Isopropyl chlorideNa−−−→Ether Cl2/hv−−−−→2 mol alc. – InterviewSolution

1.	Isopropyl chlorideNa−−−→Ether Cl2/hv−−−−→2 mol alc. KOH−−−−−−→2 mol CH2=CH2−−−−−−→Δ The end product is:
Answer» $Isopropyl chloride N a - -- \to E t h e r C l_{2} / h v - --- \to 2 m o l a l c . K O H - ----- \to 2 m o l C H_{2} = C H_{2} - ----- \to Δ$ The end product is:

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