(4104)dpp_32_50_b.pdf

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 32 & 33

This DPP is to be discussed in the week (28-09-2015 to 03-10-2015) 1. 2. 3. 4.

Course of the week as per plan : Inductive Effect, Resonance. Course covered till previous week : Hydrogen Bonding. Target of the current week : Inductive Effect, Resonance. DPP Syllabus : Inductive Effect, Resonance.

DPP No. # 32 (JEE-ADVANCED) Total Marks : 79

Max. Time : 48 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.10 Integer type Questions ('–1' negative marking) Q.11 to Q.14 Match the Following (no negative marking) Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

1.

(3 (4 (4 (8 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.) marks, 2 min.)

[15, [20, [16, [08, [20,

10] 10] 12] 06] 10]

In which of the following compounds the direction of Inductive-effect is not correct ? fuEu ;kSfxdksa esa ls fdlesa izsjf.kd izHkko dh fn'kk lgh ugha gS ?

CH = CH2 COONa (A)

(B)

(C*)

(D)

OH

C  CH

Sol.

2.

–CH2–CH2–O –O –CH2–O    Among these groups, which of the following orders is correct for the magnitude of their + effect ? fuEufyf[kr lewgksa ds + izHkko ds ifjek.k dk lgh Øe dkSulk gS ? (A) > >  (B) > >  (C) > >  (D*) > > 

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PAGE NO.- 1

3.

Which of the following statement is CORRECT regarding the inductive effect? (A) electron-donating inductive effect(+I effect) is generally more powerful than electron-withdrawing inductive effect(-I effect) (B) it implies the shifting of  electrons from more electronegative atom to the lesser electronegative atom in a molecule (C*) it implies the shifting of  electrons from less electronegative atom to the more electronegative atom in a molecule (D) it increases with increase in distance. izsjf.kd izHkko ls lEcfU/kr dkSulk dFku lR; gS \ [General Organic Chemistry] (A) bysDVªksu

nsus okyk izsjf.kd izHkko (+ I izHkko) lkekU;r;k vf/kd izHkkoh gksrk gS bysDVªkWu [khpus okys izsjf.kd (–I izHkko ls) (B) bl ?kVuk esa v.kq esa vf/kd fo|qr_.kh ijek.kq ls de fo|qr_.kh ijek.kq dh vksj  bysDVªkWu dk LFkkukUrj.k gksrk gSA (C*) bl ?kVuk esa v.kq esa de fo|qr_.kh ijek.kq ls vf/kd fo|qr_.kh ijek.kq dh vksj  bysDVªkWu dk LFkkukUrj.k gksrk gSA (D) ;g nwjh c<+us ds lkFk c<+rk gSA 4.

What is the % s character in hybridisation of carbon when it exerts strongest –I effect ? tc dkcZu ijek.kq lcls izcy –I izHkko n'kkZrk gS rks blds ladj.k esa % s y{k.k D;k gksxk \ (A) 25% (B*) 50% (C) 75% (D) 100%

5.

–OH



 SR2

–Br

–CN

I II III IV Among these groups, which of the following orders is correct for the magnitude of their – effect ? fuEufyf[kr lewgksa ds – izHkko ds ifjek.k dk lgh Øe dkSulk gS ? (A) V> > I (B) > V> III (C) > > VI (D*) >IV > >I 6.

Which order of I effect is/are incorrect. (fuEu 

7.

8.

esa ls I izHkko dk xyr Øe gS@gSa%)



(A*)  NH3 > – S(CH3 )2

[–I]

(B*) –NH2 > –NHCH3

[–I]

(C*) –OH > –Cl

[–I]

(D) –CD3 > –CH3

[+I]

Maximum–I effect group than the group –C  C – CH3 are fuEu esa ls fdldk –I izHkko –C  C – CH3 gS \ (A*) OH (B*) –OCH3 (C*) –NO2

(D) –NH2

Which of the following statements is correct about inductive effect ? (A*) Inductive effect is distance dependent and decreases drastically on increase in distance. (B*) Inductive effect is transmitted through -bond. (C) Inductive effect is transmitted through -bond (D*) Inductive effect is permanent effect

izsjf.kd izHkko ds fo"k; esa fuEu esa ls dkSulk dFku lgha gS \ (A*) izsjf.kd izHkko nwjh ij fuHkZj djrk gS rFkk nwjh c<+us ij blesa vR;f/kd deh vkrh gSA (B*) izsjf.kd izHkko -ca/k }kjk LFkkukUrfjr gksrk gSA (C) izsjf.kd izHkko -ca/k }kjk LFkkukUrfjr gksrk gSA (D*) izsjf.kd izHkko LFkk;h izHkko gksrk gSA 9.

In which of the following species, correct direction of inductive effect are shown ?

fuEu esa ls dkSulh lajpukvksa esa izsjf.kd izHkko dh fn'kk dks lgh iznf'kZr fd;k x;k gS\

(A*)

Sol.

(B)

(C*)

Case B has incorrect direction of I effect. (fodYi B esa I izHkko

(D*)

dks xyr fu:fir fd;k x;k gSA½

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PAGE NO.- 2

10.

Which of the following are + I group fuEu esa ls dkSulk@dkSulsa + I lewg gS@gSa





(A*)  NH 2



(C*)  CH2

(B) 3

11.

& (D)  N  N

1

On which C of CH3 – CH2 – CH2 – Br , the inductive effect is expected to be the maximum. 3

2

1

CH3 – CH2 – CH2 – Br

ds dkSuls dkcZu ij izsjf.kd izHkko vkisf{kd :i ls vf/kdre gksrk gsSA

Ans.

1

12.

How many groups show – effect? fuEu esa ls fdrus lewg – izHkko n'kkZrs

[General Organic Chemistry]

gSa\



–CH3 ,  NH3 , –OH, –O , –N(CH3)2 , –SO3H, –CHO, –Cl, –COO Ans.

6

13.

How many of the following molecules are polar?

[Ref. SM Sir] [M]

fuEu esa ls fdrus v.kq /kqzoh; gSa\

Ans.

(i) CO2 (ii) SO2 (iii) NO2 (iv) SOCl2 (v) COCl2 (vi) BeCl2(g) (vii) TeCl4 (ix) ClO2 06

Sol.

CO2

NO2

14.

O= C=O

= 0

;

sp

=O

 0

;

SO2

(viii) CCl4

SO2

SOCl2

COCl2

 0

;

BeCl2

TeCl4

 0

;

CCl4

ClO2

0

0

0

Cl – Be – Cl

= 0

0

How many of the following will exhibit Hydrogen bonding in water ?

[E] (CBO)

fuEu esa ls fdrus ty esa gkbMªkstu vkca/k n'kkZ;sxsa \ Ans. 15.

CH3CN, C6H5OH, D2O, H3PO3, SO3, CO2, F2, KF, CH3COOH, CH3OCH3. 10 Match the columnColumn-I (A) HCl < HF (B) PH3 < NH3 (C) H2 O < D2 O (D) H2 S < H2 O

(p) (q) (r) (s)

Column-II Intermolecular forces Dipole moment Boiling point Molar mass

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PAGE NO.- 3

feyku dhft,A dkWye -I

Ans. Sol.

(A)

HCl < HF

(p)

(B)

PH3 < NH3

(q)

(C)

H2 O < D2 O

(r)

dkW ye -II vUrjkvkf.od cy f}/kzqo vk?kw.kZ DoFkukad eksyj nzO;eku

(D) H2 S < H2 O (s) (A – p, q, r) ; (B – p, q, r) ; (C – p, q, r, s) ; (D –p,q,r) (A) In HF hydrogen bonding takes place so has higher b.p. and dipole moment is more than HCl due to more polarity of H — F bond. (B) In NH3 hydrogen bonding takes place so has higher b.p. which is absent in PH3. (C) In D2O dipole moment, H – bonding and b.p. is more than H2O as D is less electronegative than hydrogen. (D) In H2O hydrogen bonding is present. So has higher b.p. Also O — H bond is more polar. So H2O has higher dipole moment. (A) HF esa gkbMªkstu cU/ku izkIr gksrk gS vr% ;g mPp DoFkukad j[krk gS rFkk bldk HF ca/k dh vf/kd /kqzo.krk ds dkj.k HCl ls

vf/kd f}/kzqo vk?kw.kZ gksrk gSA (B) NH3 esa gkbMªkstu cU/ku izkIr gksrk gS vr% ;g mPp DoFkukad j[krk gS tksfd PH3 esa vuqifLFkr gSA (C) gkbMªkstu dh vis{kk D de fo|qr_.kh gksus ds dkj.k D2O esa] H2O dh vis{kk f}/kzqo vk?kw.kZ] H-ca/ku rFkk DoFkukad vf/ kd gksrk gSA (D) H2O esa gkbMªkstu cU/ku mifLFkr gksrk gSA vr% ;g mPp DoFkukad j[krk gS] H–O ca/k Hkh vf/kd /kqzoh; gksrk gSA vr% H2O mPp f}/kqzo vk?kw.kZ j[krk gSA

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PAGE NO.- 4

ChemINFO-4.2

POSITIVE CHARGE DELOCALISATION Positive Charge Delocalisation

Daily Self-Study Dosage for mastering Chemistry

1.

Positive charge is on conjugation or delocalisation when : (1) Adjacent atom should have negative charge e.g. (2)

  A  B  A  B

  NH  CH2  NH  CH2

Adjacent atom should have bond 

e.g. (3)

  X  Y  Z  X  Y  Z

  CH2  CH – CH2  CH2 – CH  CH2

Adjacent atom should have lone pair e.g.

   A  B  A  B

..





H2 N C H2  H2 N  CH2

Note : (i) If positive charge present on IInd period element which have complete octate then it is not delocalised .

;

;

;

(ii) If  bond and positive charge both are present on same atom of IInd period then it is delocalised . ;

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

In which of the following molecule positive charge is not in conjugation ?

(A)

17.

(C*)

(D*)

In which of the following molecule positive charge is delocalised ?

(A) 18.

(B)

(B)

(C)

(D*)

Which of the following molecule positive charge is not delocalised ? (A)

(B)

(C*)

(D)

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PAGE NO.- 5

19.

In which of the following molecule positive charge is not in conjugation ? (A)

20.

(B*)

(C)

(D)

(C*)

(D)

Which positive charge stabilised by resonance ? 

(A) CH2=CH– NH3

(B)

ChemINFO-4.2

POSITIVE CHARGE DELOCALISATION Positive Charge Delocalisation

Daily Self-Study Dosage for mastering Chemistry

1.

la;qXeu ds nkSjku /kukos'k dk foLFkkuhdj.k rc gksrk gS tc : (1) lehiLFk ijek.kq ij _.kkos'k gksrk gS % e.g. (2)

   X  Y – Z  X – Y  Z

  CH2  CH – CH2  CH2 – CH  CH2

lehiLFk ijek.kq ij ,dkdh ;qXe mifLFkr gksrk gS % e.g.

Note : (i) ;fn

  NH — CH2  NH  CH2

lehiLFk ijek.kq ij  cU/k mifLFkr gksrk gS % e.g.

(3)

  A — B  A  B

   A — B  A  B

..





H2 N C H2  H2 N  CH2

/kukRed vkos'k IInd vkoÙkZ RkRo] tks iw.kZ v"Bd j[krk gS ij mifLFkr gksrk gS rks ;g foLFkkuhd`r ugh gksrk gSA

;

(ii) ;fn  cU/k

;

;

o /kukRed vkos'k nksuks IInd vkoÙkZ ds leku ijek.kq ij mifLFkr gksrk gS rks ;g foLFkkuhd`r gksrs gSA ;

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

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PAGE NO.- 6

16.

fuEu es ls dkSuls v.kq esa /kukRed vkos'k la;qXeu es ugh gS \ (A)

17.

(B)

19.

(B)

(C)

(D*)

fuEu es ls dkSuls v.kq esa /kukRed vkos'k foLFkkuhd`r ugha gksrk gS \ (A)

(B)

(C*)

(D)

fuEu es ls fdl v.kq esa /kukRed vkos'k la;qXeu es ugh gS \ (A)

20.

(D*)

fuEu es ls fdl v.kq esa /kukRed vkos'k foLFkkuhd`r gksrk gS \

(A)

18.

(C*)

(B*)

(C)

(D)

(C)

(D)

fuEu es ls dkSulk /kukRed vkos'k vuqukn }kjk LFkk;h gksrk gS \ 

(A) CH2=CH– NH3

(B)

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PAGE NO.- 7

DPP No. # 33 (JEE-MAIN) Total Marks : 65

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

1.

(3 marks, 2 min.) (4 marks, 2 min.)

[45, 30] [20, 10]

Which of the following is not acceptable resonating structures of Buta-1, 2, 3-triene. fuEu esa ls dkSulh vuquknh lajpuk C;wVk-1, 2, 3-VªkbbZu ds fy, ekU; ugh gS \ [General Organic Chemistry] (A)

(B) CH2 = C = C = CH2

(C)

(D*)

Sol.

(A,B,C,) CH2 = C = C = CH2

2.

Which of the following is a conjugated system ?

fuEu esa ls dkSu ,d la;qXeh ra=k gS \ (A) CH2=C=C=CH2 (C) CH2=CH–CH=O 3.

(B) CH2=C=O (D*) All of these mijksDr

Resonance is not possible in : [General Organic Chemistry]

fuEu esa ls fdl essa vuqukn laHko ugha gS \ 

(A) CH2 = NH2

4.

(B*) CH3CH = C = CH2

(C)

(A)

(B*)

(C)

(D)

In which case the unshared pair (lone pair) of electrons is not delocalized. fuEu esa ls dkSulh lajpuk esa vlk¡f>r (unshared) ,dkdh bysDVªkWu ;qXe dk la;qXeh

(A)

(B*) H2C =

– CH3

rU=k esa foLFkkuhdj.k(delocalize)

[General Organic Chemistry]

Lone pair of electrons of H2C = H2 C =

6.

[GOC - EE]

dh vuquknh lajpuk dks iznf'kZr ugha djrh gS \

ugh gksrk gS \

Sol.

(D)

Which of the following does not represent the resonating structure of

fuEu esa ls dkSulh lajpuk ;kSfxd]

5.

lHkh

– CH3

(C) H2C = =

(D)

– CH3 is in sp2 hybrid orbital.

dk ,dkdh bysDVªkWu ;qXe sp2 ladfjr d{kd esa gSA

Which of the following statements is true about resonance. (A) In resonating structure hybridisation of atom will be change. (B*) Cannonical structures are imaginary (C) Cannonical structure explains all features of a molecule (D) In resonating structures position of nuclei change. 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.- 8

vuqukn ds fo"k; esa fuEu esa ls dkSulk dFku lgh gS \ (A) vuquknh lajpuk esa] ijek.kq dk ladj.k ifjofrZr gks tkrk gSA (B*) dSuksfudy lajpuk,sa dkYifud gksrh gSA (C) dSuksfudy lajpuk fdlh v.kq ds leLr y{k.kksa dh O;k[;k djrh gSA (D) vuquknh lajpukvksa esa] ukfHkd dh fLFkfr ifjofrZr gksrh gSA 

7.

CH2  CH  CH  CH2 I



CH2  CH  CH  CH2 II



III

Among these which are cannonical structures ? (A) and  (B) and  (C) and  

CH2  CH  CH  CH2 I



CH2  CH  CH  CH2 II

fuEu esa ls dkSulh lajpuk,sa dSuksfudy lajpuk,sa gS \ (A) rFkk  (B) rFkk 



CH2  CH  CH  CH2

(D*) all of these 



CH2  CH  CH  CH2

(C) rFkk 

8.

III

(D*) mi;qZDr

lHkh

[General Organic Chemistry] I II III The correct statement about the above structures is : (A) II is the minor contributor to the real hybrid. (B) III is most stable structure (C) I contributes more to the real hybrid than that of II. (D*) I and II are equal contributors and III is a minor contributor.

mijksDr lajpukvksa ds fy, lgh dFku gS % (A) II lajpuk dk ;ksxnku okLrfod ladfjr esa lcls de gSA (B) III lcls LFkk;h lajpuk gSA (C) okLrfod ladfjr lajpuk esa I dk ;ksxnku II ls vf/kd gSA (D*) I ,oa II dk ;ksxnku cjkcj rFkk III dk ;ksxnku de gSA 9.

Which of the following is not correctly orderd for resonance stability.

[General Organic Chemistry]

fuEu esa ls dkSuls fodYi vuquknh lajpuk ds LFkkf;Ro ds fy;s lgh ugha n'kkZ;s x;s gSA

Sol.

(A)

(II > I > III)

(B)

(I > III > II)

(C*)

(I > II)

(D)

(II = I)





CH3 — C  O  CH3 – C  O (I) (II) Stability II > I due to No. of  bonds.

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PAGE NO.- 9

10.

The least stable cannonical structure among these is :

mijksDr lajpukvksa esa ls dkSulh lajpuk U;wure LFkk;h dSuksfudy lajpuk gS % (A) I

11.

(B*) II

(C) III

(D) IV

The most stable resonating structure is :

[General Organic Chemistry]

lcls vf/kd LFkk;h vuquknh lajpuk gS %

12.

(A)

(B*)

(C)

(D)

Which of the following compound is not a resonance stablized ?

fuEu esa ls dkSulk ;kSfxd vuqukn }kjk LFkk;h ugha gS \

(A)

(B)

(C*)

(D)

13. In above carbanion electron is transfered in :

mijksDr dkcZ_.kk;u Lih'kht ds fdl d{kd esa bysDVªkWu dk LFkkukUrj.k gksrk gS \ (A) p-orbital (A) p-d{kd 14.

Sol.

15.

(B*) d-orbital (B*) d-d{kd

(C) s-orbital (C) s-d{kd

Which of the following molecule donot exhibit resonance ? fuEu esa ls dkSulk v.kq vuqukn ugha n'kkZrk gS \ (Made by RGP SIR on April2014)(Resonance(O)) (A*) H2C=C=CH2 (B) H2C=C=C=CH2 (C) H2C=C=O (D) NC-HC=CH-CN In H2C=C=CH2 multiple bonds are not in conjugation. H2C=C=CH2 esa cgqy ca/k esa la;qXeu ugha ik;k tkrk gSA Which of the following pairs does not represent resonating structures ? fuEu esa ls dkSulk ;qXe vuquknh lajpuk ugha n'kkZrk gS \ (Made by RGP SIR on April2014) (Resonance(O)) 









(A) CH3– C  N – O and ¼rFkk½ CH3– C = N – O (B) CH2=N









(C) CH2= N = N and ¼rFkk½ C H2 – N  N Sol.

(D) f-orbital (D) f-d{kd

O O





and ¼rFkk½ CH2–N

O O 

O

(D*) C6H5–C



NH3

and ¼rFkk½ C6H5–C

O–H NH2

In resonating structures position of atoms should not change.

vuquknh lajpukvksa esa ijek.kqvksa dh fLFkfr ifjofrZr ugha gksrh 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.- 10

ChemINFO-4.3

MESOMERIC EFFECT

Daily Self-Study Dosage for mastering Chemistry

Mesomeric effect

Mesomeric effect : Mesomeric effect is defined as permanent effect of  electron shifting from multiple bond to atom or from multiple bond to single bond or from lone pair to single bond. This effect mainly operates in conjugated system of double bond. So that this effect is also known as conjugate effect.

Types of Mesomeric effects : (a) Positive Mesomeric effect (+m effect) : When the group donates electron to the conjugated system it shows + M effect.

Relative order of +m groups (usually followed) : – O > –NH2 > –NHR > –NR2 > –OH > –OR > –NHCOR > –OCOR > –Ph > –F > –Cl > –Br > –I > –NO

(b) Negative Mesomeric effect (–m effect) : When the group withdraws electron from the conjugated system, it shows – M effect

Relative order of –m groups (usually followed) :

Note : 1. Identification of +m & –m groups : If the first atom of the group has lone pair or negative charge shows

2.

+m effect. If the group has vacant p-orbital or vacant d-orbital on first atom and also a multiple bonded group in which second atom is more electronegative than the first then it shows –m effect. The following group can show both +m & –m : –Cl, –Br, –SR, –NO, –NC

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

17.

18.

19.

20.

Identify which of the following does not show + m effect. (A) –NHR (B) –OR (C) –F

(D*) –COOH

Identify which of the following cannot show – m effect. (A) –NO2 (B) –CONH2 (C*) –OCOR

(D) –COO—

Decreasing + m effet of given group is : (I) – NR2 (II) –OCOR (A) I > III > IV > II (C) III > I > II > IV

(III) –NHCOR (B*) I > III > II > IV (D) II > I > IV > III

(IV) –Ph

Decreasing – m effet of given group is : (I) – COOH (II) –NO2 (A) I > III > IV > II (C*) II > III > I > IV

(III) –CHO (B) I > III > II > IV (D) II > I > IV > III

(IV) –CONH2

Which of the following can not show both ± m effect. (A) –Cl (B*) –F (C) –NO

(D) –NC

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PAGE NO.- 11

ChemINFO-4.3

felksesfjd izHkko

Daily Self-Study Dosage for mastering Chemistry

felkse sf jd izH kko

felksesfjd izHkko % felksesfjd izHkko ,d LFkk;h izHkko gSA ftlesa bysDVªkWuksa dk xeu cgqycU/k ls ijek.kq ij ;k cgqycU/k ls ,dy cU/k ij ;k ,dkdh ;qXe ls ,dy cU/k esa gksrk gSA ;g izHkko eq[;r% f}cU/k ds la;Xq eh rU=k esa mifLFkr gksrk gSA blfy, ;g izHkko la;Xq eh izHkko Hkh dgykrk gSA

felksesfjd izHkko ds izdkj : (a) /kukRed felksesfjd izHkko (+ m izHkko) : tc lewg la;qXeh ra=k dks bysDVªkWu nku djrk gS rks ;g + M izHkko n'kkZrk gSA Relative order of +M groups (usually followed) : + m lewgksa dk vkisf{kd Øe % – O > –NH2 > –NHR > –NR2 > –OH > –OR > –NHCOR > –OCOR > –Ph > –F > –Cl > –Br > –I > –NO (b)

_.kkRed felksesfjd izHkko (– m izHkko) : os lewg ftudh l;qXeh rU=k ls bysDVªkWu dks viuh vksj [khapus dh izo`fr gksrh gS] – M izHkko n'kkZrs gSA – m lewgksa dk vkisf{kd Øe % –NO2 > –CHO >

— – C=O > –C–O–C–R > –C–O–R > –COOH > –CONH2 > –C–O — || || || ||

O

O

O

O

uksV : 1.

2.

+m vkSj –m lewg

dh igpku : ;fn lewg ds izFke ijek.kq ij ,dkdh bysDVªkWu ;qXe ;k _.kkRed vkos'k mifLFkr gks rks og +m izHkko iznf'kZr djrk gSA ;fn lewg ds izFke ijek.kq ij fjDr p-d{kd ;k fjDr d-d{kd mifLFkr gks vkSj cgqcaf/kr lewg ftlesa f}rh; ijek.kq dh fo|qr_.krk izFke ijek.kq ls vf/kd gks rks og –m izHkko iznf'kZr djrk gSA fuEu lewg +m o –m nksuks izHkko n'kkZrs gS : –Cl, –Br, –SR, –NO, –NC

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

fuEu esa ls dkSulk lewg + m izHkko iznf'kZr ugha djrk gS \ (A) –NHR

17.

(B) –CONH2

(C*) –OCOR

(D) –COO—

(II) –OCOR

(III) –NHCOR (B*) I > III > II > IV (D) II > I > IV > III

(IV) –Ph

uhps fn;s x;s fuEUkfyf[kr lewgksa dh – m izHkko dk ?kVrk gqvk Øe gS % (I) – COOH (A) I > III > IV > II (C*) II > III > I > IV

20.

(D*) –COOH

uhps fn;s x;s fuEUkfyf[kr lewgksa dh + m izHkko dk ?kVrk gqvk Øe gS % (I) – NR2 (A) I > III > IV > II (C) III > I > II > IV

19.

(C) –F

fuEu esa ls dkSulk lewg – m izHkko iznf'kZr ugha djrk gS \ (A) –NO2

18

(B) –OR

(II) –NO2

(III) –CHO (B) I > III > II > IV (D) II > I > IV > III

(IV) –CONH2

fuEu esa ls dkSulk lewg ± m nksuks izHkko ugh n'kkZrk gS \ (A) –Cl

(B*) –F

(C) –NO

(D) –NC

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PAGE NO.- 12

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 34

This DPP is to be discussed in the week (05.10.2015 to 10.10.2015) 1. 2. 3. 4.

Course of the week as per plan : Stability of Resonating structure, Mesomeric effect. Course covered till previous week : Inductive Effect, Resonance, Resonance (Drawing Structure). Target of the current week : Stability of Resonating structure, Mesomeric effect. DPP Syllabus :

DPP No. # 34 (JEE-ADVANCED) Total Marks : 79

Max. Time : 48 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.10 Integer type Questions ('–1' negative marking) Q.11 to Q.14 Match the Following (no negative marking) Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

(3 (4 (4 (8 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.) marks, 2 min.)

[15, [20, [16, [08, [20,

10] 10] 12] 06] 10]

ANSWER KEY DPP No. # 34 (JEE-ADVANCED) 1.

(C)

2.

8*.

(ABCD) 9.*

15.

(A) – p,q,t ; (B) – p,q,s (C) – p (D) – p,q,r,s

19.

(A)

1.

+ m and + I both effects are shown by : fuEu esalsd kSulh Lih'kht + m rFkk + I nksuksaizHkko

20.

(A) 2.

4.

3.

(C)

4.

(AC)

10.*

(ACD) 11.

(B)

5.

(C)

6.*

(ABD) 7*.

(ACD)

3

12.

4

13.

6

14.

2

16.

(A)

17.

(C)

18.

(B)

(D)

iznf'kZr d jrh gS%

(B)

(C*)

(D) – C (CH3)3

(C)

(D*)

Identify which of the following shows – m effect ? fuEu esalsd kSulk lewg –m izHkko iznf'kZr d jrk gS\ (A) –NH–CH3

3.

(D)

(B) –NH

Decreasing + m power of given group is : uhpsfn;sx;sfuEUkfy f[kr lewgksad h + m {kerk d k ?kVrk gqv k Ø e gS% (I) –O –CH3 (II) –F (III) –CH2 (A) I > III > IV > II (B) III > II > I > IV (C*) III > I > II > IV

[General Organic Chemistry] (IV) –Cl (D) II > I > IV > III

The correct decreasing order of stability following resonating structure :

uhpsnh x;h vuquknh lajpuk d sLFkkf;Ro d k lgh ?kVrk gqv k Ø e gS%

Sol.

(A) I > II > III > IV (B*) I > II > IV > III (C) I > IV > III > IV (D) I > IV > II > III Non polar > Complete octate > opposite charge > same charge (-ve and lone pair) v/kqzfo; > v"Vd iw.kZ > foifjr vkos'k > leku vkos'k (–ve rFkk ,d kd h by sDVªkWu ;qXe). 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

5.

More stable resonating structure of the given cation is :

uhpsfn;sx;s/kuk;u d h lokZf/kd LFkk;h vuquknh lajpuk gS% (Made by RGP SIR on April2014) (Resonance(O) 

CH2

:

:OMe





(A)

(B)

(C*)



OMe

OMe

(D) OMe

OMe



Sol.

Follow the rules for stability of resonating structures (structure with more number of -bonds is more stable) vuquknhlajpukd sLFkkf;Ro d sfu;e d kvuql j.k d hft , ¼vf/kd re -ca/kj[kusoky hlajpukvf/kd LFkk;hgkrhgSA½

6.*

Which of the following carbanions is not resonance stabilized?

fuEu esalsd kSulk d kcZ_ .kk;u vuqukn }kjk LFkk;hd `r ugh gksrk gS\ (A*) O

Sol.

In S

S

7*.

O

(B*) H—N

N—H

(C) S

(By RGP Sir, May, 2014)

S

(D*)

S carbanion present in conjugation with vacant d-orbital of sulpher atom.

S

d kcZ_ .kk;u esalYQ j ijek.kqd sfjDr d-d {kd d h mifLFkfr d sd kj.k la;qXeu ik;k t krk gSA

In which case(s) the second structure is more stable :

d kSulsfod Yiksaesaf}rh; lajpuk vf/kd LFkk;hgS% (A*)

,

(B)

,

(C*)

,

(D*)

,

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PAGE NO.- 2

8*.

9.*

The correct number of p-electrons have been mentained in : fd lesalgh p-by sDVªkWuksad k vad u fd ;k x;k gS:

(A*)

; (4)

(B*)

(C*)

; (3)

(D*)

; (6)

; (6)

Which of the following molecule shows cross conjugation. (Made by RGP SIR on April2014) (Resonance(O)

fuEu esalsd kSulk v.kqØ kWl la;qXeu n'kkZrk gS\

Sol.

(A*) Ph – C – Ph || O

(B)

(C*)

(D)

Ph – C – Ph and || O Ph – C – Ph || O

10.*

shows cross conjugation.

rFkk

Ø kWl la;qXeu n'kkZrsgSA

Identify the correct statements

(A*) All C – C bonds in

(Made by SHK SIR ) (Resonance(O)

are equal.

(C*) All C – O bonds in

are equal.

(B) All C – C bonds in CH2 = CH – CH = CH2 are equal.

(D*) All C – O bond in

are equal.

fuEu esalsd kSulk d Fku lgh gS\ (A*)

(C*) Sol. Sol.

esalHkh C – C ca/k leku gSA

esalHkh C–O ca/kleku gSA

(B) CH2 = CH – CH = CH2 es alHkh C–C ca/k leku

(D*)

gSA

esalHkh C–O ca/k leku gSA

(i), (iii) & (iv) have equal bond length due to equivalent resonating structure. In (ii) both resonating structure are different. (i), (iii) rFkk (iv) d h ca /k y EckbZ;k¡lerqY; vuquknhlajpuk d sd kj.k leku gSA (ii) es anksuksavuquknh lajpuk,safHkUu

gSA

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PAGE NO.- 3

11.

Find the total + m groups attached to the benzene ring in the given compound?

(GOC-I)

uhpsfn;sx;s;kSfxd esafd rus+ m izHkko n'kkZusoky slewg csUt hu oy ; lst qM +sgq, gSa\

Ans. Sol.

3 –O–COCH3 , –OCH3, –N(CH3)2

12.

In how many compound lone pair of nitrogen is involved in resonance ?

fuEu esalsfd rus;kSfxd ksad sukbVªkst u d k ,d kd h ;qXe vuqukn esaHkkx y srk gS\

(1)

(2)

(3)

(5)

(6)

(7)

(4)

Ans. 13.

4 Number of p electrons in resonance in the following structure is. (Made by SHK SIR ) (Resonance(O) fuEufy f[kr lajpuk esavuqukn esalfEefy r  by sDVªkWuksad h la[;k fd ruh gksxh\

Ans. Sol.

6 It has only 6 p-electron in conjugation. blesala;qXeu esad soy 6 -by sDVªkWu mifLFkr

gksrsgSA

14.

In the given compound how many lone pair of e– are delocalised ? mijksDr ;kSfxd esafd rus,d kd h e– ;qXe d k foLFkkuhd j.k gksrk gS\

Ans.

2

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PAGE NO.- 4

15.

Match the column : Column-I

Column-II

(A) (B) H2N–CH=CH2

(p) (q)

Resonance possible Even number of p-electrons

(C)

(r)

localized lone pair of e–.

(D)

(s)

Delocalized lone pair of e–.

(t)

2 e– in p orbitals

l gh fey ku d hft ,& d kWy e-I

d kWy e-II

(A)

(p)

vuq ukn lEHko

(B) H2N–CH=CH2

(q)

p-by s DVªkWuks ad hlela[;k

(C)

(r)

by sDVªkWu d k LFkkuhd `r ,d kd h by sDVªkWu ;qXe

(D)

(s)

by sDVªkWu d k foLFkkuhd `r ,d kd h by sDVªkWu ;qXe

(t)

p d {kd

Ans.

(A) – p,q,t ; (B) – p,q,s (C) – p (D) – p,q,r,s

Sol.

(A)

(B)

(C)

(D)

Lecutre No.

Sub-topic(s) Nam e (No. of Lectures)

L60

Stability of Resonating structure

L61

Mesomeric effect (Nature & strength)

Hom e Work Sheet Sec : C Ex.1 (S) Sec : C Ex.1 (O) Ex.2 (O)

6 to 11

Ex.2 (I) Ex.2 (M)

8, 9 13 to 15

Ex.1 (S)

Sec : D

Ex.1 (O) Ex.2 (O) Ex.2 (I)

Sec : D 12 to 14 10 to 14 16 to 18, 20 to 22

Ex.2 (M)

Hom e Work NCERT Pg.N XI Th. XII 346 XI 362 Prob.

esa2 by sDVªkWu

Chem . INFO

W.S.

Hand out

Type of reagent

XII Th.

XI

346 394 to 395

XII XI Prob. XII

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.- 5

ChemINFO-5.1

HYDROCARBON Type of Reagent

Daily Self-Study Dosage for mastering Chemistry

An organic reaction can be represented as solvent

 Product Reactant (substrate) + Reagent 

Types of Bond dissociation : All reactions are initiated with bond dissociation. There are two types of bond dissociation. (a) Homolytic bond dissociation : A bond dissociation in which a bond pair electron is equally distributed to the bonding atoms. e.g.

A–B  A• + B•

a homolytic bond dissociation generates radicals. (b) Hetrolytic bond dissociation : A bond dissociation in which a bond pair electron is shifted to one atom only. e.g.

A–B  A + B

A hetrolytic bond dissociation always generate a cation and an anion.

Types of Reagents : A reagent generates three type of attacking species. Which are nucleophile, electrophile and radical. (a) Electrophiles : Electrophiles are electron deficient species. 

e.g.









Cl , Br , NO2 , CH3 (positively charged species), PCl5, SO2, SO3, BF3(species with vacant orbital

H

at central atom, carbenes) etc. (b) Nucleophiles : It is the electron rich species having atleast one unshared pair of electron. It can be neutral or negativetely charged it is always a lewis base. e.g. CN–, OH–, Br – , I – , NH3 , H2O etc. (c) Free radicals : It is electron deficient species with seven electrons around an atom. e.g.

C2H5, C2H5O, CH3COO, X etc.

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

17.

Electrophiles are : (A*) Electron deficent species

(B) having atleast one pair of electron

(C) Electron rich species

(D) negatively charged species

Which of the following is an electrophilic reagent ?

(B) OH–

(A) H2O 18.

(B*) SH–

(C) CH3+

(D) AlCl3

Which one of the following has maximum nucleophilicity ? –

(B) NH2–

(A*) CH3 20.

(D) None

Which of the following is a nucleophile ? (A) BF3

19.

(C*) NO2+

(C) OH–

(D) F

–

Which of the followings are free radicals ? 

(A) CH3 – CH2



(B) Cl



(C) CH3O

(D*) All of these

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PAGE NO.- 6

ChemINFO-5.1

HYDROCARBON v fHkd eZd ksa d s izd kj

Daily Self-Study Dosage for mastering Chemistry

,d d kcZfud vfHkfØ ;k d ksfuEukuql kj iznf'kZr fd ;k t k ld rk gS% foy k;d vfHkd kjd (inkFkZ) + vfHkd eZd  mRikn caèk fo[k.Mu d s izd kj % lHkh vfHkfØ ;k caèk fo[k.Mu gksusij gh izkjEHk gksrh gSA nksizd kj d scaèk fo[k.Mu gksrsgS%

l eka'k caèk fo[k.Mu : bl izd kj d scaèk fo[k.Mu esacaèk ;qXe by sDVªkWu caèk fo[k.Mu d si'pkr~cafèkr ijek.kqv ksa ij cjkcj : i lsforfjr gkst krsgSA (a)

mnk- A–B  A• + B• leka'k caèk fo[k.Mu d s d kj.k ewy d cursgSA fo"keka'k caèk fo[k.Mu : bl izd kj d scaèk fo[k.Mu esacaèk ;qXe by sDVªkWu caèk fo[k.Mu d si'pkr~d soy ,d ijek.kq ij gh forfjr gksrsgSA (b)

mnk- A–B  A + B fo"keka'k caèk fo[k.Mu d sd kj.k lnSo èkuk;u rFkk _ .kk;u cursgSA v fHkd eZd ksa d s izd kj % vfHkd eZd rhu izd kj d h vkØ e.kd kjh Lih'kht cukrsgS] t ksukfHkd Lusgh] by sDVªkWuLusgh rFkk ewy d gSA (a) by s DVªkWuLusgh : by sDVªkWuLusgh by sDVªkWu 

mnk-







U;wu Lih'kht gksrh gSA



f'kr Cl , Br , NO2 , CH3 (èkukos

H

Lih'kht ), PCl5, SO2, SO3, BF3(d s fUæ; ijek.kqij fjDr d {kd j[kusokyh

Lih'kht ]d kchZ u) bR;kfnA ukfHkd Lusgh : ;g ,d by sDVªkWu èkuh Lih'kht gSft uesad e lsd e ,d vlgHkkft r by sDVªkWu ;qXe ik;k t krk gSA ;g mnklhu vFkok +_ .kkosf'kr gksld rsgSA ;g lnSo ,d y wbZl {kkj Hkh gksrsgSA mnk- CN–, OH–, Br – , I – , NH3 , H2O bR;kfnA (b)

(c)

eqDr ewy d : ;g ,d by sDVªkWu U;wu Lih'kht gS ft ud s ijek.kq esa lkr by sDVªkWu gksrs gSA mnk- CH3 C2H5, C2H5O, CH3COO, X bR;kfnA

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

17.

by sDVªkuLusgh gSa% (A*) by s DVªkWu U;wu Lih'kht (C) by s DVªkWu /kuhLih'kht

(B) OH–

(B*) SH–

(D) d ks bZugha

(C) CH3+

(D) AlCl3

(C) OH–

(D) F

fuEu esad kSulklokZf/kd ukfHkd Lusfgrkj[krkgS\ –

(B) NH2–

(A*) CH3 20.

(C*) NO2+

fuEu esalsd kSulk ,d ukfHkd Lusgh gS\ (A) BF3

19.

lsd e ,d ,d kd h by sDVªkWu ;qXe mifLFkr gksa (D) _ .kkRed vkos f'kr Lih'kht

fuEu esalsd kSulk ,d by sDVªkWuLusgh vfHkd eZd gS? (A) H2O

18.

(B) d e

–

fuEu esalsd kSu eqDr ewy d gS\ 

(A) CH3 – CH2



(B) Cl



(C) CH3O

(D*) mijks Dr

lHkh

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.- 7

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 35

This DPP is to be discussed in the week (12.10.2015 to 17.10.2015) 1.

Course of the week as per plan : Mesomeric effect and SIR, Hyperconjugation, Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

2. 3.

Course covered till previous week : Stability of Resonating structure, Mesomeric effect. Target of the current week : Mesomeric effect and SIR, Hyperconjugation, Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

4.

DPP Syllabus : Mesomeric effect and SIR, Hyperconjugation, Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

DPP No. # 35 (JEE-MAIN) Total Marks : 65

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

(3 marks, 2 min.) (4 marks, 2 min.)

[45, 30] [20, 10]

ANSWER KEY DPP No. # 35 (JEE-MAIN) 1.

(D)

2.

(B)

3.

(C)

4.

(A)

5.

(C)

6.

(D)

7.

(D)

8.

(B)

9.

(B)

10.

(A)

11.

(C)

12.

(A)

13.

(D)

14.

(C)

15.

(C)

16.

(A)

17.

(C)

18.

(C)

19.

(C)

20.

(B)

1.

Hyperconjugation phenomenon is possible in :

fuEu esalsfd l v.kqesavfrla;qXeu lEHko gS\

(A)

(B) CH2 = CH2

Sol. Sol.

(C) C6H5 – CH = CH2 (D*) CH3 – CH2 – CH = CH2 CH3 – CH2 – CH = CH2 has two -hydrogen for hyperconjugation. CH3 – CH2 – CH = CH2 vfrla ;qXeu d sfy , nks-gkbMªkst u ijek.kqmifLFkr gSA

2.

Hyperconjugation is possible in which of the following species ?

fuEu izt kfr;ksaesalsfd lesavfrla;qXeu lEHko gSa\

(A)

(B*) C6H5—CH3

(C) CH2=CH2

CH3 |  (D) CH3 — C — C H2 | CH3

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3.

Number of  electrons in conjugation for these compounds ,

,

and

will be respectively :

v|ksfy f[kr ;kSfxd ksad sla;qXeu esaHkkx y susoky sby sDVªkWuksad h la[;k Ø e'k%gSa\ ,

(A) 8, 6, 6, 6 4.

rFkk

,

(B) 6, 4, 6, 6

(C*) 6, 6, 6, 6

(D) 6, 6, 8, 6

Which of the following alkenes will show maximum number of hyperconjugation forms ?

fuEu ,Yd huksaesad kSu lokZf/kd vfrla;qXeu : i n'kkZ;sxk \ (A*)

5.

(B) CH3–CH=CH–CH3

(C)

(D)

Hyperconjugation is not present in :

[SHK Sir 2012]

fuEu esalsfd lesavfrla;qXeu lEHko ughagS%

(A) Sol.

6.

(B)

(C*)

C has no sp3 carbon and H for hyperconjugation. fod Yi C esa];kSfxd esavfrla;qXeu d sfy , H rFkk sp3

(D)

d kcZu mifLFkr ughagSA

Which of the following species will not show hyperconjugation :

[General Organic Chemistry]

fuEu esalsd kSulh lajpuk vfrla;qXeu çnf'kZr ughad jsxh % (A) C6H5—CH3 7.

(B)

(C)

(D*)

(III)

(IV)

(C) II > IV > III > I

(D*) IV > III > II > I

The correct stability order of following is :

fuEu d sLFkkf;Ro d k lgh Ø e gS%

(I)

Sol.

gy -

(II)

(A) I > II > III > IV (B) III > IV > II > I Stability  resonance  Hyperconjugation length of conjugation is equal in all I, II, III & IV. LFkkf;Ro  vuqukn  vfrla ;qXeu lHkh I, II, III o IV esala;qXeu d h y EckbZleku gSA

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PAGE NO.- 2

8. Among these compounds, the correct order of resonance energy is :

fuEu ;kSfxd ksd h vuquknh Å t kZd k lgh Ø e d kSulk gS% (A) > >  9.

(B*) > > 

(C) > > 

(D) > > 

The correct decreasing order of electron density in aromatic ring of following compounds is :

fuEufy f[kr ;kSfxd kasesa,jkseSfVd oy ; ij by sDVªkWu ?kuRo d k lgh ?kVrk gqv k Ø e gksxk %

(I)

Sol.

10.

Sol.

(II)

(III)

(IV)

(A) (II) > (III) > (IV) > (I) (B*) (III) > (II) > (IV) > (I) (C) (IV) > (I) > (III) > (II) (D) (III) > (II) > (I) > (IV) + m group increases electron density and – m group decreases electron density in aromatic ring. ,sjkseSfVd oy ; esa+ m lewg by sDVªkWu ?kuRo c<+k nsrk gSrFkk – m lewg by sDVªkWu ?kuRo esad eh d j nsrk gSA Which of the following molecule has longest C=C bond length ? fuEu esalsfd l v.kqesaC=C ca/k y EckbZlokZf/kd gS\ (A*) CH3–CH=CH–CH=CH–CH3 (B) CH2=CH–CH=CH2 (C) CH3–CH=CH–CH3 (D) CH2=CH2 CH3 — CH = CH — CH = CH — CH3 (A) reso + H.C.

[General Organic Chemistry]

bond order as, bond lenght  Sol.

CH3 — CH = CH — CH = CH — CH3 (A) v uq ukn + v frl a;qXeu

ca/k Ø e , ca/k y EckbZ NH2

NH2

NH2

CH2=NH

11.

IV CHO III II I Among these compounds the correct order of C–N bond length is : mijksDr ;kSfxd kasd slUnHkZesaC–N ca/k d h y EckbZd k lgh Ø e d kSulk gS% (A) IV > I > II > III (B) III > I > II > IV (C*) III > II > I > IV

12.

(D) III > I > IV > II

The most stable cannonical structure of this molecule is :

uhpsfn;sx;sv.kqd h lokZf/kd LFkk;h d Suksfud y lajpuk d kSulh gS\ O `



O (A*)



O (B) single bond

(D) All structures are equal contribution ¼mijks Dr



O

(C)

lajpuk,sacjkcj ;ksxnku iznku d jrh gS½

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PAGE NO.- 3

O Sol.

13.

O Both ring are aromatic. ¼nks uksaoy ;



Which of the following has shortest C-Cl bond ? fuEu esalsfd l v.kqesaC-Cl ca/k lclsNksVk gS\ (A) CH3–Cl

,sjkseSfVd gSa½

(B) CH2=CH–l Cl

(C) CH2=CH–CH=CH–Cl

(D*)

Sol.

As conjugation in the molecule increases partial double bond character between C-Cl increases and bond length decreases. t Sl sgh fd lh v.kqesala;qXeu c<+rk gS]rksC-Cl ca/k d se/; vkaf'kd f}ca/k vfHky {k.k c<+rk gSrFkk ca/k y EckbZ?kVrh gSA

14.

Total number of hyperconjugative hydrogen atom in the given compound is: (By RGP Sir, May, 2014)

uhpsfn;sx;s;kSfxd esad qy fd rusvfrla;qXeh gkbMªkst u ijek.kqmifLFkr gS\

(A*) 6

15.

(B) 5

(C) 7

(D) 8

How many alkenes, from followings are more stable than

fuEu esalsfd ruh ,Yd husa

(A) 2 Lecutre No.

Sub-topic(s) Nam e (No. of Lectures)

L62

Hyperconjugation

L63

SIR + Aromaticity

L64

Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

(By RGP Sir, May, 2014)

d h vis{kk vf/kd LFkk;h gS

(B) 3

(C*) 4

Hom e Work Sheet Sec : F Ex.1 (S) Sec : F Ex.1 (O) 15 to 21, 23 Ex.2 (O) 15 to 18 Ex.2 (I) 23, 24 Ex.2 (M) Sec : E, G Ex.1 (S) Sec : E, G Ex.1 (O) 27, 30 Ex.2 (O) 19 to 21 Ex.2 (I) 25 to 27 Ex.2 (M) Sec : H Ex.1 (S) Sec : H Ex.1 (O) Ex.2 (O)

22, 24 to 26, 31 to 34

Ex.2 (I) Ex.2 (M)

22, 23 29 to 31

Hom e Work NCERT Pg.N 347 XI Th. XII 348 XI Prob. XII 391 XI Th. XII XI Prob. XII XI Th. XII 348 264 XI 363 Prob. 397

(D) 5 Chem . INFO

W.S.

Hydrocarbon

XII

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PAGE NO.- 4

ChemINFO-5.2

HYDROCARBON Type of Reagent

Daily Self-Study Dosage for mastering Chemistry

Generally an organic reaction is basically of three types (a) Substitution reaction : In a substitution reaction a group is replaced by other group. R–Y + Z  R–Z + Y (b) Addition reaction : In addition reaction one bond is broken and two new  bonds are formed.

| | YZ    C  C  | | Y Z (c) Elimination reaction : In an elimination reaction two atoms or groups (YZ) are removed from the substrate and generally resulting into formation of  bond.

| | CC | | Y Z

Elimination  YZ

 

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

Substitution reactions involve : (A*) Cleavage of a –bond and formation of a new –bond (B) Cleavage of two –bond and formation of a new –bond (C) Cleavage of a –bond and formation of two new –bond (D) None of these

17.

Which of the following reaction is a substitution reaction ? Ni / H2  CH3–CH3 (A) CH2 = CH2  

  (C*) CH3 – I + OH  CH3 OH  I

(B)

CH 2– CH2 Br

Br

Zn   CH2 = CH2 + ZnBr2

(D) CH3 – CHO

KCN    H

H CH3 – C –OH CN

18.

Addition reactions involve (A) Cleavage of a -bond and formation of a new -bond (B) Cleavage of two -bond and formation of a new -bond (C*) Cleavage of a -bond and formation of two new -bond (D) None of these

19.

Which of the following reaction is an elimination reaction ? PCl 5  CH3 – CH2 – CH2 – Cl (A) CH3 – CH2 – CH2 – OH  HCl (B) CH3 – CH = CH2  

(C*)

(D)

20.

Alc . KOH   CH3 – CH = CH2

CH OH

3   

The given reaction is an example of CH3–CH2–CHO + HCN  (A) Elimination reaction (C) Substitution reaction

(B*) Addition reaction (D) None of these 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.- 5

ChemINFO-5.2

HYDROCARBON Type of Reagent

Daily Self-Study Dosage for mastering Chemistry

lkekU;r;k,d d kcZfud vfHkfØ ;krhu izd kj d hgksrhgSA (a) iz frLFkkiu v fHkfØ ;k: izfrLFkkiu vfHkfØ ;k esa],d lewg d k nwl jslewg }kjk izfrLFkkiu gksrk gSaA R–Y + Z  R–Z + Y

(b) ;ks xkRed v fHkfØ ;k : ;ksxkRed vfHkfØ ;k esa,d caèk VwVrk gSrFkk nksu;s caèk cursgSA | | YZ    C  C  | | Y Z

(c) foy ks iu v fHkfØ ;k: foy ksiu vfHkfØ ;k esa]vfHkd kjd lsnksijek.kq;k lewg (YZ) fu"d kflr gksrsgSA ft ld s

ifj.kkeLo: i lkekU;r;k caèkcurkgSA | | CC | | Y Z

E lim ination  YZ

   

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

izfrLFkkiu vfHkfØ ;kesagksrkgS: (A*) –ca /k d k fo[k.Mu ¼fony u½ rFkk u;s–ca/k d k fuekZ.k (B) nks–ca /k d k fo[k.Mu ¼fony u½ rFkk ,d u;s–ca/k d k fuekZ.k (C) ,d –ca /k d k fo[k.Mu¼fony u½ rFkk nksu;s–ca/k d k fuekZ.k (D) bues alsd ksbZugha

17.

fuEu esalsd kSulhvfHkfØ ;kizfrLFkkiu vfHkfØ ;kgS? Ni / H2  CH3–CH3 (A) CH2 = CH2  

  (C*) CH3 – I + OH  CH3 OH  I

(B)

CH 2– CH2 Br

Br

Zn   CH2 = CH2 + ZnBr2

(D) CH3 – CHO

KCN    H

H CH3 – C –OH CN

18.

;ksxkRed vfHkfØ ;kesagksrkgS% (A) -ca /k d k fo[k.Mu ¼fony u½ rFkk u;s-ca/k d k fuekZ.k (B) nks-ca /k d k fo[k.Mu ¼fony u½ rFkk u;s-ca/k d k fuekZ.k (C*) -ca /k d k fo[k.Mu ¼fony u½ rFkk nksu;s-ca/k d k fuekZ.k (D) bues alsd ksbZughaA

19.

fuEu esalsd kSulhvfHkfØ ;kfoy ksiu vfHkfØ ;kgS\ PCl 5 HCl  CH3–CH2–CH2–Cl (B) CH3–CH=CH2  (A) CH3–CH2–CH2–OH  

(C*)

20.

Alc . KOH   CH3–CH=CH2

(D)

CH OH

3   

v|kfsyf[kr vfHkfØ ;kfuEu es alsfd l vfHkfØ ;kd kmnkgj.kgS%CH3–CH2–CHO + HCN  (A) foyks iu vfHkfØ ;k (C) iz frLFkkiu vfHkfØ ;k

(B*) ;ks xkRed vfHkfØ ;k (D) bues alsd ksbZugha 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.- 6

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 36

This DPP is to be discussed in the week (19.10.2015 to 24.10.2015) 1. 2.

Course of the week as per plan : Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring), Aromaticity, Carbanion and It’s Stability. Course covered till previous week : Mesomeric effect and SIR, Hyperconjugation, Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

3. 4.

Target of the current week : Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring), Aromaticity, Carbanion and It’s Stability. DPP Syllabus :

DPP No. # 36 (JEE-ADVANCED) Total Marks : 79

Max. Time : 48 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.5 Multiple choice objective ('–1' negative marking) Q.6 to Q.10 Integer type Questions ('–1' negative marking) Q.11 to Q.14 Match the Following (no negative marking) Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

(3 (4 (4 (8 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.) marks, 2 min.)

[15, [20, [16, [08, [20,

10] 10] 12] 06] 10]

ANSWER KEY DPP No. # 36 (JEE-ADVANCED) 1.

(C)

2.

8.*

(ACD) 9.*

(A)

3.

(C)

4.

(D)

5.

(C)

6.*

(BCD) 7.*

(AC)

(AB)

11.

3

12.

5

13.

8

14.

8

15.

(A - p,s) ; (B - p,s) ; (C - p,s) ; (D - r)

16.

(C)

17.

(C)

18.

(B)

19.

(D)

20.

(D)

1.

The least stable resonating strucutre is

(ACD) 10.*

lclsd e LFkk;h vuquknh lajpuk gS%

2.

(A)

(B)

(C*)

(D)

The correct decreasing order of electron density on pyridine nucleus :

fifjMhu ukfHkd ij by sDVªkWu ?kuRo d k ?kVrk gqv k Ø e gS%

(A*) II > III > I > IV

(B) II > I > III > IV

(C) II > IV > III > I

(D) II > IV > I > III

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PAGE NO.- 1

3. The most stable canonical structure of this molecule is

bl v.kqd h lclsvfèkd LFkk;h vuquknh lajpuk gS% (A)

(B)

(C*)

(D) All are equally stable

lHkhleku : i lsLFkk;hgSA 4.

Which of the following compounds is aromatic ?

fuEu esalsd kSulk ;kSfxd ,sjkseSfVd gS\ (A)

5.

(B)

(C)

(D*) all of these mijks Dr

lHkh

Lone-pair of electrons on nitrogen atom of pyridine is not delocalized with  – electrons because (A) it is in p orbital (B) it is in sp3 orbital 2 (C*) it in sp orbital which cannot over lap with p orbitals on adjacent carbon atoms. (D) it is in sp orbital fijhfMu d sukbVªkst u ijek.kqij by sDVªkWu ;qXe -by sDVªkWu d slkFkfoLFkkuhd `r ughagksrk gSD;ksafd (A) ;g p d {kd

gS (B) ;g sp3 d {kd gS (C*) ;g sp2 d {kd gSt kslehiLFk d kcZ u ijek.kqv ksaij p d {kd ksad slkFkvfrO;kfir ughagksrk gSA (D) ;g sp d {kd gS A 6.*

Which of the following are Aromatic in nature.

fuEu esalsd kSuls,jksesfVd gSA

(A)

7.*

(B*)

[General Organic Chemistry]

(C*)

Which of the following are Aromatic compound.

[Ref_RSS Sir_2013] [General Organic Chemistry]

fuEu esalsd kSuls,jksesfVd ;kSfxd gS%

(A*)

8.*

(B)

(D*)

(C*)

(D)

What is/are correct about Guanidine base (A*) Resonance possible (C*) localized lone pair of e–.

[Ref. DRM Mam] (B) Odd number of p-electrons (D*) Delocalized lone pair of e–.

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PAGE NO.- 2

XokfuMhu {kkj

d sfy , lgh d Fku gS@ gSa&

(A*) vuq ukn

laHko (C*) e d sLFkkuhd ` r ,d kd h;qXe

(B) p by s DVªkWu

d h fo"ke la[;k (D*) e d sfoLFkkuhd ` r ,d kd h;qXe

–

9.*

[Ref. DRM Mam]

–

Identify the correct statement (A*) All C–O bond length in carbonate ion is equal

(B) All C–C bond length in naphthalene is similar.

(C*) All N–O bond length in

(D*) All C–O bond length in

is similar

is equal.

lghd Fku d ksigpkfu;saA (A*) d kcks ZusV vk;u esalHkhC–O caèky EckbZcjkcj gSA (B) uS¶Fksfy u esalHkh C–C caèk y EckbZleku gSA esalHkh N–O caèk y EckbZleku gSA

(C*)

10.*

esalHkhC–O caèky EckbZcjkcj gSA

(D*)

Hyperconjugation is possible in

fuEu esalsfd lesavfrl;qXeu lEHko gSA

(A*)

11.

(B*)

(C)

How many species out of the following are aromatic ?

(D)

[General Organic Chemistry]

fuEu esafd ruh Lih'kht ,sjksesfVd gS no cyclic resonance ,

Ans.

,

,

,

3

Sol.

12.

,

,

,

are Aromatic species. ¼,s jksesfVd

Lih'kht gSA½

At how many carbons the negative charge of the following anion can reach through resonance (including the carbon at which – ve charge is present in the given structure).

fuEu _ .kk;u Lih'kht d svuqukn d sQ y Lo: i fd rusd kcZu ijek.kqij _ .kkos'k vkrk gSA ¼nh xbZlajpuk esa_ .kkRed vkosf'kr d kcZu ijek.kqlfgr½

Ans.

5

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PAGE NO.- 3

13.

How many p-electrons are involved in resonacne in the given structure? nh xbZlajpuk esavuqukn esaHkkx y susoky sp-by sDVªkWUkksad h la[;k fd ruh gS\

Ans.

8

14.

How many hyperconjugable H atoms are in

[Ref. DRM Mam]

esavfrla;qXeh H ijek.kqv ksad h la[;k fd ruh gS\ Ans.

8

15.

Match the following Column – I

Column – II

(A)

(p) Aromatic

(B)

(q) Non aromatic

(C)

(r) Anti aromatic

(D)

(s) Heterocyclic

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PAGE NO.- 4

Ans.

fuEu d kslqesfy r d hft ;s d kWy e– I

d kWy e– II

(A)

(p) ,s jks eS fVd

(B)

(q) ukW u

(C)

(r) ,UVh,s jkseSfVd

(D)

(s) fo"kepØ h;

(A - p,s) ; (B - p,s) ; (C - p,s) ; (D - r)

Lecutre No.

Sub-topic(s) Nam e (No. of Lectures)

L65

Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring)

L66

L67

,s jks eSfVd

Carbanion & its stability

Carbocation Stability

Hom e Work Sheet

Ex.1 (S) Ex.1 (O) Ex.2 (O) Ex.2 (I) Ex.2 (M) Ex.1 (S) Ex.1 (O) Ex.2 (O) Ex.2 (I) Ex.2 (M)

Sec : J Sec : J 47 to 49 27, 28 35, 36 I1 to I3, Sec : K I1 to I7, Sec : K 35 to 43, 45, 50 24, 26, 29 to 32 33, 34, 37, 38

Hom e Work NCERT Pg.N

Th.

XI XII XI XII XI XII XI

W.S.

Chemical reactions of alkane

Prob.

Th.

Chem . INFO

364

Prob.

Cation & radical stability order (20 Ques.)

XII

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PAGE NO.- 5

ChemINFO-5.3

Hydrocarbon Alkane :- Chemical Reaction of Alkane

Daily Self-Study Dosage for mastering Chemistry

Chemical Reaction of Alkane Halogenation of Alkane :- It is example of free radical substitution reaction CH3 –CH3 + Cl2 CH3 –CH2 Cl + HCl It is found that the rate of reaction of alkanes with halogens is F2 > Cl2 > Br2 > I2 Alkane which give more stable free radical will be more reactive towards halogenation > CH3CH2CH3 > CH3CH3 > CH4 Fluorination is too violent to be controlled.Iodination is very slow and a reversible reaction. It can be carried out in the presence of oxidising agents like HIO3 or HNO3 Mechanism  Three Step 1.

Chain Initiation Step Cl –Cl Homolysis 2

2.

Chain Propagation Step : (a) CH4 +

3.



+ HCl

(b)

+Cl2  CH3–Cl +

(b)

+

Chain Termination Step : (a) (c)

+ +

 Cl2

 CH3 – CH3

 CH3 –Cl

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

Which of the following step is chain propagation step of ethane. (A) CH3 –

+ CH3 – CH2 

(C*) CH3 –CH3+ 17.

 CH3– CH2 + HCl

(B) CH3 –CH3  CH3 – CH2 + H (D) CH3 – CH2 +

 CH3–CH2–Cl

Chlorination of an alkane involves (A) Cl

(B) Cl

(C*)

(D)

18.

The reactivity of hydrogen atom in an alkane towards substitution by bromine atom is : (A) 1°H > 2°H > 3°H (B*) 3°H > 2°H > 1°H (C) 3°H > 1°H > 2°H (D) 2°H > 3°H > 1°H

19.

Reactivity order of halogens towards free radical substitution reaction is (A) Br2 > F2 > Cl2 > I2 (B) F2 > I2 > Br2 > Cl2 (C) I2 > Br2 > Cl2 > F2 (D*) F2 > Cl2 > Br2 > I2

20.

CH3 –CH2 –CH2 –CH3  major product, major product is : (A) CH3 –CH2 –CH2 –Cl

(B) Cl

(C)

(D*) Cl

Cl

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PAGE NO.- 6

ChemINFO-5.3

Hydrocarbon Alkane :- Chemical Reaction of Alkane

Daily Self-Study Dosage for mastering Chemistry

,Yd suksad k gSy kst uhd j.k %;g eqDr ewy d çfrLFkkiu vfHkfØ ;k d k mnkgj.k gSA CH3 –CH3 + Cl2

CH3 –CH2 Cl + HCl

;g ik;k x;k gSfd gSy kst uksad slkFk ,Yd suksad h vfHkfØ ;k d h nj F2 > Cl2 > Br2 > I2 gksrh gSA ,Yd su t ksvf/kd LFkk;h eqDr ewy d nsrk gS]og gSy kst uhd j.k vfHkfØ ;k d sçfr T;knk fØ ;k'khy gksrk gSA > CH3CH2CH3 > CH3CH3 > CH4

¶yks jhuhd j.kfu;a f=kr ifjfLFkfr;ks ad svUrxZ r Hkhcgq r d fBukbZiw oZ d gks rkgS A vk;kMshuhd j.k,d cgq r /khehrFkkmRØ e.kh; vfHkfØ ;k gSA ;g vkWDlhd kjd inkFkZt Sl sHIO3 ;kHNO3 d hmifLFkfr esalEiUu gksrkgSA fØ ;kfof/k rhu in 1. J`a[ky k çkjEHkd in Cl –Cl 2.

2

J`a[ky k l ap j.k in : (a) CH4 +

3.



+ HCl

(b)

+Cl2  CH3–Cl +

(b)

+

J`a[ky k l ekiu in : (a) (c)

+ +

 Cl2

 CH3 – CH3

 CH3 –Cl

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

17.

fuEu esalsd kSulk in ,Fksu d k J`a[ky k lap j.k in gS\ (A) CH3 – CH2 + CH3 – CH2 

(B) CH3 –CH3  CH3 – CH2 + H

(C) CH3 –CH3+

(D) CH3 – CH2 +

 CH3– CH2 + HCl

,Yd su d sDy ksjhuhd j.k esafuEu esalsd kSulh Lih'kht ç;qDr gksrh gS\ (A) Cl

18.

(B) Cl

20.

(C)

(D)

czksehu ijek.kq}kjk çfrLFkkiu d sçfr ,Yd su esamifLFkr gkbMªkst u ijek.kqv ksad h fØ ;k'khy rk d k lgh Ø e fuEu gS% (A) 1°H > 2°H > 3°H

19.

 CH3–CH2–Cl

(B) 3°H > 2°H > 1°H

(C) 3°H > 1°H > 2°H

(D) 2°H > 3°H > 1°H

eqDr ewy d çfrLFkkiu vfHkfØ ;k d sçfr gSy kst uksad h fØ ;k'khy rk d k lgh Ø e d kSulk gS\ (A) Br2 > F2 > Cl2 > I2

(B) F2 > I2 > Br2 > Cl2

(C) I2 > Br2 > Cl2 > F2

(D) F2 > Cl2 > Br2 > I2

CH3 –CH2 –CH2 –CH3  (A) CH3 –CH2 –CH2 –Cl

eq[; mRikn]eq[; mRikn gS% (B) Cl

(C)

(D) Cl

Cl 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.- 7

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 37

This DPP is to be discussed in the week (26.10.2015 to 31.10.2015) 1. 2. 3. 4.

Course of the week as per plan : Carbocation and It’s Stability, Carbocation Rearrangement, Discusion. Course covered till previous week : Application of I.R.,M HC effects (Bond Length & Electron density on benzene ring), Aromaticity, Carbanion and It’s Stability. Target of the current week : Carbocation and It’s Stability, Carbocation Rearrangement, Discusion. DPP Syllabus :

DPP No. # 37 (JEE-MAIN) Total Marks : 65

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15 ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.20

(3 marks, 2 min.) (4 marks, 2 min.)

[45, 30] [20, 10]

ANSWER KEY DPP No. # 37 (JEE-MAIN) 1.

(C)

2.

(A)

3.

(A)

4.

(B)

5.

(B)

6.

(A)

7.

(D)

8.

(B)

9.

(B)

10.

(B)

11.

(C)

12.

(C)

13.

(D)

14.

(C)

15.

(A)

16.

(B)

17.

(D)

18.

(C)

19.

(B)

20.

(A)

1.

Select the correct stability order of the following carbanions.

fuEu d kcZ_ .kk;uksad sLFkkf;Ro d k lgh Ø e D;k gS\

2.

(i) CH2 = CH –

(ii) CH3 – C – CH2

(iii) CH3 – C –

(A) (i) > (ii) > (iii)

O (B) (ii) > (iii) > (i)

O (C*) (iii) > (ii) > (i)

– C –CH3 O

(D) (ii) > (i) > (iii)

Select the correct stability order of the following carbanions.

fuEu d kcZ_ .kk;uksad sLFkkf;Ro d k lgh Ø e D;k gS\

3.

(i) CH  C ..

(ii)

(iii)

(A*) (i) > (ii) > (iii)

(B) (ii) > (iii) > (i)

(C) (iii) > (ii) > (i)

(D) (ii) > (i) > (iii)

Select the correct stability order of the following carbanions.

fuEu d kcZ_ .kk;uksad sLFkkf;Ro d k lgh Ø e D;k gS\ .. CH2 (i)

(ii)

(A*) (i) > (ii) > (iii) > (iv) (B) (ii) > (iii) > (i)> (iv)

..

..

(iii)

(iv)

(C) (iii) > (iv) > (ii) > (i)

(D) (iv)> (ii) > (i) > (iii)

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PAGE NO.- 1

4.

Select the correct stability order of the following carbanions.

fuEu d kcZ_ .kk;uksad sLFkkf;Ro d k lgh Ø e D;k gS\ (ii)

.. CH

–

–

.. CH –

(iii)

–

(i)

.. CH – 2

–

–

CHO

NO2

(A) (i) > (ii) > (iii) 5.

(B*) (ii) > (iii) > (i)

(C) (iii) > (ii) > (i)

(D) (ii) > (i) > (iii)

The correct stability order of following carbocations is

fuEu d kcZ/kuk;uksad sLFkkf;Ro d k lgh Ø e gS%

Sol.

(A) I > II > III > IV (B*) IV > II > III > I (C) IV > III > II > I (D) IV > I > III > II The correct stability order of the following carbocations is IV > II > III > I

fuEu d kcZ/kuk;uksad sLFkkf;Ro d k lgh Ø e IV > II > III > I gS

>

>

>

Stability of carbocation depend upon conjugation > Hyperconjugation

d kcZ/kuk;uksad k LFkkf;Ro fuHkZj d jrk gS% la;qXeu > vfrla;qXeu 6.

Which of the following is the most stable carbocation intermediate.

fuEu esalslclsvf/kd LFkk;h e/;orhZd kcZ/kuk;u d kSulk gS\ (A*) Sol.

(C)

(D)

Carbocation stablized by + m effect. + m iz Hkko

7.

(B)

d sd kj.kd kcZ/kuk;u LFkk;hgksrkgSA

In which of the following Ist is more stable than IInd :

fuEu esalsfd lesaIst , IInd lsvf/kd LFkk;h gS\ (A)

, 

(B) 

(C) CH3– C H –CH3, CH3–CH2– C H2

,

(D*) All of these mijks Dr

lHkh

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PAGE NO.- 2

8.

Which of the following is correct order of stability of carbocation?

[Ref_RGP Sir]

bu d kcZ/kuk;u d sLFkkf;Ro d k lgh Ø e d kSulk gS\

(A) I > II > III 9.

(B*) III > I > II

(C) I > III > II

(D) III > II > I

Which of the following carbocation is most stable :

fuEu esalsd kSulkd kcZ/kuk;u lokZf/kd LFkk;hgS\ CH2 (A)

CH2 (B*)

CH3 10.

CH2

CH2 (C)

(D)

Cl

NO2

(C) C6H5–CH2

(D) CH –CH–CH

OH

Which of the following carbocation is most unstable :

fuEu esalsd kSulkd kcZ/kuk;u lokZf/kd vLFkk;hgS\ (A)

11.

(B*)

3

3

Which of the following carbocation is most unstable ?

fuEu esalsd kSulkd kcZ/kuk;u lokZf/kd vLFkk;hgS\

(A)

12.

(B)

(C*)

Maximum number of atoms in same plane in benzene are : (A) 6 (B) 10 (C*) 12

(D)

[Made by DRM Mam] (D) None of these

csat hu easvfèkd re fd rusijek.kqleku ry esagksrsgS\ (A) 6 13.

(B) 10

(C*) 12

(D) bues alsd ksbZugha

In which of the following maximum SIR.

[Made by DRM Mam]

fuEu esalsfd lesavfèkd re SIR gksxk \

(A)

(B)

(C)

(D*)

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PAGE NO.- 3

14.

In which of the following positive charge of carbocation is not in conjugation with –OH group.

fuEu esalsfd lesad kcZèkuk;u d k èkukos'k –OH d slkFk la;qXeu esaugh gksrk gS\

(A)

15.

(B)

[Made by DRM Mam]

(D) None of these bues alsd ksbZugha

(C*)

Which of the following is most stable cation.

[Made by DRM Mam]

fuEu esalsd kSulklokZfèkd LFkk;h èkuk;u gS\ (A)

(B*)

Lecutre Sub-topic(s) Nam e No. (No. of Lectures)

L68

Carbocation Rearrangement

Hom e Work Sheet Ex.1 (S)

I4

Ex.1 (O)

I8 to I12

Ex.2 (O)

44

Ex.2 (I)

25

Ex.2 (M)

32

(D)

Hom e Work NCERT Pg.N Th.

Chem . INFO

W.S.

Carbene

Chemical reaction of alkane

Hand out

XI XII XI

Prob. XII Discussions

L69

L70

(C)

Discussion of DPP's of Hydrocarbon

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PAGE NO.- 4

ChemINFO-5.4

HYDROCARBON

Daily Self-Study Dosage for mastering Chemistry

Chemical reaction of Alkane :- Nitration and sulphonation

Nitration and sulphonation of alkane Nitration :- Replacement of H-atom of alkane by nitro group is known as nitration. When nitration is carried out in vapour phase between 420 K and 720 K, a mixture of all possible mononitro derivatives is obtained (directly or after chain fission) Reactivity order = 30 > 20 > 10 For example :

R–H

R–NO2

;

CH3–CH3

CH3–CH2–NO2

Sulphonation :- When alkane is treated with oleum (H2S2O7) or fuming sulphuric acid, H-atom is replaced by – SO3H group . Reactivity order = 30 > 20 > 10 For example :

CH3–CH2–CH3

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

CH3–CH2–CH3 (A)

major product is (B*)

17.

(C) CH3–CH2–NO2

(D) CH3–NO2

(C) CH3–CH2–NO2

(D*)

major product is

(A)

(B)

18.

major product is

(A)

(B)

19.

(C*)

(D)

major product is CH3

CH2–SO3H (A)

CH3

CH3

SO3H

(B*)

SO3H (C)

(D) SO3H

20.

CH3–CH2–CH3 (A*)

major product is. (B)

(C) CH3–CH2–SO3H

(D) CH3–SO3H

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PAGE NO.- 5

ChemINFO-5.4

HYDROCARBON Chemical reaction of Alkane :- Nitration and sulphonation

Daily Self-Study Dosage for mastering Chemistry

,Yd su d k ukbVªhd j.k rFkk l YQ ksuhd j.k ukbVªhd j.k: ukbVªkslewg }kjk ,Yd su d sH-ijek.kqd sfoLFkkiu d ksukbVªhd j.k d grsgSA t c ukbVªhd j.k420 K rFkk720 K rki d se/; ok"i voLFkkes agkrskgS ]rkslHkhlEHkkfor ekuskuskbVª ksO;q RiUukasd sfeJ.k¼çR;{k: i ls;kJ` [akykfo[k.Mu d si'pkr~ ½çkIr gkrssgS A fØ ;k'khy rk d k Ø e = 30 > 20 > 10 mnkgj.k d sfy , : R–H

R–NO2

;

CH3–CH3

CH3–CH2–NO2

l YQ ksuhd j.k: t c ,Yd su d h vfHkfØ ;k vksfy ;e (H2S2O7) vFkok la/kwezlY¶;wfjd vEy d slkFk d jkrsgS]rksSO3H lewg }kjk H-ijek.kqd kfoLFkkiu gks rkgSA fØ ;k'khy rk d k Ø e = 30 > 20 > 10 mnkgj.k d sfy , :

CH3–CH2–CH3 H2SO4 + SO3

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

eq[; mRikn gS%

CH3–CH2–CH3 (A)

(B*)

(C) CH3–CH2–NO2

(D) CH3–NO2

(C) CH3–CH2–NO2

(D*)

eq[; mRikn gS%

17.

(A)

(B)

eq[; mRikn gS%

18.

(A)

(B)

(C*)

(D)

eq[; mRikn gS%

19.

CH3

CH2–SO3H (A)

CH3

CH3

SO3H

(B*)

SO3H (C)

(D) SO3H

20.

eq[; mRikn gS%

CH3–CH2–CH3 (A*)

(B)

(C) CH3–CH2–SO3H

(D) CH3–SO3H

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PAGE NO.- 6

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 38 & 41

Important Note : The Chapter of hydrocarbon will be coursed in study-XI from Board point of view only. Read the theory and momerise it, then answer the question given below that portion of theory. This is to be done during PSA holiday as Self-study. 1. 2. 3.

Course of the week as per plan : Discusion, Discusion of DPPs of hydrocarbon Course covered till previous week : Carbocation and It’s Stability, Carbocation Rearrangement, Discusion. Target of the current week : Discusion, Discusion of DPPs of hydrocarbon

4.

DPP Syllabus : Discusion of DPPs of hydrocarbon

This DPP is to be discussed in the week (02.11.2015 to 07.11.2015)

DPP No. # 38 (JEE-MAIN) Total Marks : 68

Max. Time : 34 min.

ChemINFO Questions ('–1' negative marking) Q.1 to Q.17

(4 marks, 2 min.)

[68, 34]

ANSWER KEY DPP No. # 38 (JEE-MAIN) 1. 8. 15.

(C) (B) (B)

2. 9. 16.

(B) (A) (D)

3. 10. 17.

(A) (C) (B)

4. 11.

(C) (D)

5. 12.

(D) (B)

6. 13.

(C) (C)

7. 14.

(B) (B)

3. 10. 17.

(B) (B) (C)

4. 11. 18.

(B) (B) (D)

5. 12. 19.

(A) (C) (C)

6. 13. 20.

(A) (D) (C)

7. 14.

(B) (B)

3. 10. 17.

(D) (D) (C)

4. 11. 18.

(C) (B) (D)

5. 14. 19.

(A) (A) (B)

6. 12. 20.

(C) (C) (A)

7. 13.

(D) (D)

3. 10. 17.

(A) (B) (C)

4. 11. 18.

(B) (B) (D)

5. 12. 19.

(C) (A) (C)

6. 13. 20.

(D) (A) (D)

7. 14.

(D) (D)

DPP No. # 39 (JEE-MAIN) 1. 8. 15.

(C) (B) (C)

2. 9. 16.

(C) (A) (A)

DPP No. # 40 (JEE-MAIN) 1. 8. 15.

(B) (D) (C)

2. 9. 16.

(A) (D) (A)

DPP No. # 41 (JEE-MAIN) 1. 8. 15.

(C) (C) (B)

2. 9. 16.

(D) (A) (A)

Isomerism in Alkane & Alkene

,Yd su rFkk ,Yd hu esal eko;ork % (1)

Geometrical Isomerism : Two compounds having same structural formula but different orientation of atoms or groups in space due to restricted rotation of bonds are known as geometrical isomers. e.g.

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PAGE NO.- 1

(1)

T;kferh; l eko;ork : ,sl s;kSfxd ft ud slajpukRed lw=k leku gksrsgSay sfd u ca/kksad sizfrcaf/kr ?kw.kZu d sd kj.k f=kfoe esaijek.kqv ksavFkok lewgksad k vfHkfoU;kl fHkUu&fHkUu ik;k t krk gS] mUgsaT;kferh; leko;oh d grsgSrFkk bl leko;ork d ksT;kferh; leko;ork d grsgSaA mnk-

Nomenclature of Geometrical isomers : Cis-Trans naming : If similar group present at once side of double bond/ring known as cis isomer and if present at opposite side of double bond/ring known as trans isomer. Cis and trans are G.I. of each other. e.g.

T;kferh; l eko;oh;ksa d k uked j.k % l ei{k&foi{k uked j.k % ;fn f}ca/k@oy ; d s,d rjQ leku lewg ik;st krsgS] rksblslei{k leko;oh d grs gSrFkk ;fn f}ca/k@oy ; d h foijhr lrgksaij leku lewg ik;st krsgS] rksblsfoi{k le;ko;oh d grs gS A lei{k rFkk foi{k leko;oh ,d &nwl jsd sT;kferh; leko;oh gksrsgSA mnk-

1.

Among the given compounds identify the pair of geometrical isomers :-

fn;sx;s;kSfxd ksesT;kferh; leko;oh;ksad s;qXe d ksigpkfu;s%

(A) I & II (A) I rFkkII

(B) I & III (B) I rFkkIII

(C*) II & IV (C*) II rFkkIV

(D) III & IV (D) III rFkkIV

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PAGE NO.- 2

2.

Which of the following is cis-isomer \

fuEu esalsd kSu lk lei{k& leko;oh gS\

3.

(A)

(B*)

(C)

(D)

Which of the following is trans-isomer \

fuEu esalsd kSu lk foi{k& leko;oh gS\

(A*)

(B)

(C)

(D)

Preparation of Alkene (1)

from partially hydrogenation of alkyne The hydrogenation with poisoned Pd + H2 + BaSO4 + S or quinoline (Lindlar's catalyst) reduces triple bond to double bond stereo specifically via syn addition (addition of both H atom from the same side) where as the reduction with alkali metal in liquid NH3 (Birch reduction) reduces triple bond to double bond stereo specifically via anti addition. (addition of H-atom from two different sides, above and below the plane) Poisoned R  C  C  R  H2    Pd

R

R C=C H

Na / liq NH

3 R  C  C  R  H2  

;

H

(cis alkene)

H

R C=C

R

H

(Trans alkene)

,Yd hu d k fojpu % (1)

,Yd kbZu d sv kaf'kd gkbMªkst uhd j.k }kjk % fo"kkDr Pd + H2 + BaSO4 + S ;kfDouksy hu ¼fy .My kj mRizsjd ½ d hmifLFkfr esagkbMªkst uhd j.kij f=kca/kf=kfoe fof'k"V : i lsflu ;ksx }kjkf}ca/kesavipf;r gkst krkgS¼flu ;ksx %leku lrg ij nksuks aH ijek.kqv ks ad k;ksx gksrkgSA½ t cfd SCl NH3 es a{kkj /kkrq¼cpZvip;u½d hmifLFkfr esaf=kca /kf=kfoe fof'k"V : i ls,s .Vh&;ksx }kjkf}ca/kes avipf;r gkst krk gSA ¼nksfofHkUu lrgksa]Å ij ,oeauhpsd h lrg ij H-aijek.kqd k ;ksx gksrk gSA½ Poisoned R  C  C  R  H2   

;

Pd

4.

nzo NH3 R  C  C  R  H2  

Poisoned

CH3  C  C CH3  H2  Product is Pd fo"kkDr

CH3  C  C CH3  H2   mRikn Pd

(A) CH3  CH2  CH2  CH3

(C*)

CH3

CH3 C

H

C H

gS% (B)

(D)

CH3  CH2  CH  CH2

CH3

H C

C

H

CH3

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PAGE NO.- 3

5.

Na / liq NH

3 CH3 — C  C — CH3   Product is.

Na / nz o NH

3 CH3 — C  C — CH3  

mRikn gS%

(A) CH3  CH2  CH2  CH3 (C)

CH3

CH3 C

(D*)

CH3

H C

C

H 6.

(B) CH3  CH2  CH  CH2

C

H

H

Poisoned Pd CH3 — CH2 — C  C — CH3  H2 

fo"kkDr Pd CH3 — CH2 — C  C — CH3  H2 

(A) CH3 — CH2 — CH2 — CH2 — CH3

CH3

Product is.

mRikn gS% (B)

CH3 -CH2

H C

C

H (C*)

CH3-CH2

CH3 C

7.

(D) CH3 — CH2 — CH2 — CH  CH2

C

H

CH3

H

Na / liq.NH3 CH3 — CH2 — C  C — CH3   Product is. Na / nz o NH

mRikn gS%

(A) CH3 — CH2 — CH2 — CH2 — CH3

(B*)

3 CH3 — CH2 — C  C — CH3  

CH3 -CH2

H C

C

H (C)

CH3-CH2

CH3 C

(2)

(D) CH3 — CH2 — CH2 — CH  CH2

C

H

CH3

H

Dehydrohalogenation of alkyl halide Alkyl halide on heating with alcoholic potash ( Potassiam hydroxide dissolved in alcohol) eliminate one molecule of halogen acid to form alkene. This reaction is known as dehydrohalogenation. This is example of  - elimination reaction since  -hydrogen atom is eliminated from the  - carbon atom.

H

H

H

C

C

H alc.KOH H   C  H

H X ( X = Cl, Br, I )

H C

H

Note(1) Order of ease of dehydrohalogenation of R – X : 3o > 2o > 1o (2) Reactivity order of R–X : R–I > R – Br > R– Cl (3) More stable alkene is formed as major product. (2)

,fYd y gSy kbM d sfogkbMªksgSy kst uhd j.k }kjk % t c ,fYd y gSy kbM d ks,Yd ksgkWfy d ikSVk'k¼,Yd ksgkWy esa?kqfy r iks VSf'k;e gkbMªksDlkbM½ d slkFkxeZd jrsgS]rksgSy kst u vEy d k,d v.kqfoy ksfir gksusij ,Yd hu curhgSA bl vfHkfØ ;kd ksfogkbMªksgSy kst uhd j.kd grsgSA ;g  - foy ksiu vfHkfØ ;k d k mnkgj.k gSD;ksafd  - d kcZu ijek.kqls - gkbMªkst u ijek.kqd kfoy ksiu gksrk gSA

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PAGE NO.- 4

H

H

H

C

C

,Yd k-sKOH

H



H

 H 

H C

C

H

H X ( X = Cl, Br, I )

uksV& (1) R – X d sfogkbMª ksgSy kst uhd j.k d k ljy rk lsgksusd k Ø e : 3o > 2o > 1o (2) R–X d h fØ ;k'khy rk d k Ø e : R–I > R – Br > R– Cl (3) vfèkd LFkk;h ,fYd u eq [; mRikn d s: i esaizkIr gksrh gSA 8.

alc.KOH

CH3 —CH2 —CH2 —CH2 —CH2  

,sYd k-s KOH

  Product is.

Cl (A) CH3—CH2—CH2—CH2—CH3

(B*) CH3—CH2—CH2—CH=CH2

(C) CH3— C

(D) CH3— C

CH

CH3

alc.KOH

CH3—CH2—CH2—CH—CH3  

,sYd k-s KOH 

Cl (A*) CH3—CH2—CH = CH—CH3

(B) CH3—CH2—CH2—CH=CH2

(C) CH3— C

(D) CH3— C

CH

CH3

CH

CH3

CH3 alc.KOH

CH— CH—CH—CH3  3 

,sYd k-s KOH

  Major Product is.

eq[; mRikn gS%



CH3 Cl (A) CH3—CH2—CH = CH—CH3

(B) CH3—CH2—CH2—CH=CH2

(C*) CH3— C

(D) CH3— C

CH

CH3

CH

CH3

CH3

CH3 11.

CH3

mRikn gS%

  Product is.

CH3 10.

CH

CH3

CH3 9.

mRikn gS%



alc.KOH

,sYd k-s KOH CH3—CH—CH2—CH2—Cl    Product is.mRikn gS%  

(3)

CH3 (A) CH3—CH2—CH = CH—CH3

(B) CH3—CH2—CH2—CH=CH2

(C) CH3— C

CH—CH=CH2 (D*) CH— 3

CH3

CH3 CH3 From vicinal dihalide Dihalide in which two halogen atoms are attached to two adjacent carbon atoms are known as vicinal dihalides. Vicinal dihalide on treatment with Zinc metal lose a molecule of ZnX2 to form an alkene. This reaction is known as dehalogenation.

Br

(3)

CH

Br

fofl uy MkbgSy kbM }kjk % MkbgSy kbM ft uesanksgSy kst u ijek.kqnksfud VorhZd kcZu ijek.kqv ksalst qM +sgksrsgS]mUgsafofluy MkbgSy kbM d grsgSA fofluy MkbgS y kbM d hvfHkfØ ;kZn /kkrqd slkFkd jkusij ZnX2 d sfu"d kflr gks usij ,Yd hu curhgS A bl vfHkfØ ;kd ks fogSy kst uhd j.kd grsgSA Br

Br

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PAGE NO.- 5

12.

Product is

– (A) CH2 = C = CH2

(B*) CH3— CH = CH2

Zn pw .kZ  

(A) Ph — C  CH

(D)CH3—CH2— CH = CH2

(C*) Ph — CH  CH2

(D) Ph — CH3

mRikn gS%

(B) Zn dust

  Product is

Zn pw .kZ  

(4)

(C) CH2 = CH2

Zn dust   product is

13.

14.

mRikn gS%

mRikn gS%

(A) CH2  CH — CH2 — CH3

(B*) CH3 — CH  CH — CH3

(C) CH2  CH — CH  CH2

(D)

From alcohol by acidic dehydration Alcohols on heating with concentrated sulphuric acid form alkene with the elimination of one water molecule. Since a water molecule is eliminated from the alcohol molecule in the presence of an acid ,this reaction is known as acidic dehydration of alcohols. This reaction is example of  - elimination reaction since –OH group takes out hydrogen atom from the  carbon atom.

Conc.H SO

2 4  CH –CH=CH–CH



3

3

(1) Ease.of dehydration of alcohol  3o > 2o > 1o (2) More stable alkene is formed as major product (4)

,Yd ksgkWy ksad sv Ey h; fut Zy hd j.k }kjk % ,Yd ksgkWy ksad kslkUnzlY¶;w fjd vEy d slkFkxeZd jusij t y d s,d v.kqd kfoyks iu gksusij ,Yd hu curhgSA pw fd vEy ¡ d hmifLFkfr es a,Yd ks gkW y v.kqlst y v.kqd kfoyks iu gks rkgS ]blfy, bl vfHkfØ ;kd ks,Yd ks gkW y ks ad kvEyh; fut Z y hd j.k d grsgSA ;g vfHkfØ ;k  - foy ksiu vfHkfØ ;kd kmnkgj.kgSpw¡ fd  – d kcZ u ijek.kqls–OH lewg gkbMª ks t u ijek.kqxzg.k d jrkgS A l kUn-zH2SO4   CH –CH=CH–CH 

3

3

(1) ,Yd ks gkWy d sfut Zy hd j.k d h ljy rk lsgksusd k Ø e  3o > 2o > 1o (2) vfèkd LFkk;h ,fYd u eq [; mRikn d s: i esaizkIr gksrh gSA

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PAGE NO.- 6

Conc.H2SO 4    Major product is 

15. OH

lkUn-zH2SO4   eq [; mRikn gS%  OH

(A) CH3 — CH2 — CH  CH2

(B*) CH3 — CH  CH — CH3

(C) CH3 — CH2 — CH2 — CH3

(D)

Conc.H2SO 4    product is 

16.

OH

l kUn-zH2SO4   

mRikn gS%

OH (A) CH3 — CH2 — CH  CH2

(B) CH3 — CH  CH — CH3

(C) CH3 — CH2 — CH2 — CH3

(D*)

Conc.H SO

17.

2 4       Major product is



OH

l kUn-zH2SO4   

eq[; mRikn gS%

OH

(A) Ph — CH2 — CH  CH2

(B*) Ph — CH  CH — CH3

(C) Ph

(D)

DPP No. # 39 (JEE-MAIN) Total Marks : 80

Max. Time : 40 min.

ChemINFO Questions ('–1' negative marking) Q.01 to Q.20

(4 marks, 2 min.)

[80, 40]

Chemical reactions of alkene

,Yd hu d h jkl k;fud v fHkfØ ;k (1)

Addition of halogens : Halogens like bromine or chlorine add up to alkene to form vicinal dihalides.Addition of halogen to alkene is an example of electrophilic addition reaction. This reaction is used as a test for unsaturation. CCl4 CH2  CH2  Br2  

gSy kst uksad k ;ksx % gSy kst u t Sl sfd czksehu ;kDy ksjhu],Yd hu lst qM +usij fofluy MkbgSy kbM cukrhgSA ,Yd hu ij gSy kst uksad k;ksx ,d by s DVª kW uLus gh;ks xkRed vfHkfØ ;kd kmnkgj.kgSA bl vfHkfØ ;kd kmi;ksx vla r` Irrkd kijh{k.kd jusd sfy, fd ;kt krk gS A 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.- 7

CCl4 CH2  CH2  Br2  

1.

2.

CCl

4 CH3 — CH  CH2  Br2   Product is. mRikn

gS%

(A)

(B)

(C*)

(D)

CCl4 CH3 — CH  CH — CH2 — CH3  Br2   Product is. mRikn

(A)

(B)

(C*)

(D)

Br2 / CCl4   Product is. mRikn

3.

gS%

gS%

(A)

(B*)

(C)

(D)

(2)

Addition of Hydrogen halides : Addition of HX to unsymmetrical alkene place according to Markovinikov rule. Markovnikov rule : The rule states that the negative part of the attacking species add on the carbon atom containing less number of hydrogen atom.

(2)

gkbMªkst u gSy kbMksad k ;ksx % vlefer ,Yd hu ij HX d k;ksx ekjd ksuhd kWQ fu;e d svuql kj gksrkgSA

eksjd ksuhd kWQ fu;e % vfHkd kjd d k _ .kkRed Hkkx ml d kcZu ijek.kqlst qM +rkgS]ft l ij gkbMªkst u ijek.kqv ksad h la[;kd e gksrhgSA 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.- 8

d e LFkk;h d kcZ /kuk;u ¼vYi mRikn½

vf/kd LFkk;h d kcZ/kuk;u eq[; mRikn Major Product. eq [;

4.

(A)

(B*)

(C)

(D)

5.

+ HBr

(A*)

6.

(3)

mRikn gS%

Ph — CH  CH2  HBr

Major Product. eq [;

(B)

mRikn gS%

(C)

Product is.

(D)

mRikn gS%

(A*)

(B)

(C)

(D) Ph — C  CH

Addition of HBr in Presence of peroxide or Kharash effect or Anti Markovnikov effect : In the presence of peroxide such as benzoyl peroxide [ ] and light, the addition of HBr to unsymmetrical alkene occurs contrary to Markovnikov’s rule. Peroxide CH3 — CH  CH2  HBr  

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PAGE NO.- 9

(3)

ijkWDlkbM d h mifLFkfr esaHBr d k ;ksx ;k [kjk'k izHkko ;k ,s.Vh&eksjd ksuhd kWQ izHkko iz d k'krFkkijkW DlkbM d hmifLFkfr es at Sl scs Ut kW ;y ijkW DlkbM [

] vlefer

,Yd hu ij HBr

d k;ksx ekjd ksuhd kWQ fu;e d sfoijhr gksrkgSA ijkWDl kbM CH3  CH  CH2  HBr   

7.

Peroxide Ph — CH  CH2  HBr   Product is

Ph — CH  CH2  HBr

ijkWDlkbM   mRikn gS%

(A)

(B*)

(C)

(D) Ph — C  CH

Peroxide   Product is

8.

ijkWDlkbM   mRikn gS%

(A)

(B*)

(C)

(D)

peroxide + HBr   Major product

9.

+ HBr

(A*)

(4)

CH3

ijkWDlkbM   eq[;’mRikn gS%

(B)

(C)

(D)

Addition of sulphuric acid : Cold concentrated sulphuric acid add to alkene in accordance with Markovnikov rule to form alkyl hydrogen sulphate.

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PAGE NO.- 10

(4)

lY¶;wfjd v Ey d k ;ksx % ,Yd hu ij B.MslkUnzlY¶;w fjd vEy d k;ksx eksjd ksuhd kWQ fu;e d svuq l kj gksusij ,fYd y gkbMª ks t u lYQ sV curkgSA

10.

Ph–CH=CH–CH3 + Conc.H2SO4  Major Product is :

eq[; mRikn gS%

(B*) Ph–CH–CH 2–CH 3

(A) Ph–CH 2–CH–CH 3

O–SO 2OH

O–SO 2OH

(C) Ph–CH 2–CH 2–CH 2

(D) None of these bues alsd ksbZugha

O–SO 2OH

11.

+ Conc.H2SO4 Major product is :

eq[; mRikn gS%

(A)

(B*)

(C)

(D)

(5)

Addition of water : Alkene react with water to form alcohols, in accordance with the markovinikov rule.

(5)

t y d k ;ksx % ekjd ksuhd kWQ fu;e d svuql kj ,Yd hu]t y d slkFkvfHkfØ ;kd jd s,Yd ksgkWy cukrhgSA

12.



H Ph–CH=CH–CH3+H2O  

Major product is :

eq[; mRikn gS%

(A) Ph–CH 2–CH2–CH2

(B) Ph–CH 2–CH–CH 3

OH (C*) Ph–CH–CH 2–CH 3

OH (D) None of these

buesalsd ksbZugha

OH



13.

H + H2O   Major product is :

(A)

OH

(B)

eq[; mRikn gS%

(C)

(D*)

OH

OH

(6)

OH Oxidation of alkene : Acidic potassium premanganate or acidic potassium dichromate oxidise alkene to ketone and/or acids depending upon the nature of the alkene.

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PAGE NO.- 11

CH3–CH=CH–CH3 2CH3COOH Alkene having = CH2 will give CO2 as one of the oxidation product by oxidising agent KMnO4 and K2Cr2O7. (6)

,Yd hu d k v kWDlhd j.k : vEy h; iks Vs f'k;e ijeS aXusV ;kvEyh; iksVS f'k;e MkbØ ks es V],Yd hu d ksbld hiz d` fr d svk/kkj ij d hVksu rFkk@;kvEyksaesa vkWDlhd `r d j nsrkgSA

CH3–CH=CH–CH3

2CH3COOH

,Yd hu t ks= CH2 j[krsgSt ksvkW Dlhd kjhvfHkd eZd KMnO4 rFkkK2Cr2O7 d s}kjkCO2 d s: i es a,d vkW Dlhd kjhmRikn ns rsgSA mRikn gS

14. (A) CH3–COOH and CH3–CHO

(B*) CH3–C–CH3 and CH3COOH

(C) CH 3–C–CH 3 and CH 3CHO

(D) CH3–CHO and CH3–C–COOH

O (A) CH3–COOH rFkkCH3–CHO

(B*)

(C)

(D)

O

15.

O

CH3–CH=CH–CH2–CH3

Products is :

mRikn gS

(A) CH3–C–CH3 and CH3COOH

(B) CH3–CH2–CHO and CH3–COOH

O (C*) CH3–CH2–COOH and CH3COOH

(D) CH 3–C–CH 3 and CH 3CHO

(A) CH3–C–CH3 and CH3COOH

O (B) CH3–CH2–CHO rFkkCH3–COOH

O (C*) CH3–CH2–COOH rFkkCH3COOH

(D) CH 3–C–CH 3 and CH 3CHO O

16.

CH3–CH2–CH=C–CH2–CH3

KMnO4 /H

Products is-

CH O 3 (A*) CH 3–CH 2–COOH and CH 3–C–CH 2–CH 3 O

(B) CH3COOH and CH3–C–CH2–CH3

O (D) CH 3–CH 2COOH and CH 3–C–COOH

(C) CH3–C–CH3 and CH3CH2COOH

O

O

CH3–CH2–CH=C–CH2–CH3

KMnO4 /H

mRikn gS

CH O 3 (A*) CH3–CH2–COOH

rFkk CH3–C–CH2–CH3

(B) CH3COOH

rFkk CH3–C–CH2–CH3 O

O

(C) CH3–C–CH3

rFkk CH3CH2COOH

(D) CH3–CH2COOH

rFkk CH3–C–COOH O

O

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PAGE NO.- 12

(7)

Oxidation of alkene with dil. KMnO4 (Bayer’s reagent) :

H.A.



H.A. = Hydroxylating agent. It may be any one of these : (a) Bayer’s reagent: cold dilute 1% alkaline KMnO4 (pink color) or (b) (i) OsO4 (ii) NaHSO3 Note : Double bond of Aromatic ring cannot react with these reagents. It is syn addition. Both OH groups add on the same side of pi-bond.

H.A.

Ex.

(7)



ruqKMnO4 ¼cs;j v fHkd eZd ½ }kjk ,Yd hu d k v kWDlhd j.k %

H.A.



H.A. = gkbMª ksfDly hd j.kvfHkd eZd A buesalsd ksbZHkhgksld rkgSA (a) cs ;j v fHkd eZd % B.Mk ruq1% {kkjh; KMnO4 ¼xqy kch jax½ ;k (b) (i) OsO4 (ii) NaHSO3 uksV :

,sjkseSfVd oy ; d sf}ca/kbu vfHkd eZd ksad slkFkvfHkfØ ;kughad jrsgSA ;g flu;ksx gksrk gSA nksuksaOH lewg ikbZca/k d sleku lrg ij t qM +rsgSA H OH

H.A.

Ex.



OH H

lkbDyksgs Dl hu

lei{k&lkbDyks gs Dls u-1,2-MkbvkWy Baeyer's reagent

17.

       Product ; Product will be :

cs;j vfHkd eZd   mRikn ; mRikn gksxk %

OH

HO

(A)

(B)

OH

(C*)

(D) HO

OH

OH

18. Reagent x will be : (A) 1% alkaline KMnO4 (Bayer’s reagent) (C) Peracid/H3O+

(B) OsO4 /NaHSO3 (D*) A and B both

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PAGE NO.- 13

vfHkd eZd x gksxk : (A) 1% {kkjh; KMnO4 (cs ;j vfHkd eZd ) + (C) ij&vEy /H3O

(B) OsO4 /NaHSO3 (D*) A rFkk B nks uksa

Baeyer 's Re agent

19.

      Product ; Product will be : CH3 – CH=CH – CH3 or OsO / NaHSO 4

CH3 – C  C – CH3

20.

3

cs;j vfHkd eZd   mRikn ; mRikn gksxk % ;k OsO4 / NaHSO3

(A)

(B)

(C*)

(D)

Which one of the following can not undergo in hydroxylation.

fuEu esalsfd l ;kSfxd d k gkbMªksfDly hd j.k ughagksld rk gS\ (A)

(B)

(C*)

(D)

DPP No. # 40 (JEE-MAIN) Total Marks : 80

Max. Time : 40 min.

ChemINFO Questions ('–1' negative marking) Q.01 to Q.20

(4 marks, 2 min.)

[80, 40]

Preparation of Alkyne 1. Dehydrohalogenation of gem and vic dihalide : Dihalide in which two halogen atoms are attached to two adjacent carbon atoms are known as vicinal dihalides. Dihalide in which two halogen atoms are attached to same carbon atoms are known as geminal dihalides. H H |

|

2 NaNH

2  R  C  C  R  2NaBr R  C C R    |

|

Br Br A vic  dibromide

,Yd kbZu d k fojpu 1.

t Se ,oafol MkbgSy kbMksad sfogkbMªksgSy kst uhd j.k }kjk % MkbgSy kbM ft uesanksgSy kst u ijek.kqnksfud VorhZd kcZu ijek.kqv ksalst qM +sgksrsgS]mUgsafofluy MkbgSy kbM d grsgSA MkbgSy kbM ft uesanksgSy kst u ijek.kq,d d kcZu ijek.kqv ksalst qM +sgksrsgS]mUgsat Sfeuy MkbgSy kbM d grsgSA H H |

|

2 NaNH

2  R  C  C  R  2NaBr R  C C R    |

|

Br Br A vic  dibromide

1.

NaNH

2  Product is.mRikn CH3–CH2–CH2–CHCl2   

gS%

Ä

(A) CH3–C  C–CH3 (C) CH2=CH–CH=CH2 2.

(B*) CH3–CH2–C  CH (D) CH3–CH2–CH = CH2

NaNH

2  Product is. mRikn Ph–CCl2–CH3   

gS%

Ä

(A*) Ph–C  CH

(B) CH3–C  CH

(C) Ph–CH = CH2

(D) Ph–CH = CHCl

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PAGE NO.- 14

3.

NaNH

2  Major product is eq CH3–CHCl–CH2Cl    [;

mRikn gS%

Ä

(A) CH2=C = CH2

2. 2.

(B) CH3–C = CH2

(C) CH3–CH = CCl2

(D*) CH3–C  CH

Dehalogenation of tetrahaloalkane :

VsVªkgSy k,Yd su d sfogSy kst uhd j.k }kjk % R – C  C – R + 2Zn X2

4.

Zn

 Product is mRikn CH3–CBr2–CHBr2 Ä

(A) CH2 = C = CH2

   Product is 

(A*) Ph–C  C–CH3

3. 3.

(B) CH3— CH = CH2 Zn

5.

gS% (C*) CH3–C  CH

(D) CH3–CBr = CBr2

mRikn gS%

(B) Ph–CH2–C  CH

(C) Ph–C=C=CH2

(D) Ph–CBr2–C  CH

Kolbe's electrolytic synthesis :

d ksYcsoS| qrv i?kVu la'y s"k.k }kjk: Electrolysis o| S qr viV?kVu     R – C  C – R + 2CO2 + 2KOH + H2 + H2O Current fo| qr /kkjk

If R = H, product will be CH  CH ; If R = CH3, product will be CH3 – C  C – CH3. ;fn R = H gksxk rksmRikn CH  CH gksxk; ;fn R = CH3 gksxk]mRikn CH3 – C  C – CH3 gksxkA is Electrolys    product is

6.

fo| qr vi?kVu  

(A) Ph–CC–CH3

mRikn gS%

(B) Ph–CH=CH–Ph

(C*) Ph–CC–Ph

(D) Ph–CH2–CH2–Ph

is Electrolys    product is :

7.

fo| qr vi?kVu  

(A) CHCH

mRikn gS%

(B) CH3–CH=CH2

(C) CH3–CH2–CH3

(D*) CH3–CCH

4.

Hydrolysis of carbides :

4.

d kckZbMksd st y v i?kVu }kjk %

8.

CaC2 + 2HOH  C2H2 + Ca(OH)2 ; Mg2C3 + 4HOH  CH3 – C  CH + 2Mg(OH)2 The caribide which gives propyne on hydrolysis is

fd l d kckZbZM +d st y vi?kVu lsizksikbZu izkIr gksxhA (A) Al4C3

(B) CaC2

(C) Fe3C

(D*) Mg2C3

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PAGE NO.- 15

Chemical reactions of alkyne (1)

Addition of halogens : Halogens like bromine or chlorine add up to alkyne to form trans dihalides and further addition of halogen give tetrahalo alkane. It is an example of electrophilic addition reaction. This reaction is used as a test for unsaturation.

(1)

– –

–

–

– –

Br Br Br Br2 (1eq.) Br2 (1eq.) R – C  C – R    R – C = C – R      R – C – C – R Br Br Br (Trans-dihalide) (Tetrahalide)

,Yd kbZu d h jklk;fud v fHkfØ ;k gSy kst uksad k ;ksx % gS y ks tu tS l sfd cz ks ehu ;kDykjshu],Yd kbZ u lst q Mu+sij Vª ka l MkbgS y kbM cukrhgS A t ksiq u%gS y ks t u d s;ks x lsVs VkªgS y ks ,Yd s u cukrkgSA ;g ,d bysDVªkWuLusgh;ks xkRed vfHkfØ ;kd kmnkgj.kgSA bl vfHkfØ ;kd kmi;ksx vlar`Irrkd kijh{k.kd jus d sfy , fd ;k t krk gSA Br2 (1eq.) R – C  C – R   

10.

11.

(2)

CCl4 CH3–C  CH + Br2   Product is.mRikn

gS%

(A)

(B)

(C)

(D*) CH3–CBr2–CHBr2

CCl

CH3–C  C–CH2–CH3 + Cl2  4 Product is.mRikn

gS%

(A)

(B)

(C)

(D*) CH3–CH2–CCl2–CCl2–CH3

CCl

CH3–C  C–CH3 + Cl2 (1 eq.)  4 Product is. mRikn (A)

(B*)

(C)

(D)

gS%

Addition of Hydrogen halides : Addition of HX to unsymmetrical alkyne place according to Markovinikov rule. Markovnikov rule : The rule states that the negative part of the attacking species add on the carbon atom containing less number of hydrogen atom. Br HBr HBr ( dark ) R – C  C – H     R–CBr=CH2   R – C – CH 3 Br

– –

9.

Br (1eq.) 2   

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PAGE NO.- 16

(2)

gkbMªkst u gSy kbMksad k ;ksx % vlefer ,Yd kbZu ij HX d k;ksx ekjd ksuhd kWQ fu;e d svuql kj gksrkgSA eks jd ks uhd kW Q fu;e %vfHkd kjd d k_ .kkRed Hkkx ml d kcZu ijek.kqlst qM+rkgS]ft l ij gkbMªkst u ijek.kqv ksad hla[;k d e gksrh gSA –

– –

–

Br Br HBr ( va/ksjsesa ) HBr R – C  C – H  R – C = C – H   R – C – CH 3 H Br (MK)

14.

(3)

CH3–C C – CH3 + HBr  Product is. mRikn

gS%

(A*)

(B)

(C)

(D)

Addition of HBr in Presence of peroxide or Kharash effect or Anti Markovnikov effect : In the presence of peroxide such as benzoyl peroxide [Ph–CO–O–O–CO–Ph] and light, the addition of HBr to unsymmetrical alkyne occurs contrary to Markovnikov’s rule. Peroxide CH3 – C  CH + HBr    CH3– CH2– CHBr2

(3)

ijkWDlkbM d h mifLFkfr esaHBr d k ;ksx ;k [kjk'k izHkko ;k ,s.Vh&eksjd ksuhd kWQ izHkko izd k'krFkkijkWDlkbM d hmifLFkfr es at S l scs Ut kW;y ijkWDlkbM [Ph–CO–O–O–CO–Ph] vlefer ,Yd kbZ u ij HBr d k ;ksx ekjd ksuhd kWQ fu;e d sfoijhr gksrkgSA Peroxide CH3 – C  CH + HBr    CH3– CH2– CHBr2

12.

13.

(4)

Ph–C  CH + HBr  Product is. mRikn

gS%

(A)

(B)

(C*) Ph–CBr2–CH3

(D) Ph–CH2–CHBr2

Ph–C  CH + HBr Peroxide   Product is. mRikn

gS%

(A)

(B)

(C) Ph–CBr2–CH3

(D*) Ph–CH2–CHBr2

Addition of water : Alkyne react with water to form carbonyl compounds, in accordance with the markovinikov rule.

R  C  C  H  H2 O ( Markoniko ff rule)

H | R CCH || | O H ketone ( stable )

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PAGE NO.- 17

(4)

t y d k ;ksx % ekjd ksuhd kWQ fu;e d svuql kj ,Yd kbZu]t y d slkFkvfHkfØ ;kd jd sd kcksZfuy ;kSfxd cukrhgSA

15.

Major Product is : eq [; mRikn (B) CH3–CH2–CHO

CH3– C  CH (A) (C*)

16.

gS%

(D)

Ph–CC–CH3+H2O

Major product is :

(A*)

eq[; mRikn gS%

(B) Ph–CH 2–CH–CH 3 OH

(C) Ph–CH–CH 2–CH 3

(D)

OH

(5)

Oxidation of alkyne : Acidic potassium premanganate or acidic potassium dichromate oxidise alkyne to acids . O.A.   2RCOOH

O.A.  

+ CO2

Alkyne having  C–H will give CO2 as one of the oxidation product by oxidising agent KMnO4 and K2Cr2O7. O.A.

O.A.

 2CH COOH 3

(5)



+ CO2

,Yd kbZu d k v kWDlhd j.k : vEy h; iksVsf'k;e ijeSaXusV ;k vEy h; iksVSf'k;e MkbØ ksesV],Yd kbZu d ksvEy ksaesavkWDlhd `r d j nsrk gSA O.A.   2RCOOH

O.A.  

+ CO2

,Yd kbZ u t ks CH j[krsgSt ksvkW Dlhd kjhvfHkd eZ d KMnO4 rFkkK2Cr2O7 d s}kjkCO2 d s: i esa,d vkW Dlhd kjhmRikn ns rsgSA O.A.   2CH COOH 3

17.

CH3–C  C – CH2–CH3

O.A.  

+ CO2

Products is :

(A) CH3–C–CH3 and CH3COOH

(B) CH3–CH2–CHO and CH3–COOH

O (C*) CH3–CH2–COOH and CH3COOH

(D) CH 3–C–CH 3 and CH 3CHO O

mRikn gS%

CH3–C  C – CH2–CH3 (A)

(B) CH3–CH2–CHO

(C*) CH3–CH2–COOH rFkkCH3COOH

(D)

rFkkCH3–COOH

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PAGE NO.- 18

18.

CH3–CH2–C  CH

Products is :

mRikn gS%

(A) CH3–CH2–COOH & HCOOH (C) & CO2

mRikn gS%

CH3–CH2–C  CH (A) CH3–CH2–COOH (C)

(6)

(B) CH3–COOH & CH3–COOH (D*) CH3–CH2–COOH & CO2

o HCOOH

o CH3–COOH (D*) CH3–CH2–COOH o CO2 (B) CH3–COOH

o CO2

Oxidation of alkyne with dil. KMnO4 (Bayer’s reagent) : H.A.

—CC—  H.A. = Hydroxylating agent. It may be any one of these : (a) Bayer’s reagent: cold dilute 1% alkaline KMnO4 (pink color) or (b) (i) OsO4 (ii) NaHSO3 (6)

ruqKMnO4 ¼cs;j v fHkd eZd ½ }kjk ,Yd kbZu d k v kWDlhd j.k % H.A.

—CC—  H.A. = gkbMª ksfDly hd j.kvfHkd eZd A buesalsd ksbZHkhgksld rkgSA (a) cs ;j v fHkd eZd % B.Mk ruq1% {kkjh; KMnO4 ¼xqy kch jax½ ;k (b) (i) OsO4 (ii) NaHSO3 19.

Baeyer 's Re agent

Ph–C  C–CH3       Product is : or OsO4 / NaHSO3

cs; j vfHkd eZd Ph–C  C–CH3  ;kOsO4 / NaHSO3

20.

mRikn gS%

(A)

(B*)

(C)

(D)

cs; j vfHkd eZd Baeyer 's Re agent CH3–C  CH        Product is : mRikn ;kOsO4 / NaHSO3 or OsO4 / NaHSO3 (A*)

(B)

(C)

(D)

gS%

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PAGE NO.- 19

DPP No. # 41 (JEE-MAIN) Total Marks : 80

Max. Time : 40 min.

ChemINFO Questions ('–1' negative marking) Q.01 to Q.20

(4 marks, 2 min.)

[80, 40]

Preparation of Benzene

csUt hu d k fuekZ.k 1.

By polymerisation of Acetylene : Re d hot iron tube 3HC  CH     

,flfVy hu d scgqy hd j.k }kjk : jDr rIr y kSg ufy d k 3HC  CH        1.

On heating a mixture of sodium benzoate and sodalime, the following is obtained (A) Toluene (B) Phenol (C*) Benzene (D) Benzoic acid

lksfM;e csUt ks,V rFkk lksM k y kbZe d sfeJ.k d ksxeZd jusij]fuEu esalsd kSulk d kcZfud ;kSfxd izkIr gksrk gSA (A) VkW y wbZu (B) fQ ukW y (C*) cs Ut hu (D) cs Ut ksbd vEy 2.

Re d hot iron tube jDr rIr y kSg ufy d k product is mRikn 3CH3–C  CH            

(A)

2.

(B)

(C)

(D*)

By decarboxylation of Benzoic acid : csUt ksbd v Ey d sfod kcksZfDlfy d j.k }kjk : NaOH   

NaOH   

NaOH / CaO

(A*)

+ Na2CO3

( CaO )

   product is mRikn 

3.

3.

gS:

gS:

(B)

(C)

(D)

By catalytic reforming of n-Hexane : Pt, 873 K CH3 – (CH2)4 – CH3   

 H2

Pt, 873 K   

 3H2

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PAGE NO.- 20

n-gs Dlsu d h mRizsjd h; v fHkfØ ;k }kjk : Pt, 873 K   

Pt, 873 K CH3 – (CH2)4 – CH3   

 3H2

 H2

4.

Pt  product is mRikn CH3 – (CH2)5 – CH3 873 K

(A)

4.

gS:

(B*)

(C)

(D)

(C*)

(D)

(C)

(D*)

By reduction of Phenol : fQ ukWy d sv ip;u }kjk:

Zn

 

Zn

5.

 

product is

(A)

(B)

Zn

6.

 

(A)

mRikn gS:

product is mRikn

(B)

gS:

Chemical reactions of Benzene

csUt hu d h jkl k;fud v fHkfØ ;k Benzene has clouds of pi electrons above and below its sigma bond framework. Although benzene’s pi electrons are in a stable aromatic system still they are available to attack a strong electrophile to give a carbocation. This resonance-stabilized carbocation is called a sigma complex because the electrophile is joined to the benzene ring by a new sigma bond. The sigma complex (also called an arenium ion) is not aromatic because the sp3 hybrid carbon atom interrupts the ring of p orbitals. This loss of aromaticity contributes to the highly endothermic nature of thus first step. The sigma complex regains aromaticity either by a reversal of the first step (returning to the reactants) or by loss of the proton on the tetrahedral carbon atom, leading to the substitution product. 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.- 21





The overall reaction is the substitution of an electrophile (E ) for a proton (H) on the aromatic ring: electrophilic aromatic substitution.

csUt hu d sflXekca/kd sÅ ij rFkkuhpsd hvksj ikbZbysDVª kWu vHkzik;kt krkgS A ;|fi cs Ut hu d sikbZby sDVªkW u ,d LFkk;h ,s jks eS fVd ra =kes aik;kt krkgSfQ j Hkh;g izcy bysDVªkWuLus ghd svkØ e.kij miyC/kgksrsgSarFkkft llsd kcZ/kuk;u curk gSA bl vuqukn LFkk;hd `r d kcZ/kuk;u d ksflXekl ad q y d grsgSaD;ksa fd by sDVªkW uLus gh,d u;sflXekca/k}kjkcsUt hu oy ; lst qM +rkgSA flXeklad q y (ft ls,fjfu;e vk;u Hkhd grsgSa A) ,jkseSfVd ughagks rkgSD;ka sfd sp3 la d fjr d kcZ u ijek.kqp d {kd ksad hbl oy ; d ksvo: ) d j nsrk gSA ,sjkseSfVd rk d k gklzgksusij izFke in vR;ar Å "ek'kks"kh izd `fr d k in gksrk gSA flXek ca/k ,s jks eS fVd rkiq u%rc iz kIr d j ys rkgSt c iz Fke in foijhr fn'kkes a(vfHkd kjd fuekZ .kd hrjQ ) gks rkgS;kfQ j prq "Q yd h; d kcZu ijek.kqlsizksVkWu d k fu"d klu gksrk gSft ld sQ y Lo: i izfrLFkkiu mRikn curk gSA 



lEiw.kZvfHkfØ ;kesa,sjkseSfVd oy ; ij izksVkW u (H) d sfy , by sDVªkWuLusgh(E ) d kizfrLFkkiu gksrkgS ]blfy , bl vfHkfØ ;k d ksby sDVªkWuLusgh ,sjkseSfVd izfrLFkkiu v fHkfØ ;k d grsgSA Step 1 : Attack of an electrophile on benzene ring forms the sigma complex H

E

H

E

Resonance hybrid [  – complex] Arenium Ion

in 1 : csUt hu oy ; ij by sDVªkWuLusgh d sv kØ e.k }kjk fl Xek l ad qy curk gSA H

E

H

E

vuqukn lad fjr [  – la d qy ] ,sjsfu;e vk;u Step 2 : Loss of a proton gives the substitution product. in 2 : izksVkWu d sfu"d kl u }kjk izfrLFkkiu mRikn curk gSA H

E

E 

Nu :  

7.

+ Nu – H

Electrophilic substitution reaction is characterstics reaction of : (A) Alkane (B) Alkyl halide (C) Alkene (D*) Aromatic compound

by sDVªksuLusghizfrLFkkiu vfHkfØ ;kfuEu esalsfd ld hvfHky k{kf.kd vfHkfØ ;kgS\ (A) ,Yd s u (B) ,fYd y gS y kbM (C) ,Yd hu (D*) ,jkS es fVd ;kS fxd

(1) Halogenation

AlCl3 + Cl2   

Electrophile is Cl

(1) gS y kst uhd j.k

AlCl3 + Cl2   

by sDVªkWuLusghClgSA

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PAGE NO.- 22

8.

The following reaction is : FeBr

3 C 6H6  Br2    C 6H5Br  HBr (A) An electrophilic addition reaction (B) A nucleophilic substitution reaction (C*) An electrophilic substitution reaction (D) A free radical substitution reaction fuEu vfHkfØ ;k gS:

FeBr

3 C 6H6  Br2    C 6H5Br  HBr

(A) ,d by s DVªkWuLusgh;ksXkkRed vfHkfØ ;k (C*) ,d by s DVªkW uLus ghiz frLFkkiu vfHkfØ ;k 9.

10.

ukfHkd Lus ghizfrLFkkiu vfHkfØ ;k eqDrew y d izfrLFkkiu vfHkfØ ;k

Electrophile in the case of chlorination of benzene in the presence of FeCI3 : FeCI3 d h mifLFkfr es acsUt hu d sDy ksjhuhd j.k d k by sDVªkWuLusgh fuEu gS% (A*) Cl+ (B) Cl– (C) Cl (D) FeCl3 (2) Nitration

conc. HNO3  H2 SO 4      

Electrophile is NO2

(2) ukbVª hd j.k

lkUnzHNO3  H2SO4      

bysDVª kWuLusghNO2gSA

Which of the following reactions takes place when a mixture of concentrated HNO3 and H2SO4 reacts on benzene at 350 K (A) Sulphonation (B*) Nitration (C) Hydrogenation (D) Dehydration t c 350 K rki ij cs Ut hu d hvfHkfØ ;klkUnz HNO3 rFkkH2SO4 d sfeJ.kd slkFkd jkrsgS ]rksfuEu es alsd kS ulhvfHkfØ ;k

gks rhgSA (A) lYQ ku shd j.k

11.

(B) ,d (D) ,d

(B*) ukbVª hd j.k (C) gkbMª ks t uhd j.k

(D) fut Z y hd j.k

conc. H SO

(3) Sulphonation

2 4      

Electrophile is SO3

(3) l YQ ks uhd j.k

lkUnzH2SO4     ; k H2S2O7

by sDVªkWuLusghSO3 gSA

or H2 S 2O 7

Which of the following reactions is not an electrophilic substitution ? fuEu esalsd kSulhvfHkfØ ;k,d by sDVªkuLusghizfrLFkkiu vfHkfØ ;kughagS?

Cl

OH 

(A)

+ OH

(B*)

NO2

(C)

(D)

(4) Friedel–crafts alkylation

(4) fÝ My -Ø k¶V

,Yd y hd j.k

NO2

+ NO2

AlCl

3  + R–Cl  

AlCl

3  + R–Cl  

Electrophile is R

by sDVªkWuLusghRgSA

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PAGE NO.- 23

12.

13.

Benzene reacts with CH3CI in the presence of anhydrous AlCl3 to form (A*) Toluene (B) Chlorobenzene (C) Benzylchloride (D) Xylene fut Zy AlCl3 d h mifLFkfr esacsUt hu d h vfHkfØ ;k CH3Cl d slkFk d jkusij fuEu esalsd kSulk ;kSfxd (A*) VkW y wbZu (B) Dy ks jks cs Ut hu (C) cs fUt y Dyks jkbM (D) t kbyhu C6H6 + CH3 Cl

anhydrous

   C6H5CH3 +HCI is an example of : AlCI 3

(A*) Friedel-Craft's reaction (C) Wurtz reaction

(B) Kolbe's synthesis (D) Grignard reaction

fut Zy C6H6 + CH3 Cl    C6H5CH3 +HCI A lCI

fuEu vfHkfØ ;kd kmnkgj.kgS%

(A*) fÝ My&Ø k¶V vfHkfØ ;k (C) oq VZ t vfHkfØ ;k

(B) d ks Ycsla'y s "k.k (D) fxz U;kj vfHkfØ ;k

3

3 + CH3–CH2–Cl AlCl   product is mRikn

14.

(A)

(B)

gS:

(C)

acid y wbZl vEy + CH3–CH–CH3 Lewis      product is mRikn

15.

izkIr gksrk gS%

(D*)

gS:

Cl CH2–CH2–CH3 (A)

CH2–CH=CH2 (B*)

(C)

(D)

CH3 3 + CH3–C–Cl AlCl   product is mRikn gS:

16.

CH3 CH3

CH3

CH3–C–CH3

CH2–CH=CH2

CH—CH CH3

(A*)

(B)

(C)

(D)

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PAGE NO.- 24

Cl 17.

3 AlCl   product is mRikn gS:

+

CH2–CH2–CH3 (A)

(B)

(C*)

(D)

CH2–CH2–CH3

AlCl3 + CH 3 COCl   

(5) Friedel–crafts acylation

(5) fÝ My –Ø k¶V

AlCl3 + CH3COCl  

,fl y hd j.k

Electrophile is

by sDVªkWuLusgh

gSA

3 + CH3–C–Cl AlCl   product is mRikn gS:

18.

O

O (A)

CH3

(B)

O

(C)

CH3

(D*)

Cl

AlCl 3 + Ph–C–Cl   product is mRikn gS:

19.

O Ph

O

O (A)

Ph

O

(B) Cl

Ph

(C*)

(D)

Cl

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PAGE NO.- 25

3 AlCl   product is mRikn gS:

20.

O O

(A)

(B)

Lecutre Sub-topic(s) Nam e No. (No. of Lectures)

L70

(C)

Hom e Work Sheet

Hom e Work NCERT Pg.N

Hand out

Discussion of DPP's of Hydrocarbon Ex.1 (S)

L71

(D*)

Discussion of DPP's of Hydrocarbon

Ex.1 (O)

Th.

Ex.2 (O) Ex.2 (I) Ex.2 (M)

XI

370

XII XI

397

Prob. XII

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PAGE NO.- 26

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 42

This DPP is to be discussed in the week (16.11.2015 to 21.11.2015) 1. 2. 3.

Course of the week as per plan : Discusion of DPPs of hydrocarbon Course covered till previous week : Revision of GOC-I Target of the current week : Discusion of DPPs of hydrocarbon

4.

DPP Syllabus : Revision of GOC-I

DPP No. # 42 (JEE-ADVANCED) Total Marks : 45

Max. Time : 30 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15

(3 marks, 2 min.)

[45, 30]

ANSWER KEY DPP No. # 42 (JEE-ADVANCED) 1. 8.* 15.

(A) 2. (D) 3. (B) 4. (BCD) 9.* (AD) 10.* (ABD) 11. (A) - (p,r) ; (B) - (p,s) ; (C) - (p,q,r) ; (D) - (p,q,r)

(D) 6

5. 12.

(D) 8

6.* 13.

(BC) 5

7.* 14.

(ACD) 5

1.

Which of the following species have all equally contributing resonating structures ? (Electronic effect(O))

fuEu esalsd kSulh Lih'kht +esalHkh leku : i lslg;ksx iznku d jusoky h vuquknh lajpuk,sagS\ (A*) 2.

(B) CH2=CH–CH=CH2

– + – (C) N =N=N

The species which is stable at room temperature is :

(D) [Topic-GOC-II(O)]

fuEUk esalsd kSulh Lih'khT+k d ejsd srki ij LFkk;h gS%

(A)

3.

(B)

Consider the following carbocation,

(C)

(D*)

[Electronic Effects / Carbocation/T(O))]

If we assume that all possible rearrangements are taking place by various shifts, then which of the following rearranged carbocation cannot be formed?

fuEufyf[kr d kcZèkuk;u ij fopkj d hft ,%

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PAGE NO.- 1

;fn ,sl keku fy ;kt k;s]fd fofHkUu foLFkkiuksad s}kjkmijks Dr d kcZ/kuk;u esalHkhlEHko iqufoZU;kl (rearrangements) d jk;sx,]rksfuEu esalsd kSulkiquZfoU;kflr (rearranged) d kcZèkuk;u ughacusxk\

(A)

(B*)

(C)

(D)

Sol. Sol.

The +ve charge of carbocation is not possible at bridge-head (B.H.) position in bicyclic systems. f}pØ h; ;kSfxd ksa(Bicyclo compounds) d sd kcZ/kuk;u esalsrqcU/k (bridge-head) (B.H.) ij /kukos'klEHko

4.

Which is correct for carbocation stability?

ughagSaA

[Electronic Effects / Carbocation/M(O)]

fuEu esalsd kSulk d kcZ/kuk;u d sLFkkf;Ro d k lgh Ø e gS\

(A)

(C)

5.

+

CH2

<

(B)

<

<

(D*)

<

The correct order of electron density in aromatic ring of following compounds is :

[Topic-GOC-II(O)]

fuEufy f[kr ;kSfxd kasesacsUt hu oy ; ij by sDVªkWu ?kuRo d k lgh Ø e gksxk %

I

II

III

IV

Sol.

(A) IV > III > II > I (B) I > II > III > IV (C) IV > II > I > III On the basis of electronic effect. ¼by s DVªkWfud izHkko d svk/kkj ij½

(D*) IV > II > III > I

6.*

Which of the following carbocation would not likely rearrange to a more stable carbocation ?

fuEUkesalsd kSulkvf/kd LFkk;hd kcZ/kuk;u esaiqufoZU;kfLkr ughagksxk\ (A)

7.*

(B*)

(C*)

(D)

Which of the following carbocations is/are expected to undergo rearrangement.

(Carbocation(O))

fuEu esalsd kSulk@d kSulsd kcZ/kuk;u esaiquZfoU;kl lEHko gS\ CH2

(A*)

CH2

Sol.

H CH3 | CH3–CH2–C–C CH3 (B) | CH2–CH2–CH3

(C*) O

CH2 – CH3 | HO – C – C – CH3 (D*) | | CH3 CH3

CH3

(A) (B) No rearrangement

iq uZ foU;kl ughagksrkgS A 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.- 2

(C)

O Complete Octet

O

v"Vd iw .kZ Et CH3 | | HO – C – C – CH3 | CH3 Complete Octet

Et | HO – C – C – CH3 (D) | | CH3 CH3

v"Vd iw.kZ 8.*

In which of the following first carbocation is more stable than second one ? fuEu esalsd kSulsfod Yiksaesaigy k d kcZ/kuk;u nwl jslsvf/kd LFkk;h gS\ [General Organic Chemistry]

(A)

,

(B*)

(C*)

Sol.

,

(D*)

,

,

(B)

has extended conjugation. ¼foLrkfjr

la;qXeu gSA½

(C)

has +M effect of –OCH3. (–OCH3 lew g

(D)

after delocalisation gets +M effect of –OMe.

d k +M izHkko gSA½

¼foLFkkuhd j.k d jd s–OMe d k +M izHkko fey t krk gSA½ 9.*

Observe each pair of cations. In which case (s) first is more stable than the second :

fuEu /kuk;u d s;qXeksad k izs{k.k d hft ,A buesalsfd l ;qXe esaigy k /kuk;u nwl jsd h vis{kk vf/kd LFkk;h gS:

(A*)

(C)

,

(B)

,

(D*)

,

,

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PAGE NO.- 3

Sol.

In (W) since C–H bond is weaker than C–D bond so hyperconjugation stability is more in I. In (X) only +I effect is present which is more for –C(CD3). In (Y) only +I which is more for –CD3 In (Z) –I effect of –CCl3 group will make II cation highly unstable. (W) es aC–H ca/kC–D ca/k lsnqcZy gksrk gS]blfy , vfrla;qXeu LFkkf;Ro I esavf/kd gksxkA (X) es ad soy +I izHkko mifLFkr

gS]blfy , –C(CD3) d kizHkko vf/kd gksxkA (Y) es ad soy +I izHkko mifLFkr gS]blfy , –CD3 d kizHkko vf/kd gksxkA (Z) es a–CCl3 d k–I izHkko gS ]blfy , II /kuk;u vR;f/kd vLFkkbZgksxkA 10.*

Which of the given following stability order is correct.

[Ref. RGP sir]

fuEu esalsd kSulk LFkkf;Ro d k lgh Ø e gSA (A*)

O

O

<

H 2C

<

(B*)

O

>

(C)

(D*)

11.

CH2 > Cl

H2N

CH2

CH2 F

How many carbocations given below are more stable than sec. butyl carbocation t-butyl carbocation Benzyl carbocation Allyl carbocation Cyclopropenyl cation Tropylium cation n- butyl carbocation cyclopropylmethyl carbocation

fuEu esalsfd rusd kcZ/kuk;u f}rh;d C;wfVy d kcZ/kuk;u lsvf/kd LFkk;hgSA t-C;w fVy d kcZ/kuk;u csfUt y d kcZ/kuk;u ,fyy d kcZ/kuk;u lkbDyksçks is fuy d kcZ /kuk;u Vªksikbfy;e d kcZ /kuk;u n- C;w fVy d kcZ/kuk;u lkbDyks çks ikbyes fFky d kcZ /kuk;u Ans.

6

Sol. 12.

The total number of hyperconjugable hydrogen atoms in the given two species are : [Topic-GOC-II(O)]

uhpsnh xbZnksuksaLih’kht ksaesavfrla;qXeh gkbMªkst u ijek.kqv ksad h d qy la[;k fuEu gksxh %

(i) Ans.

(ii)

8

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PAGE NO.- 4

13.

How many groups (each attached with benzene ring) show + M (Mesomeric) effect? (Electronic effect(O)) fuEu esalsfd ruslewg (t kscsUt hu oy ; lst qM +sgSa) + M (esl ksesfjd ) izHkko n’’'kkZrsgSa\

N(CH3)2

Ans.

O || C — OCH3

SH

O || O — C — CH3

5

Sol.

have + M group. ;slHkh + M lewg j[krsgSaA 14.

How many species out of the following are aromatic ?

[Topic-GOC-II(O)]

fuEu esalsfd ruh Lih’'kht ,sjksesfVd gS\

Ans. Sol.

5 Aromatic species are

fuEu ,sjksesfVd Lih’'kht gS%

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PAGE NO.- 5

15.

Match the characteristics of Column-II with the compounds given in Column-I. Column-I Column-II (Electronic effects(O))

(A)

(p) Resonance possible

(B)

(q) Rearrangement possible

(C)

(r) Hyperconjugation possible

(D)

(s) Steric inhibition of resonance possible (SIR)

d kWy e-II d sy {k.kksad ksd kWy e-I d s;kSfxd kslsfey ku d hft ,A d kWy e-I d kWy e-II

Sol.

(A)

(p) vuq ukn

laHko

(B)

(q) iq ufoZ U;kl

(C)

(r) vfrla ;qXeu

(D)

(s) vuq ukn

la Hko [Rearrangement]

laHko

d kf=kfoe ckèkk (SIR) lEHko

(A) - (p,r) ; (B) - (p,s) ; (C) - (p,q,r) ; (D) - (p,q,r)

Lecutre Sub-topic(s) Nam e (No. of Lectures) No.

Hom e Work Sheet Ex.1 (S)

L72

Revision of GOC

Hom e Work NCERT

Th.

Ex.1 (O) Ex.2 (I)

Handout

XII

Ex.2 (O) Ex.2 (M)

XI

Pg.N 379 to 388

XI

397

Prob. XII

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PAGE NO.- 6

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 43

This DPP is to be discussed in the week (23.11.2015 to 28.11.2015) 1. 2. 3. 4.

Course of the week as per plan : Introduction of s-block elements, Physical Properties of s-block elements, Chemical Properties of s-block elements. Course covered till previous week : Revision of GOC-I Target of the current week : Introduction of s-block elements, Physical Properties of s-block elements, Chemical Properties of s-block elements. DPP Syllabus : Introduction of s-block elements, Physical Properties of s-block elements, Chemical Properties of s-block elements.

DPP No. # 43 (JEE-MAIN) Total Marks : 61

Max. Time : 38 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15

(3 marks, 2 min.)

[45, 30]

ChemINFO : 5 Questions ('–1' negative marking) Q.16 to Q.19

(4 marks, 2 min.)

[16, 08]

(A) (C) (A)

7. 14.

ANSWER KEY DPP No. # 43 (JEE-MAIN) 1. 8. 15.

(B) (B) (D)

2. 9. 16.

(D) (C) (D)

3. 10. 17.

(D) (C) (B)

4. 11. 18.

(D) (C) (B)

1.

Which of the following gives propyne on hydrolysis ?

5. 12. 19.

6. 13.

(A) (D)

(C) (B)

[s-block Elements]

fuEu esalsd kSu]t y vi?kVu }kjk izksikbu cukrk gS\ (A) Al4C3 2.

(B*) Mg2C3

(C) B4C

(D) La4C3

Which of the following compounds on thermal decomposition yields a basic as well as an acidic oxide ?

fuEu esalsfd l ;kSfxd d srkih; vi?kV~u ij ,d {kkjh; vkWDlkbM d slkFk&lkFk,d vEy h; vkWDlkbM HkhizkIr gks rkgS\ (A) KClO3 Hint : 3.

(B) NaNO3

(C) K2CO3

[s-block Elements] (D*) MgCO3

MgCO3  MgO + CO2 (Basic) (Acidic) Which of the following statements is incorrrect ? [s-block Elements] (A) The superoxide ion (i.e. O2–) is stable only in presence of larger cations such as K+, Rb+, Cs+. (B) Alkali metals are normally kept in kerosene oil. (C) All the alkali metal hydrides are ionic solids with high melting points. (D*) The concentrated solution of alkali metals in liquid ammonia is paramagnetic in nature. 

fuEu esalsd kSulk d Fku vlR; gS% (A) lq ijvkWDlkbM vk;u (O2–), cM+s/kuk;uksat Sl sK+, Rb+, Cs+ d slkFkgh LFkk;hgksrk gSA (B) {kkjh; /kkrq v ksad kslkekU;r%d Sjksl hu esaj[kkt krkgSA (C) lHkh{kkjh; /kkrq v ksad sgkbMªkbM]vk;fud Bksl gksrsgSrFkkmPp xy ukad j[krsgSaA (D*) {kkjh; /kkrq v ksad k nzo veksfu;kesalkUnzfoy ;u]çd `fr esavuqp qEcd h; gksrkgSA Sol.

(A) Bigger anion is stabilised by bigger cation through lattice energy effect. (B) Because of their high reactivity towards air and water. (C) True Statement 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

(D) In concentrated solution, unpaired electrons with opposite spins paired up – forming the solution diamagnetic. (A) cM+ s_ .kk;u t ky d Å t kZizHkko lscM+s/kuk;u }kjk LFkk;hd `r jgrk gSA (B) D;ks afd bud h ok;qo t y d sizfr mPpfØ ;k'khy rk gksrh gSA (C) lR; d Fku (D) lkfUnz r foy ;u esafopfjr pØ .k ;qDr v;qfXer by sDVªkWu ;qfXer gksd j fopy u d ksizfrpqEcd h; cukrsgSA 4.

Identify the correct statement : [s-block Elements] (A) Sodium metal can be prepared by the electrolysis of an aqueous solution of NaCl. (B) Sodium metal can be kept under ethyl alcohol. (C) Sodium metal is insoluble in liquid NH3 at low temperature. (D*) Elemental sodium is easily oxidised.

lgh d Fku d h igpku d hft , % (A) NaCl d st y h; foy ;u d sfo|q r vi?kV~u d s}kjk lksfM;e /kkrqcuk;h t k ld rh gSA (B) lks fM;e /kkrqd ks,fFky ,Yd ksgkWy esaj[kkt k ld rkgSA (C) fuEu rkieku ij lks fM;e /kkrq]nzo NH3 esavfoy s; gSA (D*) rkfRod lks fM;e]vklkuhlsvkWDlhd `r gkst krkgSA 5.

Sol.

gy 6.

Sol.

gy 7.

The pair of compounds which cannot exist together in solution is : [s-block Elements][JEE-1986, 1 M] (A*) NaHCO3 and NaOH (B) Na2CO3 and NaHCO3 (C) Na2CO3 and NaOH (D) NaHCO3 and NaCI ;kSfxd ksd k d kSulk ;qXe]foy ;u esa,d lkFk ughajg ld rk gS% [JEE-1986, 1 M] (A*) NaHCO3 o NaOH (B) Na2CO3 o NaHCO3 (C) Na2CO3 o NaOH (D) NaHCO3 o NaCI Since NaHCO3 is an acid salt of H2CO3. it reacts with NaOH to form Na2CO3 and H2O. NaHCO3 + NaOH  Na2CO3 + H2O. pwafd NaHCO3, H2CO3 d k,d vEy y o.kgSA ;g NaOH d slkFkvfHkfØ ;kd j Na2CO3 o H2O cukrkgSA NaHCO3+ NaOH  Na2CO3 + H2O. Calcium is obtained by : (A*) electrolysis of molten CaCl2 (C) chemical reduction of CaCl2

[s-block Elements] [JEE-1980] (B) electrolysis of solution of CaCl2 in water (D) roasting of lime stone. d SfY'k;e fuEu d s}kjk çkIr gksrk gS% [JEE-1980] (A*) xfy r CaCl2 d sfo?kq rvi?kVu }kjk (B) t y es aCaCl2 foy ;u d sfo?kqrvi?kVu }kjk (C) CaCl2 d sjklk;fud vip;u }kjk (D) y kbe LVks u d sHkt Zu }kjk Ca is obtained by electrolysis of molten mixture of CaCl2 mixed with CaF2. Ca, CaF2 d slkFk fefJr CaCl2 d sxfy r feJ.kd sfo?kq rvi?kVu }kjk çkIr gksrk gSA Which of the following statements is true for all the alkali metals ? [s-block Elements] (A) Their nitrates decompose on heating to give the corresponding nitrites and oxygen. (B) Their chlorides are deliquescent and crystallise as hydrates. (C*) They react with water to form hydroxide and hydrogen. (D) They readily react with halogens to form ionic halides, M+X–.

lHkh {kkjh; /kkrqv ksad sfy , fuEu esalsd kSulk d Fku lR; gS\ (A) bud sukbVª sV]xeZd jusij fo?kfVr gksd j lacaf/kr ukbVªkbV rFkk vkWDlht u nsrsgSA (B) bud sDy ks jkbM]izLosn~d ¼deliquescent½ gksrsgaSrFkk t y ;ksft r : i esafØ LVy hd `r gksrsgSaA (C*) ;st y d slkFkfØ ;kd jd sgkbMª ksDlkbM rFkkgkbMªkst u cukrsgaSA (D) ;sgS y kst u d slkFkrqjUr fØ ;kd jd svk;fud gSy kbM , M+X– cukrsgaSA Sol.

(A) 4 LiNO3  2Li2O + 4NO2 + O2 2NaNO3  2NaNO2 + O2 (similar decomposition with the nitrates of K, Rb and Cs) (B) Only LiCl is deliquescent and crystallises as a hydrate LiCl.2H2O (C) 2M + 2H2O  2M+ + 2OH– + H2 (M = an alkali metal) (D) Halides of Li are covalent in nature.

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(A) 4 LiNO3  2Li2O + 4NO2 + O2 2NaNO3  2NaNO2 + O2 (ukbVª kbV K, Rb rFkk Cs d slkFklkeu fo?kVu) (B) d os y LiCl izLos| ; gSrFkk gkbVªsV LiCl.2H2O d s: i esafØ LVy hd `r jgrk gSA (C) 2M + 2H2O  2M+ + 2OH– + H2 (M = ,d {kkj /kkrq ) (D) Li d sgS y kbM lgla;kst hizd `fr d sgksrsgSA 8.

Which of the following reactions of potassium superoxide supply oxygen gas in the breathing equipments used in space and submarines ? [s-block Elements] (1) reaction of superoxide with nitrogen in the exhaled air (2) reaction of superoxide with moisture in the exhaled air (3) reaction of superoxide with carbon dioxide in the exhaled air (A) (1), (2) and (3) (B*) (2) and (3) only (C) (2) only (D) (1) and (2) only

varfj{krFkkiaM qfCc;ksaesafuEu esalsiksVsf'k;e lqij vkWDlkbM d hd kSulhvfHkfØ ;k]'olu ;a=kesavkWDlht u xSl d sçokg d sfy , iz;qDr d h t krh gS\ [s-block Elements] (1) fu"d kflr ok;qes aukbVªkst u d hlqij vkWDlkbM d slkFkvfHkfØ ;kA (2) fu"d kflr ok;qes auehd hlqij vkWDlkbM d slkFkvfHkfØ ;kA (3) fu"d kflr ok;qes ad kcZuMkbZv kWDlkbM d hlqij vkW DlkbM d slkFkvfHkfØ ;kA (A) (1), (2) rFkk(3) (B*) d s oy (2) rFkk(3) (C) d s oy (2) (D) d s oy (1) rFkk(2) Sol. 9.

Sol.

10.

(2) KO2 + 2H2O  KOH + H2O2 + 1/2O2 (3) 4KO2 + 2CO2  2K2CO3 + 3O2 STATEMENT-1 : Lithium is the most powerful reducing agent and sodium is the least powerful reducing agent amongst the alkali metals in aqueous solutions. [s-block Elements] STATEMENT-2 : Lithium has the highest hydration enthalpy and the sodium the least value. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True d Fku-1 : t yh; foy;u es ]{kkjh; /kkrq a v ks aes ayhfFk;e lclsçcy vipk;d d kjd gSrFkklks fM;e lclsnq cZ y vipk;d gS A d Fku-2 : y hfFk;e mPpre foy k;d u ,UFkSYih j[krk gSrFkk lksfM;e]lclsd eA (A) d Fku& 1 lR; gS ]d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k gSA (B) d Fku& 1 lR; gS ]d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k ughagSA (C*) d Fku& 1 lR; gS ]d Fku& 2 vlR; gSA (D) d Fku& 1 vlR; gS ]d Fku& 2 lR; gSA  + (i) E Li /Li = –3.04 ; Na+ / Na = – 2.71 which is least among the alkali metals. (ii) Hydration enthalpy / KJ mol–1 Li = – 506 ; Na = – 406 ; Cs has the least Hhyd = – 276 (i) E Li+/Li = –3.04 ; Na+ / Na = – 2.71 t ks{kkjh; /kkrq v ksaesalclsd e gSA –1 (ii) foy k;d u Å t kZ/ KJ mol Li = – 506 ; Na = – 406 ; Cs , Hfoy k;d u = – 276 d k eku d e j[krk gS A Which of the following is correct order regarding properties mentioned.

(Electronic effect(O))



(A) CH3–CH2+ > CH3 – CH– CH3 –

(Stability order)

(B) CH3– CH2 < CH3 – CH– CH3

(Stability order)

(C*) HCOOH > CH3CH2COOH (D) CH3–CH2–F > CH3–CH2–NO2

(Acidic strength) (Dipole moment)

uhpsn'kkZ;sx;sxq.kksad slUnHkZesad kSulk fod Yi lgh Ø e iznf'kZr d jrk gS\ 

(A) CH3–CH2+ > CH3 – CH– CH3 –

(LFkkf;Ro d kØ e)

(B) CH3– CH2 < CH3 – CH– CH3

(LFkkf;Ro d kØ e)

(C*) HCOOH > CH3CH2COOH (D) CH3–CH2–F > CH3–CH2–NO2

(vEy h;rkd kØ e) (f}/kq o vk?kw z .kZd kØ e)

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PAGE NO.- 3

11.

Which of the following statements is not true about the dilute solutions of alkali metals in liquid ammonia ? (A) They are deep blue coloured solutions. [s-block Elements] (B) They are highly conducting in nature. (C*) They are diamagnetic in nature. (D) Ammoniated cation and solvated electron are formed in the solution.

nzo veksfu;k esa{kkjh; /kkrqv ksad sruqfoy ;u d sfy , fuEu esalsd kSulk d Fku lR; ughagS\ (A) ;g xgjsuhy sja x d k foy ;u gSA (B) ;g iz d `fr esavR;f/kd pky d gksrk gSA (C*) ;g iz d `fr esaizfrpqEcd h; gksrkgSA (D) foy ;u es aveksfud `r /kuk;u rFkk foy ;hd `r by sDVªkWu curk gSA Sol.

M + (x + y)NH3  [M(NH3)x]+ + [e(NH3)y]– It is paramagnetic due to the presence of the unpaired electrons

v;qfXer by sDVªkWu d h mifLFkfr d sd kj.k ;g izfrpqEcd h; gksrk gSA 12.

Which of the following has the highest solubility in water ?

[s-block Elements]

fuEu esalsfd ld h]t y esafoy s;rk lokZfèkd gksrh gS\ (A) LiOH 13.

(B) KOH

(C*) CsOH

(D) RbOH

Which is correct order of ionic mobility in aqueous medium – (A) Li(aq )

 Na(aq )



Rb(aq )

(B)

(C) Li(aq )  Na(aq )  K (aq )

 Al(3aq )



[E] (SBC (I))  Mg(2aq )



Na(aq )

(D*) Both (A) & (B)

Tky h; ek/;e esvk;fud xfr'khy rk d k lgh Ø e gS%

Sol.

(A) Li(aq )  Na(aq )  Rb(aq )

2   (B) Al(3aq )  Mg( aq )  Na( aq )

(C) Li(aq )  Na(aq )  K (aq )

(D*) (A) o (B) nks uksa

Smaller the ion, higher the hydration, lower the mobility.

vk;u ft ruk NksVk gksrk gS]mruk gh t y &vi?kVu mPp gksrk gS]xfr'khy rk mruh gh d e gksrh gSA 14.

The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order. K2CO3 (), MgCO3 (), CaCO3 (), BeCO3(V) [s-block Elements]

fuEufy f[kr ;kSfxd ksad ksmud srkih; LFkkf;Ro d sc<+rsgq, Ø e esaO;ofLFkr fd ;k x;k gSA lgh Ø e d ksigpkfu,A Sol.

K2CO3 (), MgCO3 (), CaCO3 (), BeCO3(V) (A)  <  <  < V (B*) V <  <  <  (C) V <  <  <  (D)  < V <  < . As the size of cation decreases, the extent of polarisation increases so covalent character  and stability 

gy &

t Sl s& t Sl s/kuk;u d kvkd kj ?kVrk gS]/kzqo.k d sc<+uslslgla;kst h y {k.k c<+rsgSrFkkLFkkf;Ro ?kVrk gSA

15.

Which of the following statement is incorrect about the alkali metals ? [s-block Elements] (A) All alkali-metal salts impart a characteristic colour to the Bunsen flame. (B) The correct order of increasing thermal stability of the carbonates of alkali metals is Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 . (C) Among the alkali metals, cesium is the most reactive. (D*) All are incorrect.

{kkjh; /kkrqv ksad sfy , fuEu esslsd kSulk d Fku vlR; gS\ (A) lHkh {kkjh; /kkrq v ksad sy o.k]cqUlsu Toky k d ks,d y kf{.kd jax iznku d jrsgaSA (B) {kkjh; /kkrq v kasd sd kckuZsVs d srkih; LFkkf;Ro d klghc<+ rkgq v kØ e Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 gS A (C) lHkh{kkjh; /kkrq v ksaesalslhft +;e lokZfèkd fØ ;k'khy gSA (D*) lHkhxy r gS aA

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PAGE NO.- 4

ChemINFO-6.1

s-BLOCK ELEMENT Calcium oxide & Hydroxide

Daily Self-Study Dosage for mastering Chemistry

Reactions Charts (Quick Revision) : Cl2

CaOCl2 (Bleaching powder)

Na2CO3 + H2O

Ca(OH)2 (slaked lime)

SiO2

NaOH (Castic-soda) 1 part slaked lime + 3 or 4 parts silica + water (mortar) used as a building material

CaO Quicklime

Absorbs CO2 gas

+ Slaking with NaOH solution

+ Coke Heated in electric 0 furnace at 2000 C

NaOH + Ca(OH)2 (Soda lime)

CaC2

N2

CaCN2 + C (Nitrolium) use as a fertilizer

H2O

NH3

H2O C2H2 (Acetylene)

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

17.

18.

19.

Which hydrocarbon obtained by the hydrolysis of calcium carbide? (A) Methane (B) Ethane (C) Ethene

(D*) Ethyne

The formula of slaked lime is : (A) CaO (B*) Ca(OH)2

(D) None of these

(C) Ca2O2

What will obtained when slaked lime react with Na2CO3 ? (A) Na2O2 (B*) NaOH (C) NaNH2

(D) NaNO3

Which product will obtain when CaC2 react with N2 ? (A*) CaCN2 + C (B) Ca(CN)2 (C) Ca(OH)2

(D) CaO

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PAGE NO.- 5

ChemINFO-5.1

s-BLOCK ELEMENT Calcium oxide & Hydroxide

Daily Self-Study Dosage for mastering Chemistry

v fHkfØ ;k pkVZ ¼'kh?kz iquZjkoy ksd u½ : Cl2

CaOCl2 (fojatd pw.kZ)

Na2CO3 + HO 2

Ca(OH)2 (cw>k pwuk)

NaOH (dkWfLVd&lksMk) 1

SiO2

CaO cw>k gqvk pwuk) (fcuk

CO2

NaOH

foy;u ds lkFk cw>kuk

+

dksd

Hkkx cw>k pwuk + 3 ;k 4 Hkkx flfydk + ty (eksVkZV) bldk mi;ksx Hkou fuekZ.k djus okys inkFkksZ esa fd;k tkrk gSA xSl dk vo’kks"k.k

NaOH + Ca(OH)2 (lksMkykbZe)

CaC2

N2

2000ºC

rki ij oS|qr HkV~Vh esa xeZ djuk

CaCN 2+ C (ukbZVªksfy;e)

HO 2

NH3

moZjd dh rjg iz;qDRk HO 2 C2H2

¼,sflfVyhu½

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

d SfY'k;e d kckZbM d st y vi?kVu ij fuEu esalsd kSulkgkbMªksd kcZu IkzkIr gksrkgS? (A) eS Fksu (B) ,s Fks u (C) ,s Fkhu (D*) ,s FkkbZ u

17.

cw> sgq,spwusd k lw=k gS: (A) CaO

18.

(C) Ca2O2

(D) bues alsd ksbZugha

fuEu esalsd kSulk mRikn izkIr gksrk gSt c cw> sgq,spwusd h vfHkfØ ;k Na2CO3 d slkFk d jkrsgS? (A) Na2O2

19.

(B*) Ca(OH)2

(B*) NaOH

(C) NaNH2

(D) NaNO3

fuEu esalsd kSulk mRikn izkIr gksrkgSt c CaC2 d hvfHkfØ ;kN2 d slkFk d jkrsgS? (A*) CaCN2 + C

(B) Ca(CN)2

(C) Ca(OH)2

(D) CaO

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PAGE NO.- 6

Lecutre Sub-topic(s) Name (No. of Lectures) No.

Home Work NCERT

Home Work Sheet

Pg.N

Handout

291 to 292 Ex.1 (S) L73

Introduction of s-block elements

A-1 to A-4

Ex.1 (O)

A-1 to A-7

Ex.2 (O)

1

XI 298 to 299 XII XI

305

Prob.

Ex.2 (I) Ex.2 (M)

Th.

1

XII 292 to 293

Ex.1 (S)

L74

Physical Properties of s-block elements

Th.

XI 299 to 300

Ex.1 (O)

XII

Ex.2 (O)

XI

Ex.2 (I)

Prob.

Ex.2 (M)

XII

Ex.1 (S)

XI

302, 305 306

293 to 296

L75

Chemical Properties of s-block elements

Th.

300 to 302

Ex.1 (O)

XII

Ex.2 (O)

XI

Ex.2 (I) Ex.2 (M)

Prob.

305, 294 295

XII

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PAGE NO.- 7

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 44

This DPP is to be discussed in the week (30.11.2015 to 05.12.2015) 1. 2. 3.

Course of the week as per plan : Important compounds of s-block elements, Discussion. Course covered till previous week : Introduction of s-block elements, Physical Properties of s-block elements, Chemical Properties of s-block elements. Target of the current week : Important compounds of s-block elements, Discussion.

4.

DPP Syllabus : Important compounds of s-block elements, Discussion.

DPP No. # 44 (JEE-ADVANCED) Total Marks : 71

Max. Time : 43 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.6 Multiple choice objective ('–1' negative marking) Q.7 to Q.9 Comprehension ('–1' negative marking) Q.10 to Q.12 Integer type Questions ('–1' negative marking) Q.13 Match the Following (no negative marking) Q.14 ChemINFO : 5 Questions ('–1' negative marking) Q.15 to Q.19

(3 (4 (3 (4 (8 (4

marks, 2 min.) marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.) marks, 2 min.)

[18, [12, [09, [04, [08, [20,

12] 06] 06] 03] 06] 10]

ANSWER KEY DPP No. # 44 (JEE-ADVANCED) 1. 8. 13.

(A) 2. (C) 3. (C) 4. (C) 5. (D) 6. (D) 7. (ABCD) (ABCD) 9. (AD) 10. (A) 11. (A) 12. (A) (i) Na2S + 4Na2O2  Na2SO4 + 4Na2O. (ii) MnSO4+ 2Na2O2  Na2MnO4+ Na2SO4 (iii) 2NaOH + 2NO2  NaNO2 + NaNO3 + H2O (iv) 6NaOH+3Br2  5NaBr + NaBrO3 + 3H2O (v) 6NaOH + 4S  2Na2S + Na2S2O3 + 3H2O. (vi) 4NaOH + 2F2  4NaF + O2 + 2H2O. (vii) PbO + 2NaOH  Na2PbO2 + H2O (viii) Form insoluble hydroxides CrCl3 + 3NaOH  Cr(OH)3  (Green) + 3NaCl. (ix) HgCl2 + 2NaOH  Hg(OH)2  + 2NaCl ; Hg(OH)2  HgO  (yellow or brown) + H2O. 150  200 º C

  HCOONa. (x) NaOH + CO  5 10 atm 14. 19.

(A – p) ; (B – q, s, t) ; (C – t) ; (D – q, r) 15. (D)

(C)

16.

(C)

17.

(B)

18.

(C)

1.

Which of the following statements are true about the alkali metals ? [s-block Elements] (1) All alkali-metal salts impart a characteristic colour to the Bunsen flame. (2) The correct order of increasing thermal stability of the carbonates of alkali metals is Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 . (3) Among the alkali metals, cesium is the most reactive. (4) The reducing character of the alkali metal hydrides follow the order : LiH > NaH > KH > RbH > CsH. (A*) (1), (2) and (3) (B) (1), (3) and (4) (C) (2), (3) and (4) (D) (1), (2), (3) and (4)

{kkjh; /kkrqv ksad sfy , fuEu esslsd kSulsd Fku lR; gS\ (1) lHkh {kkjh; /kkrq v ksad sy o.k]cqUlsu Toky k d ks,d y kf{.kd jax iznku d jrsgaSA (2) {kkjh; /kkrq v kasd sd kckuZsVs d srkih; LFkkf;Ro d klghc<+ rkgq v kØ e Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 gS A (3) lHkh{kkjh; /kkrq v ks aesalslhft +;e lokZfèkd fØ ;k'khy gSA (4) {kkjh; /kkrqgkbMª kbM d svipk;d xq.k d k Ø e LiH > NaH > KH > RbH > CsH 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.- 1

Hint :

(A*) (1), (2) rFkk(3) (B) (1), (3) rFkk(4) (C) (2), (3) rFkk(4) (D) (1), (2), (3) rFkk(4) (4) Reducing nature increases down the group as their stability decreases down the group CsH > RbH > KH > NaH > LiH (4) oxZes auhpst kusij vipk;d izd `fr c<+rh gSD;ksafd oxZesauhpst kusij bud k LFkkf;Ro ?kVrk gSA CsH > RbH > KH > NaH > LiH

2.

HCl is added to following oxides. Which one would give H2O2? [s-block Elements] [JEE-1980] (A) MnO2 (B) PbO2 (C*) BaO2.8H2O (D) NO2 HCl d ksfuEu es alsd kSulsvkWDlkbM esafey kusij H2O2 izkIr gksrk gS? [JEE-1980] (A) MnO2 (B) PbO2 (C*) BaO2.8H2O (D) NO2

Sol.

BaO2.8H2O + 2HCl  BaCl2 + H2O2 + 8H2O

3.

S1 : Plaster of paris is a hemihydrate of calcium sulphate obtained by heating the gypsum above 393 K. S2 : Sodium carbonate is used in water softening. S3 : The order of mobilities of the alkali metal ions in aqueous solutions is Li+ > Na+ > K+ > Rb+ > Cs+. S1 : ft Ile d ks393 K d sÅ ij xeZd jusij d S fYl;e lYQ sV d kgsehgkbMªsM izkIr gksrkgS]mlsIy kLVj vkWQ isfjl d grs

ga AS S2 : lks fM;e d kcksZusV d k mi;ksx]t y d kse`nwcukusesafd ;k t krk gSA S3 : t y h; foy ;u es a{kkjh; /kkrqvk;u d hxfr'khy rkd kØ e Li+ > Na+ > K+ > Rb+ > Cs+ gSA [s-block Elements] (A) T T F (B) T T T (C*) F T F (D) F F F Sol.

393K S1 : (2 CaSO4.2H2O)   2 (CaSO4). H2O + 3H2O ; above 393 K dead burnt plaster is obtained.

S2 : Ca2+ + Na2CO3  CaCO3 + 2Na+ S3 : Li+ < Na+ < K+ < Kb+ < Cs+ Bigger hydrated ion moves slower in aqueous solution. 393K S1 : (2 CaSO4.2H2O)  r  2 (CaSO4). H2O + 3H2O ; 393 K d sÅ ij e`

t fy r Iy kLVj izkIr gksrk gSA

S2 : Ca2+ + Na2CO3  CaCO3 + 2Na+ S3 : Li+ < Na+ < K+ < Kb+ < Cs+

t y h; foy ;u esacM+k t y k;ksft r vk;u /khjslsxfr d jrk gSA 4.

STATEMENT-1 : Lithium is the most powerful reducing agent and sodium is the least powerful reducing agent amongst the alkali metals in aqueous solutions. [s-block Elements] STATEMENT-2 : Lithium has the highest hydration enthalpy and the sodium the least value. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True d Fku-1 : t y h; foy ;u esa]{kkjh; /kkrqv ksaesay hfFk;e lclsçcy vipk;d d kjd gSrFkklksfM;e lclsnqcZy vipk;d

gS A d Fku-2 : y hfFk;e mPpre foy k;d u ,UFkSYih j[krk gSrFkk lksfM;e]lclsd eA (A) d Fku& 1 lR; gS ]d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k gSA (B) d Fku& 1 lR; gS ]d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k ughagSA (C*) d Fku& 1 lR; gS ]d Fku& 2 vlR; gSA (D) d Fku& 1 vlR; gS ]d Fku& 2 lR; gSA Sol.

(i) E Li+/Li = –3.04 ; Na+ / Na = – 2.71 which is least among the alkali metals. (ii) Hydration enthalpy / KJ mol–1 Li = – 506 ; Na = – 406 ; Cs has the least Hhyd = – 276 (i) E Li+/Li = –3.04 ; Na+ / Na = – 2.71 t ks{kkjh; /kkrq v ksaesalclsd e gSA –1 (ii) foy k;d u Å t kZ/ KJ mol Li = – 506 ; Na = – 406 ; Cs , Hfoy k;d u = – 276 d k eku d e j[krk gS A

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PAGE NO.- 2

5.

Which of the following compounds on thermal decomposition yields a basic as well as an acidic oxide ?

fuEu esalsfd l ;kSfxd d srkih; vi?kV~u ij ,d {kkjh; vkWDlkbM d slkFk&lkFk,d vEy h; vkWDlkbM HkhizkIr gks rkgS\ (A) KClO3 (B) KNO3  Hint : CaCO3  CaO + CO2 (Basic) ({kkjh;)

(SBC-CPM (I)) (D*) CaCO3

(C) Na2CO3

(Acidic) (vEy h;)

6.

An inorganic compound (X) which produces brick red coloration as flame when (X) dissolves in water produces alkaline solution and a combustible gas (Y). (X) and (Y) are respectively. ,d d kcZfud ;kSfxd (X) Toky kesabZV t S l ky ky jax mRiUu d jrkgS A t c (X) d kst y es afoys; fd ;kt krkgSrksnks{kkjh; foy ;u rFkk ,d ngu’khy xSl (Y) curh gSA (X) rFkk(Y) Ø e'k%gSA [SBC-CAM] (Made by SCS SIR ON Nov.2014) (SBC-CAM-(I)) (A) CaO, O2 (B) Ca3N2, NH3 (C) CaCO3, CO2 (D*) CaH2, H2

Sol.

CaH2 + H2O  Ca(OH)2 + H2

7.

Select correct matching. (A*) Hydrolith = CaH2

[SBC-CAEM] (B*) Bleaching powder = CaCl2. Ca(OCl)2

(C*) Epsom salt = MgSO4. 7H2O

(D*) Plaster of Paris = CaSO4.

lgh fey ku pqfu, & (A*) gkbMª ks fyFk= CaH2

(B*) fojt a d

(C*) ,ile

(D*) Iy kLVj vkW Q

8.

9.

y o.k = MgSO4. 7H2O

1 HO 2 2

pw .kZ= CaCl2. Ca(OCl)2 1 2

isfjl = CaSO4. H2O

Which of the following is/are correct? (A*) Ionization energy = Ca > K (C*) Stability = MgCO3 < CaCO3

(B*) Melting point = Ca > K (D*) Hydration energy = Ca2+ > K+

[SBC-PP]

fuEu esalsd kSulk@d kSulslgh gS@ gSa\ (A*) vk;uu Å t kZ= Ca > K (C*) LFkkf;Ro = MgCO3 < CaCO3

(B*) xy uka d = Ca > K (D*) t y ;ks t u Å t kZ= Ca2+ > K+

Which s-block element shows crimson red coloration in flame ? (A*) Lithium salt (B) Rubidium salt (C) Barium salt fuEu esalsd kSulk s-Cy kWd rRo Toky k esafØ elu y ky jax n'kkZrk gS\ (A*) fy fFk;e y o.k (B) : fcfM;e y o.k (C) cs fj;e y o.k

[SBC-PP] (D*) Strontium salt (D*) LVª kWfU'k;e

yo.k

Comprehension Alkali metals burns vigorously in oxygen forming various oxides like monoxide, peroxide and superoxide. The basic character of various oxides of alkali metals increases with increasing metallic character .The increasing stability of the peroxide or super oxide, as the size of the metal ion increases, is due to stabilisation of larger anions by larger cation through lattice energy effects.

v uqPN sn {kkj /kkrq,Wok;qes arhoz: i lst y d j fofHkUu izd kj d svkDlkWbM t Sl seks uksv kWDlkbM]ijkDlkWbM o lqijkDlkWbM cukrhgSA {kkj /kkrqv ksad sfofHkUu vkDlkbMksad s{kkjh; xq.k/kkfRod vfHky {k.kc<+usd slkFkc<+rsgSA t kyd Å t kZiz Hkko d sd kj.kcM+ s/kuk;u }kjkcM+ s_ .kk;u d sLFkk;hd j.kd sd kj.kijkDWlkbM vFkoklq ijkDWlkbM d kLFkkf;Ro c<+t krk gS]D;ksafd /kkrqvk;u d k vkd kj c<+t krk gSA 10.

Principal ( i.e main) compound formed upon combustion of sodium metal in excess air is :

ok;qd svkf/kD; esalksfM;e /kkrqd sngu lseq[; ;kSfxd curk gS]og gS% (A*) Na2O2

(B) NaO2

(C) NaO3

(D) NaN3

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PAGE NO.- 3

Sol.

1 4Na + O2  2Na2O ; Na2O + O 2  Na2O2 2

11.

Among the following, which one of the superoxide is thermally most stable ?

fuEu esalsd kSulklqijkWDlkbM rkih; : i lslokZf/kd LFkk;h gS\ (A*) CsO2 12.

(B) NaO2

(C) KO2

(D) RbO2

Which of the following is most basic oxide of alkali metal ?

fuEu esalsd kSulk{kkj /kkrqd klokZf/kd {kkjh; vkDlkbM gS\ Sol.

(A*) Rb2O (B) K2O (C) Li2O (D) Na2O Basic strength of oxides increases in the order Li2O < Na2O < K2O < Rb2O. This is due to the increase in metallic character down the group.

vkW DlkbMksad h{kkjh; lkeF;ZLi2O < Na2O < K2O < Rb2O Ø e esac<+ rhgSA oxZesuhpst kusij /kkfRod vfHky {k.keao` f) d sd kj.k,sl kgksrkgSA 13.

[s-block Elements]

COMPLETE THE FOLLOWING REACTIONS :

fuEu v fHkfØ ;v ksad ksiw.kZd hft , % (i) Ans. (ii) Ans.

Na2S + Na2O2  Na2S + 4Na2O2  Na2SO4 + 4Na2O. MnSO4 + Na2O2  MnSO4 + 2Na2O2  Na2MnO4 + Na2SO4 .

(iii) Ans.

NaOH + NO2  2NaOH + 2NO2  NaNO2 + NaNO3 + H2O

(iv) Ans.

NaOH (hot & conc.) (xeZrFkk lkUnz ) + Br2    6NaOH + 3Br2 5NaBr + NaBrO3 + 3H2O.

(v) Ans.

NaOH + S  6NaOH + 4S  2Na2S + Na2S2O3 + 3H2O.

(vi) Ans.

NaOH (hot & conc.) (xeZrFkk lkUnz ) + F2  4NaOH + 2F2  4NaF + O2 + 2H2O.

(vii) Ans.

PbO + NaOH  PbO + 2NaOH  Na2PbO2 + H2O

(viii) Ans.

CrCl3 + NaOH  Form insoluble hydroxides. v?kq y u'khy

gkbMª kWDlkbM curkgSaA

CrCl3 + 3NaOH  Cr(OH)3  (Green) + 3NaCl. (ix) Ans.

HgCl2 + NaOH  HgCl2 + 2NaOH  Hg(OH)2  + 2NaCl

(x)

  NaOH + CO  5 10 atm

Ans.

  HCOONa. NaOH + CO  5 10 atm

14.

Match the compounds listed in column-I with the characteristic(s) listed in column-II. Column-I Column-II (A) BeO (s) (p) Amphoteric in nature (B) NaHCO3 (crystalline) (q) Imparts characteristic colour to Bunsen flame. (C) BeCl2(s) (r) Produce H2O2 and O2 on reaction with water. (D) CsO2(s) (s) Show hydrogen bonding (t) Has a chain structure

;

Hg(OH)2  HgO  (yellow or brown) + H2O.

150  200 º C

150  200 º C

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PAGE NO.- 4

LrEHk -I esafn;sx;s;kSfxd ksad ksLrEHk -II esafn;sx;svfHky {k.kksad slkFklqesfy r fd ft ,A LrEHk -I LrEHk -II (A) BeO (s) (p) mHk;/kehZiz o`fr (B) NaHCO3 (fØ LVyh;) (q) cq ulsu Toky k esavfHky k{k.khd jax n'kkZrsgSA (C) BeCl2(s) (r) t y d slkFk vfHkfØ ;k d j H2O2 o O2 ns rsgSA (D) CsO2(s) (s) gkbMª kst u ca/kn'kkZrsgSA (t) J` a[ky k ljap uk j[krk gSA Ans. Sol.

(A – p) ; (B – q, s, t) ; (C – t) ; (D – q, r) (A) BeO is amphoteric in nature becuase it reacts with acid as well as base. (B) Hydrogen bonding ; HCO3– ions are linked into an infinite chain through H-bonding. (C) (D) 2CsO2 + 2H2O  2Cs+ + 2OH– + H2O2 + O2 (A) BeO mHk;/kehZiz d `fr d kgksrkgSD;ksafd ;g vEy lkFk ghlkFk{kkj d slkFkvfHkfØ ;kd jrkgSA (B) gkbMª kst u ca/k ; HCO3– vk;u H-ca/k }kjk vUur%J`[kay kesat qM +sgksrsgSA (C) (D) 2CsO2 + 2H2O  2Cs+ + 2OH– + H2O2 + O2

ChemINFO-7.1

HYDROGEN

Daily Self-Study Dosage for mastering Chemistry

Preparation of H2

There are three main sources from which hydrogen may be prepared. These are (i) Water (ii) Acids (C) Alkalines (i) Hydrogen from water : (a) Cold water react with alkali and alkaline earth matals to evolve hydrogen 2Na – (Hg) + 2H2O  2NaOH + H2  Sodium amalgam

Note : The reaction are vigorous to minimum the rate of reaction, alkali metals are used in the form of amalgam. (b) Hot water or steam, when passed over metals like Zn, Fe, Mn, Co, Cr, Sn etc is decompose to librate hydrogen. Zn + H2O  ZnO + H2  Note: When steam, when passed over hot coke, hydrogen is produced in the form of water gas. C + H2O  H2 + CO Steam

water gas

(II) Hydrogen from Acid : Many reactive metals such as alkali metals, alkaline earth metals, Zn, Mg, Fe etc. react with dil HCl or dil H2SO4 to evolve hydrogen. Mg + 2HCl  MgCl2 + H2 Note : Mn and Mg react with dil HNO3 to evolve hydrogen. (III) Hydrogen from alkali : Zn, Al, Sn, Pb, Si (Amphoteric metals) reacts with boiling NaOH or KOH to evolve hydrogen. Zn + 2NaOH  Na2ZnO2 + H2 

Sodium Zincate Sn + 2NaOH + H2O  Na2SnO3 + 2H2 

Sodium stannate Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice.

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PAGE NO.- 5

15.

Which of the following metal give H2 after reaction with cold water. (A) Cu (B) Ag (C*) K (D) Pt

16.

Metals on reaction with water or dilute mineral acids can give (A) Monohydrogen (B) O2 (C*) Dihydrogen (D) Tritium

17.

Water gas is (A) CO + N2 (C) CO2 + H2

18.

Which of the following metal gives H2 gas after reaction with acid as well as alkali. (A) Cu (B) Na (C*) Zn (D) Fe

19.

Which of the following metal does not give H2 after reaction with dil H2SO4 (A) Mg (B) Fe (C) Zn (D*) Cu

(B*) CO + H2 (D) CO2 + N2

ChemINFO-7.1

HYDROGEN Preparation of H2

Daily Self-Study Dosage for mastering Chemistry

;gk¡rhu eq[; L=kksr gSft ulsgkbMªkst u cuk;h t k ld rh gS& ;g gS% (i) t y (ii) vEy (C) {kkj (i) t y l sgkbMª kst u izkIr d juk : (a) B.Mkt y ]{kkj rFkk{kkjh; e` nk/kkrqv ksad slkFkfØ ;kd jd sgkbMªkst u fu"d kflr d jrkgSA” 2Na – (Hg) + 2H2O  2NaOH + H2 

lksfM;e vey xe uksV : vfHkfØ ;k rhozgksrh gS]vkSj vfHkfØ ;k d h nj d ksd e d jusd sfy , {kkjh; /kkrq,¡vey xe d s: i esaiz;qDr gksrh gSA (b) xeZt y ;kHkki d kst c /kkrqt S l sZn, Fe, Mn, Co, Cr, Sn bR;kfn ij iz okfgr d jrsgS ]rks;g fo?kfVr gkds j gkbMª ks t u fu"d kflr d jrkgSA Zn + H2O  ZnO + H2  uksV : t c Hkki d ksxeZd ksd ij izokfgr d jrsgS]rksgkbMªkst u t y xSl d s: i esamRiUUk gksrh gSA C + H2O  H2 + CO

Hkki

t y xSl

(II) v Ey

l sgkbMªkst u izkIr d juk : {kkjh; /kkrq,¡]{kkjh; e``nk/kkrq,¡]Zn, Mg, Fe bR;kfn d sleku vf/kd fØ ;k'khy /kkrq,¡ruqHCl ;kruqH2 SO4 d slkFkfØ ;kd jd s gkbMª kst u fu"d kflr d jrhgS aA Mg + 2HCl  MgCl2 + H2 uksV : Mn rFkk Mg ruqHNO3 d slkFk fØ ;k d jd sgkbMªkst u fu"d kflr d jrh gSA (III) {kkjkslsgkbMª kst u

iz kIr d juk: Zn, Al, Sn, Pb, Si (mHk;/kehZ/kkrq ,¡) mCky rsgq, NaOH ;kKOH d slkFkfØ ;k d jd sgkbMªkst u fu"d kflr d jrsgSaA Zn + 2NaOH  Na2ZnO2 + H2 

lks fM;e

ft ad sV

Sn + 2NaOH + H2O  Na2SnO3 + 2H2 

lksfM;e LVsusV Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 15.

fuEu esalsd kSulh/kkrqB.Mst y d slkFkvfHkfØ ;kd s 17. i'pkRk~H2 nsrh gS\ (A) Cu (C*) K

16.

(B) Ag (D) Pt

/kkrq,¡]t y ;k ruq[kfut vEy d h fØ ;k ij nsrh gS& (A) eks uks gkbMª ks tu (B) O2 (C*) MkbZ gkbMª kts u (D) Vª kbZfV;e

18.

t y xSl gS& (A) CO + N2 (C) CO2 + H2

fuEu esalsd kSulh/kkrqvEy d slkFk&lkFk{kkj d slkFk vfHkfØ ;k d si'pkRk~H2 nsrh gS\ (A) Cu

19.

(B*) CO + H2 (D) CO2 + N2

(B) Na

(C*) Zn

(D) Fe

fuEu esalsd kSulh /kkrqruqH2SO4 d slkFk vfHkfØ ;k d si'pkRk~H2 ugh nsrh gS\ (A) Mg (B) Fe

(C) Zn

(D*) Cu

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PAGE NO.- 6

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 45

This DPP is to be discussed in the week (07.12.2015 to 12.12.2015) 1. 2. 3.

Course of the week as per plan : Boron Family, Compound of Boron (B2O3, H3BO3, Borax), Compounds Al2O3, AlCl3, Alum Course covered till previous week : Important compounds of s-block elements, Discussion. Target of the current week : Boron Family, Compound of Boron (B2O3, H3BO3, Borax), Compounds Al2O3, AlCl3, Alum

4.

DPP Syllabus : Boron Family, Compound of Boron (B2O3, H3BO3, Borax), Compounds Al2O3, AlCl3, Alum

DPP No. # 45 (JEE-MAIN) Total Marks : 59

Max. Time : 36 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.13 ChemINFO : 5 Questions ('–1' negative marking) Q.14 to Q.18

(3 marks, 2 min.) (4 marks, 2 min.)

[39, 26] [20, 10]

ANSWER KEY DPP No. # 45 (JEE-MAIN) 1. 8. 15.

(B) (C) (C)

2. 9. 16.

(D) (B) (B)

3. 10. 17.

(A) (C) (C)

4. 11. 18.

1.

Moderate electrical conductivity is shown by : (A) silica (B*) graphite

(D) (C) (A)

5. 12.

(D (D)

(C) diamond

fuEu esalsfd ld s}kjk e/;e fo|qrpky d rk n'kk;h t krh gS% (A) flfyd k (B*) xz sQ kbV (C) ghjk(diamond)

6. 13.

(C) (A)

7. 14.

(C) (D)

[JEE-1982, 1 M] (D) (BN)X [JEE-1982, 1 M] (D) (BN)X

2.

Which of the following halides is least stable and has doubtful existence? [JEE 1996] (A) C4 (B) Ge4 (C) Sn4 (D*) Pb4 fuEu esalsd kSulk gSy kbM lclsd e LFkk;h rFkk langsiw.kZvfLrRo j[krk gS\ [JEE 1996] (A) C4 (B) Ge4 (C) Sn4 (D*) Pb4

3.

Which one of the following oxides is neutral ?

[JEE 1996, 1 M]

fuEu esalsd kSulk mnklhu vkWDlkbM gSA (A*) CO 4.

(B) SnO2

(C) ZnO

(D) SiO2

Which of the following is bauxite?

fuEu esalsd kSulk ckWDlkbV gS\ (A) Al(NO3)3

(B) AlCl3

(C) Al2(SO4)3. xH2O

(D*) Al2O3. xH2O

5.

Name the type of the structure of silicate in which one oxygen atom of [SiO4]4– is shared ? (A) Three – dimensional (B) Linear chain silicate (C) Sheet silicate (D*) Pyrosilicate flfy d sV d h lajpuk d sizd kj d k uke D;k gSft lesa[SiO4]4– d s,d vkWDlht u ijek.kqlslkf>r gksrk gS? (A) f=kfofe; (B) js [kh; J`a[ky kflfy d sV (C) 'khV flfy d s V (D*) ik;jkf slfyd s V

6.

When orthoboric acid (H3BO3) is heated, the residue left could be (A) metaboric acid (B) Boron (C*) Boric anhydride

(D) Borax

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PAGE NO.- 1

t c vkFkksZcksfjd vEy (H3BO3) d ksxeZfd ;k t krk gS]rks'ks"k cpk vof'k"V fuEu gksxk % (A) es Vkcks fjd vEy (B) cks jks u (C*) cks fjd ,ugkbMª kbM (D) cks jDsl 7.

Boron carbide , B4C, is widely used for (A) making acetylene (C*) as a hardest substance after diamond cksjksu d kckZbM B4C, d k mi;ksx eq[; : i lsfd l (A) ,s l Vhy hu

d sfuekZ.k esa (C*) ghjsd sckn lclsd Bks j inkFkZd s: i esa 8.

9.

(B) making plaster of paris (D) making boric acid

gsrqfd ;k t krk gSA (B) Iy kLVj vkW Q isfjl d sfuekZ.k esa (D) cks fjd vEy d sfuekZ.k esa

An aqueous solution of borax is (A) Neutral (B) Amphoteric

(C*) Basic

(D) Acidic

cksjsDl d k ,d t y h; foy ;u gS% (A) mnklhu (B) mHk;/kehZ

(C*) {kkjh;

(D) vEyh;

Boric acid is polymeric due to (A) its acidic nature (C) its monobasic nature

(B*) the presence of hydrogen bonds (D) its geometry.

cksfjd vEy fuEu d sd kj.k cgqy d h; gS(A) bld hvEy h; iz d `fr d sd kj.k (C) bld h,d y {kkjh; iz d `fr d sd kj.k 10.

(B*) gkbMª kst u

cU/kd hmifLFkfr d sd kj.k (D) bld hT;kferh; d sd kj.k

The type of hybridization of boron in diborane is

Mkbcksjsu esacksjksu d slad j.k d k izd kj gS% (A) sp 11.

(B) sp2

(D) dsp2

Which of the following is formed when aluminium oxide and carbon is strongly heated in dry chlorine gas ? (A) Aliminium chloride (B) Hydrated aluminium chloride (C*) Anhydrous aluminium chloride (D) None of these fuEu es alsD;kfufeZ r gkxskt c ,Y;q fefu;e vkDWlkbM rFkkd kcZ u d ks'k"qd Dykfsju xS l es siz cyrj : i lsxeZfd ;kt krkgS ? (A) ,Y;q fefu;e

Dy ksjkbM (C*) fut Z y ,Y;qfefu;e Dy ksjkbM 12.

(C*) sp3

(B) gkbMª sV

,Y;qfefu;e Dy ksjkbM (D) bues alsd ksbZugha

Which of the following statements about H3BO3 is not correct ? (A) It is prepared by acidifying an aqueous solution of borax. (B) It has a layered structure in which planar B(OH)3 units are joined by hydrogen bonds. (C) It does not act as proton donor but acts as well as Lewis acid by accepting hydroxyl ion. (D*) It is a strong acid. fuEu esalsd kSulk d Fku H3BO3 d sckjsesavlR; gS? (A) blscks jsDl

d svEy hd `r t y h; foy ;u }kjk cuk;k t krk gSA (B) ;g ,d ijrh; la jpuk j[krkgSft lesalery h; B(OH)3 bd kb;k¡gkbMªkst u cU/kksa}kjk t ksM +st krsgSaA (C) ;g iz ksVkWu nkrk d h rjg gh d k;Zughad jrk gSy sfd u gkbMªksfDly vk;u d ksxzg.k d j y qbZl vEy d h rjg Hkh d k;Z d jrkgSA (D*) ;g ,d iz cy vEy gSA 13.

Which gas is liberated when Al4C3 is hydrolyzed ? d kSulh xSl eqDr gksxh t c Al4C3 t y vi?kfVr gksrk gS? (B) C2H2 (C) C2H6 (A*) CH4

(D) CO2

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PAGE NO.- 2

ChemINFO-7.2

HYDROGEN Properties of Hydrogen

Daily Self-Study Dosage for mastering Chemistry

Properties of Hydrogen (1) It is colourless, odourless and taste less gas. (2) It is only slightly soluble in water. (3) It is lightest of all elements. (4) certain metals like palladium, platinum etc can adsorb large quantities of hydrogen. The adsorbed hydrogen is called occluded hydrogen and is more active than ordinary hydrogen. Combustion : Hydrogen is flammable or combustible gas it burn with blue flame in oxygen atmosphere. 2H2 + O2  2H2O Combination reaction : Due to very high bond dissociation energy (436 KJ mol–1) hydrogen is not a very active. (A) Except Beryllium all alkali and alkaline earth metal directly combine with hydrogen and ionic hydride is formed. 2Li + H2  2LiH 2Na + H2  2NaH

Ca + H2  CaH2 Ba + H2  BaH2

(B) Halogens directly combine with Hydrogen to form covalent compounds. X  F , Cl , Br , I

H2 + X2  2HX

The reactivity of Halogen : F2 > Cl2 > Br2 > I2 (C) Synthesis of methyl alcohol : In presence of ZnO and CrO3 catalyst at 200 atmosphere CO and H2 combine to form CH3OH. catalyst

  CH3OH CO + 2H2  300 C (D) Unsaturated fats are changed to saturated fats in presence of nickel .

Unsaturated fat + H2

Ni

  Saturated Fat catalyst

(Oil)

(vanaspati ghee)

(E) Synthesis of Ammonia (Haber's Process) : A mixture of N2 and H2 in ratio of 1 : 3 at 200 atm, 500°C in presence of Fe catalyst NH3 is formed. N2 + 3H2

2NH3

H = –22.4 Kcal mol–1

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 14.

15.

16.

17.

18.

The colour of hydrogen is : (A) Yellow (B) Orange

(C) Red

(D*) None

Which of the following is the lightest gas ? (A) Nitrogen (B) Helium

(C*) Hydrogen

(D) Oxygen

Which of the following metal adsorb hydrogen ? (A) Zn (B*) Pd (C) Al

(D) K

Which of the following react with Hydrogen very fast : (A) Br2 (B) Cl2 (C*) F2

(D) I2

H2 cannot combine directly with which metal : (A*) Be (B) Ca

(D) Na

(C) Ba

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PAGE NO.- 3

ChemINFO-7.2

HYDROGEN gkbMªkst u d sxq.k/keZ

Daily Self-Study Dosage for mastering Chemistry

gkbMªkst u d sxq.k/keZ (1) ;g ja xghu]xa/kghu rFkk Loknghu xSl gSA (2) ;g d s oy t y esad e ?kqy u'khy gksrh gSA (3) ;g lHkhrRoks alsgYd hgksrhgSA (4) fuf'pr /kkrq ,sat S l sis y sfM;e]IysfVue bR;kfn gkbMªkst u d ksvf/kd re la [;kes avf/k'kks f"kr d j ld rhgSA vf/k'kksf"kr gkbMª ks tu vo: ) gkbMªkst u (occluded hydrogen) d gy krhgSrFkk;g lkekU; gkbMªkst u d hvis{kkvf/kd fØ ;k'khy gksrhgSaA ngu : gkbMªkst u Toy u'khy ;k ngu'khy xSl gS;g vkWDlht u;qDr ok;qe.My esauhy h Toky k d slkFk t y rh gSA 2H2 + O2  2H2O l a;kst u v fHkfØ ;k : cgqr vf/kd cU/kfo;kst u Å t kZ(436 KJ mol–1) d sd kj.kgkbMªkst u vf/kd lfØ ; ughgksrhgSA (A) cs jhfy;e d ksNks Md+ j lHkh{kkj rFkk{kkjh; e` nk/kkrq ,s agkbMª ks t u d slkFkiz R;{k: i lst q MrhgSrFkkvk;fud gkbMª kbM cukrhgS A 2Li + H2  2LiH Ca + H2  CaH2 2Na + H2  2NaH Ba + H2  BaH2 (B) gs y kst u gkbMªkst u d slkFkizR;{k: i lst qM d j lgla;kst d ;kSfxd cukrsgSA H2 + X2  2HX X  F , Cl , Br , I

gsy kst u d h fØ ;k'khy rk : F2 > Cl2 > Br2 > I2 (C) es fFky ,Yd ks gy d kla 'ys "k.k: ZnO rFkkCrO3 mRiz jd d hmifLFkfr es s a200 ok;q e.Myh; nkc ij CO rFkkH2 t Mq+ d j CH3OH cukrsgSA mRizsjd

CO + 2H2  CH3OH 300 C (D)

vlar`Ir olk fud y d h mifLFkfr esalar`Ir olk esacny rh gSA

Ni vlar`Ir olk + H2  r`Ir olk  la mRizsjd (rs y) (ouLifr ?kh) (E) veks fu;kd kla 'ys "k.k(gs cj iz Ø e) : N2 rFkkH2 d kfeJ.k200 atm rFkk500°C ij Fe mRis jd d hmifLFkfr es z a1 : 3 d svuq ikr esafØ ;k d jd sNH3 cukrkgSA

N2 + 3H2

2NH3

H = –22.4 Kcal mol–1

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 14. 15.

16.

gkbMªkst u d kjax gksrkgS& (A) ihyk (B) ukja xh (C) y ky

17. (D*) d ks bZugh

fuEu esalsd kSu gYd h xSl gS? (A) ukbVª kst u (B) ghfy;e (C*) gkbMª ks tu (D) vkD Wlht u

(A) Br2 (C*) F2 18.

(B*) Pd

(C) Al

Lecutre Sub-topic(s) Nam e No. (No. of Lectures)

Ex.1 (S) L78

Boron Family

Hom e Work NCERT Pg.N Th.

Ex.2 (O) Ex.2 (I)

L79

Compound of Boron (B2O3, H3BO3, Borax)

Ex.1 (O)

L80

Compounds -Al2O3, AlCl3, Alum

Ex.1 (O)

Th.

Ex.2 (M)

Handout

323, 311

Chemical reaction of D2O

XI

312 to 314

XII XI

323, 324, 325

Prob. XII Th.

Ex.2 (O) Ex.2 (I)

Chem INFO

XII

Ex.2 (M) Ex.1 (S)

(B) Ca (D) Na

307 to 312

Prob.

Ex.2 (O) Ex.2 (I)

XI XII XI

Ex.2 (M) Ex.1 (S)

fuEu esalsfd l /kkrqd slkFk gkbMªkst u izR;{k : i lsugh t qM +rhgS&

(D) K

Hom e Work Sheet Ex.1 (O)

(B) Cl2 (D) I2

(A*) Be (C) Ba

fuEu eslsd kSulh/kkrqgkbMªkst u d ksvf/k'kksf"kr d jrhgS? (A) Zn

gkbMªkst u fuEu eslsfd l d slkFklokZf/kd fØ ;k'khy gS&

XI

314

XII XI

323, 324, 325

Prob. XII

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PAGE NO.- 4

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 46

This DPP is to be discussed in the week (14.12.2015 to 19.12.2015) 1.

Course of the week as per plan : Carbon family, Compounds-CO,CO2,C3O2, SiO2 Silicate & Silicone, Discurssion of p-block and Basic Strength.

2. 3.

Course covered till previous week : Boron Family, Compound of Boron (B2O3, H3BO3, Borax), Compounds Al2O3, AlCl3, Alum Target of the current week : Carbon family, Compounds-CO,CO2,C3O2, SiO2 Silicate & Silicone, Discurssion

4.

of p-block and Basic Strength. DPP Syllabus : Carbon family, Compounds-CO,CO2,C3O2, SiO2 Silicate & Silicone, Discurssion of p-block and Basic Strength.

DPP No. # 46 (JEE-ADVANCED) Total Marks : 48

Max. Time : 28 min.

Multiple choice objective ('–1' negative marking) Q.1 to Q.5 Integer type Questions ('–1' negative marking) Q.06 to Q.09 ChemINFO : 5 Questions ('–1' negative marking) Q.10 to Q.14

(4 marks, 2 min.) (4 marks 3 min.) (4 marks, 2 min.)

[12, 06] [16, 12] [20, 10]

ANSWER KEY DPP No. # 46 (JEE-ADVANCED) 1. 8.

(AB) 4

2. 9.

(ACD) 3. 1 10.

(AB) (D)

4. 11.

(ABC) 5. (D) 12.

(C) (A)

6. 13.

5 (D)

1.

Which of the following are correct [M] (A*) B2H6 reacts with excess of ammonia at low temperature to form an ionic compound. (B*) Borax, when heated with oxides of certain transition metals, forms coloured beads. (C) One mole borax in aqueous solution will require one mole HCl for titration (D) B2H6 can be methylated completely to give B2(CH3)6

7. 14.

4 (A)

(PBC(INO))

fuEu esad kSulsd Fku lR; gSa\ (A*) B2H6 fuEu rki ij veks fu;k d svkf/kD; lsfØ ;k d jd s,d vk;fud ;kSfxd cukrk gSA (B*) t c cks jsDl d ksd qN laØ e.k /kkrqvkWDlkbMksad slkFk xeZfd ;k t krk gSrksjaxhu eud k curk gSA (C) t y h; foy ;u es a,d eksy cksjsDl d svuqekiu d sfy , 1 eksy HCl d h vko';d rk gksxhA (D) B2H6 d ksiw .kZr%esfFky hd `r d jd sB2(CH3)6 mRiUu fd ;kt kld rkgSA Sol.

gy 2.

(C) One mole borax will require two moles of HCl for complete reaction. (D) The bridging hydrogens can not be methylated in B2H6. (C) iw .kZvfHkfØ ;kd sfy , ,d eksy cksjsDl d ks2 eksy HCl d hvko';d rk gksxhA (D) B2H6 es alsrqgkbMªkst uksad ksesfFky hÑ r ughad j ld rsgSaA Which of the following can produce B2O3 ? (A*) Heating borax with conc. H2SO4 (C*) Combustion of diborane, B2H6. fuEu esalsd kSu B2O3 d ksmRikfnr d j ld rk gS\ (A*) lkUnzH2SO4 d slkFkcks jsDl d ksxeZd j (C*) Mkbcks jsu, B2H6 d kngu d j

(p-Block)

[By SM Sir, Oct 2013] [M] (B) Passing CO2 through aq. NaBO2 (D*) Warming H3BO3 crystals till red hot. (B) t y h; NaBO2 es als CO2 izokfgr d j (D*) H3BO3 fØ LVy d ksy ky rIr rd xeZd jd s

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PAGE NO.- 1

3.

Sol. 4.

Which of the following are correct about B2H6 ? (A*) each ‘B’ atom is sp3 hybridised (B*) It consists of two-”3 centre 2 electron bonds” (C) The two bridging H-atoms are in the plane of the molecules (D) All of these B2H6 d sckjses afuEu esalslgh gS\ (A*) iz R;sd ‘B’ ijek.kqsp3 lad fjr voLFkk esagSA (B*) bles anks]rhu (C) nksls rqH-ijek.kqv.kqd sry esagh gksrsgSaA (D) mijks Dr lHkh ‘A’ and ‘B’ are correct on the basis of structure of B2H6. B2H6 d h la jpuk d svk/kkj ij ‘A’ o ‘B’ lgh gSaA Which of the following hydrides react with water ?

[M]

d sUnz-nks-by sDVªkWu cU/k gksrsgSaA

[M]

fuEu esalsd kSulsgkbMªkbM t y d slkFk vfHkfØ ;k d jrsgS\ (A*) B2H6 Sol.

(B*) NH3

(Inorganic)

(C*) NaH

(D) C2H6

(C*) B2O3

(D*) CO2

B2H6 + H2O  H3BO3 + H2 NH3 + H2O  NH4OH NaH + H2O  NaOH + H2 C2H6 + H2O  no reaction. vfHkfØ ;kughagks rhA

5.

Which of the following oxides are acidic :

fuEu esalsd kSulsvkWDlkbM vEy h; gSa% (A) CaO 6.

(B) Na2O

In borax ion how many B–O–B bonds present : cksjsDl vk;u esafd rusaB–O–B caèk gS% 5

[Made by DRM Mam]

[Made by DRM Mam]

Ans.

In diborane maximum how many 2c – 2e bonds present : MkbZcksjsu esavf/kd re fd rus2c – 2e caèk gS% 4

8.

How many oxides in the following are basic in water:

[Made by DRM Mam]

Ans. 7.

fuEu esalsfd rusvkWDlkbM t y esa{kkjh; gksrsgS%+ Ans. 9. Ans.

B2O3, Al2O3, CO2, SO2, NO2 CaO, Na2O, K2O, Cs2O, CaO 4 Basicity of H3BO3 is : H3BO3 d h {kkjd rk gS% 1

[Made by DRM Mam]

Lecutre Sub-topic(s) Nam e (No. No. of Lectures)

Ex.1 (S) Ex.1 (O) L81

Carbon family

Hom e Work NCERT

Hom e Work Sheet

Pg.N Th.

Ex.2 (O) Ex.2 (I)

L82

Compounds-CO, CO2, C3O2

Ex.1 (O)

Ex.1 (O) L83

SiO2 Silicate , Discussion

Th.

Ex.2 (I)

XI

319 to 320

XII XI

324, 325

Prob. XII Th.

Ex.2 (O) Ex.2 (M)

324

XII

Ex.2 (M) Ex.1 (S)

314 to 319

XII XI

Ex.2 (O) Ex.2 (I)

Handout

Prob.

Ex.2 (M) Ex.1 (S)

XI

Chem INFO

XI

320 to 322

XII XI

325

Alums

Prob. XII

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PAGE NO.- 2

ChemINFO-8.1

p-BLOCK ELEMENT Group 13 & 14 Elements

Daily Self-Study Dosage for mastering Chemistry

ALUMS ; M2SO4. M2 (SO4)3. 24H2O or MM (SO4)2. 12H2O Alums are transparent crystalline solids having the above general formula where M is almost any univalent positive cation (except Li+ because this ion is too small to meet the structural requirements of the crystal) and M’ is a trivalent positive cation (Al3+, Ti3+, V3+, Cr3+, Fe3+, Mn3+, Co3+, Ga3+ etc.). Alums contain the ions [M(H2O)6]+, [M’(H2O)6]3+ and SO42– in the ratio 1 : 1 : 2. Some important alums are : (i) Potash alum K2SO4 . Al2(SO4)3 . 24H2O (ii) Chrome alum K2SO4 . Cr2(SO4)3 . 24H2O (iii) Ferric alum K2SO4 . Fe2(SO4)3. 24H2O (iv) Ammonium alum (NH4)2SO4 . Al2(SO4)3 . 24H2O Alums are double salts which when dissolved in water produce metal ions (or ammonium ions) and the sulphate ions.

Preparation : A mixture containing solutions of M2SO4 and M’2(SO4)3 in 1 : 1 molar ratio is fused & then the resulting mass is dissolved into water. From the solution thus obtained, alums are crystallised.

Uses : It is used 1.

2. 3.

As a mordant in dye industry. The fabric which is to be dyed is dipped in a solution of the alum and heated with steam. Al(OH)3 obtained as hydrolysis product of [Al(H2O)6]3+ deposits into the fibres and then the dye is absorbed on Al(OH)3. as a germicide for water purification. As a coagulating agent for precipitating colloidal impurities from water.

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 10.

Sol. 11.

Sol.

12. 13. 14.

Which mixed sulphate is not an alum: (A) K2SO4.Al2(SO4)3.24H2O (B) K2SO4.Cr2(SO4)3.24H2O (C) Na2SO4.Fe2(SO4)3.24H2O (D*) CuSO4.Al2(SO4)3.24H2O M is divalent, it should be monovalent according to the formula of alum. Alum is used by dyer for : (A) for fire-proofing fabrics (B) as first aid for cuts (C) for softening hard water (D*) as mordant As a mordant in dye industry. The fabric which is to be dyed is dipped in a solution of the alum and heated with steam. Al(OH)3 obtained as hydrolysis product of [Al(H2O)6]3+ deposits into the fibres and then the dye is absorbed on Al(OH)3. Alums contain the ion [M(H2O)6]+, [M'(H2O)6]+3 and SO4–2 in the ratio : (A*) 1 : 1 : 2 (B) 1 : 1 : 1 (C) 2 : 1 : 1

(D) 1 : 2 : 1

In question no. 12, M ion can not be : (A) Li+ (B) K+

(C) NH4+

(D*) Al+3

Alums are : (A*) double salt

(C) Triple salt

(D) none of these

(B) single salt

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PAGE NO.- 3

ChemINFO-8.1

p-BLOCK ELEMENT Group 13 & 14 Elements

Daily Self-Study Dosage for mastering Chemistry

,sy e~(M2SO4. M2 (SO4)3. 24H2O v Fkok MM (SO4)2. 12H2O) : ,y e~] ikjn'khZfØ LVy h; Bksl gS] ft ld k mijksDr lkekU; lw=k gSrFkk ;gk¡M ,d la;kst h /kkrqvFkok /kukRed ewy d gS(Li+ d s vfrfjDr D;ksafd ;g vk;u]fØ LVy lajpukd hvko';d rklscgqr NksVsvkd kj d kgS) rFkkM f=kla;kst h/kkrqgS(Al3+, Ti3+, V3+, Cr3+, Fe3+, Mn3+, Co3+, Ga3+ etc.). ,y e~[M(H2O)6]+, [M’(H2O)6]3+ rFkkSO42– vk;uks ad ks1 : 1 : 2. vuqikr esaj[krk gSA d qN izeq[k ,sy e~fuEu gS& (i) iks Vk'k,y e~K2SO4 . Al2(SO4)3 . 24H2O

(ii) Ø ks e

,y e~K2SO4 . Cr2(SO4)3 . 24H2O (iii) Q S fjd ,y e~K2SO4 . Fe2(SO4)3. 24H2O (iv) veks fu;e ,s y e~(NH4)2SO4 . Al2(SO4)3 . 24H2O ,s y e~f}d ~yo.kgS A t c bud kst y es afoys ; fd ;kt krkgSrks/kkrqvk;u (vFkokvekfsu;e vk;u) rFkklYQ Vs vk;u cukrsgS Aa

fojpu % ,sy e~d k 1 : 1 eksy vuqikr esaM2SO4, M’2(SO4)3 d ksxfy r d jd sçkIr ifj.kkeh nzO;eku d kst y esafoy ; fd ;k t krk gSrFkk bl çd kj çkIr foy ;u ls,sy e~d ksfØ LVy hd `r fd ;k t krk gSA

mi;ksx : 1.

2. 3.

jat d m|ks x es aca/kd d s: i es a]Q S fcz d (d iM+s @ /kkxs a) ft ud ksja ft r fd ;kt kukgSd ks ],sy e~d sfoy;u es afoy; fd ;kt krk gSrFkkHkki d slkFkxeZfd ;kt krkgSA /kkxks aij [Al(H2O)6]3+ d st y vi?kVu lsiz kIr Al(OH)3 fu{ks fir gkst krkgSrFkk fQ j jat d Al(OH)3 ij vo'kksf"kr gkst krkgSA t y 'kqf) d j.k d sfy , d hVk.kquk'kd d s: i esa t y lsd ksy kbMy v'kqf) ;ksad ksvo{ksfir d jusd sfy , Ld Und kjd d s: i esa

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 10.

fuEu esalsd kSulk fefJr lYQ sV],sy e~ughagS%

Sol.

(A) K2SO4.Al2(SO4)3.24H2O (B) K2SO4.Cr2(SO4)3.24H2O (C) Na2SO4.Fe2(SO4)3.24H2O (D*) CuSO4.Al2(SO4)3.24H2O M f}la ;kst hgS];g fQ Vd jhd slw=kd svuql kj ,d y la;kst hgksukpkfg,A

11.

Sol.

12.

oL=kkasd ksjaxusoky sO;fDr (dyer) ,sy e~d ksfd l mís'; lsiz;qDr d jrsgSa% (A) oL=kks ad ksvfXuizfrjks/khcukusd sfy , (B) d Vusij iz kFkfed mipkj d sfy , (C) d Bks j t y d kse`nqd jusd sfy , (D*) ca /kd d s: i esa jat d m|ksx esac/kad d s: i esa]Q sfczd ¼d iM+s;k /kkxsa½ ft Ugsajaft r fd ;k t krkgSd ks],sy e~d sfoy ;u esafHkxks;k t krkgS rFkk Hkki d slkFkxeZfd ;k t krkgSA /kkxksaij [Al(H2O)6]3+ d st y vi?kVu lsizkIr mRikn Al(OH)3 fu{ksfir gkst krkgS rFkkfQ j jat d Al(OH)3 ij vo'kksf"kr gkst krkgSA ,y e esavk;u [M(H2O)6]+, [M'(H2O)6]+3 rFkkSO4–2 fuEu vuqikr esagksrsgS% (A*) 1 : 1 : 2

13.

14.

(B) 1 : 1 : 1

(C) 2 : 1 : 1

(D) 1 : 2 : 1

(C) NH4+

(D*) Al+3

(C) f=kd

(D) bues alsd ksbZugha

iz'u la- esa12 esaM vk;u ughagksld rk gS& (A) Li+

(B) K+

,y e gksrsgS& (A*) f}d y o.k

(B) ,d y

y o.k

yo.k

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PAGE NO.- 4

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 47

This DPP is to be discussed in the week (21.12.2015 to 26.12.2015) 1. 2.

Course of the week as per plan : Basic Strength, Basic Strangth and Acidic Strength Course covered till previous week : Carbon family, Compounds-CO,CO2,C3O2, SiO2 Silicate & Silicone,

3. 4.

Discurssion of p-block and Basic Strength. Target of the current week : Basic Strength, Basic Strangth and Acidic Strength DPP Syllabus : Basic Strength, Basic Strangth and Acidic Strength

DPP No. # 47 (JEE-MAIN) Total Marks : 39

Max. Time : 26 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.13

(3 marks, 2 min.)

[39, 26]

ANSWER KEY DPP No. # 47 (JEE-MAIN) 1. 8.

(B) (A)

2. 9.

(D) (B)

3. 10.

(B) (D)

4. 11.

(B) (A)

5. 12.

(C) (B)

6. 13.

(B) (C)

7.

(D)

1.

The following structures represent various silicate anions. Their formulas are respectively [M] (PBC(INO))

fuEu lajpuk,¡fofHkUu flfy d sV _ .kk;uksad ksiznf'kZr d jrh gSA mud slw=k Ø e'k%fuEu gSa& O O = Oxygen = Silicon

O

O

(A) SiO44– & Si3O88– (A) SiO44– o Si3O88– 2.

(B*) SiO44– & Si3O108– (B*) SiO44– o Si3O108–

(C) SiO42– & Si3O92– (C) SiO42– o Si3O92–

(D) SiO34– & Si3O108– (D) SiO34– o Si3O108–

Select the correct statements (A) Oxides of boron (B2O3) and silicon (SiO2) are acidic in nature. (B) Oxides of aluminium (Al2O3) and gallium (Ga2O3) are amphoteric in nature. (C) Oxides of germanium (GeO2) and tin (SnO2) are acidic in nature. (D*) both (1) and (2)

[E]

(PBC(I))

lghd Fkuksad kp;u d hft ;sA (A) cks jksu (B2O3) rFkkflfy d kWu (SiO2) d svkWDlkbM vEy h; izd `fr d sgksrsgSA (B) ,y q ehfu;e (Al2O3) rFkk xsfy ;e (Ga2O3) d svkWDlkbM mHk;/kehZizd `fr d sgksrsgSA (C) t es Zfu;e (GeO2) rFkkfVu (SnO2) d svkWDlkbM vEy h; izd `fr d sgksrsgSA (D*) (1) rFkk(2) nks uks a Sol.

S1 and S2 are correct statements. S3 : GeO2 is acidic while SnO2 is amphoteric in nature.

gy %

S1 o S2 lgh d Fku

gSaA S3 : GeO2 vEy h; gSt cfd SnO2 mHk;/kehZiz d `fr d k 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.- 1

3.

BC3 is more stable than TC3 because : [p-Block] [SA Mam_Jan_2013] [M] (A) The difference between electronegativity of B and C is larger than that of T and C. (B*) The higher oxidation state of T (+ 3) is less stable than that of B (+ 3). (C) Thallium being larger in size is able to accomodate more C atoms around it. (D) B3+ is more easily formed than T3+. BC3, TC3 d hvis {kkvf/kd LFkk;hgksrkgS]D;kasfd : [p-Block] [SA Mam_Jan_2013] (A) B rFkkC d h fo|q r_

Sol. 4.

.krk d se/; vUrj T rFkkC d h vis{kk vf/kd gksrk gSA (B*) T (+ 3) d h mPpre vkW DLkhd j.k voLFkkB (+ 3) d h vis{kk d e LFkk;h gksrh gSA (C) FkS fy ;e vkd kj esacM+kgksrk gS]rFkk bld spkjksvksj vf/kd C ijek.kqfLFkr gksrsgSA (D) B3+, T3+ d h vis {kk vf/kd ljy rk lscurk gSA Refer notes (uks V~l ns[ksa) The straight chain polymer is formed by : (A) hydrolysis of (CH3)3 SiCl followed by condensation polymerization. (B*) hydrolysis of (CH3)2 SiCl2 followed by condensation polymerization. (C) hydrolysis of (CH3) SiCl3 followed by condensation polymerization. (D) hydrolysis of (CH3)4 Si followed by condensation polymerization.

[M]

(PBC(INO))

lh/kh J`a[ky h; cgqy d d k fuekZ.k fd ;k t k ld rk gS% (A) (CH3)3 SiCl d kt y &vi?kVu rRi'pkr~la ?kuu cgqy d hd j.k}kjk (B*) (CH3)2 SiCl2 d kt y &vi?kVu rRi'pkr~la ?kuu cgqy d hd j.k}kjk (C) (CH3) SiCl3 d kt y &vi?kVu rRi'pkr~la ?kuu cgqy d hd j.k}kjk (D) (CH3)4 Si d kt y &vi?kVu rRi'pkr~la ?kuu cgqy d hd j.k}kjk 5.

Sol.

Which gas is formed when CaC2 is allowed to react with dilute HCl ? (A) Ethane (B) Ethene (C*) Ethyne t c CaC2 d ksruqHCl d slkFk vfHkd `r fd ;k t krk gS]rksd kSulh xSl curh gS\

[Change by CJ Sir] [E] (D) Methane

(A) ,s Fks u

(D) es Fks u

(B) ,Fkhu

(C*) ,Fkkbu

CaC2 + 2HCl  CaCl2 + HCCH Ethyne

,Fkkbu 6.

The function of fluorspar in the electrolytic reduction of alumina dissolved is fused cryolite (Na3AlF6) is (A) as a catalyst (B*) to lower the temperature of the melt and to make the fused mixture very conducting (C) to decrease the rate of oxidation of carbon at the anode (D) none of these above laxfy r Ø k;ksy kbV (Na3AlF6) esa?kqfy r ,sY;qfeuk d soS| qr vi?kVuh; viPk;u esa¶y ksjLikj d k d k;Zgksrk gSA (A) mRiz sjd

d s: i esagSA (B*) xfy r d k rki d e d juk rFkk la xfy r feJ.k d ksvfrpky d cukuk (C) ,uks M +ij d kcZu d hvkWDlhd j.kd hnj d ks?kVkuk (D) bues alsd ksbZugha 7.

Which of the following is bauxite?

fuEu esalsd kSulk ckWDlkbV gS\ (A) Al(NO3)3 8.

(B) AlCl3

(C) Al2(SO4)3. xH2O

(D*) Al2O3. xH2O

Which of the following oxidation states are the most characterstics for lead and tin, respectively ?

y SM o fVu d sfy , fuEu esalsd kSulh vkWDlhd j.k voLFkk Ø e'k%lokZf/kd : i lsvfHky k{kf.kd gSa\ (A*) + 2 , + 4

(B) + 4 , + 4

(C) + 2 , + 2

(D) + 4 , + 2

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PAGE NO.- 2

9.

Which of the following anions is present in the chain structure of silicates ?

flfy d sV d h J`a[ky k lajpuk esafuEu esalsd kSulk _ .kk;u mifLFkr gksrk gS\ (A) (Si2O 52– )n 10.

11.

(B*) (SiO 23 – )n

Glass reacts with HF to produce Xy kl]HF d slkFk fØ ;k d j fuEu mRikn curk gSA (A) H2SIO3 (B) SiF4 The oxide which is not a reducing agent is :

(C) SiO 44 –

(D) Si2O76 –

(C) Na3AlF6 (D*) H2SiF6 [p-Block Element (Boron & Carbon Family)]

og vkWDlkbM t ksfd vipk;d ughagS]fuEu gS%

Sol.

gy 12.

(A*) CO2 (B) CO (C) SO2 (D) Both (A) & (C) nks uks(A) (C) CO2 can not act as reducing agent because carbon is in its highest oxidation state, i.e., +4. CO2 vipk;d d s: i es ad k;Zughad jrk gSD;ksafd d kcZu bld h mPpre vkWDlhd j.k voLFkk +4 esagSaA Aqueous solution of potash alum is : (A) alkalline (B*) acidic

rFkk

[p-Block Element (Boron & Carbon Family)] (C) neutral (D) soapy

iksVk'k ,sy e~d k t y h; foy ;u fuEu gksrk gS% (A) {kkjh; (B*) vEyh; Sol.

(C) mnklhu (D) lkcq uhd ` r It is acidic because of the hydrolysis of Al2(SO4)3 according to the following reaction. Al2 (SO4)3 + 6H2O   2Al(OH)3 + 3H2SO4.

gy -

fuEufy f[kr vfHkfØ ;kd svuql kj Al2(SO4)3 d st y vi?kVu d sd kj.k;g vEy h; gSA Al2 (SO4)3 + 6H2O   2Al(OH)3 + 3H2SO4.

13.

From B2H6, all the following can be prepared except : [p-Block Element (Boron & Carbon Family)] B2H6 lsfuEu es alsfd l ,d d svfrfjDr lHkh d kscuk;k t k ld rk gS% (A) H3BO3 (B) [BH2(NH3)2]+ [BH4]– (C*) B2(CH3)6 (D) NaBH4 (C) CH3 group being larger can not form a bridge between two small sized boron atoms. (C) CH3 lew g d k cM+k vkd kj gksusd sd kj.k ;g nksNksVsvkd kj d scksjkWu ijek.kqv ksad se/; lsrqughacuk ld rk gSA z

Sol.

gy -

Lecutre Sub-topic(s) Nam e (No. No. of Lectures) L84

Pg.N

Handout

Basic Strength Ex.1 (S) Ex.1 (O)

L85

Hom e Work NCERT

Hom e Work Sheet

Basic Strength

Th.

Ex.2 (O) Ex.2 (I) Ex.2 (M)

XI XII

388 to 391

XI Prob. XII

401, 392, 396, 400

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PAGE NO.- 3

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 48 & 49

This DPP is to be discussed in the week (28.12.2015 to 02.01.2016) 1. 2. 3. 4.

Course of the week as per plan : Acidic Strength, Acidic Strength and Tautomerism. Course covered till previous week : Basic Strength, Basic Strangth and Acidic Strength Target of the current week : Acidic Strength, Acidic Strength and Tautomerism. DPP Syllabus : Acidic Strength, Acidic Strength and Tautomerism.

DPP No. # 48 (JEE-MAIN) Total Marks : 54

Max. Time : 36 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.18

(3 marks, 2 min.)

[54, 36]

ANSWER KEY DPP No. # 48 (JEE-MAIN) 1. 8. 15.

(A) (A) (D)

2. 9. 16.

(A) (D) (C)

3. 10. 17.

(D) (C) (B)

4. 11. 18.

(B) (A) (C)

5. 12.

(C) (B)

6. 13.

(C) (C)

7. 14.

(A) (C)

(A) 4. (D) 5. (C) (C) 11. (C) 12. (C) (A  p) ; (B  r) ; (C  q) ; (D  s)

6. 13.

(D) (D)

7.

(B)

DPP No. # 49 (JEE-ADVANCED) 1. 8. 14.

(B) 2. (B) 9. 2 (II & IV)

1.

Which molecule have more tendency to accept H? d kSulk v.kqH xzg.k d jusd h izo`fÙk vf/kd j[krk gS\ (A*)

2.

(D) (B)

3. 10. 15.

(B)

(C)

(D)

Which N atom can donate its electron pair more easily? d kSulk N ijek.kqby sDVªkWu ;qXe vf/kd vklkuh lsnsld rk gS\ O || NH—C—CH3

(A*)

3.

(B)

(C)

(D)

(C)

(D*)

Choose the strongest base among the following :

fuEu esaizcy re {kkj d k p;u d hft ;sA

(A) Sol.

(B)

Only in (D) the .p of N atom is not involved in resonance with benzene ring. d soy (D) esaN ijek.kqd k .p csat hu oy ; d slkFk vuqukn esaughagSA 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

4.

Which nitrogen in LSD (Lysergic acid diethylamide) is most basic ? LSD (y kblft Z d vEy Mkb,fFky ,ekbM ) esad kSulkukbVªkst u lclsvf/kd {kkjh; O || (3) C – N(C2H5)2

(1)

H–N (2)

N | CH3 (B*) 2

(A) 1 5.

gSA

(C) 3

(D) all are equally basic (D) lHkhleku

(A) 1 (B*) 2 (C) 3 Increasing order of their basic strength follows for : Aniline p-nitroaniline p-toluidine I II III

{kkjh; gSA

fuEufy f[kr d sfy , {kkjh; lkeF;Zd k c<+rk gqv k Ø e gS% ,fuyhu p-ukbVª k,ss uhfyu p-Vky s bwfMu I (A) III < I < II 6.

II

III (B) III < II < I

(C*) II < I < III

(D) I < III < II

Consider the set of three compounds, one set of three compound being (1) MeNH2 (2) Me2NH and (3) Me3N and another set of three compounds being (4) EtNH2 (5) Et2NH and (6) Et3N Which is true about basicity of these compounds in water ? (A) 2 > 1 > 3 and 5 > 4 > 6 (B) 2 > 3 > 1 and 5 > 4 > 6 (C*) 2 > 1 > 3 and 5 > 6 > 4 (D) 3 > 2 > 1 and 6 > 5 > 4

fuEu ;kSfxd ksad sleqPp; ij fopkj d hft ;sA rhu ;kSfxd ksad k ,d leqPp; gS (1) MeNH2 (2) Me2NH rFkk (3) Me3N vkSj vU; rhu ;kSfxd ksad k leqPp; gS (4) EtNH2 (5) Et2NH rFkk (6) Et3N bu ;kSfxd ksad h t y esa{kkjh;rk d k lgh Ø e gS? (A) 2 > 1 > 3 rFkk 5 > 4 > 6 (B) 2 > 3 > 1 rFkk 5 > 4 > 6 (C*) 2 > 1 > 3 rFkk 5 > 6 > 4 (D) 3 > 2 > 1 rFkk 6 > 5 > 4

(B)

NH2

NH2

NH2

–

(C) 

NH2

–

CCl3

(D)V

–

(A*) 

–

7.

In RNH2, R2NH, R3N; the order of base strength is 2º > 1º > 3º when R = Me and the order is 2º > 3º > 1º when R  Me. RNH2, R2NH, R3N; es a{kkjh; lkeF;Zd k Ø e gS2º > 1º > 3º t c R = Me rFkk ;g Ø e gksxk 2º > 3º > 1º ;fn R  Me Identify the most basic nitrogen atom. (vf/kd {kkjh; ukbVª kst u ijek.kqd ksigpkfu;s\)

–

Sol.

–

CCl3

–

8.

CCl3

Sol.

(o) (m) (p) Aniline (A*) Aniline > p > m > o (B) Aniline > m > p > o (C) p > o > Aniline > m (D) p > o > m > Aniline (A*) ,fuy hu > p > m > o (B) ,fuy hu> m > p > o (C) p > o > ,fuy hu > m (D) p > o > m > ,fuy hu –CCl3 group shows –I effect so the greatest base weakening effect will be at ortho position than at meta positive and then at para position. Unsubstituted aniline will be strongest in the above list. –CCl3 , –I iz Hkko n'kkZ rkgSvr%{kkjh; {kerkd ks?kVkusokyklewg esVkrFkkiSjkd hrqy ukes avkFkks Zij gSA vr%viz frLFkkih 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.- 2

,fuy hu vU; d h rqy uk esavf/kd {kkjh; gSA 9.

Which of the following statement is wrong about basic strength. (A) Generally electron donating group increases basic strength & electron withdrawing group decreases basic strength. (B) As we move down the group basic character decreases. (C) As we move across the period basic character decreases. (D*) Ortho effect in aniline increases basic strength.

{kkjh; lkeF;Zd slanHkZesad kSulk d Fku xy r gS\ (A) lkekU;r;k by s DVªkWu nkrk lewg {kkjh; lkeF;Zc<+krsgSarFkk by sDVªkWu vkd "khZlewg {kkfj; lkEkF;Z?kVkrsgSaA (B) ,d oxZes aÅ ij lsuhpst kusij {kkjh; xq.k ?kVrsgSaA (C) ,d vkorZd sla xr t kusij {kkjh; xq.k ?kVrsgSaA (D*) ,s fuy hu esavkWFkksZizHkko {kkjh; xq.k c<+krkgSA 10.

Which of the following acid has highest tendency to ionise in aqueous solution ?

t y h; foy ;u esad kSulsvEy d h vf/kd vk;fur gksusd h izo`fr gksrh gSA Sol. Sol. 11.

(A) HCOOH (B) CH3COOH Acidity  - I effect – F show – I effect, Acidity  as vEy h;rk - I iz Hkko – F, – I iz Hkko n'kkZrk gS]blfy , vEy h;rk c<+rh gSA

(C*) FCH2COOH

(D) BrCH2COOH

Among -nitro acetic acid (1), -fluoroacetic acid (2) -bromoacetic acid (3), -cyanoacetic acid (4), the correct order of increasing acid strength is : ukbVª ks ,lhfVd vEy (1), ¶yks jks ,lhfVd vEy (2), cz ks eks ,lhfVd vEy (3), lk;uks ,lhfVd vEy (4) d s e/;

vEy h; lkeF;Zd k lgh c
(A*) 3 < 2 < 4 < 1 (B) 1 < 2 < 3 < 4 due to – I effect – No2 > – CN > – F > – Br – I iz Hkko d sd kj.k– NO2 > – CN > – F > – Br

(C) 2 < 1 < 3 < 4

Ease of hydrogen release in decreasing order follows gkbMªkst u d sfud y us(Ease of hydrogen release) d h nj d k ?kVrk gqv k Ø e

(A) a > c > e > d > b

(B*) a > e > c > b > d

(D) 4 < 1 < 2 < 3

gksxkA

(C) a > d > c > b > e

Sol. Sol.

Due to stability of conjugate base

13.

Which of the following phenols has the largest pKa value ? fuEufy f[kr fQ ukWy esalspKa d k eku fd lesalclsvf/kd gksxk \

(D) a > c > b > d > e

le;qXeh {kkj d sLFkkf;Ro d sd kj.k

(A)

(B)

(C*)

(D)

Sol.

is lowest Ka value so this compound has largest pKa value.

Sol.

d k fuEure Ka eku blfy , ;g ;kSfxd vf/kd re pKa eku j[krk gSA

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PAGE NO.- 3

14.

What is the acidity order of the following compounds.

fn;s x;s ;kSfxd ksa d s vEy h; lkeF;Z d k Ø e crkb;sA

I

II

COOH

COOH

COOH

COOH

CH3

III

COOH CH 3

V CH3

IV

CH3

CH3

Sol. Sol.

(A) I > II > III > IV > V (B) II > III > IV > V > I (C*) V > III > I > II > IV (D) V > III > II > I > IV (C) Due to ortho effect of benzoic acid, acidity as (C) cs Ut ksbd vEy d svkFkhZizHkko d sd kj.k vEy h;rk c<+rh gSA

15.

The feasible reaction is

[VPM Madam]

lql axr vfHkfØ ;k gSA

Sol.

16.

(A) CH3COOH + NaCl 

(B) C6H5COOH + KBr 

(C)

(D*)

+ KHSO4 

Salicylic acid is more acidic than p-hydroxy benzoic acid. lsy hflfyd vEy ] p-gkbMªkWDlhcSUt ksbd vEy d hvis {kkvf/kd vEy h;

+



gSA

Select the incorrect statement order :

xy r d Fku@Ø e d k p;u d hft , %

(A)

>

(B)

>

(C*)

>

(D) Sol.

Sol.

(A) (B) (D) (A)

<

.......... (Stability)

(LFkkf;Ro)

.......... (Acidic-strength)

(vEyh;

lkeF;Z )

.......... (Acid-strength)

(vEyh;

lkeF;Z )

.......... (Acid-strength)

(vEyh;

lkeF;Z )

 Intramolecular H-bending  due to ortho effect  Due to ortho effect, ka as

vUrvkf.od H-cU/ku (B)  vkFkhZiz Hkko d sd kj.kA (D)  vkFkhZiz Hkko d sd kj.k, Ka c<+rk gSA 

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PAGE NO.- 4

17.

If statement is True mark T and for False mark F

(i)

Imidazole

is more basic than pyridine because conjugate acid of imidazole have

equal distribution of charge between both nitrogen. (ii)

Ethylamine is stronger base than aniline.

(iii)

NH2 is stronger base than C H3

(iv)

(CH3)2NH is less basic than (CH3)3N

–

–

;fn d Fku lgh gSrksT rFkk vlR; gksusij F fy f[k,A (i)

behMst kW y

(ii)

ij vkos'kd k leku forj.k ,fFky ,ehu],uhy hu lsizcy {kkj gSA

(iii)

NH2 , C H3 lsiz cy

(iv)

(CH3)2NH, (CH3)3N lsd e

–

–

(A) T T F T

, fifjfMu

lsvf/kd {kkjh; gSD;ksa fd behMst kW y d sla;qXehvEy esanks uksaukbVª ks tu

{kkj gSA {kkjh; gSA

(B*) T T F F

(C) F T F T

(D) T F T F

Sol.

Sol.

18.

STATEMENT -1 :

is stronger base than

STATEMENT -2 : The –:NH2 group of (I) is better electron pair donor to H+ ion than the –:NH2 group of (II) (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True

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PAGE NO.- 5

d k {kkjh; lkeF;Z

d Fku-1 :

lsvf/kd gksrk gSA

d Fku-2 : ;kSfxd (I) d k –:NH2 lewg ;kSfxd (II) d s–:NH2 lewg lsH+ vk;u d sfy ;svPNk by sDVªkWu ;qXe nkrk gSA (A) d Fku& 1 lR; gS ] d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k gSA (B) d Fku& 1 lR; gS ] d Fku& 2 lR; gS; d Fku& 2, d Fku& 1 d k lgh Li"Vhd j.k ughagSA (C*) d Fku& 1 lR; gS ] d Fku& 2 vlR; gSA (D) d Fku& 1 vlR; gS ] d Fku& 2 lR; gSA



H  

Sol.

Conjugate base stabilise by resonance

la;qXeh {kkj vuqukn }kjk LFkk;hd `r gksrsgSA

DPP No. # 49 (JEE-ADVANCED) Total Marks : 51

Max. Time : 35 Amin.

Single choice Objective ('–1' negative marking) Q.1 to Q.10 Comprehension ('–1' negative marking) Q.11 to Q.13 Integer type Questions ('–1' negative marking) Q.14 Match the Following (no negative marking) Q.15

1.

(3 (3 (4 (8

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 6 min.)

[30, [09, [04, [08,

20] 06] 03] 06]

[Tautomerism]

Which of the following will not show tautomerism ?

fuEu esalsd kSu py ko;ork ughan'kkZrk gS\

(A) 2.

(B*)

(C)

(D)

[Tautomerism]

Which one of the following compound does not exist in enol form ? fuEu ;kSfxd ksaesalsd kSu bZuksy : i iznf'kZr ughad jrk gS?

O O || || (A) CH3  C  CH2  C  C 2H5

(B)

(C)

(D*)

Sol.

(D) All carbon atoms are sp2 hybridized. d fjr gSA (D) lHkh C ijek.kqsp2 la

3.

Which of the following compound will not show tautomerism.

fuEufy f[kr esalsd kSulk ;kSfxd py ko;ork iznf'kZr ughad jrsgSA

(A*) Ph — C — Ph || O

(B) CH3 – CH – NO2

(C)

CH 3

(D) CH3  C  CH2  C  CH3 || || O O

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PAGE NO.- 6

Sol.

For tautomerism -H must be present. py ko;ork d sfy , -H mifLFkr gksuk pkfg,A

4.

Tautomerism will be exhibited by :

fuEu esalsfd ld s}kjk py ko;ork iznf'kZr d h t krh gSA (A) (CH3)2NH

(B) (CH3)3CNO

(C) (CH3)3C – CN

(D*) RCH2NO2

Sol.

5.

In which of the following compound. D–exchange will take palce in OD/D2O solution ? fuEu esad kSulk;kSfxd OD/D2O foy ;u esaD-fofue;(D–exchange ) d jrkgS\

(A)

(B)

(C*)

(D)

6.

Stability order among these tautomers is :

mijksDr py ko;oh;ksad sLFkkf;Ro d k lgh Ø e d kSulk gS% (A) I > II > III

(B) III > II > I

(C) II > I > III

(D*) II > III > I

(C) I = II

(D) Can not predict

(C) I = II

(D) crk;k ugh t k ld rk

(C) I = II

(D) Can not predict

(C) I = II

(D) crk;k ugh t k ld rk

7.

Which of these tautomers is more stable ? (A) I (B*) II

fuEu esalsd kSulk py ko;oh : i vf/kd LFkk;h gS\ (A) I

(B*) II

8.

Which of these tautomers is more stable ? (A) I (B*) II

fuEu esalsd kSulk py ko;oh : i vf/kd LFkk;h gS\ (A) I

(B*) II

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PAGE NO.- 7

9.

The correct acidity order is :

vEy h;rk d k lgh Ø e D;k gS\ O || CH3 – C – OH

O || H – C – OH

(Y)

(A) X > Z > Y 10.

Sol. Sol.

(B*) Z > X > Y

(Z)

(C) Y > Z > X

(D) X > Y > Z

The compound in which the CH3 –group causing increase in acid strength and decrease in basic strength are respectively : ;kSfxd ft lesaCH3 –lewg vEy h; lkeF;Zd ksc<+krk gSrFkk {kkjh; lkeF;Zd ks?kVkrk gSØ e'k%gS: (1) CH3–COOH

(2)

(3)

(4)

(5)

(6)

(A) 3 and rFkk 5 (B) 1 and rFkk 6 (C*) 3 and rFkk 4 (D) 2 and rFkk 5 Due to ortho effect –CH3 group in ‘3’ is increasing acidic strength & in ‘4’ is decreasing basic strength. vkFkhZizHkko d sd kj.k ‘3’ esa–CH3 lewg vEy h; lkeF;Zc<+rk gSrFkk ‘4’ esavEy h; lkeF;Z?kVkrk gSA

Write up : (Q.11 to Q.13) Observe the following sequence of reactions

vfHkfØ ;k vuqØ e d k izs{k.k d hft ;sA

 H

 H





1

11.

2

 H

 3

The correct stability order is

LFkkf;Ro d k l gh Ø e gSA

(A) III > II

(B) IV > II

(C*) III > IV

(D)

> IV

Sol. Sol.

III is more stable becose –OH group show – I effect but in compound IV – O  show + I effect III vf/kd LFkk;h gSD;ks afd –OH lewg – I izHkko n'kkZrk gSy sfd u ;kSfxd IV – O  lewg + I izHkko n'kkZrk gSA

12.

In compound II which of the following electronic effect is absent (A) Inductive effect (B) Delocalisation of  electrons. (C*) Hyperconjugative (D) Negative mesomeric effect

fn;sx;s;kSfxd II esafuEu esalsd kSulk by sDVªkWfud izHkko vuqifLFkr gSA (A) iz sjf.kd iz Hkko (B)  bys DVª kWuksad kfoLFkkuhd j.k (C*) vfrla ;q Xeu (D) + _ .kkRed esl kses fjd izHkko Sol. Sol.

(C) sp3 Hydrogen is absent (C) sp3 gkbMª kst u ijek.kqvuqifLFkr

gSA

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PAGE NO.- 8

13.

Which of the following is not an acid

fuEu esalsd kSulk vEy ughagSA Sol. Sol.

(A) I (B) II (D) Due to absence of acidic Hydrogen (D) v Ey h; gkbMª kst u d h v uqifLFkfr d sd kj.kA

(C) III

14.

How many of the following will show tautomerism ?

(D*) IV

fuEu esalsfd rus;kSfxd py ko;ork iznf'kZr d jsxs\ O

NH Ans.

2 (II & IV) Some important order of decreasing acidic strength (based on Ka values ) d qN egRoiw.kZ v Ey h; l keF;Z d k ?kVrk gqv k Ø e (Ka d s ekuksa ij v k/kkfjr) CF3COOH > CCl3COOH > CHCl2COOH > NO2CH2COOH > NC–CH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH > HCOOH > ClCH2CH2COOH > C6H5COH > C6H5CH2COOH > CH3COOH > CH3CH2COOH

15.

Match the column :

fey ku fd ft , & Column I Compounds d kWy e-I

Column II Kb d kWy e-II Kb

;kSfxd

Ans.

[Ref. MM Sir] [Ref. MM Sir]

(A)

(p)

5.6 × 10–7

(B)

(q)

4 × 10– 4

(C)

NH3

(r)

4.6 × 10– 9

(D)

CH3 –CH2 – NH –CH3

(s)

10 × 10– 4

(A  p) ; (B  r) ; (C  q) ; (D  s) Lecutre No.

Sub-topic(s) Nam e (No. of Lectures)

Ex.1 (S)

Acidic Strength

Ex.2 (O)

Ex.1 (S)

Sec : B, C Sec : B, C

Ex.2 (O)

6 to 23

Ex.2 (I)

5 to 11

Ex.2 (M)

5 to 15

Ex.1 (O) L91

Tautomerism

XII

Ex.1 (O)

Ex.1 (S)

Th.

Ex.2 (I)

345, 378, 379 402, 376

XI XII XI

Deutarium Exchange

Prob. XII Th.

Ex.2 (O) Ex.2 (M)

XI XII

Prob.

Ex.2 (M)

Tautomerism

Handout

XI

Ex.2 (I)

L90

Chem INFO

Pg.N Th.

Ex.1 (O) L89

Hom e Work NCERT

Hom e Work Sheet

XI XII XI

Prob.

fesibility of acid -base reaction

XII

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PAGE NO.- 9

ORG./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2017 Course : VIKAAS (JA)

NO. 50

This DPP is to be discussed in the week (04-01-2016 to 09-01-2016) 1. 2. 3. 4.

Course of the week as per plan : Acidic Strength, Acidic Strength and Tautomerism. Course covered till previous week : Basic Strength, Basic Strangth and Acidic Strength Target of the current week : Acidic Strength, Acidic Strength and Tautomerism. DPP Syllabus : Acidic Strength, Acidic Strength and Tautomerism.

DPP No. # 50 (JEE-MAIN) (Revision DPPS) Total Marks : 60

Max. Time : 40 min.

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

(3 marks, 2 min.)

[60, 40]

ANSWER KEY DPP No. # 50 (JEE-MAIN) (Revision DPPS) 1.

(C)

2.

(B)

3.

(C)

4.

(B)

5.

(D)

6.

(D)

7.

(D)

8. 15.

(C) (B)

9. 16.

(D) (B)

10. 17.

(C) (A)

11. 18.

(B) (C)

12. 19.

(A) (B)

13. 20.

(D) (C)

14.

(C)

1.

is named as : (A) 2, 3-Dimethylenebutanal

(B) 3-Methyl-2-methylenebut-3-enone

(C*) 3-Methyl-2-methylenebut-3-enal

(D) 2, 3-Dimethylenebutanone

d k uke gS% (A) 2, 3-Mkbes fFky huC;w VsuSy

(B) 3-es fFky -2-es fFkyhuC;w V-3-bZ uks u

(C*) 3-es fFky -2-es fFkyhuC;w V-3-bZ uS y

(D) 2, 3-Mkbes fFkyhuC;wVs uks u

4

O

Sol.

H

1

2

H

1

3-Methyl-2-methylenebut-3-enal

4

O

Sol.

3

2

3

3-es fFky -2-es fFkyhuC;w V-3-bZ uS y

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PAGE NO.- 1

2.

How many structural isomers could be obtained from the alkane C6H14 ?

,Yd su C6H14 d sfd ruslajpukRed leko;oh lEHko gS\ (A) 4

3.

(B*) 5

(C) 6

Reductive ozonolysis An alkene (A)  

A is

vip;kRed vkst ksuhvi?kVu ,d ,Yd hu (A) 

(A)

(D) 7

A gS%

(B)

(C*)

(D)

Sol.

4.

Acetaldehyde and Propyne can be distinguish by

,s l hVS fYMgkbM vkSj izksikbu d ksfoHksfnr fd ;kt kld rkgSA

Sol.

(A) NaHCO3

(B*) I2/NaOH

(C) Lucas reagent

D) neutral FeCl3

(A) NaHCO3

(B*) I2/NaOH

(C) Y;q d kl

(D) mnklhu FeCl3

vfHkd eZ d

Acetaldehyde and Propyne can be distinguish by Iodoform test.

,s l hVSfYMgkbM vkSj izksikbu d ksvk;ksMksQ keZijh{k.k}kjkfoHksfnr fd ;kt kld rkgSA 5.

Arrange following compounds in decreasing order of their dipole moment.

fn;sx, ;kSfxd ksad sf}/kqzo vk?kw.kZd k ?kVrk gqv k Ø e gS% CH3—CH2—NO2 I

Ans.

CH3—CH2—NH2 II

CH3—CH2—F III

(A) IV > III >I > II (B) IV > I > III > II (C) I > III > IV > II Decreasing order of dipole moment for given compounds

CH3—CH2—CN IV (D*) I > IV > III > II

;kSfxd ksad sf}/kqzo vk?kw.kZd k ?kVrk gqv k Ø e fuEu gS% I > IV > III > III

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PAGE NO.- 2

6.

Identify which of the following does not show + m effect ?

fuEu esalsd kSulk lewg + m izHkko iznf'kZr ughad jrk gS\ (A) –NH2

Sol.

Sol. 7.

O || (C)  NH  C  CH 3

(B) –O



O || (D*)  C  NH  CH 3

O ||  C  NH  CH3 do not contain lone pair or negative charge on first atom. O || aizFke ijek.kqij ,d kd h by sDVªkWu ;qXe ;k _ .kkRed vkos'k mifLFkr ughagSA  C  NH  CH3 es The correct stability order of following is :

fuEu d sLFkkf;Ro d k lgh Ø e gS%

Sol.

(I) (II) (A) I > II > III > IV (B) III > IV > II > I Stability  resonance  Hyperconjugation length of conjugation is equal in all I, II, III & IV.

gy -

LFkkf;Ro  vuqukn  vfrla ;qXeu lHkh I, II, III o IV esala;qXeu d h y EckbZleku gSA

8.

Correct order of stability of given carbocation is :

(III) (C) II > IV > III > I

(IV) (D*) IV > III > II > I

fuEu d kcZ/kuk;uksad sLFkkf;Ro d k lgh Ø e gS% O

+ I (A) I > III > II 9.

+

+ II (B) I > II > III

(C*) II > III > I

III (D) II > I > III

(C)

(D*)

Which molecule have lowest pKb value?

d kSulk v.kqU;wure pKb eku j[krk gS\

(A)

(B)

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PAGE NO.- 3

10.

Using anhydrous AlCl3 as catalyst, which one of the following reaction produce ethylbenzene (PhEt)?

mRizsjd fut Zy h; AlCl3 d h mifLFkfr esafuEu esalsfd ld h fØ ;k }kjk ,fFky csUt hu (PhEt) curk gS? [AIPMT 2004] (A) H3C – CH2OH + C6H6 (C*) C2H5Cl + C6H6 11.

(B) CH3 – CH == CH2 + C6H6 (D) H3C – CH3 + C6H6

A particular form of tribromobenzene forms three possible mononitro Bromo-benzenes the structure of compound is :

(A)

(B*)

(C)

(D) Both B and C

,d VªkbZczksekscsat hu]rhu lEHko eksuksukbVªksczksekscsat hu cukrhgSrks;kSfxd d hlajpukgksxhA

(A)

(B*)

[Ref. DRM mam]

(D) B rFkk C nks uksa

(C)

conc. HNO  conc. H SO

Sol.

3 2 4 

+

12_.

Which is more stable :

fuEu esalsd kSulk vf/kd LFkk;h gS%&

13_.

(I)

(II)

(A*) I > II (A*) I > II

(B) II > I (B) II > I

(C) II = I (C) II = I

(D) stability can’t be predicted (D) LFkkf;Ro d kvuq eku ughayxk;kt kld rk

In HCOO–, the two carbon-oxygen bonds are found to be of equal length. What is the reason for this ? (A) The anion is obtained by the removal of a proton from the acid molecule. (B) Electronic orbitals of carbon atoms are hybridised. (C) The C=O bond is weaker than C–O bond. (D*) The anion HCOO– has two equally stable resonating structures.

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PAGE NO.- 4

esanksuksd kcZu vkWDlht u ca/k leku y EckbZd sgksrsgSA bld k lgh d kj.k D;k gS\ v.kqlsizksVksu fu"d klu d si'pkr _ .kk;u izkIr gksrk gSA (B) d kcZ u v.kqd sby sDVªkfud d {kd lad fjr gksrsgSA (C) C=O ca /k C–O ca/k lsnqcZy gksrk gSA (D*) _ .kk;u HCOO– d hnksleku LFkkf;Ro oky hvuq uknhlajpuk,sagksrhgSA HCOO– (A) vEy

Sol. In HCOO– , the two carbon oxygen bonds are of equal length because the anion HCOO– has two equally stable resonating structures.

HCOO– es anks uksd kcZ u

vkW Dlht u ca /kleku yEckbZd sgks rsgSD;ks fd a_ .kk;u HCOO– d hnksleku LFkkf;Ro okyhvuq uknh

lajpuk,sagksrhgSA

14_.

Which of the following benzene ring has greater electron density than

fuEu esalsd kSulh csUt hu oy ; d k by sDVªkWu ?kuRo

(A)

15._

(B)

lsvf/kd gSA

(C*)

(D)

Which is most basic in aqueous solution ?

fuEu esalsd kSulk t y h; foy ;u esavf/kd {kkjh; gS& Sol.

(A) CH3NH2 (B*) (CH3)2NH (C) (CH3)3N (D) Ph–NH2 Secondary amine is most basic in aqueous solution among aliphatic amines.

,fy Q sfVd ,ehuksaesalsf}rh;d ,ehu t y h; foy ;u esalokZf/kd {kkjh; gksrk gSA 16._

Decreasing order of enol content of the following compound in aqueous phase.

fuEu ;kSfxd ksd k t y h; voLFkk esabZukWy ?kVd d k ?kVrk gqv k Ø e gS& (a)

(b)

(c)

(d)

(A) (a) > (b) > (c) > (d)

(B*) (c) > (b) > (a) > (d) (C) (c) > (b) > (d) > (a)

(D) (b) > (c) > (a) > (d)

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PAGE NO.- 5

17._

The correct order of decreasing acid strength of trichloroacetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is : [AIPMT 2012 (Pre.)]

Vª kbZDy ksjks ,s l hfVd ,flM (A), VªkbZ ¶yqv ks jks,sflfVd ,flM (B), ,s l hfVd ,flM (C) vkSj Q kfeZ d ,flM (D) d s?kVrsgq , vEy lkeF;Zd k lgh Ø e gS% Sol.

(1*) B > A > D > C (2) B > D > C > A (3) A > B > C > D (4) A > C > B > D CF3–COOH > CCl3–COOH > HCOOH > CH3COOH (Ka order) (Ka d kØ e)

18.

Which is most basic among the followings ?

fuEu easlsd kSulk lclsvf/kd {kkjh; gS\ (1) Ph-NH2 19._

(2) NH3

(GOC-II) (MG Sir April 2015) (3*) CH3–NH2

(4) C2H5–CN

Arrange the following compounds in order of decreasing acidity.

fuEu ;kSfxd ksd k vEy h;rk d k ?kVrk gqv k lgh Ø e gS&

Sol.

(i) (ii) (iii) (iv) (1) (i) > (ii) > (iii) > (iv) (2*) (iii) > (i) > (ii) > (iv) (3) (iv) > (iii) > (i) > (ii) (4) (ii) > (iv) > (i) > (iii) Electron withdrawing group increase acidic strength and electron relasing group decrease acidic strength.

by sDVªkWu vkd "khZlewg vEy h; lkeF;Zd ksc<+krsgSrFkk by sDVªkWu fu””"d f"kZlewg vEy h; lkeF;Zd ks?kVkrsgSA 20.

In which pairs first compound is stronger acid than the second ? (A) Adipic acid, succinic acid (B) Fumaric acid, maleic acid (C*) Pthalic acid, terepthalic acid (D) Picric acid, o-toluic acid

fd l ;qXe esaizFke ;kSfxd ]f}rh; d h rqy uk esavf/kd izcy vEy gS\ (A) ,Mhfid vEy ]lfDlfud vEy (B) ¶;w esfjd vEy ] eSy sbd vEy (C*) FkS fy d vEy ]VjFkSfy d vEy (D) fifØ d vEy , o-VkW yw bd vEy Sol.

gy %

(C) Phthalic acid is stronger acid due to intramolecular hydrogen bonding. (C) FkS fy d vEy vUr%vkf.od gkbMªkst u ca/ku d sd kj.kizcy vEy gSA

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