Match the compounds given in List-I with List-II and select the suitable optionusing the code given below :
List-I List-II
(A)Benzaldehyde (i)Phenolphthalein
(B)Phthalic anhydride (ii)Benzoin condensation
(C)Phenyl benzoate (iii)Oil of wintergreen
(D)Methyl salicylate (iv)Fries rearrangement
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Solution
Clemmensen reduction of a ketone is carried out in the presence of which of the following ?
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Solution
Following compounds are given:
Which of the above compound(s), on being warmed with iodine solution and NaOH,will give iodoform?
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Solution
Among the given compounds only CH3OH does not give iodoform reaction.
Which of the following reactions will not result in the formation of carbon-carbon bonds?
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Solution
Propionic acid with Br2/P yields adibromo product. Its structure would be:
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Solution
This reaction is an example of Hell - Volhard Zelinsky reaction. In this reaction acids containing α– H on treatment with X2/P give di-halo substituted acid.
The product formed in Aldol condensation is
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Solution
Aldehydes and ketones having at least one α-hydrogen atom in presence of dilutealkali give β-hydroxy aldehyde or β-hydroxy ketone
Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid?
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Solution
Aldehydes containing no -hydrogen atom on warming with 50% NaOH or KOH undergo disproportionation i.e., selfoxidation - reduction known as cannizzaro’s reaction.
Which of the following represents the correct order of the acidity in the given compounds?
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Solution
Electron withdrawing substituent (like halogen, —NO2,C6H5 etc.) would disperse the negative charge and hence stabilise the carboxylate ion and thus in creaseacidity of the parent acid. On the other hand, electron-releasing substituents would intensify the negative charge, destabilise the carboxylate ion and thus decrease acidity of the parent acid.Electro negativity decreases inorder
F >Cl > Br
and hence–I effect also decreases in the same order,therefore the correct option is[FCH2COOH > ClCH2COOH > BrCH2COOH >CH3COOH]
Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc. HCl is called
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Solution
Which of the following pairs can be distinguished by sodium hypoiodite?
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Solution
Oxidation of acetal aldehydes and methyl ketones with sodium hypoiodite gives this test. So, here in option(b)ketone is having (CH3CO —) group and the other is having (CH3CH2CO—) group which do not give hypoiodite test. So thus they can be distinguished.