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Practice each question, then open the answer or worked solution.
Predict the major product (s) of the following reactions and explain their formation.
Step 1: Homolysis of peroxide to from free radicals
Step 2: Formation of Bromine free radical
∙C6H5 + H-BR → C6H6 + Br∙
Step 3: reaction of hydrogen bromide with an alkyl radical
Nucleophiles and electrophiles are reaction intermediates having electron-rich and electron-deficient centres’ respectively. Hence, they tend to attack electron-deficient and electron-rich centres respectively. Classify the following species as electrophiles and nucleophiles.
(i), (vi), (vii), (viii) are nucleophiles
(ii), (iii), (iv), (v) are electrophiles
The relative reactivity of 1°, 2°, 3° hydrogen’s towards chlorination is 1: 3.8: 5. Calculate the percentages of all mono-chlorinated products obtained from 2-methyl butane.
Amount of mono-chlorinated products = No. of hydrogen reactivity
Mono-chlorinated products from 1° H = 9 1 = 9
Mono-chlorinated products from 2° H = 2 3.8 = 7.6
Mono-chlorinated products from 3° H = 1 5 = 5
Total mono-chlorinated products = 9+7.6+5 = 21.6
Yield % for 1° H chlorination = 9×100/21.6 = 41.67%
Yield % for 2° H chlorination = 7.6×100/21.6 = 35.18%
Yield % for 3° H chlorination = 5×100/21.6 = 23.15%
Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane.
When the mixture of 1-iodo-2-methylpropane and 2-iodopropane is treated with sodium it gives a mixture of three products formed by intermolecular and intramolecular reaction as follows:
Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.
The 3° free radical is stabilised by 9 hyperconjugation structures and thus has more stability as compared to 1° free radical.
An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination, this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.
The ring systems having the following characteristics are aromatic.
(i) Planar ring containing conjugated π bonds.
(ii) Complete delocalisation of the π−electrons in-ring system i.e. each atom in the ring has unhybridised p-orbital, and
(iii) Presence of (4n+2)−electrons in the ring where n is an integer (n = 0, 1, 2,………..) [Huckel rule].
Using this information classifies the following compounds as aromatic/non-aromatic.
Aromatic compounds: A, E and F
Non-Aromatic : B, C, D and G
Which of the following compounds are aromatic according to Huckel’s rule?
According to Huckel’s rule, it should satisfy (4n+π) rule
Compound B, C, D and F are aromatic.
Suggest a route to prepare ethyl hydrogen sulphate (CH3–CH2–OSO2—OH) starting from ethanol (C2H5OH)
Solution;
Ethanol when treated with sulphuric acid at around 140°C gives hydrogen sulphate. The reaction is
IV. Matching Type
40. Match the reagent from Column I which on reaction with CH3
—CH=CH2 gives some product given in Column II as per the codes given below :
| Column I
(i) O3/Zn + H2O (ii) KMnO4/H+ (iii) KMnO4/OH– (iv) H2O/H+ (v) B2H6/NaOH and H2O2 |
Column II
(a) Acetic acid and CO2 (b) Propan-1-ol (c) Propan-2-ol (d) Acetaldehyde and formaldehyde (e) Propane-1,2-diol |
(i) is d
(ii) is a
(iii) is e
(iv) is c
(v) is b
Match the reagent from Column I which on reaction with
CH3 – CH=CH2 gives some product given in Column II as per the codes given below:
Column I | Column II |
(i) O3/Zn + H2O | (a) Acetic acid and CO2 |
(ii) KMnO4/H+ | (b) Propan-1-ol |
(iii) KMnO4/OH- | (c) Propan-2-ol |
(iv) H2O/H+ | (d) Acetaldehyde and formaldehyde |
(v) B2H6/NaOH and H2O2 | (e) Propan-1,2-diol |
Ans:
(i) O3/Zn + H2O | (d) Acetaldehyde and formaldehyde |
(ii) KMnO4/H+ | (a)Acetic acid and CO2 |
(iii) KMnO4/OH- | (e) Propan-1,2-diol |
(iv) H2O/H+ | (c) Propan-2-ol |
(v) B2H6/NaOH and H2O2 | (b) Propan-1-ol |