Question of 90
Q.(a) What is Lindlar's catalyst? (½)
(b) Identify A, B and C. (1½)
CH3 – CH = CH2 + O3 → A —[Zn, H2O]→ B + C.
CH3 – CH = CH2 + O3 → A —[Zn, H2O]→ B + C.
(c) Complete the reaction.
(1)
[benzene ring] + Cl2 —[anhydrous AlCl3]→ _____
[benzene ring] + Cl2 —[anhydrous AlCl3]→ _____
Kerala DhseKerala DHSE Plus One Board 2020Subjective· 3mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →Lindlar's catalyst gives selective partial hydrogenation of alkynes; ozonolysis of propene cleaves the C=C bond into acetaldehyde and formaldehyde; benzene undergoes Friedel-Crafts-type halogenation with Cl2/anhydrous AlCl3.
- Lindlar's catalyst Palladium metal deposited on calcium carbonate (or barium sulphate) and partially deactivated ("poisoned") with sulphur compounds such as quinoline. It selectively hydrogenates alkynes to cis-alkenes, without over-reducing them further to alkanes.
- Identify A, B, C CH3–CH=CH2 (propene) reacts with ozone (O3) to form A = propene ozonide (the cyclic ozonide intermediate). Treating the ozonide with Zn/H2O (reductive workup) cleaves the C=C bond: CH3–CH=CH2 + O3 → A --[Zn, H2O]--> B + C
- B = CH3CHO (acetaldehyde/ethanal)
- C = HCHO (formaldehyde/methanal) (B and C together, i.e. their identities, may be interchanged in order — one product retains the CH3 group, the other is formaldehyde from the terminal =CH2.)
(c) Benzene + Cl2 (anhydrous AlCl3) …
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