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Following compounds are given to you:2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane (i) Write the compound which is most reactive towards S_(N)2reaction. (ii) Write the compound which is optically active. (iii) Write the compound which is most reactive towards beta-elimination reaction. |
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Answer» Solution :`underset("2-Bromopentane") (CH_(3)- CH_(2) - CH_(2) - underset(H) underset(|)overset(Br) overset(|) C -CH_(3)) underset("2 -Bromo-2-methylbutane") (CH_(3) - CH_(2) - underset(CH_(3))underset(|)overset(Br)overset(|)C - CH_(3)) underset("1-Bromopentane") (CH_(3) - CH_(2) - CH_(2) - CH_(2) - CH_(2) - Br) ` (i) 1-Bromopentane is most REACTIVE towards SN2 reaction because it is a primary ALKYL halide. (II) Compound 2-Bromopentane is optically active because it contains an asymmetric carbon atom. `CH_(3) - CH_(2) - CH_(2) - underset(H) underset(|) overset(Br) overset(|)(C^(***)) - CH_(3) ` (iii)2-Bromo-2-methylbutane is most reactive towards b-elimination because it gives the alkene with greatest number of alkyl GROUPS attached to the double bond. `CH_(3) - CH_(2) - underset(CH_(3))underset(|) overset(Br) overset(|)C - CH_(3) overset("Alc.KOH") to CH_(3) - CH = underset(CH_(3)) underset(|) C - CH_(3) ` |
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