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CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

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Question : 25 of 26
Marks: +1, -0
Write the structures of A, B, C, D and E in the following reactions:
OR
(a) Write the chemical equation for the reaction involved in Cannizzaro reaction.
(b) Draw the structure of the semicarbazone of ethanal.
(c) Why pKa\mathrm{p}K_a of F−CH2−COOH\mathrm{F}-\mathrm{CH}_2-\mathrm{COOH} is lower than that of Cl−CH2−COOH\mathrm{Cl}-\mathrm{CH}_2-\mathrm{COOH} ?
(d) Write the product in the following reaction
(e) How can you distinguish between propanal and propanone?
Solution:  
OR
(a) 2HCHO→50% NaOHCH3OH+HCOONa2 \mathrm{HCHO} \xrightarrow[50\%\ \mathrm{NaOH}]{} \mathrm{CH}_3 \mathrm{OH} + \mathrm{HCOONa}
(b) CH3−CH=N−NH−CO−NH2\mathrm{CH}_3-\mathrm{CH}=\mathrm{N}-\mathrm{NH}-\mathrm{CO}-\mathrm{NH}_2
(c) In FCH2−COOH\mathrm{FCH}_2-\mathrm{COOH}, fluorine is more electron withdrawing than chlorine in ClCH2−COOH\mathrm{ClCH}_2-\mathrm{COOH}, so FCH2−COOH\mathrm{FCH}_2-\mathrm{COOH}, flurine is more acidic than ClCH2COOH\mathrm{ClCH}_2 \mathrm{COOH} hence its pKa\mathrm{p}K_a value is lesser than ClCH2COOH\mathrm{ClCH}_2 \mathrm{COOH}.
(c) CH3−CH=CH−CH2−CNPent-3-enenitrile→ (i) H2O (i) DIBL−H\underset{\text{Pent-3-enenitrile}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CN}} \xrightarrow[\text{ (i) } \mathrm{H}_2\mathrm{O}]{\text{ (i) } \mathrm{DIBL-H}} CH3−CH=CH−CH2−CHOPent-3-ene-1-al\underset{\text{Pent-3-ene-1-al}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO}}
(e) Propanal and propanone can be differentiated by Tollens' reagent i.e., propanal will give silver mirror but propanone will not.
CH3−CH2−CHO+2[Ag(NH3)2]+→CH3−CH2\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHO} + 2 [\mathrm{Ag}(\mathrm{NH}_3)_2]^{+} \rightarrow \mathrm{CH}_3-\mathrm{CH}_2 −COO−+2Ag- \mathrm{COO}^{-} + 2 \mathrm{Ag} +H2O+4NH3Silver mirror\underset{\text{Silver mirror}}{+ \mathrm{H}_2\mathrm{O} + 4 \mathrm{NH}_3}
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