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CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

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Question : 36 of 37
Marks: +1, -0
(a) Write the products formed when benzaldehyde reacts with the following reagents :
(i) CH3CHO\mathrm{CH}_3 \mathrm{CHO} in presence of dilute NaOH\mathrm{NaOH}
(iii)Conc. NaOH\mathrm{NaOH}
(b) Distinguish between following:
(i) CH3−CH=CH−CO−CH3\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-\mathrm{CH}_3 and CH3−CH2\mathrm{CH}_3-\mathrm{CH}_2 −CO−CH=CH2-\mathrm{CO}-\mathrm{CH}=\mathrm{CH}_2
(ii) Benzaldehyde andBenzoic acid
OR
(a) Write the final products in the following:
(iii) CH2=CH−CH2−CN→(b) H3O+(a) DIBAL−H\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CN}\xrightarrow[\text{(b)}\,\mathrm{H}_3\mathrm{O}^{+}]{\text{(a)}\,\mathrm{DIBAL}-\mathrm{H}}
(b) Arrange the following in the increasing order of their reactivitytowards nucleophilic addition reaction:
(c) Draw the structure of 2, 4 DNP derivative of acetaldehyde.
Solution:  
(a)
(b) (i) CH3−CH=CH−CO−CH3\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-\mathrm{CH}_3 gives iodoform test while CH3−CH2−CO−CH=CH2\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}-\mathrm{CH}=\mathrm{CH}_2 does not give.
CH3−CH=CH−C∥O−CH3\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\underset{\mathrm{O}}{\underset{\parallel}{\mathrm{C}}}-\mathrm{CH}_3 +3 NaOI→Yellow ppt.CH3I+2 NaOH+3\,\mathrm{NaOI}\xrightarrow[\text{Yellow ppt.}]{\mathrm{CH}_3\mathrm{I}}+2\,\mathrm{NaOH} +CH3−CH=CHCOONa+\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHCOONa}
CH3CHCOCH=CH2+3 NaOI→No ppt\mathrm{CH}_3\mathrm{CHCOCH}=\mathrm{CH}_2+3\,\mathrm{NaOI}\rightarrow\text{No ppt}
(ii) ● Benzaldehyde reacts with tollen's reagent to form silver mirror. Benzoic acid does not give this reaction.
● With NaHCO3\mathrm{NaHCO}_3 benzaldehyde does not react while benzoic acid produces brisk effervescences.
OR
(a)
(iii) CH2=CH−CH2−CN→(b) H3O+(a) DIBAL−H\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CN}\xrightarrow[\text{(b)}\,\mathrm{H}_3\mathrm{O}^{+}]{\text{(a)}\,\mathrm{DIBAL}-\mathrm{H}} CH2=CH−CH2−C∥O−H\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{H}
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