10

Q1. Complete the following reactions:
(i)      (ii)       


(iii) CH₃CH₂Br + NaI

Solution

(i)                    + SO₂ + HCl   (ii)               (iii) CH₃CH₂Br + NaI   CH₃CH₂I + NaBr

Q2. Why is sulphuric acid not used during the reaction of alcohols with KI?

Solution

H₂SO₄ converts KI to corresponding HI and then oxidizes it to I₂. Hence it cannot be used during the reaction of alcohols with KI. Reaction:   KI_(s) + H₂SO_{4(l)  →}        KHSO_4(s) + HI_(g) 8H⁺ + 8I^- + H₂SO_{4 (l) →} H₂S _(g)    + 4H₂O _(l) + 4I_2(s)

Q3. What are freons? Give one example.

Solution

The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. Example: Freon-12( CCl₂F₂)

Q4. What are enantiomers? Give an example.

Solution

The stereoisomers related to each other as nonsuperimposable mirror images are called enantiomers. For example: 2-chloro butane (CH₃CH₂ C^*H (Cl) CH₃). Here the carbon atom marked is the chiral centre and posseses four different groups.

Q5. Give the IUPAC name of the compound CH₃CH (Cl) CH (Br) CH₃.

Solution

2-bromo-3-chloro-Butane

Q6. Define racemisation.

Solution

A mixture containing two enantiomers in equal proportions is known as racemic mixture and the process of conversion of enantiomers into a racemic mixture is known as racemisation.

Q7. Give the three uses of each of the following: (i) Dichloromethane (ii) Tetrachloromethane

Solution

Dichloromethane: It can be used (i) As solvent (ii) As paint remover (iii) As aerosol propellant Tetrachloro methane: It can be used (i) As a solvent (ii) As a fire extinguisher (iii) As a degreasing agent

Q8. Which alkyl halides show colour on exposure to light?

Solution

Alkyl Bromide and Alkyl iodide.

Q9. Which compound in each of the following pairs will react faster in S_N2 reaction with -OH? (i) CH₃Br or CH₃I (ii) (CH₃)₃CCl or CH₃Cl

Solution

(i) CH₃I will react faster than CH₃Br because the order of reactivity is as follows:     R-I > R-Br > R-Cl > R-F (ii) CH₃Cl will react faster than (CH₃)₃CCl because the order of reactivity is as follows in S_N2 reaction.      Primary halide > Secondary halide > Tertiary halide

Q10. What is DDT and why its use is banned in United States?

Solution

DDT is p, p'-Dichlorodiphenyltrichloroethane and is used as insecticide. It is banned for the following reasons:(i) Its high toxicity towards fish.(ii) DDT is not metabolized very rapidly by animals instead it gets deposited and stored in the fatty tissues.

Q11. Why the use of chloroform as anesthetic is decreasing?

Solution

Inhaling chloroform vapors depresses the central nervous system and its chronic exposure may cause damage to the liver and kidneys due to metabolism of chloroform to phosgene gas. Hence the use of chloroform as anesthetic is decreasing.

Q12. Why chloroform is kept closed in dark colored bottles?

Solution

Chloroform is oxidized slowly by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as Phosgene                  Hence it is kept closed in dark colored bottles.

Q13. Give reasons for the following:         i)     Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanide as the chief products.         ii)     Alcohols do not react with NaBr, but when H₂SO₄ is added, they form alkyl bromides.

Solution

i) KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms of CN can donate electron pairs but  the attack takes place mainly through carbon atom and not through nitrogen atom because C-C bond is more stable than C-N bond. However AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product. ii) Br^- is a weak base. So it cannot displace the strong base OH^- when H₂SO₄ is added, it leads to protonation of alcohol, as a result of which water molecule is formed. Since water molecule is a very weak base it is easily replaced by Br^-. 

Q14. A primary alkyl halide (A) C₄H₉ reacts with alcoholic KOH to give a compound (B). The compound (B) reacts with HBr to give the compound C which is an isomer of A. When A reacts with sodium metal it give a compound (D) whose molecular formula is C₈H₁₈. The compound D is different from the compound formed when n-butyl bromide reacts with sodium. Give the structural formula of A and write all the equations involved in the reaction.

Solution

For C₄H₉Br, two primary halides are possible CH₃CH₂CH₂CH₂Br (n-butyl bromide) and (CH₃)₂CHCH₂Br (iso-butyl bromide). Since A is not normal –butyl bromide, it must be iso-butyl bromide.  

Q15. Write the structure of the major organic product in each of the following reactions: (i) CH₃CH₂CH₂OH + SOCl₂ (ii) CH₃CH₂CH = CH₂ + HBr (iii) CH₃CH = C (CH₃)₂ + HBr

Solution

(i)                  CH₃CH₂CH₂Cl (ii)                CH₃CH₂CH₂CH₂Br (iii)               CH₃CH₂ CBr(CH₃)₂

Q16. Give reasons for the following: (i) Grignard reagent should be prepared under anhydrous conditions. (ii) Neo-pentyl bromide undergoes nucleophilic substituting reactions very slowly. (iii) p-Methoxybenzyl bromide reacts faster than p-nitrobenzyl bromide with ethanol to form an ether product.

Solution

(i) Grignard reagent reacts with water and gets decomposed, so it is produced in anhydrous conditions.(ii) Neo-pentyl bromide being a primary halide reacts slowly through SN₁, and being a sterically hindered halide reacts slowly even through SN₂ mechanism. (iii) Methoxy (-OCH₃) group being an electron releasing group stabilizes the intermediate carbocation. On the other hand nitro group (-NO₂) is an electron withdrawing group and hence it destabalises the intermediate carbocation. Thus p-methoxy benzyl bromide reacts faster than p-nitrobenzyl bromide because  the reaction proceeds via more stable intermediate carbocation.

Q17. Write the equations for the preparation of 1-iodobutane from:         (i)    1-butanol         (ii)    1-chlorobutane         (iii)    But-1-ene

Solution

(i)  CH₃CH₂CH₂CH₂OH     CH₃CH₂CH₂CH₂I                1-butanol                                       1-iodobutane   (ii) CH₃CH₂CH₂CH₂Cl + NaI    CH₃CH₂CH₂CH₂I + NaCl        1-chlorobutane                             1-iodobutane     (iii) CH₃CH₂CH = CH₂ + HI   CH₃CH₂CH₂CH₂I          But-1-ene.                                 1-iodobutane