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Chemistry Question on Aldehydes, Ketones and Carboxylic Acids

Acetaldehyde can be prepared by

A

CH3CH2Br\ce−>[KCN(alc)][Δ]  \ce−>[(i)SnCl2/HCl,ether][(ii)H2O/Δ]CH_3CH_2Br \ce{->[KCN(alc)][\Delta]} \,\, \ce{->[(i) SnCl_2/HCl, {\text{ether}}][(ii) H_2O/\Delta]}

B

CH≡CH+H2O\ce−>[H2SO4]CH \equiv CH +H_2O \ce{->[H_2SO_4]}

C

\ceCH2=CH2−>[CO, H2, (Co(CO)4)2][High Temperature, pressure]\ce{CH2 = CH2 ->[CO, \,H_2,\, (Co(CO)_4)_2][{\text{High Temperature, pressure}}] }

D

\ceCH3COOH−>[CaCO3]  −>[Δ]\ce{CH3COOH ->[CaCO3]\,\, ->[\Delta]}

Answer

CH≡CH+H2O\ce−>[H2SO4]CH \equiv CH +H_2O \ce{->[H_2SO_4]}

Explanation

Solution

CH3CH2Br−>[KCN(alc)][Δ]CH_3CH_2Br {->[KCN(alc)][\Delta]} CH3CH2CNPropane nitrile\underset{\text{Propane nitrile}}{ {CH3CH2CN }} −>[(i)snCl2/HCl,Ether][(ii)H2O/Δ] {->[(i) snCl_2/HCl,{\text{Ether}}][(ii)H_2O/\Delta]} CH3CH2CHOPropanaldehyde\underset{\text{Propanaldehyde}}{ {CH3CH2CHO }} CH≡CH+H2O−>[H2SO4]CH3CHOCH \equiv CH +H_2O {->[H_2SO_4]} CH_3CHO CH2=CH2+CO+H2−>[(Co(CO)4)2][HighT,P]CH_2 = CH_2 +CO +H_2 {->[(Co(CO)_4)_2][{\text{High}}T,P]} CH3CH2CHOOxoprocess\underset{\text{Oxoprocess}}{ {CH3CH2CHO }} CH3COOH−>[CaCO2](CH3COO)2Ca−>[Δ][−CaCO3](CH3)2COCH_3COOH {->[CaCO_2]}(CH_3COO)_2 Ca {->[\Delta][-CaCO_3]}(CH_3)_2CO