Chemistry-2.PART TEST-1-MAILING-Solutions

CHEMISTRY SOLUTIONS 1. a) p-chlorophenol will have high pKa value, as its phenoxide is not much stabilized as compared to phenoxide ion of m-chlorophenol. The electron withdrawing Cl group stabilizes m-chlorophenol due to close proximity position to phenoxide ion. [4] b) i) ii) [3 + 3] c) [6] 2. a) [2 + 2 + 2] Reaction mechanism of Claisen condensation intramolecular i.e. Dieckmann condensation [2] b) [3 + 3] 3. a) i) ii) [2 + 2] b) Since compound B reacts with HIO4 to give benzaldehyde so it should contain one benzene ring and two adjacent – OH groups. Moreover since compound (A) gives iodoform test so it must contain group. The structures of compounds and reactions are as [2 × 3] c) [2 × 3] 4. a) There are 5 isomeric hexane (C6H14) of which 2,2-diemthylbutane and n-hexane produces 3 isomers (monochloro derivatives). [8] b) Tollen’s test given by three types of compounds i) Aldehydes (CnH2nO) ii) Formic acid (HCO2H) iii) α-hydroxyketone. Molecular formula of compound C3H6O2 indicates that it must be a α-hydroxyketone. [2 × 3] 5. a) Neopentyl chloride (CH3)3CCH2Cl, a 1° RCl, does not participate in typical SN2 reactions because the bulky (CH3)3C – group sterically hinders backside attack by a nucleophile. [3] b) The H’s on the C attached to the ring (the benzylic H’s), although they are in this case 2°, are nevertheless more reactive toward Brx than are ordinary 3° H’s like a C = C group in the allylic system, the Ph group can stabilize the free radical by electron donation through extended p orbital overlap. [3] 6. a) The compound A must be a ketone as it forms oxime but does not reduce Tollen’s reagent. The compound D must be an aldehyde. Its structure does not include the fragment CH3 – – as it does not respond idoform test. The compound E must be a ketone containing CH3CO – fragment. Let the compounds D and E be RCH2CHO and R′COR″ respectively. Where R, R″, R′ are all alkyl groups. From these we get. Since the M.F. of A is C6H12O, it follows that R = R′ = R″ = CH3. Hence, the structure of molecules (A) to (E) and the reactions are as follows. [2 × 5] b) [2 + 2] ‘α’ hydroxy acids from lactides, these are 6 membered ring compounds formed by reaction between two molecules of the hydroxy acid. [2] 7. a) i) Due to delocalisation of lone pair electron of nitrogen in benzene ring. ii) In the SN1 reaction mechanism carbocation are formed as intermediates. The carbonium ion is sp2 hybridized, the structure of which permits the attack of nucleophile from above and below the plane, consequently, racemisation occurs. iii) This is in accordance with Saytzeff rule. iv) Because Cl– is good leaving group in comparison to –OCOCH3 v) Due to rearrangement of intermediate carbonium ion. vi) group lone pair electrons is involved in delocalisation with of [2 × 6] b) [10] 6