Answer
Verified
399.9k+ views
Hint: Conversion of an ammonia like odour organic compound to the sweet smelling organic compound. In between silver is also produced. A has an ammonia like smell and B has an alcoholic smell. On controlled oxidation of B we get a carbonyl compound C and this compound can reduce tollen’s reagent and gets converted to an acid D. On esterification of D and B we get a sweet smelling compound E.
Complete step-by-step answer:
It is given that the molecular formula of A is \[{{C}_{2}}{{H}_{7}}N\] .
Let us consider A as ethyl amine \[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\].
A reaction with nitrous acid gives ethanol or B.
\[\underset{A}{\mathop{{{C}_{2}}{{H}_{5}}N{{H}_{2}}}}\,\xrightarrow{HN{{O}_{3}}}\underset{B}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\]
Next step is B (ethanol) on controlled oxidation gives C (acetaldehyde).
\[\underset{B}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{controlled\quad oxidation}\underset{C}{\mathop{C{{H}_{3}}CHO}}\,\]
C, that is acetaldehyde reduces Tollen’s reagent. The produces are silver mirror and D (acetic acid)
\[\underset{C}{\mathop{C{{H}_{3}}CHO}}\,+2{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}+3O{{H}^{-}}\xrightarrow{heat}\underset{D}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+2Ag\downarrow +2{{H}_{2}}O+4N{{H}_{3}}\]
D or acetic acid and ethanol reacts in the presence of sulphuric acid to give the sweet smelling compound of an ester ethyl acetate.
\[\underset{D}{\mathop{C{{H}_{3}}COOH}}\,+{{C}_{2}}{{H}_{5}}OH\xrightarrow{conc{{H}_{2}}S{{O}_{4}}}\underset{E}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}\,\]
This is the conversion of a primary amine to an ester.
Thus, A-\[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\], B-\[{{C}_{2}}{{H}_{5}}OH\], C-\[C{{H}_{3}}CHO\], D-\[C{{H}_{3}}COOH\], E-\[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\]
C reacts with ammonia to give acetaldehyde ammonia.
Ammonia attacks the double bond O of the aldehyde and converts the aldehyde to acetaldehyde ammonia.
Additional Information: Conversions are the easy steps or hacks to convert one organic compound to another. It can be single step or multi step. Conversion schemes cannot be used as such in organic synthesis, when actually synthesising various parameters is involved. We should have a general idea of all the organic reactions while developing a conversion scheme.
Note: Whenever nitrogen is present in the given molecular formula without oxygen, then the only possibilities are derivatives of amines or nitrile compounds. Only aldehydes reduce Tollen’s reagent to silver, ketones do not.
Complete step-by-step answer:
It is given that the molecular formula of A is \[{{C}_{2}}{{H}_{7}}N\] .
Let us consider A as ethyl amine \[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\].
A reaction with nitrous acid gives ethanol or B.
\[\underset{A}{\mathop{{{C}_{2}}{{H}_{5}}N{{H}_{2}}}}\,\xrightarrow{HN{{O}_{3}}}\underset{B}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\]
Next step is B (ethanol) on controlled oxidation gives C (acetaldehyde).
\[\underset{B}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\xrightarrow{controlled\quad oxidation}\underset{C}{\mathop{C{{H}_{3}}CHO}}\,\]
C, that is acetaldehyde reduces Tollen’s reagent. The produces are silver mirror and D (acetic acid)
\[\underset{C}{\mathop{C{{H}_{3}}CHO}}\,+2{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}+3O{{H}^{-}}\xrightarrow{heat}\underset{D}{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+2Ag\downarrow +2{{H}_{2}}O+4N{{H}_{3}}\]
D or acetic acid and ethanol reacts in the presence of sulphuric acid to give the sweet smelling compound of an ester ethyl acetate.
\[\underset{D}{\mathop{C{{H}_{3}}COOH}}\,+{{C}_{2}}{{H}_{5}}OH\xrightarrow{conc{{H}_{2}}S{{O}_{4}}}\underset{E}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}\,\]
This is the conversion of a primary amine to an ester.
Thus, A-\[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\], B-\[{{C}_{2}}{{H}_{5}}OH\], C-\[C{{H}_{3}}CHO\], D-\[C{{H}_{3}}COOH\], E-\[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\]
C reacts with ammonia to give acetaldehyde ammonia.
Ammonia attacks the double bond O of the aldehyde and converts the aldehyde to acetaldehyde ammonia.
Additional Information: Conversions are the easy steps or hacks to convert one organic compound to another. It can be single step or multi step. Conversion schemes cannot be used as such in organic synthesis, when actually synthesising various parameters is involved. We should have a general idea of all the organic reactions while developing a conversion scheme.
Note: Whenever nitrogen is present in the given molecular formula without oxygen, then the only possibilities are derivatives of amines or nitrile compounds. Only aldehydes reduce Tollen’s reagent to silver, ketones do not.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE