BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI - HYDERABAD CAMPUS 2nd Semester, 2011-2012 CHEM C142: Chemistry II Test-1 (Regular) Closed Book Max. Marks: 40 Time: 50 minutes Date: 03.02, 2012 Note: There are two questions in all. Ensure that both sides of the paper are printed. Start answering each question on a fresh page and answer all parts of the question together. Pencil should not be used. Answer to the point in short.

1. (a) When is chlorinated, among the products found are traces of chloroethane (CH 3CH 2Cl). How is it formed? Show with a mechanism. (3)

(b) When 2-methylpropane is monochlorinated in the presence of light at room temperature, 36% of the product is 2-chloro-2-methylpropane and 64% is 1-chloro-2-methylpropane. From these data, calculate how much easier it is to remove a from a tertiary than from a primary carbon under these conditions. (4)

(c) Write only the major product for the following reaction reactions. [NBS= N-bromosuccinamide] (2+2)

CH3 NBS, CCl NBS 4 (i) (ii) hv Peroxides S

(d) The formation of free radicals from is easier than chlorination but still iodination of alkane is slower than chlorination. Why? (2)

(e) Identify the alkane in each of the following pairs that has the lower carbon-carbon bond dissociation energy and explain the reason for your choice. (i) propane or 2-methyl propane (ii) 2-methyl propane or 2,2-dimethylpropane (4) (f) What is the order of reactivity for the carbon centres 2, 3 and 4 towards free radical halogentation for the following molecule? Answer with explanation. (3) 3 1 F . 4 2 2. (a) Comment on the chirality of , and bromochlorofluoromethane. ( 3) (b) Mention whether the following molecules are chiral or achiral? (4) t-Bu t-Bu Br Br Br Cl Me (A) Cl (B) (C) H Cl (D) Me H t-Bu I I I (c) The following compound has only one asymmetric center Why, then, does it have four stereoisomers. Draw the structures of all ? (2) H H3C C CH C CH3 H

OH (d) Assign R/S wherever applicable (4x1 )

H Cl H H Me OH H OMe (ii) (iii) (i) H (iv) H CO H H H 2 Cl O H H OH (e) Optically pure 2-methyl-1-butanol of specific rotation (-) 5.9 0 was treated with HCl to obtain 1-chloro-2- methyl butane of specific rotation (+) 1.253 0. Optically pure 1-chloro-2-methyl butane has a specific rotation of (+) 1.67 0. What are the percentages of (+) and (-) isomer in a sample of 1-chloro-2-methyl butane of specific rotation (+) 1.253 0 ? (3) (f) Are they meso compounds (Yes/NO)? (2)

Br CH3

H H Br Cl (i) (ii) H Br Cl Br

CH CH H 2 3 (g) How can you separate (+) and (-) isomer from (±)-acid? Write in one sentence. (2)

Model Solution:

1. (a) Steps leading to chloroethane are

(i) Propagation

(ii)

(iii)

(iv) Termination

(b) Number of 1 0 hydrogen= 9, Number of 3 0 hydrogen= 1

Contribution per 1 0 hydrogen is 64/9=7.11

Contribution per 3 0 hydrogen is 36/1=36

So 3 0 are 36/ 7.11= 5 times more reactive then 1 0 hydrogens. Thus, it is 5 times easier to remove than 1 0 hydrogens.

(c) (i) CH 3 CH2Br

NBS Peroxides S S

(ii) Br

NBS, CCl4 hv

(d) Although the initiation step is faster, the chain propagation step (rate determining step) is not favorable for iodination because this step-I need very high activation energy and also step-II is highly endothermic. (e) (i) The carbon-carbon bond dissociation energy is lower for 2-methylpropane because it yields a more stable (secondary) radical; propane yields a primary radical. (ii) The carbon-carbon bond dissociation energy is lower for 2,2-dimethylpropane because it yields a still more stable tertiary radical. (f) 3>2>4 1 C1 attached to and C2 attached to carbon attached to fluorine shows less reactivity because of the electron withdrawing effect of fluorine, whereas C3 and C4 shows normal reactivity difference between primary and secondary carbon undergoing halogenation. 2

2. (a) bromochloromethane (one plane of symmetry) and dichloromethane (two plane of symmetry) are achiral whereas bromochlorofluoromethane does not have any plane of symmetry. Hence bromochlorofluoromethane is chiral and exist as enantiomer

(b) (A) achiral (B) achiral (C) achiral (D) chiral

(c) Possibility of cis-trans isomers due to double bond as well as enantiomers.

H3C H H H

CH3 CH3

H H3C H OH H OH

H3C H H H

CH3 CH3

H H3C HO H HO H

(S) (R)

(d) (i) S (ii) S (iii) (S) at both (iv) O (e) Specific Rotation Purity

(+) 1.67 0 100 % (+) isomer

(+) 1.253 0 75 % (+) isomer (2)

Remaining 25 % is racemic modification hence (-) isomer is 25 / 2 = 12.5%

Hence (+) isomer is 75 + 12.5 = 87.5% (1) and (-) isomer is 12.5% ( 1)

(f) (i) NO (ii) Yes

(g) by using an optically active base.