10
10.5 The following two compounds I and II can be distinguished by using reagent
OH
COOH
HO
COOH
(1) aq. NaHCO3 (2) Neutral FeCI3 (aq.) (FfeCI3 + NH4OH + H2O)
(3) Blue litmus solution (4) Na metal (5) HCI (ZnCI2 anhydrous) (A) 1 or 3 (B) 2 or 5 (C) 4 or 5 (D) 3 or 4
10.6 Tollen's regent (AgNO3 + NH4OH) can be distinguished between
(A) H–C–H and Ph–C–H
O O
(C)
C O D
Me C=O and
Me C=O
O
O H
O H H
O and
Ph–C–Ph and Me–C–Me
O O
(B)
(D)
10.7 Observe the4 following compound and select +Ve & –Ve tests respectively
C CH
COOH OHC
HO
(i) Na metal (ii) NaHCO3 (iii) 2,4-DNP (iv) Lucas reagent
(A) + + + – (B) + + + + (C) + – + – (D) + – – +
10.8 Consider following compounds and decide as to which of the following statements are true ?
CH = CH – C – COOH2 H
CH3
CH – C – OCH3 3 CH3 CH3
CH – C – CH OCH3 2 3 OH
CH3 CH O–C 3 C – H
(I) (II) (III) (IV)
(A) (II) gives no reaction with Na metal, however, 1 mole of (IV) on reaction with Na metal willliberate 22.4 litres of H2 gas at STP
(B) (I) will give brisk effervescence on addeition of NaHCO3 but will not bring any change in the colour of Br2 water
(C) (III) Iiberates H2 gas with Na metal, gives white precipitate with Tollen's reagent but does not respond towards lucas reagent or 2, 4-DNP test.
(D) (IV) gives turbidity with anhydrous ZnCI2
10.9 An aromatic compound 'X' (C9H8O3) turns blue litmus to red.It gives yellow precipitate with I2/ NaOH and forms Y (C8H6O4).Y forms three mononitro isomeric products. Identify X.
COOH COCH3 (A)
COOH COCH3 (B)
COOH COCH3
(C) (D) none of these
10.10 Which will not give iodoform reaction with I2/OH–?
(A) CH3COCH2CH3 (B) CH3CONH2 (C) C6H5COCH3 (D) CH3CHO
10.11 Compound Y1C7H8O is insoluble in water, dil HCI and aqueous NaHCO3.It dissolves in dilute NaOH. When Y is treated with bromine water it is converted rapidly into a compound of formula C7H5OBr3.Identfy the structure of Y
(A) OH
CH3
(B) (C) (D)
OH O – H O – CH3
CH3
CH3
10.12 Compounds I and II can be distinguished by using reagent
(I) (II)
4-Amino-2-methlbut-3-en-2-ol 4–Amino–2, 2-dimethylbut-3-yn-1-ol.
(A) NaNO2/HCI (B) Br2/H2O
(C) HCI/ZnCI2 (anhydrous) (D) Cu2CI2 + NH4OH
10.13 Compound P(C6H10) does not have any geometrical isomer. ON ozonolysis, two products R(C3H4O) and Q(C3H6O) are formed. R gives negative iodoform test while Q responds positively towards I2/NaOH solution. S, another isomer of P is an unsyumetrical alkene and on ozonolysis produces T(C6H10O2) which also gives a yellow precipitate with I2/NaOH solution and also gives positive test with Tollen's reagent. Which of the following does not represent any of the molecules amongst P,Q,R,S&T.
(A) O (B) (C) (D)
H OO H H OO H
10.14 A set of reagents (1 to 8) are successively reacted with the followit compound
OH
OH O
1. NaHCO3 2.2, 4, DNP 3.Na metal 4. AgNO3 +OH
5. Fehling's solution 6.Cu2CI2 + NH4OH 7. Br2/H2O 8. NaNO2 + HCI The reagents which give positive test with the given compound are :
(A) 1, 2, 3, 4, 5 (B) 3, 4, 5, 6, 8 (C) 1, 2, 3, 4, 8 (D) All reagents except 1 and 8
10.15 Compounds (C8H8O) X will give following laboratory tests.
Isomers FeCl3 AgNO / NH OH3 4 Na metal X Coloured solution Negative 1/2 H 2
X can be :
10.16 Which of the following alcohol will show positive iodoform test ?
CH – CH – CH3 2 – NO2
(D) none of these 10.17 The compound A gives following reactions.
Na metal
2, 4-DNP
O3
yellow orange ppt H 2gas
10.18 In compound A (C30H60O) following tests are observed negatively, A can be :
-Ve
(A) an unsaturated ether(B) an epoxide (C) a cyclic ketone (D) a cycloalkanol
10.19 A mixture of two orgnaic compound gives red coloured precipitate with cuprous chloride (ammonical) and silver mirror on heating with Zn dust and NH4CI followed by AgNO3 + NH4OH solution. The mixture contains
COOH COOH
SECTION - II : MULTIPLE CORRECT ANSWER TYPE
10.20 2D, 3D, 4D, 5D, 6-Pentahydroxy hexanal can give.
(A) Tollen's Test (B) Lucas Test (C) 2, 4-DNP Test (D) FeCI3 Test
10.21 Compound X and Y both have the same molecular formula (C4H8O). They give following observation in some lab test.
Test X Y
Br2–water positive negative
Na–metal positive negative
Lucas reagent turbidity after some time negative
lodoform test negative negative The compound X and Y are
(A) OH (B) OH (C) OH (D) O
10.22 Compound 'P' (C10H12O) evolves H2 gas with Na metal. It reacts with Br2/CCI4 to give 'Q' (C10H12Br2O). With I2/NaOH it forms iodoform and an acid 'R' (C9H8O2). 'P' has geometrical and optical isomers. The structure of 'P and R' should be
(A)
COOH
CH=CH2
(C)
CH = CH – CH – CH3 OH
(B) PH – CH = CH – COOH
(D)
CH = CH – CH3 OH
10.23 compound (X) C9H10O is inert to Br2/CCI4.Vigorous oxidation with hot alkaline KMnO4/OH yields C6H5COOH. (X) gives precipitate with 2,4-dinitrophenyl hydrazine. How can these isomers be distinguished by the usual chemical tests?
Following are possible isomkers of X : (l)
(lv) (l)
(lll)
C H – CH – CH – CHO6 5 2 2
C H – CH – C – CH6 5 2 3 O
C H – CH – CHO6 5
C H – C6 5 – CH – CH2 3 O
CH3
(I) C6H5 – CH2 – CH2– CHO
(A) I gives red ppt. with Fehling solution and II & III cand be distinguished by iodoform test (B) I & II can be distinguished by simple chemical method
(C) I & II give red ppt. with Fehling solution and III & IV can be distinguished by iodoform test (D) II give red ppt. with Fehling solution and I & IV can be distinguished by iodoform test.
10.24 Which is/are the correct method for separating a mixture of benzoic acid, p-methylaniline &
phenol.
(A) aq.NaHCO3 aq.NaOH (B) aq.HCI aq.NaHCO3 (C) aq.NaOH aq.NaHCO3 (D) aq.NaOH aq.HCI
SECTION - III : ASSERTION AND REASON TYPE
10.25 Statement -1 : Only one Aldehyde 'X' responds positively with all the tests of carbonyl compounds like Tollen's test, Fehling test, 2, 4-DNP test, as well as iodoform test.
Statement-2 : All aldehydes respond all the four tests given in assertion.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Satement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
10.26 Statement-1 : A mixture os p-methylbenzoic acid and picric acid is separated by NaHCO3 solution .
Statement -2 : p-Methylbenzoic acid is soluble in NaHCO3 because it give effervesence of CO2 (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Satement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
SECTION - IV : TRUE AND FALSE TYPE
10.27 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false. X (molecular formula, C7H6O2) is and aromatic white solid which liberates colourless, odourless gas on rescting with NaHCO3.
S1 : Only three of th3e five functional isomers of X (including 'X' itself) will give positive 2, 4-DNP test.
S2 : The liberated colourles, odourless gas will contaning radioactive 14C.
S3 : Except 'X', no other functional isomer will liberate colourless odourless gas with NaHCO3.
S4 : The DU of higher homolog of 'X' will be four.
(A) TTTF (B) FTTF (C) FTTT (D) TTFF
SECTION - V : COMPREHENSION TYPE
Comprehension # 1
Observe the following sequence of reactions P (C H Br)9 9 R + Q
(s) + Ag
reductive O / Zn,H O3 2
(i) Tollen's Reagent (ii) H+
P shows geometrical isomersm. Q gives positive Tolen's test and the oxidation product of Tollen's test followed by acidification is the strongest acid among its all position isomers. R gives positive lab tests with 2,4-DNP, Fehling solution and I2/NaOH reagents.
10.28 The compound P can be
(A) Br
CH – CH = CH2 2
(B) Br
CH= CH – CH3
(C) Br CH= CH – CH3
(C)
CH= CH – CH3
Br 10.29 What could be the structure of Q ?
(C)
CHO
CH3 (B)
(A) (D)
CHO Br COOH
CH3 CHO
Br
10.30 Identify the structure of R
(A) HCOOH (B) CH3CHO (C) BrCH2 – CHO Comprehension # 2
An aromatic compound T (C10H10O2) give 2 moles of CHI3 and compound U (C8H4O4Na2) On treatment with I2 and NaOH. After acidification U gives two mononitro produicts on nitration.
10.31 Compound (T) can aoso be obtained by ozonolysis of V, in this ozonolysis one mole of OHC – CHO is obtained alongwith (T). Possible structure for Compound V could be
CH3 CH3 (A)
(C)
CH3
CH3 CH3
CH3 (B)
CH3 (D) CH3
10.32 Which of the following statement is true
CH3 C CH3 O O
(A) T is C
(B) Compound (V) decolourises pink colour of diluted solution lkof KMnO4.
(C) All isomers (only acidic) of U after acidification gives one mole of CO2 with NaHCO3 (D) After acidification of (U), it is most acidic in its all other isomers.
10.33 Compound U is
O (A)
COONa
O C H
||
(B)
COONa COONa
(C)
COONa
COONa
(D)
COONa
COONa
SECTION - VI : MATRIX - MATCH TYPE
10.34 X(C8H14) by ozonolysis forms Y[C8H14O2]. Y on reaction with NaOI followed by acidification gives CHI3 and compound Z on strong heating forms W.
Column-I Column-II
(A) Compound X (p) Bayer's Test
(B) Compound Y (q) NaHCO3
(C) Compound Z (r) 2, 4-DNP
(D) Compound W (s) Iodoform Test
(t) Na Metal
10.35 Match the following : (More than one option in column - II may match with single option in column-I)
Column - I Column - II
(A) Sodium metal
(B) Sodium bicarbonate
(C) 2, 4-Dinitrophenylhydrazine
(D) Lucas reagent
(X)
O
C C – H
(Y)
CH3 CH3 OH C –
(Z) OHC
C CH COOH
(W)
CH = CH2 OH H CO3
C O
SECTION - VII : SUBJECTIVE ANSWER TYPE SHORT SUBJECTIVE :
10.36 A water insoluble organic mixture contained following compounds
(1) = Benzoic acid (2) = salicylaldehyde
(3) = p-Hydroxybenzaldehyde (4) =a-Naphthylamine (5) = Naphthalene
The following sequence of reagents are uksed to separate this mixture
1 + 2 + 3 + 4 + 5 mixture (Step-X) aq. HCI
[Insoluble] [Soluble]*
(Step-Y) aq. NaHCO3
[Insoluble] [Soluble]*
(Step-Z) aq. NaOH
[Insoluble]* [Soluble]
(Step-W) Steam distillation
[distilled]* [left behind]
Fill up the serial number of starred compound ontained in the steps X, Y, Z and W reapectively.
X Y Z W
10.37 How many acidic H is present in given compound.
O OH
O
OH
10.38 How many CHI3 will be released from the given compound.
O
||
O
|| l / NaOH2
10.39 How many molecule of phenyIhydrazine is used to form osazone from glucose.
Hint :
CHO
H OH
HO
CH OH2 OH OH H H
H
3Ph–NH–NH2
CH OH2 N––NH
N––NH HO
H OH
H
H
OH
+ Ph––NH + NH2 3
10.40 How many CHI3 will be released from given compound.
|| ||
O O
H N2 OCH3
O O