Alcohols Essay

Published: 2021-07-06 06:33:50
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Category: Chemistry

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1. isomers are compounds with the same molecular formula but differ in structural conformation. In C4H10O, the carbons can be arranged in a straight chain to form, but i-ol and but-2ol.we can also have a chain with three carbon atoms and a branched methyl group in the middle to give 2-methyl propanol. Compounds formed can be ethers or alcohols Three isomers with the formula C4H10O hence can be butan-1-ol, butan-2- and 2 methyl-propan-1-ol which are the compounds C, D, E, respectively. CRO3/ pyridine(Collins reagent) oxidizes primary alcohols to form an aldehyde. The compounds formed are butanal, but-2-nal and 2 methylpropanal (F, G, H )Primary alcohols are first oxidized to aldehydes, and further oxidation will form carboxylic acids. Unlike primary alcohols, secondary alcohols are oxidized to give ketones. Upon further oxidation, ketones cannot be oxidized further to form a carboxylic acid. Tertiary alcohols cannot be oxidized. Further oxidation of C and E will lead to formation of I and J which are butanoic and 2-methyl butanoic acid respectively.When alcohol is dehydrated, both the hydroxyl group and the hydrogen atom adjacent to the carbon atom in the chain, are removed. C and E give single compounds ( k and L) which are but-1-ene and 2methyl butene. While D will give rise to ( KMN). In D, the hydrogen atom either to the left or the right of the carbon can be removed, will give rise to bute-1-ene and but- 2-ene. Consequently, bute-2-ene shows isomerism, hence cis bur-2-ene and cis but-2ene will be formed. Usually, trans isomers are more stable than cis isomers .trans but-2-ene will be M while cis but-2-ene will be N. The overall result will be a mixture of three compounds but-1-ene, trans bute-2-ene and cis but-2-ene which are the compounds K, M N in that orderWhen alkenes are hydrated in the presence of aqueous acid, it leads to the formation of alcohols. L will form 2methyl propan 2-ol (compound O)Methoxypropane is another isomer of C4H10O. It is an ether and can be prepared from alkoxide and haloalkane. In the Williamson synthesis. Alkoxide are conjugated bases of alcohol .groip while haloalkane are group of compounds derived from alkane and they contain a halogen group. The two pairs of reactant that can be used in the synthesis of the compound are:N-propyl iodide and sodium methoxide which reacts to form methoxide propane and sodium iodide.Sodium ethoxide and chloromethane which react to form methoxide propane and sodium chloride.Higher yields can be obtained in the second pair (R+S) compared to the first pair (P+Q)2. the reaction proceeds via SN2mechanism. In this reaction, one bond is broken, and another one is formed all in a one-step reaction. The nucleophile(Ro-) attacks the carbon with the best leaving group to form C-O bond and breaking the c-halogen bond. In the two pairs of reactants, chloride in 1-propyl chloride is easily lost hence there the reaction is faster giving more yields.Presentation of the mechanism3. 2-methoxypropan-1-ol is a primary alcohol. Grignard is a simple way of preparing primary, secondary and tertiary alcohols. A Grignard reagent is an organometallic reagent of magnesium halide. It reacts with formaldehyde, aldehydes or ketones to form primary, secondary or tertiary alcohols.A pair of reagent that can be used used to prepare compound E is:Methyl magnesium bromide –Grignard reagentProp aldehyde- carbonyl. Compound

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