Identify the correct reagents that would bring about the transformation C6H5-CH2-CH=CH2 -> C6H5-CH2-CH2-CHO (allylbenzene to 3-phenylpropanal).

Correct answer: (A) (A) (i) BH3, (ii) H2O2/OH-, (iii) PCC. Answer: (A) (i) BH3, (ii) H2O2/OH-, (iii) PCC Solution: Hydroboration-oxidation (BH3 then H2O2/OH-) adds water across the terminal C=C in anti-Markovnikov fashion, giving the primary alcohol C6H5-CH2-CH2-CH2OH. PCC then mildly oxidises this primary alcohol only as far as the aldehyde. Alkaline KMnO4 would over-oxidise to the acid, and acid-catalysed hydration would give a Markovnikov secondary alcohol, so option (A) is correct.

NEET 2024 · ChemistryRosenmundPrevious Year Question

Identify the correct reagents that would bring about the transformation $C X_{6} H X_{5} - CH X_{2} - CH = CH X_{2} C X_{6} H X_{5} - CH X_{2} - CH X_{2} - CHO$ (allylbenzene to 3-phenylpropanal).

Answer: (A) (A) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) PCC. \textbf{Answer:} (A) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) PCC \textbf{Solution:} Hydroboration-oxidation ( $BH X_{3}$ then $H X_{2} O X_{2} / OH X^{-}$ ) adds water across the terminal $C = C$ in anti-Markovnikov fashion, giving the primary alcohol $C X_{6} H X_{5} - CH X_{2} - CH X_{2} - CH X_{2} OH$ .

A.(A) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) PCC✓
B.(B) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) alk. $KMnO X_{4}$ , (iv) $H X_{3} O X^{+}$
C.(C) (i) $H X_{2} O / H X^{+}$ , (ii) PCC
D.(D) $H X_{2} O / H X^{+}$

Correct Answer

(A) (A) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) PCC

Solution & Explanation

\textbf{Answer:} (A) (i) $BH X_{3}$ , (ii) $H X_{2} O X_{2} / OH X^{-}$ , (iii) PCC \textbf{Solution:} Hydroboration-oxidation ( $BH X_{3}$ then $H X_{2} O X_{2} / OH X^{-}$ ) adds water across the terminal $C = C$ in anti-Markovnikov fashion, giving the primary alcohol $C X_{6} H X_{5} - CH X_{2} - CH X_{2} - CH X_{2} OH$ . PCC then mildly oxidises this primary alcohol only as far as the aldehyde. Alkaline $KMnO X_{4}$ would over-oxidise to the acid, and acid-catalysed hydration would give a Markovnikov secondary alcohol, so option (A) is correct.

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ChemMatch List-I with List-II: List-I (Reaction) (a) C6H6 -> C6H5-CO-C6H5 (benzophenone) (b) Alkene -> carbonyl (oxidative cleavage) (c) C6H5-OH -> carbonyl (oxidation) (d) C6H5-CH2-CH3 -> C6H5-COOH List-II (Reagents/Condition) (i) C6H5COCl, anhyd. AlCl3 (ii) CrO3 (iii) KMnO4/KOH, (iv) (1) O3, (2) Zn-H2O Choose the correct option:ChemThe following conversion is known as: C6H5-COCl ->[H2][Pd-BaSO4] C6H5-CHO ChemThe intermediate compound X in the following reaction is: C6H5-CH3 ->[CrO2Cl2][CS2] X ->[H3O+] C6H5-CHO ChemIdentify compound X in the following reaction: C6H5-CH3 ->[Cl2][h ] X ->[H2O][373 K] C6H5-CHO ChemWhen the vapours of a secondary alcohol are passed over heated copper at 573 K, the product formed is:ChemIdentify the major products P, Q and R in the following sequence of reactions: C6H6 + CH3CH2CH2Cl ->[anhydrous AlCl3] P P ->[(i) O2][(ii) H3O+] Q + R ChemThe correct order of decreasing acidity of the following aliphatic acids is:ChemWhich one of the following reactions does NOT belong to Lassaigne's test?

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Identify the correct reagents that would bring about the transformation CX6​HX5​−CHX2​−CH=CHX2​​CX6​HX5​−CHX2​−CHX2​−CHO (allylbenzene to 3-phenylpropanal).

Solution & Explanation

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Identify the correct reagents that would bring about the transformation $C X_{6} H X_{5} - CH X_{2} - CH = CH X_{2} C X_{6} H X_{5} - CH X_{2} - CH X_{2} - CHO$ (allylbenzene to 3-phenylpropanal).