Which one among the following reaction products gives the iodoform test?

  • A
    $CH_3CH_2CH_2CN \xrightarrow[\text{(ii) } H_3O^{+}]{\text{(i) } C_6H_5MgBr} CH_3CH_2CH_2COC_6H_5$
  • B
    $(C_6H_5)_2Cd \xrightarrow{CH_3COCl} C_6H_5COCH_3$
  • C
    $CH_2=CHCH_2CH(OH)C_6H_5 \xrightarrow{PCC} CH_2=CHCH_2COC_6H_5$
  • D
    $C_6H_5CH_2CN \xrightarrow[\text{(ii) } H_3O^{+}]{\text{(i) } C_2H_5MgBr} C_6H_5CH_2COC_2H_5$

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Similar Questions

Write down functional isomers of a carbonyl compound with molecular formula $C_3H_6O$. Which isomer will react faster with $HCN$ and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction conditions? If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why?

Assertion : Benzaldehyde is more reactive than ethanol towards nucleophilic attack.
Reason : The overall effect of $-I$ and $+R$ effect of phenyl group decreases the electron density on the carbon atom of $>C=O$ group in benzaldehyde.

Compound '$A$' (molecular formula $C_3H_8O$) is treated with acidified potassium dichromate to form a product '$B$' (molecular formula $C_3H_6O$). '$B$' forms a shining silver mirror on warming with ammoniacal silver nitrate. '$B$' when treated with an aqueous solution of $H_2NCONHNH_2 \cdot HCl$ and sodium acetate gives a product '$C$'. Identify the structure of '$C$.

Acetyl bromide reacts with excess of $CH_3MgI$ followed by treatment with a saturated solution of $NH_4Cl$ gives:

$CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO \xrightarrow{KOH} ?$ The number of intramolecular aldol condensation products is:

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