Reaction Mechanism Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE COCH2COH3 1 NaOH P Br D H Which of the following statement is correct O COCH2COH3 (A) P = H C (B) P = OH D D O O O CH3 (C) P = H H CH3 (D) P = DO H D Br 2 NaOH  Product, Product is OH OH (A) (B) (C) O OH 3 4 CH3 – CH2SH (i) CH3O (ii) ethylene oxide (D) OH Product, Product is : (A) CH3 – CH2 – S – CH2 – CH2 – OH  (B) CH3  CH 2O (C) CH3 – CH2 – O – CH2 – CH2 – OH (D) Read the following road map carefully CH3 K I Ph — CH2COH 1-Phenyl-2-propanol S.R = + 33.02° CH3 Ph—CH2 CHOK + gas II EtOTs CH3 Ph—CH2—CH—OC2H5 ethyl–1–phenyl–2–propylether III TsCl CH3 Ph—CH2—CHOTs IV C2H5OK CH3 Ph—CH2 —CH—OC2H5 Ethyl–1–phenyl–2–propylether O REACTION MECHANISM 5 (A) Both the ethers obtained by the two routes have opposite but equal optical rotation. (B) One of the either is obtained as a recemic mixture. (C) Step II & III both are SN2 reaction and both have inversion (D) Step II has inversion but step III has retention. A compound A has the molecular formula C5HgCl. It does not react with bromine in carbon tetrachloride. On treatment with strong base it produces a single compound B. B has a molecular formula C5H8 and reacts with bromine in carbon tetrachloride. Ozonolysis of B produces a compound C which has a molecular formula C 5H8O2. Which of the following structures is that of A? Cl Cl (A) Cl (B) Cl (C) (D) CH3 CH3 6 Me2SO4 NaOH PhOH P, P is (A) Ph–O–SO2OMe (B) PhOMe (C) PhOSO2OPh (D) PhMe Br 7  EtO + Ph – CH2 – NH2   Product, Product is Ph O OEt (B) Ph (A) Ph O NH Ph NH Ph O Br (C) Ph NH Ph (D) Ph OH CH3 8 In the reaction conc.H 2SO 4 CH3   A. The product A is – OH OH CH3 (A) O CH3(CH2)6 CH3 (B) CH3 CH3 CH3 CH3 (C) (D) COCH3 CH3 O H KI, DMSO TsCl  B (Major) ; ‘B’ has stereochemistry –  A  C—OH  9 C2H5 10 11 (A) R (B) S (C) R, S both (D) None of these – +   Product (CH3)3CCl + (CH3)3CO K (A) SN Product will be more (B) E2 Product will be more (C) both will be same (D) None of these Neopentyl iodideis treated with aq. AgNO3 solution, a yellow precipitate is formed along with other compound which is NO3 OH (A) (CH3)3CCH2NO3 (B) (CH3)2C CH2CH3 (C) (CH3)3CCH2OH (D) (CH3)2C CH2CH3 * CH 2Cl 12 The major end product of the following reaction is OH H3C CH3 * CH2 — OH OH (A) aq.AgNO3   * (B) H3C * O (C) H 3C * O (D) H3C O CH3 NO2 NO2 13 CH3 C  H A; Product ‘A’ is CH3  C OH OH CH3 NO2 O (A) CH3 C (B) C NO2 C C NO2 O CH3 NO2 CH3 CH3 (C) O N 2 C CH(OH) CH3 (D) None of these NO2 14  The products of hydrolysis of O C2H5 O (A) HOCH2CH2CH2CH2CHO + CH3CHO H3O   , is (B) HOCH2CH2CH2CH2CH2OH + CH3CHO (C) HOCH2CH2CH2CH2CHO + C2H5OH (D) HOCH2CH2CH2CH2CH2OH + C2H5OH CH2 — OH 15 A. The produt is : CH2 — HSO4 (A) CH2 (B) CH3 CH3 (C) (D) OH REACTION MECHANISM 16 The major product P of the following reactionis NH2 CH3 CH3 CH CH CH NH2 Cl NH2 CH3 C (A) NH2 CH3 CH2 CH NH2 CH3 CH (B) CH3 CH2 CH3 CH (C) NH2 OCH3 OCH3 NH2 C CH3 CH NH 2 CH NH2 CH 3 (D) CH OCH3 CH 3 CH C NH 2 O 17  The product of following reaction is H3O   14 O Ar OH O 14 (A) (B) O 14 OH Ar O OH + ArOH 14 (C) (D) 14 O 18 + ArOH + ArOH O H /H 2O18 CH 2   H 2C 14 O Which can not be the product (A) 14 H 2C CH2 OH CH2 14 OH CH2 (C) 18 OH CH 2 (B) 18 OH 18 OH CH2 18 OH (D) A and B both SECTION - II : MULTIPLE CORRECT ANSWER TYPE 19 Which of the following reaction(s) is/are possible  OH  CH3CH(NEt2)CH2 – OH (A) CH3 – CHClCH2NEt2  (B) Either CH3 CH CH2 HCl  CH3CHClCH2SEt SEt or CH3CH(sEt)CH2OH  OH (C) Treatment of either epoxide I or epoxide II with aqueous  OH gives the same product III O CH3 CH CH O I OH CH3Br or CH3 Br OH EtO EtOH (D) Cl CH CH II OH + OEt OEt  OH CH2   CH3 CH CH O III CH2 OH Cl OH 20 OH CH2 C Cl NaOH 1 eq P O O NaOH Q   R O O (A) R = (B) Q = O OH O OH O OH C=O OHCH2Cl (C) Q = CH2 C Cl (D) P = OH CH3C6H4 21 Ph OH H NH3 HNO 2  ? Ph Ph (A) Ph O pCH3C6H5 (B) Ph H (C) 22 pCH3C6H5 H pCH3C6H5 O Ph Ph O Ph pCH3C6H5 O Ph (D) H Ph H The correct of SN / E2 ratio for the % yield of product of the following halide is – 2 Ph (A) CH3 CH C (B) CH3 CH3 I (D) CH3 (C) CH3 – CH2 – I CH CH Ph I CH CH CH3 CH3 I 23 The product in the given reaction is Me H Br ONa + H Ph I Me (A) H P Ph II H Me Me (B) H O Ph H O Ph (C) H Ph O Me Me (D) H Ph O Me Me H Ph H Ph Me The correct statement(s) about solvent effect is/are : (A) Decreasing soslvent polarity causes a large increase in the rate of the SN2 attack by ammonia on an alkyl halide Ph 24 Me Ph REACTION MECHANISM   ..  RNH 3  X RX  NH3  (B) Increasing solvent polarity causes a large causes a large decrease in the rate of the SN 2 attack by hydroxide ion on trimethylsulfonium ion.   HO +  CH 3 )3  S   CH3OH + (CH3)3S Trimethyl DMS sulphonium ion (C) Increasing solvent polarity causes a small decrease in the rate of the SN2 attack by trimethylamins on trimethylsulfornium ion.  ..   CH 3 N(CH 3 )3 + (CH3)2S (CH 3 )3 N + (CH 3 )3 S  SECTION - III : ASSERTION AND REASON TYPE Read the following question and choose the correct answer (A) Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1. (B) Statement-2 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is Fasle (D) Statement-1 is False, Statement-2 is True. 25 H2SO4 Statement-1 : (CH3)3 CCH (CH3)3CCH = CH2 + CH3 OH Statement-2 : It follows Hofmann’s elimination by E2. 26 Statement-1 : CH3 OH H2SO4 Ph Heat CH3 CH3 C=C CH3 CH3 + CH2=C CH3 CH(CH3)2 Ph Statemen-2 : Elimination follows E1 mechanism. 27 Statement-1: Ethers are prepared from alcohols acid medium through SN1 or SN2 mechanisms depending upon nature of alcohol. Statement - 2 : ROH + H+ ƒ R or O H + slow R OH2  R – OH + H+ ƒ R+O H  R – OH 2 R slow R OH 2 R O+ H + R O H R H2O fast  ROR + H O+ R + H 2O  3 R+ + H2O R OR + H3O+ OH 28 Statement - 1 :  OH OH H  + H Statement - 2 : OH + etc. + H SECTION - IV : MATRIX - MATCH TYPE 29 Match the column I with column II. Column (I) Reactions H CH3 KNH 2 C=C (A)   Ph Cl Column (II) Reactions (P) -elimination CH3 SOCl2  OH  (B) H (Q) SN2 Ph (C) 30 NaOH   (R) -elimination (D) (S) SNi Substrate stereochemistry of product CH3CH2CH2 C Br + H2O H3C CH CH 2 3 (B) CH3CHDCl + OH– (p) Retention (C) (r) Inversion (A) (q) Racemisatioin C Br + H2O H3C CH CH 2 3 (D) CH2=CH CH Cl + SH / (CH3COCH3) (s) Mixture of inversion & product with -bond shift CH3 SECTION - V : SUBJECTIVE TYPE REACTION MECHANISM Ph H 31 + KNH2 Ph Br Identify the product 32 Give products Product excess NH3 O HC CH2 A CH3CH2SH base B LiAlD4 then H2O C aq. NaCN aq. NaN3 KI D E F SECTION - VI : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension A hydrocarbon (X) of the formula C6H12 does not react with bromine water but reacts with bromine in presence of light, forming compound (Y).Compound (Y) on treatment with Alc. KOH gives compound [Z] which on ozonolysis gives (T) of the formula C6H10O2. Compound (T) reduces Tollens reagent and gives compound (W). (W) gives iodoform test and produce compound (U) which when heated with P 2O5 forms a cyclic anhydride (V). 33 Compound V is O CH3 O O (A) O (B) O O O (C) CH3 C CH2 CH2 CH = O 34 Compound W is (A) COOH – (CH2)2 – COOH (C) 35 (D) CHO – CH = CH – CHO COCH3 (B) COOH COOH (D) CH3 COOH CH2 CH COOH CH = O Compound ‘X’ is CH3 (A) CH3 (B) Comprehension Read the following reaction (C) CH3 CH3 CH3 (D) O  H  P 36 NaEt/ NBS NaOH Q   R   S  T Compound T is OH OH (A) (B) (D) HO OH Compound ‘Q’ is OH OH (A) OH OH (B) (C) Cl 38 OH (C) OH OH 37 OH (D) Cl Cl When treating ‘S’ with strong base, product obtained is OH OH (A) (B) OH OH (C) (D) Br Comprehension Two optically active isomeric alcohols X and Y (M.F. = C9H12O) on oxidation with KMnO4 give benzoic acid. X gives positive iodoform test but Y does not X looses optical activity on oxidation with CrO 3/H+ but Y does not. 39 ‘Y’ can be CH2 CH2 OH (A) (C) 40 CH3 CH3 C CH2 OH H (B) (D) CH3 C CH3 OH H C CH2 CH3 OH The rate of reaction of ‘X’ with the following acids will be in the following order is REACTION MECHANISM (I) NO2 COOH (III) CH3 COOH (A) I > II > III > IV 41 (II) COOH NO2 (IV) CH3O (B) II > I > III > IV COOH (C) III > I > II > IV (D) IV > II > III > I The possible product(s) on reaction of ‘X’ with NaBr + H2SO4 (along with appropaite mechanism) will be (A) PhCH2 CH CH3 by SN2 mechanism (B) PhCH2 Br CH CH3 SN1 mechanism (C) Ph Br CH CH2CH3 by SN1 mechanism (D) Ph Br CH CH2CH3 by SN2 mechanism Br MOCK TEST 1 C 2 B 3 A 4 A 5 B 6 B 7 B 8 D 9 14 C 15 C 16 B 17 C 18 C 19 A,B,C,D 20 A,C,D 25 D 26 A 27 A 28 A 33 B 34 C 36 B 35 A A 10 C 21 A,C 37 A 38 B 11 B 12 D 13 B 22 A 23 A 24 (B,C) 39 C 40 A 41 C