Complete the following: i) 2-butyne xrightarrowLindlar Catalyst ii) CH2=CH2 xrightarrowI₂ iii) CH2(Br)-CH2(Br)

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Q. Complete the following:

i) 2-butyne

\xrightarrow{\text{Lindlar Catalyst}}

ii) CH2=CH2

\xrightarrow{I_2}

iii) CH2(Br)-CH2(Br)

\xrightarrow{Zn/C_2H_5OH}

iv) CaC2

\xrightarrow{H_2O}

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Step 1 (i). 2-Butyne + H2 over Lindlar's catalyst (Pd on CaCO3, partially poisoned) is a stereospecific SYN addition that stops at the alkene stage: CH3-C $\equiv$ C-CH3 + H2 --Lindlar--> cis-2-butene (CH3-CH=CH-CH3, both methyls on the same side).

Step 2 (ii). Ethene + I2 is a straightforward (if slow and somewhat reversible) halogen addition across the double bond: CH2=CH2 + I2 --> CH2I-CH2I (1,2-diiodoethane).

Step 3 (iii). The vicinal dibromide CH2Br-CH2Br treated with zinc in ethanol undergoes DEhalogenation (loss of both bromines as ZnBr2, regenerating the double bond): CH2Br-CH2Br + Zn --C2H5OH--> CH2=CH2 (ethene) + ZnBr2. …

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