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Published on: 13/05/2022
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Take MCQ Chemistry Test1.
An organic compound 'A' is a sodium salt of phenolic acid with molecular formula C7H5O3Na. 'A' on heating with soda lime gives compound 'B' of molecular formula C6H6O. 'B' gives violet colour with neutral ferric chloride. 'B' on treatment with C6H5COCI in the presence of NaOH gives an ester 'C'. Identify 'A', 'B' and 'C'. Explain the reactions.
2.
An organic compound (A) C6H6O gives violet colour with neutral FeCl3 solution. With NH3 in the presence of anhydrous ZnCI2, (A) gives (B) (C6H7N). (A) with dimethyl sulphate gives (C) (C7H8O). What are (A), (B) and (C)? Explain the reactions.
3.
An organic compound (A) of molecular formula C3H8O2 is obtained as by-product in the manufacture of soap. Compound (A) on heating with P2O5 gives an unsaturated compound (B) of molecular formula C3H4O. Compound (A) with well cooled mixture of Conc.H2SO4 and fuming HNO3 form compound (C) which is an explosive. Identify A, B and C and explain the rection.
4.
An organic compound A of molecular formula C3H6O on reduction with LiAlH4 gives B. Compound B gives blue colour in Victor Meyer's test and also forms a chloride C with SOCl2. The chloride on treatment with alcoholic KOH gives D. Identify A, B, C and D and explain the reactions.
5.
An organic compound (A) C2H6O liberates hydrogen on treatment with metallic sodium. (A) on mild oxidation gives (B) C2H4O which answers iodoform test. (B) when treated with cone. H2SO4 undergoes polymerisation to give (C) a cyclic compound. Identify (A) (B) and (C) and explain the reactions.
1.
(i) The organic compound (A) which is sodium salt of phenolic acid is sodium salicylate.
(ii) (A) on heating with soda lime gives Compound (B) phenol. Phenol gives violet colouration with neutral perchloride.
(iii) Phenol on treatment with C6H5COCl in the presence of NaOH gives on ester 'C':
C6H5OH + C6H5COCl \(\overset { NaOH }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }OCO{ C }_{ 6 }{ H }_{ 5 }+HCl } \)
| Compound | Compound Name | Formula |
| A | Sodium benzoate' |  |
| B | Phenol | C6H5OH |
| C | Phenyl benzoate | C6H5OCO C6H5 |
2.
(i) An organic compound (A) C6H6O gives violet colour with neutral FeCl3 solution.
(ii) With NH3 in the presence of anhydrous ZnCI2,(A) gives (B) (C6H7N).
| Compound | Compound Name | Formula |
| A | Phenol | C6H5OH |
| B | Aniline | C6H5NH2 |
| C | Anisole | C6H5OCH3 |
3.
(i) Compound (A) with molecular formula C3H8O3 is glycerol.
(ii) Glycerol on heating with P2O5 gives an unsaturated compound (B).
(iii) Glycerol with cooled mixture of cone. H2SO4 and fuming HNO3 form compound (C).
| Compound | Compound Name | Formula |
| A | Glycerol | \(\overset { CH_{ 2 }OH }{ \underset { \overset { CHOH }{ \underset { { CH }_{ 2 }OH }{ | } } }{ | } } \) |
| B | Acrolein | \(\overset { CH_{ 2 } }{ \underset { \overset { CH }{ \underset { { CH }O }{ | } } }{ || } } \) |
| C | Nitroglycerine | \(\overset { CH_{ 2 }O{ NO }_{ 2 }\quad }{ \underset { \overset { CHO{ NO }_{ 2 } }{ \underset { { CH }_{ 2 }O{ NO }_{ 2 } }{ | } } }{ | } } \) |
4.
(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH4 gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\)
(iii) (B) reacts with SOCl2to give (C).
(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \)
| Compound | Compound Name | Formula |
| A | Acetone | \({ CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\) |
| B | Isopropyl alcohol | \({ CH }_{ 3 }-\underset { \overset { | }{ OH } }{ CH } -{ CH }_{ 3 }\) |
| C | Isopropyl chloride | \({ CH }_{ 3 }-\underset { \overset { | }{ Cl } }{ CH } -{ CH }_{ 3 }\) |
| D | Propylene | CH3-CH=CH2 |
5.
(i) \( \underset { (A) }{2 \mathrm{C}_{2} \mathrm{H}_{5}} \mathrm{OH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2} \uparrow \)
(ii) Ethyl alcohol on mild oxidation gives acetaldehyde (B) It answers idoform test.
\({ CH }_{ 3 }-\underset{(A)}{{ CH }_{ 2 }OH}\overset { Mild\left( O \right) }{ \longrightarrow } \underset { (B) }{ CH_{ 3 }CHO } +{ H }_{ 2 }O\)
\( \underset {(B)}{\mathrm{CH}_{3} \mathrm{CHO}+}3 \mathrm{I}_{2}+\mathrm{NaOH} \overset { B }{ \underset { Iodoform }{ \longrightarrow } } \mathrm{CHI}_{3}+\mathrm{HCOONa}+3 \mathrm{HI}\)
(iii) Acetaldehyde when treated with Cone H2SO4 undergoes polymerisation and the product obtained is paraldehyde (C).
| Compound | Compound Name | Formula |
| A | Ethyl alcohol | CH3CH2OH |
| B | Acetaldehyde | CH3CHO |
| C | Paraldehyde |  |
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