General Chemistry III (Chem EH 301) Solved Question 2019: Section II (Organic Chemistry)

Solved Question

Solution to the third Semester Final Examination Question Paper from NEHU, Shillong.

Paper Chem EH 301, October 2019 Examination.

Section - II (Organic)

Click on Question No to view the solution: [7 (a)] [7 (b)] [7 (c)] [7 (d)] [7 (e)]

7 (a) Discuss the effect of electronegative elements on the acidity of aliphatic carboxylic acids with suitable examples. [1.5]

Solution: Carboyxlic acids undergo dissociation as follows:
\[\ce{R-COOH <=> R-COO- + H+}\] The abover reaction is in equilibrium. So, any factor that would shift the equilibrium to the forward direction, would increase the acidity of the molecule. Electronegative elements attached to the R- group would stabilise the RCOO- ion, thus would increase the acidity of aliphatic carboxylic acids in general. This is due to inductive effect. Consider the following example:
(i) \(\ce{CH3-COOH}\) pka of 4.7
(ii) \(\ce{Cl-CH2-COOH}\) pKa of 2.8
(ii) \(\ce{O2N-CH2-COOH}\) pKa of 1.68
When one of the hydrogen atom from CH3- is replaced by a more electronegative element Cl the pKa decrease which means acidity increase. Similarly, when -Cl atom is replaced -NO2 group, which is more electronegative, it further increase the acidity of the carboxylic acid.

7 (b) Write the preparation of H-COOH from oxalic acid [1]

Answer: Oxalic acid on heating alone above its melting point decompose to give formic acid and carbondioxide
\(\ce{HOOC-COOH ->[\Delta] HCOOH + CO2}\)

7 (c) How can acetyl chloride be converted into
(i) acetamide; (ii) acetic anhydride? [2]

Answer: Acetyl Chloride (\(\ce{CH3-COCl}\)) can be converted to:
(i) Acetamide by reacting with \(\ce{NH3}\).
\(\ce{CH3-COCl + NH3 -> CH3-CONH2 + HCl}\)
(ii) Acetic anhydride by reacting with \(\ce{CH3COOAg}\)
\(\ce{CH3-COCl + CH3COOAg -> (CH3CO)2O + AgCl}\) [post_ads]

7 (d) Synthesise the following using a suitable Grignard reagent: [2]
(i) CH3-CO-CH3
(ii) A secondary alcohol

Answer: When a suitable ester is treated with limited amount of a suitable Grignard reagent, a ketone is obtain. Thus, to synthesise acetone, methyl acetate is treated with methyl magnesium bromide as follows:


When and aldehyde is treated with a Gridnard reagent and then followed by hydrolysis, a secondary alcohol is obtained. Eg.


7 (e) Complete the following reactions: [3]

Molecular Diagram



Click on Question No to view the solution: [8 (a)] [8 (b)] [8 (c)] [8 (d)] [8 (e)] [8 (f)]

8 (a) Synthesise the following compounds form malonic ester (any one): [2]
(i) (CH3)2CH-CH2-COOH
(ii) Crotonic acid.

Answer: (i) (CH3)2CH-CH2-COOH can be synthesised by reacting malonic ester in presence of NaOEt and (CH3)2CH-Br, followed by hydrolysis and then decarboxylation.

(ii) Malonic ester undergo condensation reaction with Acetaldehyde in presence of a base like piperidine/pyridine which on hydrolysis and decarboxylation yield Crotonic acid. This condensation reaction is also known as Knoevenagel condensation. The reaction is as follows:

8 (b) Explain why dry ether is used in the preparation of Grignard reagents. [1]

Answer: Grignard reagent is very reactive towards protic solvents like water, alcohols etc. If any trace of water is present in the reaciton mixture, Grignard reagent that is formed will react to produce alkanes as follows: \[\ce{R-MgX + H2O -> R-H + Mg(OH)X }\] Therefore dry ether is used in the preparation of Grignard Reagents


8 (c) How will you convert the following? [2]
(i) Ethanoic acid into methyl acetate
(ii) Ethanoic acid into ethanamide

Answer (i) Ethanoic acid on reacting with thionyl chloride, yields Acetyl chloride. Acetyl chloride on treating with methanol, yields Methyl acetate.

(ii) Ethanoic acid on reacting with thionyl chloride, yields Acetyl chloride. Acetyl chloride on treating with ammonia yields ethanamide.

8 (d) Arrange the relative reactivity of the following carboxylic acid derivative towards nucleophilic substitution with justification.

Answer: The relative reactivity of the carboxylic acid derivative are as follows:
The general structure of a carboxylic acid derivative can be represented as R-CO-G. The reactivity of carboxylic acid derivatives depends on the nature of the group -G as a leaving group. If -G is a good leaving group, then the reactivity will be high and if -G is a poor leaving group then the reactivity is low. Here, Cl- is a good leaving group compared to R'COO-. R'COO- is a better leaving group than R'O-, and NH2- is a very poor leaving group compared to the rest.

8 (e) Write the different tautomeric forms of ethyl acetocetate. [1]

Answer: The following are the different tautomeric forms of ethyl acetate.

8 (f) Write the preparation of citric acid from glycerol. [1]

Answer: Citric acid can be prepared from glycerol as follows:


Click on Question No to view the solution: [9 (a)] [9 (b)] [9 (c)] [9 (d)] [9 (e)]

9 (a) Why aryl amines cannot be prepared by phthalimide syntheisis? [1]

Answer: The first step of the phathalimide synthesis involve a nucleophilic attack of the phthalimide on an aryl halide. As aryl halides does not undergo nucleophilic substituion readily, therefore, aryl amines cannot be prepared by the phthalimide synthesis.

9 (b) Outline the synthetic route of aniline from benzene. [2]


9 (c) Arrange the following molecules in order of their increasing basic strength with proper justification: [2]

Answer: The order of the increasing basic strength is as follows:

The basic strength of aromatic amines (or amines in general) depends on the availability of the lone pair of electrons on nitrogen atom for protonation. This is influenced many factors. Here, the lone pair from the N is delocalised with π electrons of the benzene ring. (1) Electron withdrawing nature of -NO2 coubpled with the extended delocalisation of electrons at para- position, would lead to decrease availability of the lone pair for protonation. (2) Electron donating nature of -OCH3 would increase the electron density on the ring and hence on the N atom, thus would lead to increased availability of the lone pair for protonation.


9 (d) Complete the following reaction by writtng a suitable mechanism: [2]

\[\ce{CH3-CH2-CH2-NH2 ->[HNO2][] ? }\]

Answer: Primary aliphatic amines reacts with nitrous acid to form an unstable diazo salt, which would then decompose to give N2 gas and carbocation. The carbocation further reacts to form alcohols and alkenes. The mechanism is as folows:

9 (e) Write the products of the following reactions: [1+1.5]

(i) \(\ce{C6H5-NO2 ->[Sn/HCl] ?}\)
(ii) \(\ce{ClCH2-COOH ->[NaNO2] ? ->[\Delta] ?}\)

Answer: (i) \(\ce{C6H5-NO2 ->[Sn/HCl] C6H5-NH2}\)
(ii) \(\ce{ClCH2-COOH ->[NaNO2] O2N-CH2-COOH ->[\Delta] CH3-NO2 + CO2 }\)


Click on Question No to view the solution: [10 (a)] [10 (b)] [10 (c)] [10 (d)] [10 (e)]

10 (a) Identify the products in the following reactions: [1.5+1]

(i) \(\ce{C6H5NH2 ->[NaNO2/HCl][{0 °C - 5 °C}] ? ->[C6H5OH][OH-] ?}\)
(ii) \(\ce{CH2N2 + HCl -> ?}\)

Answer:(i) \(\ce{C6H5NH2 ->[NaNO2/HCl][{0 °C - 5 °C}] C6H5-N2+Cl- ->[C6H5OH][OH-] C6H5-N=N-C6H4-OH}\)
(ii) \(\ce{CH2N2 + HCl -> CH3-Cl + N2}\)

10 (b) Discuss the formation of different products on the treatment of methyl amine, dimethyl amine and trimethyl amine with HNO2. [3]

Answer: Each class of amine (1° , 2° and 3°) reacts differently with HNO2
1° amines, reacts with HNO2 to form unstable diazo compounds. The diazo compounds quickly loose a N2 gas and form a mixture of compouns depending upon the reaction condition. The evolution of N2 gas, form the basis of the Van Slyke method of estimating primary -NH2 groups.

2° amines react slowly with nitrous acid yielding oily neutral N-nitrosoamines which are yellowish or greenish and characterised by Liebermann's test.

3° amines readily dissolve in cold nitrous acid forming soluble nitrite salt.


10 (c) How will you prepare \(\ce{CH3-COOCH3}\) using diazomethane and ethanoic acid? [1]

Answer: When diazomethane is reacted with ethanoic acid in presence of ether, methyl acetate is obtained.

\[\ce{CH3-COOH + CH2N2 ->[ether] CH3-COOCH3 + N2 }\]

10 (d) Why is nitromethane acidic in nature? [1]


The nitro group on nitromethane is a highly electron withdrawing group. Hence, via inductive effect it would cause a decrease in the electron density on the carbon and hydrogen atoms. Therefore, the hydrogen atoms on nitromethane can be easily abstracted by a base. Also, after the abstraction of a hydrogen ion by a base, the conjugate base of nitro methane is resonance stabilised by the nitro group as follows.

10 (e) Convert -
(i) aniline into chlorobenzene;
(ii) benzene diazonium chloride into cyanobenzene. [2]




F Diengdoh.com: General Chemistry III (Chem EH 301) Solved Question 2019: Section II (Organic Chemistry)
General Chemistry III (Chem EH 301) Solved Question 2019: Section II (Organic Chemistry)
F Diengdoh.com
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