11.5 Write structures of the products of the following reactions:

<p><strong>11.5</strong> In this reaction, when propene reacts with the given reagent then the double bond of propene breaks down with charges on them. So, H⁺ gets placed on the carbon which already has two hydrogen atom and OH^- gets substituted on center carbon because it has the more positive charge which attracts OH^-. Thus we get propene-2-or as a</p><p>&nbsp;</p><p><strong>2.</strong> In this reaction, when Methyl ( 2-oxocyclohexyl) ethanoate reacts with the given reagent then the double bond between the oxygen atom and cyclohexyl gets breaks down, such that O has a negative charge and that particular carbon will have a positive charge on it. So, to neutralize it, H⁺ gets substituted to that carbon and with O^- to form the structure of alcohol. Thus we get Methyl (2-hydroxycyclohexyl) ethanoate as a</p><p>&nbsp;</p><p><strong>3.</strong> In this reaction, when 2-Methylbutanal reacts with the given reagent NaBH₄, then the double bond between carbon and oxygen gets to break down in the above aldehyde compound, where</p><p>&nbsp;</p><p><strong>4.</strong> carbon has positive charge and oxygen will have a negative charge on So, H⁺ gets placed on the carbon to complete its octet and H⁺ gets substituted on O^- to form OH, i.e, an alcohol. Thus we get 2-Methylbutan-1-ol as a product.</p>

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