Jp Xii Organic Chemistry (27).pdf

ORGANIC CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 31

Course : VIJETA (JP)

This DPP is to be discussed in the week (05-10-2015 to 10-10-2015)

DPP No. # 31 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

(3 marks, 2 min.)

[60, 40]

(D) (C) (D)

7. 14.

ANSWER KEY DPP No. # 31 (JEE-MAIN) 1. 8. 15.

(C) (B) (C)

2. 9. 16.

(C) (B) (A)

1.

The reagent ‘A’ for the following reaction would be fuEufyf[kr vfHkfØ;k esa vfHkdeZd 'A' gS %

O || (A) CH3 – C – OC 2H5

3. 10. 17.

(B) (C) (C)

4. 11. 18.

(C) (B) (D)

(B) CH3CHO

5. 12. 19.

6. 13. 20.

(C*) HCHO

(C) (B) (B)

(B) (B)

(D)

Sol.

2.

The correct statement about the following reaction is :

(A) The product is a dicarboxylic acid. (B) The product is an -keto acid. (C*) The product is a racemic mixture of -hydroxy acids. (D) The product is a mixture of two aromatic acids. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 1

fuEu vfHkfØ;k ds fy;s lgh dFku dkSulk gSA

(A) mRikn ,d f} dkcksZfDlfyd vEy gSA (C*) mRikn ,d -gkbMªksDlh vEy dk jsflfed

(B) mRikn (D) mRikn

feJ.k gSA

,d -fdVks vEy gSA nks ,jkseSfVd vEyksa dk feJ.k gSA

Sol.

It is Benzil-benzilic acid rearrangement reaction.

;g ,d csfUty&csUtkbfyd vEy iqUkfoZU;kl vfHkfØ;k gSA

(1) Ac O,AcONa

2  (A), Identify product (mRikn (2) H O

3.

2

(A)

4.

(B*)

dks igpkfu;s)

(C)

(D)

In the following conversion, (i) MeMgBr

NaOH / I

2    X  Y   (ii) H3O

H3O

the major products X and Y, respectively, are :

(A)

(B)

(C*)

(D)

fn;s x, ifjorZu esa eq[; mRikn (i) MeMgBr

NaOH / I

(ii) H3O 

H3O

2    X   Y

X

rFkk

(A)

Y Øe'k%

gS %

(B)

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PAGE NO.- 2

(C*)

(D)

Sol.

5.

Find the suitable reagents of the following reaction

fuEufyf[kr vfHkfØ;k esa mi;qDr vfHkdeZd gS %

(A) PhCHO / NaOH, LiAlH4 (C) CrO3 / H

 ; Al

(B) CrO3 /

O3 ; HBr / R2O2 ; PhCH2MgBr

2

, LiAlH4

(D*) PhCHO / NaOH, N 2H4 / KOH, 

Sol.

6.

–

OH   (A)

+ Product (A) is : mRikn (A) gS %

(A)

7.

(B)

(C*)

(D)

The product/s of the following reaction is

fuEufyf[kr vfHkfØ;k esa izkIr mRikn fuEu esa ls gksxk ?

PhCO H

3   

Sol.

(A)

(B*)

(C)

(D)

(B) This is Baeyer Villiger oxidation and the methyl group is the weakest migrating group.

;g cs;j foyhtj vkDlhdj.k vfHkfØ;k gS vkSj esfFky lewg nqcZyre LFkkukUrfjr lewg gksrk gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 3

PhCO H

3   

8.

The following product is obtained by using reactant/s ....... in a base catalysed condensation reaction {kkj mRizsjfd; la?kuu vfHkfØ;k esa fuEu mRikn dks izkIr djus ds fy, .......... vfHkdeZd dk mi;ksx djrs gSA

(A)

+

(C)

(B*)

+

(D)

+

Sol. The is a self claisen condensation reaction.

;g Loa; Dsystu la?kuu vfHkfØ;k gSA 9.



Above conversion can be carried out by :

mijksDr vUr%ifjorZu fdlds }kjk gksrk gS % (A) LiAlH4, H+,  (C) (i) H+,  (ii) OH–

10.

(B*) (i) OH– (conc.) (lkUæ),  (ii) H + (D) (i) NaBH4 (ii) H+

CO  HCl

     P Anhydrous AlCl3





H3O Q   R  S.

Which is not true about above sequence of reaction. (A) Product R and S reacts with NaHCO3 solution with effervescence of CO2 gas. (B) P when reacts with NaCN and H2SO4, gives two optically active resolvable product. (C*) Q is formed by nucleophilic substitution reaction on P. (D) S is formed by evolution of H2O and CO2 from R. CO  HCl

        AlCl3

P





H3O Q   R  S.

mijksDr vfHkfØ;k ds lanHkZ esa dkSulk dFku xyr gSA (A) mRikn R rFkk S NaHCO3 foy;u ds lkFk fØ;k dj CO2 xSl nsxsaA (B) tc P dh NaCN ,oa H2SO4 ds lkFk fØ;k dh tkrh gS] rks nks izdkf'kd lfØ; i`FkDdj.kh; mRikn izkIr gksrs gSA (C*) mRikn Q, ;kSfxd P ij ukfHkdLusgh izfrLFkkiu vfHkfØ;k }kjk izkIr gksrk gSA (D) mRikn S, ;kSfxd R ls H2O rFkk CO2 ds fu"dklu ls izkIr gksrk gSA

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PAGE NO.- 4

CO  HCl

CO  HCl

      

Sol.

Anhydrous AlCl3

AlCl3

 Ph  CH  CH  COOH Ph  CH  CH  COOH 

Cinnamic Acid

11.

+ 2C2H5OH

–H2O – CO2

The product of following reaction is :

fuEu vfHkfØ;k dk mRikn gS& AC 2 O / ACONa /  AC2O / ACONa /       (Product) (mRikn) ( excess )

Sol.

(

(A)

(B*)

(C)

(D)

)

It is perkin reaction.

;g ifdZy vfHkfØ;k gSA

12. The compound (S) is : ;kSfxd (S) gSa %

(A)

(B)

(C*)

(D)

Sol.

13.

Which of the following compound on reaction with O3/Zn, H2O followed by aq. NaOH/ will form

fuEu esa ls dkSulk ;kSfxd O3/Zn, H2O ls fØ;k djkus ds i'pkr~ NaOH/ ls fØ;k djkus ij

(A)

(B*)

(C)

cukrk gSA

(D)

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PAGE NO.- 5

O3 / Zn   

Sol.

14.



+ (C6H5)3P = CH2  product (mRikn )

(A) Sol.



OH  

(B*)

(C)

(D)

It is witting reactions.

;g fofVZx vfHkfØ;k gSA 15.

An aldehyde X (C11H10O), which does not undergo self aldol condensation, gives benzaldehyde and two moles of Y on ozonolysis. compound Y, on oxidation with silver ion gives oxalic acid. The compounds X and Y, are respectively : ,d ,fYMgkbM X (C11H10O), tks Lor% ,YMksy la?kuu ugha n'kkZrk ysfdu vkstksuh vi?kVu ij csUtSfYMgkbM vkSj nks eksy Y nsrk gSA ;kSfxd Y, flYoj vk;u ds lkFk vkWDlhdj.k ij vkWDtSfyd vEy nsrk gSA ;kSfxd X o Y Øe'k% gS %

(A)

(B)

(C*)

(D)

Sol.

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PAGE NO.- 6

16.

1–Ethylcyclopent–1–ene on reductive ozonolysis followed by aq. NaOH/ gives 1–,fFkylkbDyksisUV–1–bZu dk vipk;d vkstksuhvi?kVu djus ds i'pkr~ tyh; NaOH/ ls

(A*)

O / Zn

Sol.

17.

(B)

3  

(C)

fØ;k djus ij nsrk gSA

(D)



(i ) H   



OH

 

(ii) 

Compound ‘A’ (molecular formula C4H10O) is treated with KMnO 4 solution to form product ‘B’ (molecular formula C4H8O). B does not form shining silver mirror on warming with amonical AgNO 3 . When B is reacted with 2,-4 DNP, yellow precipitate of C is obtained. Identify the structure of C. ;kSfxd ‘A’ (v.kqlw=k C4H10O) KMnO4 foy;u ds lkFk vfHkd`r djus ij mRikn ‘B’ (v.kqlw=k C4H8O) cukrk gSA B veksfud`r AgNO3 ds lkFk xeZ djus ij pedhyk jtr niZ.k ugha nsrk gSA tc ;kSfxd B dks 2,-4 DNP ds lkFk vfHkd`r fd;k tkrk gS ] rks ihyk vo{ksi C izkIr gksrk gSA C dh lajpuk gS %

(A) CH3CH2CH2CH = N.

(B)

(C*)

(D)

 1. . NaOH /  1 . dil NaOH /           Final product is :

18.

vafre mRikn gSA

(A)

Sol.

(B)

(C)

(D*)



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PAGE NO.- 7

 



H / Ni

NaNO2 / HCl / H2 O KCN / H  (W) 2    

19.

The compound W is ;kSfxd (W) gS

(A)

(B)

(C)

(D*)

Sol.

20.

(1) Zn   (B) PhCHO + BrCH2CO2 Et ( 2) HO (A)  ( H O) 2

Product of (B) of the above reaction is : mijksDr vfHkfØ;k esa mRikn (B) gksxk % (A) Ph – CH = CH – CH3 (C) Ph – CH = CH – CH2 – OH

2

(B*) Ph – CH = CH – CO 2Et (D) Ph – CH = CH – CH3

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PAGE NO.- 8