Q. 26 A5.0( 1 Vote )

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Answer :

a) Step 1: When CH3CN is treated with SnCl2 (stannous chloride) in the


presence of HCl, it gives CH3CHO (A)


CH3CN CH3CHO


The above reaction is called Stephen reaction.


Step 2: When CH3CHO is treated with HCN, it gives cyanohydrin.


CH3 CH3 CH3



(A) Cyanohydrin (D)


Step 3: When CH3CHO (A) is treated with dil. NaOH, it undergoes aldol condensation to form β-hydroxyacetaldehyde (B).


OH ion removes the acidic α-hydrogen to form enolate ion



The enolate ion being a strong nucleophile attacks the carbonyl group of the second molecule of acetaldehyde (CH3CHO) to anion.



Protonation of anion to form a β-hydroxy aldehyde (aldol)



Step 4: Now, when aldol(B) is heated with dilute acids undergo dehydration to from But-2-en-1-al(C).



Thus, A is CH3CHO


B is CH3—CH(OH)—CH2—CHO


C is CH3—CH=CH—CHO


D is CH3—CH(CN)(OH)


b) (i) Iodoform test:


C6H5—CH=CH—COCH3 being a methyl ketone on treatment with I2/NaOH(NaOI) undergoes iodoform test to give yellow precipitate but C6H5—CH=CH—CO-CH2CH3 being ethyl ketone does not.


C6H5—CH=CH—COCH3 + 3NaOI C6H5—COONa + CHI3 + 2NaOH


Iodoform


(yellow ppt)


C6H5—CH=CH—CO-CH2CH + NaOI No yellow ppt. of CHI3


(ii) CH3—CH2—COOH and HCOOH can be distinguished by the following test:


Tollen’s reagent test: Formic acid contains a H-atom attached to >C=O group and hence can be regarded as an aldehyde.


It can be easily oxidized to CO2 and H2O but acetic acid cannot. Therefore, HCOOH (formic acid) behaves as a reducing agent whereas CH3—CH2—COOH (ethanoic acid) does not.


Formic acid reduces Tollen’s Reagent to metallic silver but ethanoic acid does not.


Tollens’ reagent is an ammoniacal solution of silver nitrate and is prepared by adding NH4OH solution to AgNO3 solution till the precipitate of Ag2O first formed just redissolves. When propanal (aldehyde) is heated with Tollens’ reagent, the latter is reduced to metallic silver which deposits on the walls of the test tube as bright silver mirror.


During this reduction, the following reaction occur:


Formation of Tollens’ reagent:


2AgNO3 + 2NH4OH Ag2O + 2 NH4NO3 + H2O


Ag2O + 4NH4OH 2[Ag(NH3)2]+OH- + 3H2O


Tollens’ Reagent


Now, formic acid is treated with Tollens’ reagent:


HCOOH + 2Ag(NH3)2]+ + 3OH-


2Ag + CO2 + 4NH3 + 2H2O


The silver (Ag) thus deposited shines like a mirror. The formation of silver mirror confirms the presence of propanal.


On the other hand, CH3—CH2—COOH does not give this test. It does not form silver with Tollens’ reagent.


c) The increasing order of boiling points of the given organic compounds is


CH3CHO < CH3CH2OH < CH3COOH


Explanation:


The boiling point of carboxylic acids are much higher than those of aldehydes and alcohols of comparable molecular masses. This means that carboxylic acids form even stronger intermolecular H- bonding than alcohols and aldehyde.



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