Q. 265.0( 1 Vote )

a. Predict the ma

Answer :

a. (i)


(ii)


(iii)


b. Iodoform test is done to distinguish between these two compounds. On adding NaOH / I2 and heat, acetophenone forms yellow ppt. of iodoform(CHI3) whereas benzophenone does not.


c. Alpha (α) hydrogen of carbonyl compounds is acidic in nature due to resonance stabilisation of conjugate base of carbonyl compound


OR


a. (i) When propanal reacts with excess of methanol in the presence of HCl, it forms 1,1-dimethoxy propane.


C2H5CHO+2CH3OHC2H5CH(OCH3)2


(ii) Propanal having α-hydrogen atom undergo aldol condensation in presence of dil. NaOHand forms 3-hydroxy-2-methyl pentanal.


C2H5CHO+ NaOH (dil) CH3CH2CH (OH) CH (CH3) CHO


(iii) The carbonyl group of propanal is reduced to CH2 group on treatment with hydrazine followed by heating with KOHin ethylene glycol and the product formed will be propane (CH3CH2CH3)


b.


(i) increasing acid character order is


CH3COOH < HCOOH < FCH2COOH < O2N-CH2COOH


(ii) increasing order of reactivity towards addition of HCN


Acetophenone < Benzaldehyde < acetone < acetaldehyde


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