A) H2O would most likely donate the electron pair from oxygen into Carbon-5 as it is carrying a +1 formal charge (not shown in question). CH3CH2CH=CH-CH2(+) + H:O:H (which carries 2 non-bonded electron pairs) The Pi-electrons of the double bond between C-3 and C-4 would shift to C-4 and absorb the Hydrogen from the Nucleophilic water. This would then leave C-3 with a formal +1 charge which would attract another water that would dissociate leaving an Hydroxide on C-3. Yield => CH3CH2CHOHCH2CH2OH => Pentan-1,3-diol.
B. In the illustrated formula, from right to left, Carbons 2,3 & 4 need another Hydrogen. This would then be 2-Pentene (CH3CH=CHCH2CH3). This then could be Hydrogenated at the double bond in a prototypical conversion of an alkene to an alkane in the presence of an appropriate metallic catalyst. The yield in this case is Pentane.
Complete the following equations. Include names of all reactants and products
A) CH3CH2CHCH - CH2+H2O -->
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CH3
B) CH3CHC=CCH3 + H2 --->
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CH3
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