Chapterwise IIT JAM Organic Chemistry

 Chapter wise IIT JAM Organic Chemistry : 2005 - Onwards

Acidity of Particular Hydrogen Atom:

JAM 2005: Arrange the following in the decreasing order of acidity of the hydrogen indicated in italics:

CH₃COCH₃, CH₃COCH₂COCH₃, CH₃OOCCH₂COOCH₃, CH₃COCH₂NO₂

Solution: 

JAM 2005: benzyl chloride is reacted with different nucleophiles shown below; arrange them in the decreasing order of reactivity:

HO^-, CH₃COO^-, PhO^-, CH₃O^-

Solution: Nucleophilicity refers to the ability to be a nucleophile. Greater is the extent of negative charge over Oxugen greater nucleophilicity. The methyl group in methoxide ion pushes electrons towards O atom and increases its nucleophilicity compared to hydroxide ion. Now in phenoxide ion the negative charge is distributed between ring carbon and on oxygen atom but in acetate ion the negative charge is distributed between two oxygen atom which is more effective distribution compared to phenoxide. Thus acetate ion has the lowest charge over O atom.

Hence the order is:

CH₃O^- > HO^- > PhO^- > CH₃COO^-

JAM 2005: The rate of nitration of the following aromatic compounds decrease in the order:

Benzene, Pyridine, Thiophene, Toluene

Solution: Nitration is an electrophilic substitution reaction. The difference between the calculated energy of the nitro substituted structure and experimentally determined energy is called the resonance energy. Lower is the resonance energy greater is the reactivity towards nitration.

Thiophene has the lowest R.E of 29kcal/mol, hence it readily goes electrophilic substitution reaction compared to the rest three. The rest three has almost same resonance energy but higher is the electron density on the ring higher will be the reactivity. CH₃ pushes electron towards the ring in toluene hence it is next more reactive, and then comes benzene and then pyridine.

Thys the order is: Thiophene > Toluene > Benzene > Pyridine

JAM 2005: For the reaction shown below if the concentration of KCN is increased four times the rate of reaction will be:

Solution: The reactant is a tertiary alkyl halide which proceeds through SN1 mechanism. In SN1 mechanism the nucleophile (cyanide ion) has no effect on the reaction rate, as it dont appear in the slow step. Thus changing the concentration of KCN does not change the rate of reaction.

IIT JAM 2005: The major product formed in the reaction of 1,3-butadiene with bromine is:

a) BrCH₂CH(Br)CH=CH₂

b) CH₂=CH-CH₂CH₂Br

c) CH₂=C(Br)-C(Br)= CH₂

d) BrCH₂CH=CHCH₂Br

Solution: (d) At first one molecule of bromine is added to one of the double bond. The bromine on last carbon makes a cyclic transition state by pushing the next bromine out as bromide ion. That bromide ion attacks the next double bond, pi bond shifts and the product is formed as following:

IIT JAM 2005: The reaction of (+)2-iodobutane and NaI^* (I^* is radioactive isotope of iodine) in acetone was studied by measuring the rate of racemization (k_r) and the rate of incorporation of I^* (k_i)

(+)CH₃CH(I)CH₂CH₃ + NaI^* -----------> CH₃CH(I^*) CH₂CH₃ + NaI

For this reaction, the relationship between k_r and k_i is:

Solution:

Here ½ mole of (I) reacts with ½ mole of NaI^* to produce ½ mole of inverted product. ­­Thus ½ mole of (I) and ½ mole of (II) produce 1 mole of racemic mixture. Hence rate of racemization is twice of rate of incorporation. K_r = 2k_i

IIT JAM 2005:

In the scheme shown above P,Q, R and S are:

P = nucleotide (Base + Sugar + phosphate), Q = Nucleoside (sugar + Base), R = pyrimidine base, S = Purine base.

IIT JAM 2005: The products obtained from the following reaction are:

Solution:

IIT JAM 2005: The product(s) obtained in the following reaction is/are:

Solution: 

IIT JAM 2005: Match the isoelectric point with the given amino acids.

H₂NCH₂COOH, 

HOOCCH₂CH₂CH(NH₂)COOH, 

H₂N(CH₂)₄CH(NH₂)COOH 

and 9.5, 6.0, 3.1.

Solution: Neutral amino acids having equal acidic and basic group will have pH 5.5 to 6.3. Thus the lysine (first one) should get the value of 6.0

Acidic amino acids having more acidic group than basic group will have pH around 3 and accordingly the second amino acid glutamic acid should have Isoelectric point of 3.1.

Basic amino acids having less acidic group than basic group will have pH 7.6 to 10.8 and accordingly the third amino acid lysine should have Isoelectric point of 9.5.

IIT JAM 2005:  identify the product(s) formed in the following reactions:

Solution:

 

Solution:

Solution: from the attack of water you should follow  the above ester hydrolysis mechanism

IIT JAM 2005: How may the following transformation be affected? Indicate clearly the reagent / reaction condition in each step.

A. (Not involving any functional group transformation of the COOH group in the starting material)

Solution:

B. Using diethyl malonate as the only source of carbon

Solution:

C. 

Solution:

IIT JAM 2005: Write suitable mechanism

Solution:

Solution:

IIT JAM 2005: Rationalise the following observations using suitable mechanisms.

A. Nitration of 4-t-butyltoluene gives 4-nitrotoluene as one of the products.

Solution: A stable tertiary carbocation is eliminated in exchange of hydrogen ion from nitric acid to regain the aromaticity, then usual nitration of toluene occurs.

B. cis-4-t-butylcyclohexyltrimethylamoniumhydroxide undergoes Hoffman elimination to yield 4-t-butyl whereas the trans isomer does not. (Use conformation to explain).

Solution: In geometrical isomerism, a cyclohexane ring structure having upward-upward or dowm-down bisubstitution is considered Cis. In chair form the 1,2 or 1,4 bisubstitution is considered Cis if the substitutions are equatorial and axial to each other.

 The Hofmann elimination becomes successful in the cis form because a hydrogen is present anti or trans to the aminehydroxide group which is not found in trans form.

C. PhMgBr and 2 moles of PhCHO react first in dry ether and after acidic work up form PhCOPh and PhCH2OH.

Solution:

IIT JAM 2005:

a. Suggest a chemical method for the separation of the mixture containing p-N.N-dimethylaminophenol and p-aminobenzoicacid and give confirmatory test for phenol.

b. Write the structure X, Y and Z in the following reactions.