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$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{B}_2 \mathrm{H}_6 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{NaOH}}$ Product.
Product in above reaction is
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Product in above reaction is
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The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{B}_2 \mathrm{H}_6 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{NaOH}} \mathrm{HC}_2 \mathrm{CH}_2 \mathrm{OH}$
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