Jp Xii Organic Chemistry (24).pdf

ORGANIC CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 28

Course : VIJETA (JP)

This DPP is to be discussed in the week (14-09-2015 to 19-09-2015)

DPP No. # 28 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

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

1.

Sol. 2.

(3 marks, 2 min.)

[60, 40]

Acetaldehyde and Acetone can be distinguish by : (A) 2,4–DNP (B) I2/NaOH (C) Lucas reagent (D*) Ammonical AgNO3 ,slhVSfYMgkbM vkSj ,lhVksu ds e/; fdlds }kjk foHksnu dj ldrs gS : (A) 2,4–DNP (B) I2/NaOH (C) Y;wdkWl vfHkdeZd (D*) veksuhd`r AgNO3 Acetaldehyde give positive Tollen (Ammonical AgNO3) but ketones are not ,slhVSfYMgkbM VkWysu (veksuhd`r AgNO3) ds lkFk /kukRed ijh{k.k nsrs gSa] ysfdu dhVksu ugha Acetaldehyde and Propyne can be distinguish by :

,slhVSfYMgkbM vkSj izksikbu dks fdlds }kjk foHksfnr fd;k tk ldrk gS % Sol.

(A) NaHCO3 (B*) I2/NaOH (C) Lucas reagent (A) NaHCO3 (B*) I2/NaOH (C) Y;wdkWl vfHkdeZd Acetaldehyde and Propyne can be distinguish by Iodoform test.

(D) neutral FeCl3 (D) mnklhu FeCl3

,slhVSfYMgkbM vkSj izksikbu dks vk;ksMksQkeZ ijh{k.k }kjk foHksfnr fd;k tk ldrk gSA

3.

and

can be distinguished by :

(A) Iodoform test (C*) Fehling solution test

rFkk

(B) Tollen's test (D) 2,4-DNP test

dks fdlds }kjk foHksfnr dj ldrs gS %

(A) vk;ksMksQkWeZ ijh{k.k (C*) Qsgfyax foy;u ijh{k.k 4.

(B) VkWysu ijh{k.k (D) 2,4-DNP ijh{k.k

I and II are homologues. These can be distinguished by using : I rFkk II letkr gS aA fdlds mi;ksx }kjk buesa foHks n fd;k tk ldrk

(A) Tollen’s reagent (A) VkWysu vfHkdeZd

(B) 2, 4 – DNP (B) 2, 4 – DNP

gS :

(C) Baeyer’s reagent (C) cs;j vfHkdeZd

(D*) I 2 + NaOH (D*) I2 + NaOH

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5.

Which of the following compound will give black or silver ppt. with Tollen's reagent.

fuEu esa ls dkSulk ;kSfxd VkWysu vfHkdeZd ds lkFk dkyk ;k jtr vo{ksi nsrk gSA

Sol.

(A) H  C  C – CH2  C  CH3 || O

(B) CH3  C  C  C  CH3 || O

(C*) CH3 – C  C – CH2 – CHO

(D) CH2  CH  C  CH  CH2 || O

Aldehydes gives black or silver ppt. with tollen's reagent.

,fYMgkbM VkWysu vfHkdeZd ds lkFk dkyk ;k jtr vo{ksi nsrk gSA 6.

Tollen’s reagent (AgNO 3 + NH4OH) can be used to distinguish between.

(A)

(C*)

and

(B)

and

(D)

and

and

VkWysu vfHkdeZd (AgNO3 + NH4OH) ls fuEu esa foHksn dj ldrs gSA (A)

(C*)

(B)

rFkk

(D)

rFkk

rFkk

rFkk

Sol.

Aldehydes gives black or silver ppt. with tollen's reagent.

gy %

,fYMgkbM VkWysu vfHkdeZd ds lkFk dkyk ;k jtr vo{ksi nsrk gSA

7.

2-Pentanone can be distinguished from 3- Pentanone by which reagent ? (A) 2, 4- Dinitrophenyl hydrazine (B) Tollen's reagent (C*) I2 and dilute NaOH (D) Fehling solution 2-isUVsukWu vkSj 3- isUVsukWu dks fuEu esa ls fdl vfHkdeZd }kjk foHksfnr fd;k tk ldrk (A) 2, 4-MkbukbVªks (C*) I2 vkSj

Sol.

QSfuy gkbMªsthu

ruq NaOH Iodoform test. (vk;ksMksQkeZ ijh{k.k)

gSA

(B) VkWysu

vfHkdeZd (D) Qsgfyax foy;u

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8.

Which of the following compound will not react with I2 /OH–. fuEu esa ls dkSulk ;kSfxd I2 /OH– ds lkFk vfHkfØ;k ugh djrk gSaA

(A)

Sol. 9.

(B)

(C*)

group gives positive iodoform test.

gy %

(D) CH3 – CHO

lewg /kukRed vkW;ksMksQkWeZ ijh{k.k nsrk gSA

The compound A gives following reactions.

Its structure can be ;kSfxd 'A' fuEufyf[kr

vfHkfØ;k,sa nsrk gS %

mi;qDr vfHkfØ;kvksa ds vk/kkj ij ;kSfxd 'A' dh lajpuk gksxh \ (A)

(B) OHC – (CH2)2 – CH = CH – COOH

(C*)

(D)

Sol.

Sol.

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10.

A alkene on ozonolysis give only one product(x). x does not respond with Tollen’s reagent and NaOI but give yellow precipitate with 2,4-DNP, The structure of alkene can be,d ,Ydhu vkstksuhvi?kVu ij dsoy ,d mRikn (x) nsrh gSA x, VkWysu vfHkdeZd rFkk NaOI ds lkFk fØ;k ugh nsrk ijUrq 2,4-DNP ds

lkFk ihyk vo{ksi nsrk gSA ,Ydhu dh lajpuk gks ldrh gS %

(A)

11.

(B)

(C*)

(D)

Which of the following compound can not be formed in given aldol condensation reaction :

nh xbZ ,YMksy la?kuu vfHkfØ;k esa fuEu esa ls dkSulk ;kSfxd ugh cusxkA NaOH

Dil. NaOH    CH3CHO + CH3COCH3   

(B) CH3–C=CH–CHO

(A) CH3–CH=CH–CHO

CH3 (C) CH3–C=CH–C–CH3

CH3

(D*) CH3–CH2–CH=CH–CHO

O

O Dil. NaOH

12.

O

  Major product, Major product is – 

O NaOH

O

  eq[; mRikn] eq[; mRikn gS – 

O (A)

(B*)

(C)

O 13.

O

(D)

O

Which of the following can show Aldol condensation reaction :

fuEu esa ls dkSulk ;kSfxd ,YMksy la?kuu vfHkfØ;k n'kkZ ldrk gSA O (A*)

O

(B)

(C)

(D) H–C–D

O

O Dil. NaOH   Major product, Major product is : 

14.

NaOH

  eq[; mRikn] eq[; mRikn –  (A*)

(B)

(C)

(D)

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15.

OH 

O3  Y + Z  X  

Zn,H2O

Compound X in the above reaction is mijksDr vfHkfØ;k esa ;kSfxd X gSA

(A)

(B*)

(C)

(D)

O / Zn

3  

Sol.

+



OH  

16.





H  

Which of the following can not show Cannizzaro reaction :

fuEu esa ls dkSu dsfutkjks vfHkfØ;k iznf'kZr ugh dj ldrk gS %

H–C–D (A)

17.

CHO

(B)

O

O

(C)

(D*) CD3CHO

(C) C6H5COCH3

(D) C6H5COOH

(C*) CCl3CHO

(D)

Conc. NaOH C6H5CHO   C6H5COONa + X ‘X’ is :

NaOH

 C6H5COONa + X C6H5CHO   ‘X’ gS : (A*) C6H5CH2OH 18.

(B) C6H5OH

Which can not show Cannizzaro reaction :

dkSu dsfutkjks vfHkfØ;k iznf'kZr ugh dj ldrk gS%

(A)

(B) H–C–D

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19.

Conc. NaOH X  

X is : NaOH

 X   X

gSA :

(A*)

(B)

(C)

(D)

Conc. H SO

(1) Conc. NaOH 4  Product is :     2

20.

( 2 ) H

(1)



NaOH

H2 SO 4     mRikn gSA (2) H 

(A*)

(B)

(C)

(D)

NaOH

H

Sol.

  

 

Sol.

  

NaOH

 

H

Conc. H SO

4 2 



H SO

2 4  



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