Q.Write the reagent required (denoted '?') for the following reaction: benzonitrile (ring-CN) --?--> benzaldehyde (ring-CHO).

Hg$^{2+}$/H$_2$SO$_4$ hydration of propyne (Markovnikov) gives propanone via the enol.

Terminal alkynes undergo Markovnikov (mercuric-ion catalysed) hydration with dilute $H_2SO_4$ in presence of 1% $HgSO_4$, adding water across the triple bond to give an enol which tautomerises to a ketone (following Markovnikov's rule).

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2,3-Dimethylbut-2-ene, $(CH_3)_2C=C(CH_3)_2$, is symmetric, so ozonolysis cleaves it into two identical acetone molecules.

2,3-Dimethylbut-2-ene has the structure $(CH_3)_2C=C(CH_3)_2$ — both carbons of the double bond bear two identical methyl substituents. Ozonolysis (formation of the ozonide, followed by reductive cleavage with Zn dust/water) cleaves the $C=C$ bond, converting each doubly-bonded carbon into a carbonyl carbon:

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Hydration of acetylene over HgSO4/dil. H2SO4 gives acetaldehyde.

When acetylene (HC≡CH) is passed through dilute sulphuric acid in the presence of mercuric sulphate as catalyst, water adds across the triple bond (Markovnikov):

HC≡CH + H2O --(HgSO4/dil.H2SO4)--> [CH2=CH-OH] -> CH3-CHO

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A nitrile can be reduced selectively to an aldehyde by the Stephen reduction (SnCl2/HCl) or by DIBAL-H, both followed by hydrolysis of the intermediate imine.

Identifying the reaction: Nitrile (-CN) converting to aldehyde (-CHO) is a partial (controlled) reduction — the nitrile must be stopped at the imine stage rather than going all the way to the primary amine.

Method 1 — Stephen reduction: Benzonitrile is treated with stannous chloride (SnCl2) and HCl gas, which reduces it to an imine-stannic chloride complex; this is then hydrolysed with water to release the aldehyde.

$C_6H_5CN \xrightarrow{SnCl_2/HCl} C_6H_5CH=NH\cdot HCl \xrightarrow{H_2O} C_6H_5CHO$

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Dry distillation of the calcium salt of formic acid (calcium formate) yields formaldehyde, HCHO.

Dry distillation of the calcium salt of a carboxylic acid is a classic route to aldehydes/ketones. Calcium formate, (HCOO)2Ca, on dry distillation gives formaldehyde:

(HCOO)2Ca ---> HCHO + CaCO3

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Dry distillation of the calcium salt of a carboxylic acid is a classic laboratory route to a ketone — for calcium acetate, this gives acetone.

When the calcium salt of acetic acid (calcium acetate) is heated strongly in the absence of air (dry distillation), it undergoes ketonic decarboxylation: two acetate groups combine, releasing calcium carbonate and forming a ketone — in this case, the simplest ketone, acetone (propan-2-one):

$Ca(CH_3COO)_2 \xrightarrow{\Delta,\,dry\ distillation} CaCO_3 + CH_3COCH_3\ (acetone)$

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