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90501.

A compound contains atoms of three elements A,B and C. If the oxidation number of A is +2 , B is +5 and that of C is - 2 , then the possible formula of the compound is

Answer»

`A_(3)(BC_(4))_(2)`
`A_(3)(B_(4)C)_(2)`
`ABC_(2)`
`A_(3)(BC_(3))_(2)`

Answer :A
Discussion
90502.

A compound contains atoms, A ,B and C. The oxidation number of A is+2, of B is +5 and of C is -2. The possibleformula of the compound is

Answer»

`ABC_(2)`
`B_(2)(AC_(3))_(2)`
`A_(3)(BC_(4))_(2)`
`A_(3)(B_(4)C)_(2)`

Solution :N/A
Discussion
90503.

A compound contains 8% sulphur. The minimum molecular weight of the compound is

Answer»

100
200
350
400

Solution :Let .a. ATOMS of sulphur be present in the compound and y be the weight of the compound
8% of weight of compound = weight of sulphur in compound
`8/100 y = 32 xx a`
` y = 3200/8 a = 400 a .
The MINIMUM molecular weight is the ONE in whicha = 1 (i.e 1 atom of sulphur is present)
Discussion
90504.

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g. What are its emirical and molecular formulas?

Answer»

Solution :To calculate the empirical formula

Hence, the empirical formula of the compound is `CH_(2)Cl`
Step 2. CALCULATION of empirical formula mass
E.F. mass of `CH_(2)Cl=12+2xx1+35.5=49.5`
Step 3. Calculation of VALUE of n
`n=("Molecular mass")/("E.F. mass")=(98.96)/(49.5)=2`
Step 4. Calculation of molecular formula
Molecular formula `=nxxE.F. = 2XX(CH_(2)Cl)=C_(2)H_(4)Cl_(2)`
Discussion
90505.

A compound contains 34.8% oxygen, 52.2% carbon and 13.0% hydrogen. What is the empirical formula mass of the compound?

Answer»

SOLUTION :
The empirical formula is `C_(2)H_(6)O`.
Empirical formula mass `=(2xx12)+(6xx1)+16=46`
Discussion
90506.

A compound contains 28 percent nitrogen and 72 percent of a metal by weight. If 3 atoms of the metal combine with 2 atoms of nitrogen, then atomic weight of the metal is :–

Answer»

24
28
32
42

Answer :A
Discussion
90507.

A compound contains 28% of nitrogen and 72% metal by mass 3 atoms of the metal combine with 2 atoms of the nitrogen. Find the atomic mass of the metal.

Answer»

Solution :Valency of METAL=2 and valency of NITROGEN =3
Equivalent mass of nitrogen `=(14)/(3),("Eq. mass of metal")/(14//3)=(72)/(28)`
Equivalent mass of metal=12
ATOMIC mass of metal `=12xx2=24`
Discussion
90508.

A compound contains 2 dissimilar asymmetric carbon atoms. The number of optically active isomers is:

Answer»

2
3
4
5

Answer :C
Discussion
90509.

A compound contains 10^(2)% of phosphorous . If atomic mass of phosphorouus is 31, the moleular mass of the compound having one phosphorous atom per molecules is :-

Answer»

31
`3.1xx10^(3)`
`3.1xx10^(5)`
`3.1xx10^(4)`

ANSWER :C
Discussion
90510.

A compound containing two -OH groups attached with one carbon atom is unstable but which one of the following is stable:

Answer»




None

Answer :C
Discussion
90511.

A compound containing sodium, sulphur, hydrogen and oxygen gave the following results on analysis : Na=14.28%, S=9.92%, H=6.20% Calculate the molecular formula of the anhydrous compound. If all the atoms of hydrogen in the compound are present in combination with oxygen as water of crystallization, what is the structure of the crystalline salt? The molecular mass of the crystalline salt is 322.

Answer»

Solution :Step 1. To calculate the percentage of OXYGEN. The given compound contains oxygen but is percentage is not given. This can, however, be calcuated by subtracting the sum of percentages of NA, S and H from 100 as shown below:
Sum of percentage composition of Na, S and H `=14.28+9.92+6.20=30.40`
`therefore"Percentage of oxygen"=100.00-30.40=69.60`
Step 2. To calculate the empirical formula.

Hence, the empirical formula of the compound is `Na_(2)SH_(20)O_(14)`
Step 3. To calculate the empirical formula mass. Empirical formula mass of the compound `(Na_(2)SH_(20)O_(14))`
`=2xx23.0+1xx32.0+20xx1.0+14xx16.0=322`
Step 4. To calculate the value of 'n'
`n=("Molecular mass")/("Empirical formula mass")=(322)/(322)=1`
Step 5. To calculate the molecular formula of the compound.
`"Molecular formula"=nxx"Empirical formula"=1xxNa_(2)SH_(20)O_(14)=Na_(2)SH_(20)O_(14)`
Step 6. To calculate the number of molecules of water of crystallisation. SINCE all the H-atoms are present in the form of `H_(2)O`, 20 atoms of hydrogen would combine with 10 oxygen atoms to produce10 molecules of `H_(2)O`.
Thus, the number of molecules of water of crystallization present in the salt `Na_(2)SH_(20)O_(14)` is 10.
Step 7. To determine the STRUCTURE of the crystalline salt. Out of the 14 oxygen atoms, 10 are present in the form of `H_(2)O`. The remaining 4 must be a part of the salt consisting of Na and S. Hence, molecular formula for the ANHYDROUS salt is `Na_(2)SO_(4)` and that of crystalline salt will be `Na_(2)SO_(4).10H_(2)O`.
Discussion
90512.

A compound containing carbon, hydrogen and oxygen gave the following analytical data : C=40.0% and H=6.67% Calculate the molecular formula of the compound if its molecular mass is 180.

Answer»

Solution :As crystalline salt on becoming ANHYDROUS loses `51.2%` by mass, this MEANS
48.8 g of anhydrous salt contains `H_(2)O=51.2g`
`THEREFORE"120 g of anhydrous salt contains "H_(2)O=(51.2)/(48.8)xx120=126g=(126)/(18)" molecules = 7 molecules."`
Hence, molecular formula of crysatlline salt = `MgSO_(4).7H_(2)I`
Discussion
90513.

A compound, containing only carbon, hydrogen and oxygen, has a molecular weight of 44. On complete oxidation it is converted into a compound of molecular weight 60. The original compound is

Answer»

an aldehyde
an acid
an alcohol
an ether

Solution :On complete oxidation the obtained compoundshows increament in molecular weight of only 16. It means only one OXYGEN atom is added here. This condition is fulfilled by only aldehyde which on oxidation gives acid
`RCHOoverset([O])to RCOOH`
Hence, original compound MUST be
`underset(mol.wt.44)(CH_(3)CHO) OVERSET([O])tounderset(mol wt 60)(CH_(3)COOH)`
Discussion
90514.

The oxygen containing angle is maximum in

Answer»

`H_2O`
`CH_3OH`
`CH_3OCH_3`
`(CH_3)_3COC(CH_3)_3`

ANSWER :D
Discussion
90515.

A compound contain two dissimilar chiral carbon atoms. The number of optical isomes is/are

Answer»

2
3
4
1

Answer :C
Discussion
90516.

A compound contain 4% oxygen, 4% sulphur and 10% carbon by mass. How many oxygen atoms will be present in 1 molecule of that compound.

Answer»

Solution :`n_(o)=(4)/(16)=(1)/(4) n_(o):n_(s):n_(C)=(1)/(4):(1)/(8):(5)/(6)`
`n_(s)=(4)/(32)=(1)/(8) n_(o):n_(s):n_(c)=6:3:5`
`n_(C)=(10)/(12)=(5)/(6)`
Discussion
90517.

Acompound CO(en)_(2)NO_(2)Cl has been prepared in a number of isomeric form . One form undergoes no reaction with AgNO_(3) or (en) and is optically inactive. A third form is optically acitve and reacts with both AgNO_(3) and (en) complexes are bis-bis (ethylenediamine) dinitrocobalt (III) chloride. trans -bis (ethylyenediamine) dinitrocobalt (III) chloride. trans chloronitrobis (ethylendiamine)cobalt (III) nitrite. One of the complexes starch iodide paper blue in acidic medium . This comples is

Answer»

C
B
A
all of these .

ANSWER :A
Discussion
90518.

Acompound CO(en)_(2)NO_(2)Cl has been prepared in a number of isomeric form . One form undergoes no reaction with AgNO_(3) or (en) and is optically inactive. A third form is optically acitve and reacts with both AgNO_(3) and (en) complexes are bis-bis (ethylenediamine) dinitrocobalt (III) chloride. trans -bis (ethylyenediamine) dinitrocobalt (III) chloride. trans chloronitrobis (ethylendiamine)cobalt (III) nitrite. First form of the complex is

Answer»

A
B
C
none of these

ANSWER :C
Discussion
90519.

A compound, C_(7)H_(14), on ozonolysis gives ethanal and 3-pentanone. The structure of the compound is :

Answer»

`CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH = CH_(2)`
`(CH_(3))_(2) C = C (CH_(3))CH_(2)CH_(3)`
`CH_(3)CH_(2) - C (C_(2)H_(5))=CH - CH_(3)`
`CH_(3) - underset(CH_(3))underset(|)(C) = CH - CH_(2)CH_(2)CH_(3)`

ANSWER :C
Discussion
90520.

A compound C_(6)H_(14)O_(2) has two tertiary alcoholic groups . The IUPAC name of this compound is

Answer»

2, 3 - DIMETHYL - 1, 2 - butanediol
3, 3 - dimethyl - 1, 2 - butanediol
2, 3 - dimethyl - 1, 2 - butanediol
2 - METHYL - 2, 3 - pentanediol

Answer :C
Discussion
90521.

A compound C_5H_8 which gives white ppt. with ammonical AgNO_3.A give (CH_3)_2CHCOOH with hot alcoholic KOH then compound is

Answer»

`CH_3CH_2-CH_2-CH=CH_2`
`CH_3-CH_2-C-=CH`
`(CH_3)_2CH-C-=CH`
`CH_2=CH-CH_2-CH=CH_2`

ANSWER :C
Discussion
90522.

A compound C_(5)H_(10)O(A) forms a phenylhydrazone and gives negative Tollens' test and a positive iodoform reaction. It gives n-pentane on reduction. The compound (A) is

Answer»

pentanal
2-pentanone
3-pentanone
AMYL ALCOHOL

ANSWER :B
Discussion
90523.

A compound C_(3)H_(8)O is inert to sodium and when heated with excess HI yields a mixture of methyl iodide and ethyl iodide. What is the structure and name of the compound?

Answer»

Ethyl methyl ether
Diethyl ether
DIMETHYL ether
Isopropyl alcohol

Solution :The formula `C_(3)H_(2)O` represents, either the compound is an alcohol or ether, as it is saturated and contains ONE OXYGEN atom. But the compound is inert to sodium, so it is an ether, because only alcohol reacts with sodium metal. Again the ether when heated with HI yields a mixture of methyl iodide and ethyl iodide, so it is ethyl methyl ether (`C_(2)H_(5)-O-CH_(3)`)
`underset("(Ethyl methyl ether)")(C_(2)H_(5))-O-CH_(3)+2HI overset(Delta)to C_(2)H_(5)-I+CH_(3)-I+H_(2)O`
Discussion
90524.

A compound alloy of gold copper crystallies in a cubic lattice in which gold atoms occupy the lattice points at the corners of each cube and copper atoms occupy the centres of each cubic faces. Find the number of total atoms per unit cell.

Answer»


ANSWER :4
Discussion
90525.

A compound alloy of gold and Cu crystallises in a cubic lattice in which the gold atoms occupy the lattice points at the comers of a cube and the copper atoms occupy the centres of each of the cube faces. What is the empirical formula of this compound ?

Answer»

`AuCu_3`
`Au_3Cu`
`Au_2Cu_3`
AuCu

Answer :A
Discussion
90526.

A compound A_(x)B_(y) crystallises on a fcc lattice in which A occupies each corner of a cube and B occupies the centre of each face of the cube. What is the formula of the compound ?

Answer»

Solution :`A_(x)B_(y)`
No. of corner PARTICLES `(x) = (1)/(8) XX 8 = 1`
No. of centre particles `(y) = (1)/(2) xx 6 = 3`
`A_(1)B_(3)`
Discussion
90527.

A compound alloy of gold and copper crystallizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of a cube and the copper atoms occupy the centres of each of the cube faces. Determine the formula of this compound.

Answer»

SOLUTION :One-eighth of each corner atom (Au) and one-half of each face-centred atom (CU) are CONTAINED within the UNIT cell of the compound.
Thus, number of Au atoms per unit cell =`=8 xx 1/8` and number of Cu atoms per unit cell =`6 xx 1/2=3`. The FORMULA of the compound is `AuCu_(3)`.
Discussion
90528.

A compound A(C_(7)H_(5)N) on hydrolysis with strong aqueous acid gives another compound B which is a monobasic aromatic carboxylic acid. The compound B on treatment with ammonia gives a salt which on heating gives C. The compound C undergoes Hofmann/s bromamide reaction to yield aniline. name A,B andC and write the chemical reactions involved.

Answer»

Solution :(i) SINCE COMPOUND `A(C_(7)H_(5)N)` on hydrolysis with strong aqueous ACID gives compound B which is a monobasic aromatic acid, therefore, compound (A) must be benzonitrile and compound (B) must be benzoic acid.
`underset("Benonitrile (A) "M.F.C_(7)H_(5)N)(C_(6)H_(5)CN)+2H_(2)O+H^(+)to underset("Benzoic acid (B)")(C_(6)H_(5)COOH)+NH_(4)^(+)`
(ii) Since compound B on treatment with ammonia gives a salt which on heating gives C, therefore, the salt must be AMMONIUM benzoate and the compound C must be benzamide.
`underset("Benzoic acid (B)")(C_(6)H_(5)COOH) overset(NH_(3))to underset("Amm. benzoate")(C_(6)H_(5)COONH_(4)) underset(-H_(2)O)overset(Delta)to underset("Benzamide (C)")(C_(6)H_(5)CONH_(2))`
(iii) The fact that compound (C) is benzamide is confirmed by the fact that it on Hofmann mromamide reaction gives aniline.
`underset("BEnzamide (C)")(C_(6)H_(5)CONH_(2)) underset(("Hofmann bromamide reaction"))overset(Br_(2)-NAOH)to underset("Aniline")(C_(6)H_(5)NH_(2))`
Discussion
90529.

A compound A(C_(5)H_(10)Cl_(2)) on hydrolysis gives C_(5)H_(10)O which reacts with NH_(2)OH, forms iodoform but does not give fehling test. A is

Answer»

`CH_(3)-underset(Cl)underset(|)OVERSET(Cl)overset(|)(C)-CH_(2)-CH_(2)-CH_(3)`
`CH_(3)CH_(2)-underset(Cl)underset(|)overset(Cl)overset(|)(C)-CH_(2)CH_(3)`
`CH_(3)CH_(2)CH_(2)CH_(2)overset(Cl)overset(|)(C)H-Cl`
`CH_(3)-overset(Cl)overset(|)(C)H-overset(Cl)overset(|)(C)H-CH_(2)CH_(3)`

Solution :Since compound A on hydrolysis Gives `C_(5)H_(10)O`, does not REDUCE Fehling's solution but reacts with `NH_(2)OH` and forms iodoform, therefore, it must be a methyl ketone
`underset((A))(CH_(3)-underset(Cl)underset(|)overset(Cl)overset(|)(C)-CH_(2)CH_(2)CH_(3) overset("Hydrolysis")to underset("A methyl ketone")(CH_(3)-overset(O)overset(||)(C)-CH_(2)CH_(2)CH_(3))`.
Discussion
90530.

A compound A(C_(4)H_(10)O) is found to be soluble in concentrated sulphuric acid. (A) does not react with sodium metal or potassium permangante. When (A) is heated with excess of HI, it gives a single alkyl halide. Deduce the structure of compound (A) and explain all the reactions involved.

Answer»

Solution :(i) Since COMPOUND `A(C_(4)H_(10)O)` does not reat with Na metal or `KMnO_(4)`, it cannot be an alcohol.
(ii) Since compound A dissolve in conc. `H_(2)SO_(4)`, it may be an ETHER.
(iii) Since ether A on heating with excess of HI gives a single ALKYL halide, therefore, ether (A) must be symmetrical. now the only symmetrical ether having M.F. `C_(4)H_(10)O` is diethyl ether `(CH_(3)CH_(2)OCH_(2)CH_(3))`.
(iv) All the reactions CANN now be explained as follows.
`underset("Diethyl etehr (A) M.F. "C_(4)H_(10)O)((CH_(3)CH_(2)-O-CH_(2)CH_(3)) overset("conc. "H_(2)SO_(4))hArr underset("Soluble ozonium salt")([CH_(3)CH_(2)-underset(H)underset(|)overset(+)(O)-CH_(2)CH_(3)])HSO_(4)^(-)`
`underset("Diethyl ether (A)")(CH_(3)CH_(2)-O-CH_(2)CH_(3)) underset(Delta)overset("HI (excess)")to underset("Ethyl iodide")(2CH_(3)CH_(2)-I)+H_(2)O`
Discussion
90531.

A compound AB has NaCl type structure. The face diagonal of the compound is 120sqrt(2) pm. Calculate the distance between cation and anion.

Answer»

30 PM
120 pm
60 pm
45 pm

ANSWER :C
Discussion
90532.

A compound AB_2 possesses the CaF_2 type crystal structure. Write the coordination number of A^(2+) and B^(-) ions in its crystals.

Answer»

SOLUTION :COORDINATION number of `A^(2+)` ION = 8
Coordination number of `B^(-)` = 4
Discussion
90533.

Crystal has face centred cubic structure, having atomic weight 6.023y g mol^-1. If the minimum distance between two atoms is y^(1//3) nm and the observed density is 20 kg m^-3 find type of defect in crystal lattice.

Answer»

Solution :AB has a rock salt (NaCl) structure. This TYPE of crystal structure possesses fcc unit cell and contains four formula units PER unit cell, i.e., Z = 4.
In CASE of a rock salt structure, the edge length (a) of the unit cell = 2 `xx` (radius of cation + radius of anion)
Therefore, the edge length (a) of the unit cell of AB crystal = `2xxY^(1//3)nm=2Y^(1//3)xx10^(-9)m`.
We know, `rho=(ZxxM)/(Nxxa^(3))`
GIVEN: `M=6.022Yg*mol^(-1)=6.022xx10^(-3)Ykg*mol^(-1)`
`therefore" "rho=(4xx6.022xx10^(-3)Y)/(6.022xx10^(23)xx(2Y^(1//3)xx10^(-9))^(3))=5.0kg*m^(-3)`
(1) Density of the crystal = `5.0kg*m^(-3)`
(2) The observed density `(=20kg*m^(-3))` is higher than that of the calculated density. This indicates that the crystal structure of AB is likely to have non-stoichiometric defect in the form of metal excess or metal deficiency defect or to have impurity defect in the form of substitutional impurity defect or interstitial impurity defect.
Discussion
90534.

A compound AB has a cubic structure and molecular mass 99. Its density is 3.4 g cm^(-3) . What is the length of the edge of the unit cell ?

Answer»

Solution :Let the EDGE length of each edge = a cm
Volume of unit CELL `= a^3 cm^3`
Density (d) = `3.4 g cm^(-3)`
Number of ATOMS per unit cell (Z) for simple cubic = 1
Molar mass (M) = 99 g `"mol"^(-1)`
Density , d `= (ZM)/(N_A xx a^3)`
or `a^3 = (ZM)/(d xx N_A) = (1 xx 99)/(3.4 xx 6.02 xx 10^(23)) cm^3`
`= 4.84 xx 10^(-23) cm^3`
`a = 3.64 xx 10^(-8) cm = 364 `pm
Discussion
90535.

A compound AB has a cubic structure and molecular mass 99. Its density is 3*4 g cm^(-3) . What is the length of the edge of the unit cell?

Answer»

Solution :Let the edge length of each edge = a cm
Volume of unit CELL `= a^3 cm^3`
Density (d) = `3.4 g cm^(-3)`
Number of atoms per unit cell (Z) for SIMPLE cubic = 1
Molar mass (M) = 99 g `"mol"^(-1)`
Density , d `= (ZM)/(N_A xx a^3)`
or `a^3 = (ZM)/(d xx N_A) = (1 xx 99)/(3.4 xx 6.02 xx 10^(23)) cm^3`
`= 4.84 xx 10^(-23) cm^3`
`a = 3.64 xx 10^(-8) cm = 364 `PM
Discussion
90536.

A compound AB crystallized in bcc lattice with unit cell edge length of 480 pm. If the radius of B^(-) is 225pm, then the radius of A^(+) is

Answer»

225 PM
190.7pm
129.6pm
132.8pm

Solution :Body diagonal of a cube `= SQRT( 3) a `
`= 1.732 xx 480`
`=831.36 `pm
Body diagonal`= 2( r_(x^(+)) + r_(y^(-)))`
` r_(x^(+)) + r_(y^(-)) = ( 831.36)/( 2) = 415.68`pm
`:. r_(x^(+)) = 415.68 - 225`
`= 190.68 = 190.7 `pm
Discussion
90537.

A compound A with molecular formula C_(5)H_(10)O gave a positive 2,4-DNP test but a negative Tollens reagent test. It was oxidised to carboxylic acid B with molecular formula C_(3)H_(6)O_(2) when treated with alkaline KMnO_(4) under vigorous conditon. Sodium salt of B gave a hydrocarbon C on kolne's electrolytic reduction. Identify A,B and C and write the chemical equations for the reaction.

Answer»

Solution :Structures of A,B and C are derived as under:
`C_(5)H_(10)O` gives positive 2,4-DNP test ut a negative TOLLENS. reagent test, therefore it is a ketone.
Since, it givescarboxylic acid `C_(3)H_(6)O_(2)` on vigorous oxidationit should be 3- pentanone.
Discussion
90538.

A Compound (A) with molecular formula C,H,N on acid hydrolysis gives(B) which reacts with thionylchloride to give compound(C). Benzene reacts with compound (C) in presence of anhydrous AICI_3to give compound(C). Compound (C) on reduction with gives (D). Identify (A), (B), (C) and D. Write the equations.

Answer»

Solution :(i) COMPOUND (A) with molecular formula `C_2H_3N` is methyl CYANIDE. `(CH_3CN) `
(ii) Methyl cyanide (A) on HYDROLYSIS gives acetic acid (B)
Discussion
90539.

A compound AB completely decomposes into A and B on heating. 50g of AB, on strong heating gave 40g of A. how much quantity of AB should be decomposed by heating to obtain 2.5 g of B? How much quantity of A will be produced in the process?

Answer»


SOLUTION :`ABrarrA+B`
50G 40G 10G
Discussion
90540.

A compound 'A' with molecular formula C_4H_10O is unreactive towards sodium metal. It does not add Bromine water and does not react with NaHSO, solution. On refluxing 'A' with excess of HI, it gives 'B' which reacts with aqueous NaOH to form 'C'. 'C' can be converted into 'B' by reacting with red P and I_3. 'C' on treating with conc. H_2SO_4 forms 'D'. 'D' decolourises bromine water. Identify A to D and write the reactions involved.

Answer»

Solution :A is not an ALCOHOL therefore it does not react with sodium metal. .A. is ALSO not an aldehyde or a KETONE as it does not react with `NaHSO_3`. .A. is not an unsaturated hydrocarbon as it does not add `Br_2` (aq). So, it is likely to be a ether.
`underset((C_2H_10O))underset(.A.)(CH_3CH_2OC_2H_5)+ underset(("excess"))(2HI)to underset((.B.))(2C_2H_5I)+ H_2O`
`underset(.B.)(2C_2H_5I)+ underset(("excess"))(NaOH(aq)) to underset(.C.)(C_2H_5OH) + NaI`
`underset(.C.)(C_2H_5OH) OVERSET(P//I_2)to underset(.B.) C_2H_5I`
`underset(.C.)(C_2H_5OH)overset(Conc. H_2SO_4)to underset(.D.)(CH_2=CH_2)`
Discussion
90541.

A Compound (A) with molecular formula C_(2)H_(3)N on acid hydrolysis gives (B) which reacts with thionylchloride to give compound ( C). Benzene reacts with compound ( C) in presence of anhydrous AlCl_(3) to give compound (D). Compound (D) on reduction gives (E). Identify (A), (B), ( C), (D) and (E). Write the equations.

Answer»

Solution :COMPOUND with molecular formula ` C_(2)H_(3)N " is " CH_(3)CN(A)` acetonitrile or Methylcyanide.
`underset(" (A) Methyl cyanide")CH_(3)CN overset(H_(2)O) underset(H^(+))tounderset(" (B) Acetic acid")(CH_(3) COOH) `
Acidhydrolysisof nitriliesgive the correspondingcarboxylicacid.
`underset(" (B) Acceticacid") (CH_(3)COOH) + underset("Thionyl chloride")(SoCI_(2)) tounderset("Acetylchloride")( CH_(3)COCI + SO_(2) + HCI`

(A) ` CH_(3) CN `= acetonitrile
(B) ` CH_(3)COOH` =acetic acid
(C) `CH_(3) COCI ` = acetylchloride
(D) `C_(6)H_(5)COCH_(3) ` =acetylchloride
(E) Depending upon the reducing agent
(E) is ` C_(6) H_(5)CHOHCH_(3) to `Phenylmethylcarbinol
(OR)
` C_(6)H_(5)CH_(2)CH_(3) to`EthylBenzene
Discussion
90542.

A compound A with molecular formula C H Cl gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives C and D.C gives Cannizaro reaction but not aldol condensation. D gives aldol condensation but not Cannizaro reaction. A is :

Answer»

`C_6H_5-CH_2-CH_2-CH_2-CH_2-Cl`
`C_6H_5-CH_2-CH_2-underset(Cl)underset(|)CH-CH_3`

Solution :Compound A reacts with alc.KOH to give compound B which on further azonolysis gives C (does not contains `alpha`-H atom) and D (contains `alpha`-H atom). This reaction SEQUENCE can be achieved by COMPOUNDS in option (a) and (c). SINCE compound A gives white ppt. with `AgNO_3` preferable option will be (c) as tert alkyl reacts with `AgNO_3` more quickly
Discussion
90543.

A compound A with molecular formula C_(10)H_(13)Cl give a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product, B on ozonolysis gives C and D. C gives cannizaro reaction but not aldol condensation D gives aldol condensation but not Cannizaro reaction A is

Answer»

`C_(6)H_(6)-CH_(2)-CH_2-CH_(2)-CL`
`C_(6)H_(5)-CH_(2)-CH_(2)-underset(Cl)underset(|)(C)H-CH_(3)`

Answer :D
Discussion
90544.

A compound (A) whose formula is C_6 H_10 reacts with H_2/Pt in excess to give a product C_6 H_12 , which does not decolorize Br_2//C Cl_4 . Ozonolysis of A gives 1 mol of HCHO and 1 mol of . Give the structure of A .

Answer»



`CH_3 CH_2 CH_2 CH_2 C-= CH`

SOLUTION :
Discussion
90545.

A compound A with molecular formula C_(15)H_(13)Cl gives a white precipitate on adding silver nitrate solution A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives C and C gives Cannizaro reaction but not aldol condensation. D gives aldol condensation but not Cannizaro react A is :

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`C_(6)H_(5)-CH_(2)CH_(2)CH_(2)-CH_(2)-CL`
`C_(6)H_(5)-CH(2)-CH_(2)-underset(Cl)underset(|)CH-CH_(3)`

Answer :B
Discussion
90546.

A compound A when reacted with PCl_5 and then with ammonia gave B,B when treatment with bromine and caustic potash produced C . C on treatment with NaNO_2 and HCl at 0^@C and then boiling produced orthocresol. Compound A is :

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o-toluic ACID
o-chlorotoluene
o-bromotoluene
m-toluic acid

ANSWER :A
Discussion
90547.

A compound A on oxidation gives B(C_(2)H_(4)O_(2)). A reacts with dil. NaOH and on subsquent heating forms C. C on catalytic hydrogenation gives D. Identify A, B, C, D and write down the reactions involved.

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Solution :The COMPOUND A is `CH_(3)CHO` (ETHANAL). It SATISFIES all REACTIONS, which are given below:
Discussion
90548.

A compound A on oxidation gives B(C_(2)H_(4)O_(2)). A reacts with dil. NaOH and on subsequent heating forms C. C on catalytic hydrogenation gives D. Identify A,B,C,D and write down the reactions involved.

Answer»

Solution :(i) Since compound A on oxidation gives compound B with M.F. `C_(2)H_(4)O_(2)`, therefore, compound B may be acetic acid, `CH_(3)COOH and A` and may be acetaldehyde, `CH_(3)CHO`.
(ii) Since compound A, i.e., acetaldehyde reacts with dil. NaOH, therefore, it undergoes aldol condensation to AFFORD an aldol. further since this aldol on heating gives compound (C), therefore, (C) must be an `alpha,beta-`unsaturated ALDEHYDE, i.e., but-2-en-1-al (crotonaldehyde).
(iii) Since compound (C) on catalytic hydrogenation gives compound D., therefore, D may be either 1-butanal or 1-butanol depending upon the extent of hydrogenation.
All the REACTIONS INVOLVED in this question are explained below:
Discussion
90549.

A compound 'A' on oxidation gave acetaldehyde , then again on oxidation gave acid . Afterfirst oxidation it was reacted with ammoniacal AgNO_3 then silver mirror was produced. A is likely to be

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PRIMARY ALCOHOL
Tertiary alcohol
Acetaldehyde
Acetone

Solution :`{:(CH_(3)CH_(2)OH overset("Oxi.")to CH_(3)CHO overset("Oxi.")to CH_(3)CHOOH),(UNDERSET(A)(1^(@) "alcohol")""underset("Silver Mirror")darr{:("Ammoniacal"),(AgNO_(3)""):}):}`
Discussion
90550.

A compound A reacts with thionyl chloride to give compound B. B reacts with magnesium in ether medium to form a Grignard reagent which is treated with acetone and the product on hydrolysis givesIdentify A and B. Write down the chemical equations for the reactions involved.

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SOLUTION :
Discussion