈O and gives yellow precipitate with 2,4-DNP on heating with iodine in the presence of sodium hydroxide. It neither reduces tollens’ or Fehling’s reagent nor does it decolorize bromine water or Bayer’s reagent. On drastic condition with chromic acid, it gives a carboxylic acid B having molecular formula C₇H₆O₂. Identify compound A&B and explain the reaction involved.

b) An organic compound Z is known as an oil of mirbane which is prepared by nitration of benzene.

i. What product would you expect when Z is reduced with LiAlH_4.

ii. Convert Z into P hydroxy azobenzene.

Solution:

A)

Identifying Compound, A:

Compound (A) is characterized by the formation of a 2,4-DNP derivative, which suggests that it may be either an aldehyde or a ketone. However, since it does not reduce Tollens' or Fehling's reagent, it can be concluded that Compound (A) is a ketone. Additionally, it gives a positive iodoform test, indicating that it is a methyl ketone.

Although Compound (A) has a molecular formula of C₈H₈O, which suggests a high degree of unsaturation, it does not decolorize bromine water or Baeyer's reagent. This indicates that the unsaturation is due to the presence of an aromatic ring in the molecule.

Further evidence for the identity of Compound (A) comes from its reaction with NaOH and I₂, which gives a positive iodoform test. This confirms that Compound (A) is a methyl ketone with an aromatic ring.

Taking all these observations into account, it can be concluded that Compound (A) is phenyl methyl ketone, also known as acetophenone, with the molecular formula C₈H₈O.

 

Identifying Compound B:

Compound (B) with the molecular formula C₇H₆O₂ is oxidized to form a carboxylic acid.

Based on this information, it can be concluded that Compound (B) is benzoic acid.

The chemical equation for the oxidation of Compound (B) to form benzoic acid is:

C₇H₆O₂ (Compound B) + 3[O] → C₇H₆O₂ (Benzoic acid) + H₂O

Therefore, the compound (B) is benzoic acid, which can be represented by its molecular formula C₇H₆O₂.

 

b) Since Z is oil of mirbane i.e., nitrobenzene which can be easily prepared by nitration of benzene.

i) Nitrobenzene on reduction with lithium aluminum hydride (LiAIH₄) gives azobenzene.

ii) Conversion of Oil of Mirbane (Comp. Z) i.e., Nitro benzene into P Hydroxy azobenzene involves the following reaction

6. a) Arrange the following compound in increasing order of basic strength in their aqueous solution

NH₃, CH₃NH₂, (CH₃)₂NH,   (CH₃)₃N.

b) Give you a chemical test to distinguish between ethylamine and aniline.

c) How can methyl bromide be preferentially converted to methyl isocyanide?

d) Complete the following reaction equations:

i. C₆H₅N₂Cl + H₃PO₂ + H₂O →

ii. C₆H₅NH₂ + Br₂ →

Solution:

a) CH₃NH₂, > (CH₃)₂NH > (CH₃)₃N

b) Solution will be available soon.

c) CH₃​Br (Methyl Brommide)+AgCN→CH₃​NC (Methyl Isocyanide)+AgBr

d) (i)

d(ii)